Graduate Course

UNIT 1 : Theory of Equations and Expansions of Trigonometric Functions

Lesson 1 : Complex Numbers

Lesson 2 : De Moivre’s Theorem

Lesson 3 : Applications of De Moivre’s Theorem

Lesson 4 : Applications of De Moivre’s Theorem to Summation of Seric

Lesson 5 : Theory of Equations–I

Lesson 6 : Theory of Equations–II

UNIT 2 : Matrices

Lesson 1 : Matrices : Basic Concepts

Lesson 2 : Elementary Operations on a Matrix and Inverse of a Matrix

Lesson 3 : Rank of a Matrix

Lesson 4 : Systems of Linear Equations

Lesson 5 : The Characteristic Equation of a Matrix

UNIT-3 : Groups, Rings and Vector Spaces

Lesson 1 : Groups

Lesson 2 : Subgroup

Lesson 3 : Ring

Lesson 4 : Vector Spaces

Editor:

Dr. S.K. Verma

SCHOOL OF OPEN LEARNING

University of Delhi

5, Cavalry Lane, Delhi-110007

Semester-II

Paper II: Algebra

Course Objectives: Students will get conceptual understanding and the applicability of the

subject matter, helps students to see how linear algebra can be applied to real-life situations.

Modern concepts and notation are used to introduce the various aspects of linear equations,

leading readers easily to numerical computations and applications.

Course Learning Outcomes: The course will enable the students to understand:

i) Solving higher order algebraic equations.

ii) Become aware of De Moivre’s theorem and its applications.

iii) Solving simultaneous linear equations with at most four unknowns.

iv) Get an overview of abstract algebra by learning about algebraic structures namely,

groups, rings and vector spaces.

Unit 1: Theory of Equations and Expansions of Trigonometric Functions

Fundamental Theorem of Algebra, Relation between roots and coefficients of nth degree

equation, Remainder and factor theorem, Solutions of cubic and biquadratic equations, when

some conditions on roots of the equation are given, Symmetric functions of the roots for cubic

and biquadratic; De Moivre’s theorem (both integral and rational index), Solutions of equations

using trigonometry and De Moivre’s theorem, Expansion for cos nx, sin nx in terms of powers of

cos x, sin x, and cosnx , sinnx, in terms of cosine and sine of multiples of x.

Unit 2: Matrices

Matrices, Types of matrices, Rank of a matrix, Invariance of rank under elementary

transformations, Reduction to normal form, Solutions of linear homogeneous and non-

homogeneous equations with number of equations and unknowns up to four; Cayley−Hamilton

theorem, Characteristic roots and vectors.

Unit 3: Groups, Rings and Vector Spaces

Integers modulo n, Permutations, Groups, Subgroups, Lagrange’s theorem, Euler’s theorem,

Symmetry Groups of a segment of a line, and regular n-gons for n = 3, 4, 5, and 6; Rings and

subrings in the context of C[0,1] and ℤn; Definition and examples of a vector space, Subspace

and its properties, Linear independence, Basis and dimension of a vector space.

References:

1. Beachy, J. A., & Blair, W. D. (2006). Abstract Algebra (3rd ed.). Waveland Press, Inc.

2. Burnside, William Snow (1979). The Theory of Equations, Vol. 1 (11th ed.) S. Chand &

Co. Delhi. Fourth Indian Reprint.

3. Gilbert, William J., & Vanstone, Scott A. (1993). Classical Algebra (3rd ed.). Waterloo

Mathematics Foundation, Canada.

4. Meyer, Carl D. (2000). Matrix Analysis and Applied Linear Algebra. Society for Industrial

and Applied Mathematics (Siam).

Additional Readings:

i. Dickson, Leonard Eugene (2009). First Course in The Theory of Equations. The Project

Gutenberg EBook (http://www.gutenberg.org/ebooks/29785).

ii. Gilbert, William J. (2004). Modern Algebra with Applications (2nd ed.). Wiley-

Interscience, John Wiley & Sons.

1

UNIT 1

LESSON 1

COMPLEX NUMBERS

1.1 Complex Number

Definition : An ordered pair (x, y) of real numbers x, y is called complex number.

Illustration : (1, 6) (, –2), (0, 1), ( –1, 0), ( 2,1) , (1/2, 3) are complex numbers.

Two complex numbers : (x, y) and (p, q) are said to be equal if x = p, y = q.

In symbols, (x, y) = (p, q) x = p and y = q.

1.2 Addition of Complex Numbers

If = (a, b) and = (c, d) be to complex numbers, we define their sum + to be the complex

number (a + c, b + d). Symbolically

(a, b) + (c, d) = (a + c, b + d)

Illustrations : (3, 7) + (4, –1) = (7, 6)

(2p, q) + (0, –2q) = (2p, – q)

1.3 Multiplication of Complex Numbers

If = (a, b) and = (c, d) be two complex numbers, we define the product to be the complex

number (ac – bd, ad + bd).

In symbols (a, b) (c, d) = [(ac – bd), (ad + bc)]

Illustrations : (i) (2, 4) (–1, 3) = [2(–1) –4.3, 2.3 + 4 (–1)]

= (–14, 2)

(ii) (1, 0) (2, –3) = [(1.2) – 0.(–3), 1.(–3) + 0.2)]

= (2, –3)

1.4 Division

If , , be any two complex number ( 0) then by definition = –1. We shall also use the

symbols /, to denote .

1.5 Usual Notation for Complex Numbers

We shall now see at to how we can express a complex number (a, b) in the usual way as

a + ib. The mapping a → (a, 0) associates to each real number a, a complex number (a, 0) whose first

number is ‘a’ and whose second number is ‘0’. This mapping is compatible with addition and

multiplication. That is, if ‘a’ and ‘b’ be any two numbers, then

a + b → (a + b, 0) = (a, 0) + (b, 0)

ab → (ab, 0) = (a, 0) (b, 0)

The above relations mean the if we have any relation involving real numbers, then it remains true if

each real number ‘a’ in the relations is replaced by the complex number (a, 0). Because of this important

property we can write any relation involving complex numbers.

2

Also (0, 1) (0, 1) = (–1, 0) = – 1

which suggest that we may write (0, 1) = i

If we agree to write ‘a’ for (a, 0) and i for (0, 1) then

(a, b) = (a, 0) + (0, b)

= (a, 0) + (0, 1) (b, 0)

= a + ib

Thus we find that if we set up the convention of writing ‘a’ for (a, 0) and i for (0, 1), then

(a, b) will be written as a + ib (where i = 1− )

We shall give below, some examples to illustrate the Algebraic operation on complex numbers, with

usual Notation.

Example 1. Simplify (4 + 5i) + (2 – 7i)

Solution. (4 + 5i) + (2 – 7i) = (4 + 2) + (5i – 7i)

= 6 + (5 – 7) i

= 6 – 2i

We note that to add two complex numbers, we add the real parts and the pure imaginary parts

separately.

With the new notation, the same example can be solved as follows :

(4 + 5i) + (2 + 7i) = (4, 5) + (2, –7)

= (4 + 2, 5 – 7)

= (6, –2)

= 6 – 2i

Example 2. Simplify (2 + 3i) + (5 – 7i)

Solution. (2 + 3i) + (5 – 7i) = 2(5 – 7i) + 3i (5 – 7i)

= 10 – 14i + 15i – 21i2

= 10 + i (–14 + 15) –21(–1)

= 10 + 21 + i (15 – 14)

= 31 + i

With the new notation, we have

(2 + 3i) (5 – 7i) = (2, 3) (5, –7)

= [2 × 5 – 3 (–7), 2(–7) + 3 × 5]

= (10 + 21, –14 + 15)

= (31, 1)

= (31 + i)

Note : By the above two examples, it is evident that any one of the two notations can be used.

1.6 Real and Imaginary Parts of a Complex Number

Let z = x + iy be any complex number. Then x is called the real part of z and y is called the

imaginary part of z. We write R(z) = x, I(z) = y. We say that the complex number z is purely real if

I(z) = 0, and that z is purely imaginary if R(z) = 0. Also, the complex number x – iy is called the complex-

3

conjugate of x + iy, the symbol z is usually used to denote the complex-conjugate of the complex

number z. That is, if z = x + iy, then z = x – iy.

If z is the conjugate of z, then z is the conjugate of z that the relation of conjugacy between

complex number is a symmetric one. We therefore, generally say that z and z are conjugate complex

numbers instead of saying that z the complex-conjugate of z.

The following results follow directly from the definition.

(i) The sum of two conjugate complex numbers is purely real.

(ii) The difference to two conjugate complex numbers is purely imaginary.

(iii) The product of two conjugate complex numbers is a non-negative real number.

In order to see the truth of the above three statements, we have z = x + iy so that

z = x – i y i.e. z = (x, y) so that z = (x, – y), then

z z+ = 2x which is purely real [ ( , ) ( , ) (2 , 0)]z z x y x y x + = + − =

z z− = 2iy, which is purely imaginary [ ( , ) ( , ) (0, 2 )]z z x y x y y − = − − =

zz = x2 + y2, which is a non-negative real number.

2 2[ ( , ) ( , ) ( , 0]zz x y x y x y= − = +

We give below some solved examples with the usual notation which would help us at a later state.

Example 3. Simplify a bi

c di

+

+

To divide two complex numbers, multiply both numerator and denominator of the fraction by the

conjugate of the denominator.

a bi

c di

+

+ =

( ) ( )

( ) ( )

a bi c di

c di c di

+ −

+ −

= 2 2

( ) ( )ac bd bc ad i

c d

+ + −

+

= 2 2 2 2

ac bd bc adi

c d c d

+ −+ + +

Example 4. Find the value of

(2 + i) (3 – 2i) (8 + i)

Solution. Here (2 + i) (3 – 2i) = 6 + 3i – 4i – 2i2

= (6 + 2) + (3i – 4i) [ i2 = –1]

= (8 + i)

(8 – i) (8 + i) = 64 – i2 = 64 + 1 = 65

Example 5. Find the value of (1 + i)3.

Solution. (1 + i)3 = 1 + 3i + 3i2 + i3

= 1 + 3i – 3 – i

= – 2 + 2i

4

Example 6. Divide (1 + 3i) by (2 + i).

Solution. 1 3

2

i

i

+

+ =

(1 3 ) (2 )

(2 ) (2 )

i i

i i

+ −

+ −

= (2 3) (–1 6)

4 1

i+ + +

+

= 5 5

15

ii

+= +

Example 7. Express 5 2

3 4

i

i

−

− in the form x + iy

Solution. Here 5 2

3 4

i

i

−

− =

(5 2 ) (3 4 )

(3 4 ) (3 4 )

i i

i i

− +

− +

= (15 8) (20 6)

9 16

i+ −

+

= 23 14 3 4

25 25 25

ii

+ = +

Example 8. Find the roots of the quadratic equation

x2 – 6x + 10 = 0.

Solution. This can be written as (x2 – 6x + 9) + 1 = 0

or (x – 3)2 = –1 = i2

x – 3 = ± i

or x = 3 ± i

Thus 3 + i and 3 – i are the roots of the quadratic equation

x2 – 6x + 10 = 0

It is seen that the roots are conjugate. Verify that these values satisfy the equation. The same

examples can also be solved as

x = 6 36 40 6 4

2 2

− −=

= 6 2

32

ii

=

1.7 Graphical Representation of a Complex Number

The complex number z = x + iy may be representated graphically by a point P whose rectangular co-

ordinates are (x, y) i.e., we associate with each complex number z = x + iy the point of the plane which

has, with reference to a fixed rectangular system, the coordinates x and y and conversely with each point

having the co-ordinates (x, y) we associate the complex number z = x + iy.

From each theorem involving complex numbers we can deduce a definite relationship between the

geometrical points of the Cartesian plane and conversely, the diagram showing points which represent

complex numbers geometrically is called Argand Diagram.

5

It may be noted that all points on the x-axis are of the form (x, 0) and are x + oi = x and so

corresponds to real numbers x. Similarly points on the y-axis correspond to pure imaginary numbers yi.

In addition, the complex number may be repesented by the directed line segment or vector OP.

1.8 Trigonometric form of Complex Numbers

The complex number z = x + iy is represented by the vector OP. This vector and hence the complex

number is described in terms of the length r of the vector and the angle which is vector makes with the

positive direction of the x-axis (real axis) measured positively.

Thus OP = 2 2r x y= + and tan q = y/x

Thus a complex quantity can always be put in the form r (cos + i sin ) where r and are both

real.

Let x + iy = r (cos + i sin )

Equating real and imaginary parts, we get

x = r cos , and y = r sin .

By squaring and adding, we get

x2 + y2 = r2

r = 2 2x y+

6

Here, we take only positive sign of the square root.

Also by dividing y

x =

sintan

cos

r

r

=

or = 1tan ( 0)

yx

x

−

2 2r x y= + is called the modulus or absolute value of the complex number x + iy and = 1tan

y

x

−

is called amplitude (or argument or phase).

There are many values of satisfying the equation tan = y/x, the value of , such that – < is

called the principal value of the amplitude. We shall generally take the principal value of .

In symbols the modulus of a complex number z = x + iy is also denoted by | z | or | x + iy |

i.e., | z | = | x + iy |

= 2 2x y+

That amplitude is denoted by the symbols amp z or arg z.

amp (x + iy) = amp z =

here the value of is so choosen that it is the principal value i.e.,

– <

The form r(cos + i sing ) is called the standard or the polar form of the complex number

x + iy, R, y R.

Example 1. Express 1 + i in the standard form.

Solution. Let 1 + i = r (cos + i sin )

Equating real and imaginary parts, we get

1 = r cos ; 1 = r sin

Then r2 = 2 r = 2

Also tan = 1 i.e., = 4

Here cos is positive and sin is positive, and hence lies in the first quadrant.

= 4

Hence 1 + i = 2 cos sin4 4

i

+

The modulus of the complex number is 2 and the amplitude is .4

Example 2. Express 1 3i− in the polar form.

Solution. Let 1 3i− = r (cos + i sin )

so that 1 = r cos ,

3− = sin

r = 2, tan = 3−

7

Here cos is +ve and sin is –ve.

lies in the fourth quadrant

Also since – <

= 3

−

Hence 1 3i− = 2 cos sin3 3

i

− + −

Example 3. Express sin + i cos in the standard form.

Solution. Let sin + i cos = r (cos + i sin )

so that sin cos

cos sin

r

r

=

=

2 1

tan cot

r =

=

r = 1, tan = cot

= tan2

−

= 2

−

Hence sin + cos = cos sin .2 2

i

− + −

EXERCISE–1

1. Simplify the following :

(i) (2 – 5i) + (–3 + 4i) + (8 – 3i)

(ii) 5

3 2

i

i

+

−

2. Evaluate the following :

(i) (–7 – 2i) (–1 – 5i)

(ii) 2

2

(2 4 )

(3 )

i

i

+

−

3. Solve the following equations :

(i) [(x + 2y), (2x –y – 6) = (3, 2)

(ii) (x – y) + i (x + y) = 2 3

1

i

i

− +

+

4. Express the following complex number in the polar form

(i) 1 3i− +

(ii) 2 12i− −

8

5. Find the trigonometric representation of

(i) sin – i cos

(ii) 1 + cos + i sin

6. Show that | cos + i sin | = 1

7. Express the complex number 5

2

i

i

+

+ in the form (x + iy) where x and y are real. Find its modulus

and amplitude.

9

LESSON 2

DE MOIVRE’S THEOREM

In this lesson, we shall discuss an important theorem which is used in finding the roots of a complex

number, solution of equations, expansion of trigonometric function etc. This theorem is known as De

Moivre’s Theorem.

2.1 De Moivre’s Theorem

Statement :

(i) If n is an integer, positive, or negative, then (cos + i sin ) = cos n + i sin n.

(ii) If n is a rational number, then one of the values of (cos + i sing )n is

Proof : Part (i) case (a) : Let n be a positive integer

By simple multiplication

(cos + i sin ) (cos + i sin ) = (cos cos – sin sin )

+ i (sin cos + cos sin )

= cos ( + ) + i sin ( + )

Again multiplying the above results by (cos + i sin ), we have

(cos + i sin ) (cos + i sin ) (cos + i sin )

= [cos ( + ) + i sin ( + )] (cos + i sin )

= cos ( + + ) + i sin ( + + )

This process can be continued to any number of factors so that

(coa + i sin ) (cos + i sin ) (cos + i sin ) ...... to n factors

= cos ( + + + ....... to n terms)

+ i sin ( + + + ...... to n terms)

In this expression, put = = = ........ = .

So that we have (cos + i sin )n = cos n + i sin n .

Case (b) : Let n be a negative integer. Suppose n = – m where is a positive integer.

Then (cos i sin )n = (cos + i sin )–m

= 1

(cos sin )mi + (by the law of indices)

= 1

(cos sin )m i m + (by case (a))

= (cos sin )

(cos sin )(cos sin )

m i m

m i m m m

−

+ −

= 2 2

(cos sin )cos sin

(cos sin )

m i mm i m

m m

− = −

+

= cos( ) sin ( )m i m− + −

= con n + i sin n

10

Thus (cos + i sin )n = cos n + i sin n for integral values of n, whether positive or negative

Proof of (ii) : If n is (a rotational number, positive or negative), let ( 0)p

n qq

=

We shall take q as a positive integer and p as a positive or a negative integer.

We haver cos sin

q

iq q

+

= cos sinq i q

q q

+

= cos + i sin

i.e., qth power of cos siniq q

+

is cos + i sin

i.e., cos siniq q

+

is one of the qth roots of (cos + i sin )

i.e., cos siniq q

+ is one of the values of

1

(cos sin )qi +

Raising each of these quantities of pth power, we have the results that one of the values of

/(cos sin ) is cos sin

p

p qi iq q

+ +

i.e., cos sinp p

iq q

=

Hence cos n + i sin is one of the values of (cos + i sin )n if n is a rational number positive or

negative.

Corollary 1. cos n – i sin n is one of the values of (cos – i sin )n for all rational values of n.

Illustratives : With the help of the De Moivre’s Theorems, we see that

1. (cos + i sin )4 = cos 4 + i sin 4

2.

7

cos sin2 2

i

+

= 7 7

cos sin2 2

i

+

3. (cos + i sin )–7 = cos (–7) + i sin (–7)

= cos 7 – i sin 7

4. Since (cos + i sin ) (cos – i sin )

= (cos2 + sin2 ) + i (sin cos – sin cos ) = 1 + i0 = 1

1

cos sincos sin

ii

− = +

5. sin + i cos = cos sin2 2

i

− + −

6. (sin + i cos )n = cos sin2 2

n i n

− + −

11

e.g., (sin + i cos )3 =

3

cos sin2 2

i

− + −

= cos 3 sin 3 02 2

i

− + −

= 3 3

cos 3 sin 32 2

i

− + −

= – sin 3 – i cos 3

Remarks : (i) As any complex number x + iy can be put in the form r (cos + i sin ), therefore, De

Moivre’s Theorem may be written as

( )nx iy+ = (cos sin )n nr i +

= (cos sin )nr n i n +

where r is modulus of the complex number and is the amplitude.

(ii) The product and quotient of any two complex numbers can be obtained by using this method.

Let any two complex number be given by

r1 (cos

1 + i sin

1) and r

2 (cos

2 + i sin

2),

then their product is

r1r

2[cos (

1 +

2) + i sin (

1 +

2)]

i.e., the product of two complex numbers is a complex number whose modulus is the product of the

modulli and whose amplitude is the sum of the amplitudes two complex numbers.

The quotient of these two complex numbers, r1(cos

1 + i sin

1) and r

2(cos

2 + i sin

2) is

1 1 1

2 2 2

(cos sin )

(cos sin )

r i

r i

+

+ = 1

1 1 2 32

(cos sin ) (cos sin )r

i ir

+ −

= 11 2 1 2

2

[cos( ) sin ( )]r

ir

− + −

Solved Examples

Example 1. Simplify 4 5

3 7

(cos 2 sin 2 ) (cos sin ).

(cos 6 sin 6 ) (cos 4 sin 4 )

i i

i i

+ +

+ −

Solution. By De Moivre’s Theorem, we have

cos 2 + i sin 2 = (cos + i sin )2

cos – i sin = (cos + i sin )–1

cos 6 + i sin 6 = (cos + i sin )–1

cos 4 – i sin 4 = (cos + i sin )–4

Now, 4 5

3 7

(cos 2 sin 2 ) (cos sin )

(cos 6 sin 6 ) (cos 4 sin 4 )

i i

i i

+ −

+ −

= 8 5

18 –28

(cos sin ) (cos sin )

(cos sin ) (cos sin )

i

i i

− + +

+ +

= (cos + i sin )8 – 5 –18 + 28

12

= (cos + i sin )36 – 23

= cos + i sin )13 = cos 13 + i sin 13.

Example 2. Simplify 3 5

8 2

(cos 2 sin 2 ) (cos sin ).

(cos 3 sin 3 ) (cos 5 sin 5 )

i i

i i

+ −

− +

Solution. By De Moiver’s Theorem, we have

3 5

8 2

(cos 2 sin 2 ) (cos sin )

(cos 3 sin 3 ) (cos 5 sin 5 )

i i

i i

+ −

− + =

6 5

24 –10

(cos sin ) (cos sin )

(cos sin ) (cos sin )

i

i i

−

−

+ +

+ +

= (cos + i sin )6 – 5 + 24 – 10

= (cos + i sin )15

= cos 15 + i sin 15

Example 3. Obtain the quotient of 1 3

1

i

i

− +

+ by using De Moivre’s Theorem.

Solution. 1 3

1

i

i

− +

+ =

1 3 2 22 2 cos sin2 2 3 3

1 12 cos sin2

4 42 2

ii

ii

− + +

=

++

= 2 2

2 cos sin3 4 3 4

i

− + −

= 5 5

2 cos sin .12 12

i

+

Example 4. Obtain ( 3 ) .i +

Solution. Here ( 3 )i + =

43 1

2 2 cos sin2 2 6 6

i

+ = +

= 4 4

16 cos sin16 6

i

+

= 2 2

16 cos sin3 3

i

+

= 1 3

162 2

i − +

= 8 8 3 i− +

Example 5. Prove that

1 sin cos

1 sin cos

ni

i

+ +

+ − = cos sin

2 2

n nn i n

− + −

Solution. Let

1 + sin + i cos = r (cos + i sin ).

13

Then equating real and imaginary part, we have

1 + sin = r cos , cos = r sin

Hence, tan = cos

1 sin

+ =

sin2

1 cos2

−

+ −

= sin

, where1 cos 2

= −

+

= 2

2 sin cos2 2 tan

22 cos

2

=

= 2 4 2

= −

Also 1 + sin – i cos = r(cos – sin )

The given expression

= (cos 1sin ) (cos sin )

(cos sin ) (cos sin )

n n n

n n n

r i

r i i −

+ + =

− +

= (cos + i sin )n (cos + i sin )n

= (cos + i sin )2n

= (cos 2n + i sin 2n)

= cos 2 sin 24 2 4 2

n i n

− + −

= cos sin2 2

n nn i n

− −

Example 6. Prove that :

(a + bi)2m/n + (a – bi)m/n = 2 2 /2 12( ) cos tanm n m b

a bn a

− +

Solution. Here put a + bi = r (cos + i sin )

Then a = r cos and b = r sin

a2 + b2 = r2(cos2 + sin2 ) r2 = a2 + b2

and tan = b

a

Hence r = 2 1and = tan

ba b

a

−+

Also a – bi = r cos – i sin ).

Thus the given expression can be written as

(a + bi)m/n + (a – bi)m/n = / /[ (cos sin )] [ (cos sin )]m n m nr i r i + + −

= / /cos sin cos sinm n m nm m m m

r i r in n n n

+ + −

14

= /2 cosm n mr

n

= 2 2 /2 12( ) cos tanm n m ba b

n a

− +

Example 7. If x = cos + i sin , then show that

1m

mx

x+ = 2 cos m

and 1m

mx

x− = 2 i sin m.

Solution. Here xm = (cos + i sin ) = cos m + i sin m,

and 1mx

= (cos + i sin )–m

= cos m – i sin m.

Hence 1m

mx

x+ = 2 cos m

Also, 1m

mx

x− = 2 i sin m

Example 8. If 1 1

2 cos , show that 2 cos .n

nx x n

x x+ = + =

Solution. Here, it is given that

1

xx

+ = 2 cos

or x2 – 2x cos + 1 = 0

or x2 – 2x cos + (cos2 + sin2 ) = 0

or (x2 – 2x cos + cos2 ) = –sin2

or (x – cos )2 = ( i sin )2

x – cos = ± i sin .

i.e., x = cos ± i sin

Let x = cos + i sin (taking the + ve sign).

Then, 1

x =

1cos sin

cos sini

i= −

+

1n

nx

x+ = (cos + i sin )n + (cos – sin )n

= (cos n + i sin n) + (cos n – i sin n)

= 2 cos n

If x cos – i sin , then 1

cos sin .ix

= + Then also the result can be proved.

Example 9. If sin + sin + sin = 0 and cos + cos + cos = 0. Prove that

cos 3 + cos 3 + cos 3 = 3 cos ( + + )

15

sin 3 + sin 3 + sin 3 = 3 sin ( + + )

Solution. Let x = cos + i sin

y = cos + i sin

z = cos + i sin

Then x + y + z = (cos + cos + cos ) + i (sin + sin + sin )

= 0 + i0 = 0

x3 + y3 + z3 = 3xyz

(cos + i sin )3 + (cos + i sin )3 + (cos + i sin )3

= 3(cos + i sin ) (cos + i sin ) (cos + i sin )

(cos 3 + i sin 3) + (cos 3 + i sin 3) + (cos 3 + i sin 3)

= 3[cos( + + ) + i sin ( + + )]

(cos 3 + cos 3 + cos 3) + i (sin 3 + sin 3 + sin 3)

= 3 cos( + + ) + 3 i sin ( + + )

Equating real and imaginary parts, we get

cos 3 + cos 3 + cos 3 = 3 cos( + + )

and sin 3 + sin 3 + sin 3 = 3 sin ( + + )

Hence the results.

Example 10. Show that for an integer n

1 ( )(1 cos sin ) (1 cos sin ) 2 cos cos

2 2

n n n n ni i +

+ + + − =

Solution. We have (1 + cos + i sin )n + (1 + cos – i sin )n

= 2 22 cos 2 sin cos 2 cos 2 sin cos

2 2 2 2 2 2

n n

i i

+ + −

= 2 cos cos sin 2 cos cos sin2 2 2 2 2 2

n nn n n ni i

+ + −

= 2 cos cos sin 2 cos cos sin2 2 2 2 2 2

n

n n n nn ni i

+ + − + −

= 2 cos cos sin cos sin2 2 2 2 2

n n n n n ni i

+ + − + −

= 2 cos cos sin cos sin2 2 2 2 2

n n n n n ni i

+ + −

= 2 cos 2 cos2 2

n n n

= 12 cos cos

2 2

n n n+

16

Example 11. If sin + sin + sin = 0 = cos + cos + cos . Prove that

(i) cos( + ) + cos( + ) + cos( + ) = 0

(ii) sin( + ) + sin( + ) + sin( + ) = 0

Solution. Let x = cos + i sin , y = cos + i sin , z = cos + i sin

so that x + y + z = (cos + cos + cos ) + i(sin + sin + sin )

= 0 + i0 = 0

Also 1

x =

1 1cos sin , cos sin , cos sini i i

y z − = − = −

1 1 1

x y z+ + = (cos cos cos ) (sin sin sin )i + + − + +

= 0 – i0 = 0

yz + zx + xy = 0

xy + yz + zx = 0

(cos + i sin ) (cos + i sin ) + (cos + i sin ) (cos + i sin )

+ (cos + i sin ) (cos + i sin ) = 0

[cos ( + ) + i sin ( + )] + [cos ( + ) + i sin ( + )]

+ [cos ( + ) + i sin ( + )] = 0

[cos ( + ) + cos ( + )] + cos ( + )] + i [sin ( + ) + sin ( + )]

+ sin ( + )] = 0

Equating real and imaginary parts on both sides, we get

cos ( + ) + cos ( + ) + cos ( + ) = 0

sin ( + ) + sin ( + ) + sin ( + ) = 0

Hence the result.

Example 12. If Z = cos + i sin , show that

2

2

1

1

n

n

Z

Z

−

+ = i tan n, n being an integer.

Solution. 2

2

1

1

n

n

Z

Z

−

+ =

n n

n n

Z Z

Z Z

−

−

−

+ (Dividing numerator and denominator by Zn)

= (cos sin ) (cos sin )

(cos sin ) (cos sin )

n n

n n

i i

i i

−

−

+ − +

+ + +

= (cos sin ) (cos sin )

(cos sin ) (cos sin )

n i n n i n

n i n n i n

+ − −

+ + −

= 2 sin

tan proved2 cos

i ni n

n

=

Example 13. If (cos + i sin ) (cos 2 + i sin 2) ...... (cos n + i sin n) = 1 prove that

= 4

, where is an integer.( 1)

kk

n n

+

17

Solution. We have 1 = cos 0 + i sin 0

= cos 2k + i sin 2 k ......, where k is an integer. ...(1)

Also (cos + i sin ) (cos 2 + i sin 2) ....... (cos n + i sin n) = 1

L.H.S. = cos ( + 2 + ...... + n) + i sin ( + 2 + ...... + n)

= ( 1) ( 1)

cos sin2 2

n n n ni

+ + +

( 1)1 2 3 ......

2

n nn

+ + + + + =

Thus ( 1) ( 1)

cos sin2 2

n n n ni

+ + + =1 = cos 2 k + i sin 2k by (1)

This gives us ( 1)

2

n n + =

42

( 1)

kk

n n

=

+ Hence the result.

Example 14. If (1 + x)n = 20 1 2 ,n n n n n

xC C x C x C x n+ + + + is a positive integer. Prove that

(i) 2

0 4 8

22 2 cos

2 4

n n n n n nC C C − −

+ + + = +

(ii) /2

0 2 4 6 2 cos4

n n n n n nC C C C

− + − + =

(iii) /2

1 3 5 7 2 sin4

n n n n n nC C C C

− + − − =

Solution. We are given that

(1 + x)n = 2 3 40 1 2 3 4

n n n n nC C x C x C x C x+ + + + +

Putting x = 1, –1, i, – i in succession, we get

2n = 0 1 2 3 4n n n n n n

nC C C C C C+ + + + + ...(1)

0 = 0 1 2 3 4– – –n n n n nC C C C C+ + ...(2)

(1 + i)n = 0 1 2 3 4– –n n n n nC i C C C i C− + + ...(3)

= 0 2 4 1 3 5( – ) ( )n n n n n nC C C i C C C+ + − + ...(4)

(1 – i)n = 0 1 2 3 4–n n n n nC i C C i C C− + + ...(5)

Adding (1), (2), (4) and (5) we get

2 (1 ) (1 )n n ni i+ + + − = 0 4 84( .......)n n nC C C+ + + ...(6)

Now (1 + i)n = 2 cos sin4 4

n

i

+

= /22 cos sin

4 4

n n ni

+

...(7)

Similarly, (1– i)n = /22 cos sin

4 4

n n ni

−

...(8)

Adding (7) and (8) and substituting in (6), we get

0 4 84( )n n nC C C+ + + = /22 2.2 cos

4

n n n

+

18

= 1

22 2 cos4

n

n n

+

+

0 4 8n n nC C C+ + + =

2cos

2 2 42 2 Proved

n n

n

−

− +

(ii) From (4) and (7)

0 4 8 1 3 5( ) ( ......)n n n n n nC C C i C C C+ + + + − +

= /22 cos sin4 4

n n ni

+

Equating real and imaginary parts, we get

0 2 4n n nC C C− + = 2n/2 cos n/4

1 3 5n n nC C C− + = / 22 sin / 4n n

EXERCISE 2.1

1. Simplify 9

9 8

(cos sin ) (cos 2 sin 2 )

(cos 2 sin 2 ) (cos sin )

i i

i i

+ −

−

2. Evaluate:

10 10

6

cos sin cos sin15 15 15 15

cos sin3 3

i i

i

+ + −

+

3. Use De’ Moivre’s Theorem to simplify 4

3

(cos sin )

(sin cos )

i

i

+

+

4. Prove that (sin x – i cos x)n = cos sin2 2

n x i n x

− − −

5. Obtain the value of (3 + 4i)3 with the help of De Moivre’s Theorem.

6. If n is a positive integer, show that :

( ) ( ) 13 3 2 cos .6

n nn n

i i − + + − =

7. Prove that :

[cos + cos ) + i (sin + sin )]n + [cos + cos ) – i sin + sin )]n

= 1 ( )2 cos cos

2 2

n n n+ − +

(cos cos (sin sin )

2 cos cos 2 sin cos2 2 2 2

2 cos cos sin2 2 2

i

i

i

+ + + + − + −

= +

− + + = +

Hint :

19

8. Show that

[(cos – cos ) + i (sin – sin )]n + [(cos – cos ) – i(sin – sin )]n

= 12 sin cos2 2

n n n+ − + +

(cos cos (sin sin )

2 sin sin 2 sin cos2 2 2 2

2 sin sin cos2 2 2

2 sin cos sin2 2 2 2 2

i

i

i

i

− + −

+ − − + = + − + + = − +

− + + = + + +

Hint :

EXERCISE 2.2

1. If 2cos = 1

xx

+ , 2 cos =1

yy

+

Show that

1m n

m nx y

x y+ = 2 cos (m + n)

m n

n m

x y

y x+ = 2 cos (m – n)

where m and n are integers. What happens if m and n are rational numbers?

2. If x = cos + i sin , y = cos + i sin and z = cos + i sin such athat x + y + z = 0 then

prove that :

1 1 1

x y z+ + = 0

and 1

xyzxyz

+ = 2 cos ( + + )

3. Let the complex numbers x, y, z be given respectively by cos a + i sin a, cos b + i sin b and cos c

+ i sin c, then prove that for any integers p, q, r.

1p q r

p q rx y z

x y z+ = 2 cos (pa + qb + rc)

4. Let x1 + iy

1, x

2 + iy

2 ......, x

n + iy

n be any n complex number and A + iB be some other complex

number, such that

(x1 + iy

1) (x

2 + iy

2) ...... ( x

n + iy

n ) = A + iB

Then show thats

(i) 1 1 1 11 2

1 2

tan tan ...... tan tann

n

yy y B

x x x A

− − − −+ + =

(ii) 2 2 2 2 2 2 2 21 1 2 2( ) ( ) ( )n nx y x y x y A B+ + + = +

20

5. Expand (1 + i)n in two different ways and find the sum of the two series,

2 4 6 1 3 5(1 ) and )n n n n n nC C C C C C− + − + − +

[Hint : First expand (1 + i)n by Boinomial theorem and then expand it by help of De Moivre’s

Theorem.

6. If sin + sin + sin = cos + cos + cos = 0. Prove that

(i) sin 2 + sin 2 + sin 2 = 0

(ii) cos 2 + cos 2 + cos 2 = 0

(iii) cos2 + cos2 + cos2 = 3/2

21

LESSON 3

APPLICATIONS OF DE MOIVRE’S THEOREM

In this lesson, we shall discuss certain applications and uses of De Moivre’s Theorem. As mentioned

earlier, it helps in finding the roots of a number, solution of some equations and expressions of

trigonometric functions, etc.

We have seen that De Moivre’s Theorem can be stated as

cos sinq i qq q

+ = cos sin

q

iq q

+

, where q is a positive integer.

i.e., cos + i sin = cos sin

q

iq q

+

cos siniq q

+ is a qth root of cos + i sin

i.e., cos siniq q

+ is one of the values of (cos + i sin )1/q

Now the questions arises :

‘What about the other values of (cos + i sin )n for rational values of n ?

We shall now determine all the values of (cos + i sin )1/q, where is a positive integer.

We know that

(cos + i sin ) = cos (2r + ) + i sin (2r + )

(cos + i sin )1/q = [cos (2r + ) + i sin (2r + )]1/q where q is any integer.

Now by De Moiver’s Theorem one of the values of (cos + i sin )1/q is

2 2

cos sinr r

iq q

+ + +

By giving ‘r’ various values, we get different values of (cos + i sin )1/q.

Let us give r in succession the values 0, 1, 2, ...... q – 1 and we see that each of the following

quantities

cos siniq q

+

2 2

cos siniq q

+ + +

4 4

cos siniq q

+ + +

....... ....... ....... ........ ......

2( 1) 2( 1)

cos sinq q

iq q

− + − + +

22

is equal to one of the values of (cos + i sin )1/q. We note here that if we give to ‘r’ integral values

greater than (q – 1) viz., the values q, (q + 1), (q + 2), ...... etc. then we do not obtain any new values of

(cos + i sin )1/q.

For example, for r = q we obtain the value

2 2

cos sinq q

iq q

+ + +

i.e., cos 2 sin 2iq q

+ + +

i.e., cos sin .iq q

+

which is the same values as obtained by putting r = 0.

Thus the greatest values we need to assign to r is q – 1 for the values q, q + 1, q + 2 ...... will be

found to give the same result as the values, 0, 1, 2, ...... q – 1, for ‘r’.

Also no two of the quantities obtain by giving ‘r’, the values 0, 1, 2, ...... q – 1 will be the same for

all the angles involved there in differ from one another by less than 2 and no two angles differing by less

than 2 have their consines the same and also sines the same.

Thus the expression cos (2 ) (2 )

sin ,r r

iq q

+ ++ where r = 0, 1, 2, ...... (q – 1) gives q and only

q different values of (cos + i sin )1/q. We may extend the above result to (cos +

i sin )1/q where p and q are integers, q being taken as positive.

Thus (cos + i sin )1/q has q and only q different values and the these are given by

(2 ) (2 )cos , 0,1, 2, ...... ( 1).

r p r pi ain where r q

q q

+ + + = −

Then, the values are

cos sin , when 0p p

i rq q

+ =

(2 ) (2 )

cos sin , where 1p p

i rq q

+ + + =

(4 ) (4 )

cos sin , where 2p p

i rq q

+ + + =

.......................................................................

[2 ( 1) ] [2 ( 1) ]

cos sin , when 1.q p q p

i r qq q

− + − + + = −

Note. We may note here that the ‘q’ distinct values of (cos + i sin )p/q can be arranged in a

geometrical progression, for if

= 2 2

cos sin and cos sin .p p p p

i a iq q q q

+ = +

then the q values of (cos + i sin )p/q can be arranged as a, a, a2, ...... aq – 1 which is a G.P. The

student should note that

23

(2 ) (2 ) 2 2cos sin cos sin cos sin

rp r p r p p p p

i i iq q q q q q

+ + + = + +

3.1 Extraction of Roots of Complex Quantity

We know that

x + iy = r (cos + i sin ),

where r = 2 2 1and tan .

yx y

x

−+ =

Hence (x + iy)1/q = r1/q[cos + i sin ]1/q

= 1/ 1 /[cos(2 ) sin (2 )]q qr n i n + + +

= 1/ (2 ) (2 )cos sinq n n

r iq q

+ + +

when ‘n’ is given in succession the values, 0, 1, 2, ......, q – 1.

This gives the required ‘q’th roots of x + iy.

It may be noted that even if we write x + iy = [cos (2 ) sin (2 )],r k i k k + + + being an integer,

the same values are obtained.

Example 1. Find the values of (1 + i)1/6.

Solution. We know that cos4

=

1sin .

42

We can write (1 + i) = 1 1

22 2

i

+

=

1

22 cos sin4 4

i

+

=

1

22 cos 2 sin 24 4

n i n

+ + +

(1 + i)1/6 =

1

122 /4 2 /4

2 cos sin6 6

n ni

+ + +

when n is given the six values 0, 1, 2, 3, 4, and 5.

The six values of (1 + i)1/6, are therefore

1 1

12 129 9

2 cos sin , 2 cos sin ,24 24 24 24

i i

+ +

1 1

12 1217 17 25 25

2 cos sin , 2 cos sin ,24 24 24 24

i i

+ +

1 1

12 1233 33 41 41

2 cos sin , 2 cos sin24 24 24 24

i i

+ +

24

It may be observed that the 4th value, viz.,

1

1225 25

2 cos sin24 24

i

+

=

1

122 cos sin24 24

i

+ + +

=

1

122 cos sin ,24 24

i

− +

since cos ( + ) = – cos and sin ( + ) = – sin . Similarly the 5th and 6th values can simplified. The

six value of (1 + i)1/6 can then be written as

1

122 cos sin ,24 24

i

+

1

129 9

2 cos sin ,24 24

i

+

1

1217 17

2 cos sin ,24 24

i

+

or simply as

1

122 cos sin ,24 24

r ri

+

where r = 1, 9 and 17.

Example 2. Find all the values of

3/41 3

2 2i

+

and show that the continued product of all the

values is 1.

Solution. Let 1 3

2 2i+ = (cos sin ),r i +

then 1

2 = r cos

and 3

2 = r sin

so that r2 = 1 3

14 4

+ =

and tan = 3 tan3

=

or r = 1 and 3

=

Now

3/41 3

2 2i

+

=

13 4

cos sin3 3

i

+

25

= (cos + i sin )1/4

= [cos (2k + ) + i sin (2k + )]1/4

= cos (2 1) sin (2 1) , 0,1, 2, 34 4

k i k k

+ + + =

Hence the required four values are

when k = 0 cos sin ,4 4

i

+

k = 1 3 3

cos sin ,4 4

i

+

k = 2 5 5

cos sin ,4 4

i

+

k = 3 7 7

cos sin ,4 4

i

+

And the continued product of four values

= 3 3 5 7 7

cos sin cos sin cos sin cos sin4 4 4 4 4 4 4 4

i i i i

+ + + +

= 3 5 7 3 5 7

cos sin4 4 4 4 4 4 4 4

i

+ + + + + + +

= cos 4 + i sin 4

= 1 + i.0 [ cos 4 = 1, sin 4 = 0]

= 1.

Example 3. Find the seven 7th roots of unity and prove that the sum of their nth powers always

vanishes unless n be a multiple of 7, n being integer and that the sum is 7 when n is multiple of 7.

Solution. We have to find the 7 values of (1)1/7,

We know that cos 0 = 1 and sin 0 = 0

We can write 1 = cos 0 + i sin 0

= cos 2n + i sin 2n

(1)1/7 = 2 2

cos sin7 7

n ni

+

where n = 0, 1, 2, 3, 4, 5, 6.

The 7 roots are, therefore

(cos 0 + i sin 0) i.e., 1,

2 2

cos sin7 7

i

+ = (say)

224 4 2 2

cos sin . ., cos sin7 7 7 7

i i e i

+ + =

and similarly for the other roots.

Thus the seven 7th roots of unity are

1, , 2, 3, 4, 5, 6

26

where = 2 2

cos sin .7 7

i

+

The sum of the nth power of the roots

= (1)n +s ()n + (2)n + (3)n + (4)n + (5)n + (6)n

= 1 + n + 2n + 3n + 4n + 5n +6n

(which is geometric progression)

= 7 71 ( ) 1 ( ) 1 (1 ) 1 1

01 1 1 1

n n n

n n n n

− − − −= = = =

− − − −

Since 7 = 1

Also n 1, since n is not a multiple of 7.

If ‘n’ is a multiple of 7, i.e., if n = 7 m say where m is an integer, then

n =

72 2 2 2

cos sin cos sin7 7 7 7

n m

i i

+ = +

= cos 2m + i sin 2m

= 1.

Hence when n is a multiple of 7, the sum of the nth power of the 7th roots of unity is

= 1 + 1 + 1 +1 +1 +1 + 1 = 7

Example 4. Prove that n nx iy x iy+ + − has ‘n’ real values.

Solution. Let x = r cos and y = r sin . Then

x + iy = r[cos (2k + ) + i sin (2k + )],

where k is an integer.

(x + iy)1/n + (x + iy)1/n

= 1/ 1/ 1/{[cos(2 ) sin(2 )] [cos(2 ) sin(2 )] }n n nr k i k k i k + + + + + − +

= 1/ 2

2 cosn kr

n

+ , where k = 0, 1, 2, ...... (n – 1)

Here r2 = 2 2 , and tan .

yx y

x+ =

Hence (x + iy)1/n + (x – iy)1/n =

1

2 2 1/ 22 tan

2( ) cosn

yk

xx yn

− +

+

where k = 0, 1, 2, ...... (n – 1).

This is real and has n distinct values.

3.2 Solution of Equations

De Moivre’s Theorem can be used to obtain solutions of certain types of equations. This will be

illustrated by a few solved examples. Conversely De Moivre’s Theorem can be used to find the equation

whose roots are given as trigonometical functions.

27

Example 5. Solve the equation zn = 1, where n is a positive integer.

Solution. Here we write

zn = 1 = cos (2k + 0) + i sin (2k + 0)

z = 1/ 2 2(1) cos sinn k k

in n

= +

where k = 0, 1, 2, ..... (n – 1)

Hence the nth roots of unity are

(cos 0 + i sin 0), i.e., 1, 2 2

cos sinin n

+

4 4

cos sinin n

+

=

22 2

cos sinin n

+

...............................................................

...............................................................

12( 1) 2( 1) 2 2

cos sin cos sin

nn n

i in n n n

− − −

+ = +

It may be noted that if we write = 2 2

cos sin ,in n

+

then the nth root of ‘1’ can be written as 1, , 2, ...... n – 1. Thus we get an interesting result that the ‘n’

nth roots of unity form a G.P. Also it is seen that each of these roots can be expressed as a power of

another.

Exmple 6. Solve the equation z9 – z5 + z4 – 1 = 0.

Solution. Here (z9 – z5 + z4 – 1) = 0

z5 (z4 – 1) + (z4 – 1) = 0

i.e., (z5 – 1) (z4 – 1) = 0

Either (i) z4 = 1 or (ii) z5 = – 1

since 1 = cos 2 n + i sin 2 n

and – 1 = cos + i sin

= cos (2n + ) + i sin (2n + )

= cos (2n + 1) + i sin (2n + 1) .

From (i), we have

(cos 2n + i sin 2n)1/4 = 2 2

cos sin4 4

n ni

+

z = cos sin , where 0,1, 2, 3,2 2

n ni n

+ =

and form (ii), we have

z = (–1)1/5 = [cos (2n + 1) p + i sin (2n + 1) ]1/5

= (2 1) (2 1)

cos sin , when 0,1, 2, 3, 4.5 5

n ni n

+ + + =

These (4 + 5) = 9 values give the complete solution of the given equations.

28

Example 7. Solve the expression z5 + 1 = 0.

Solution. z5 = –1 = [cos (2r + 1) + i sin (2r + 1)]

where r is an integer.

z = (2 1) (2 1)

cos sin , where 0,1, 2, 3, 4.5 5

r ri r

+ + + =

When r = 0, cos sin .5 5

z i

= +

When r = 1, 3 3

cos sin5 5

z i

= +

And for r = 2, z = cos + i sin = –1.

For r = 3, the value of z is

7 7

cos sin5 5

i

+

= 3 3

cos 2 sin 25 5

i

− + −

= 3 3

cos sin5 5

i

− + −

= 3 3

cos sin5 5

i

−

For r = 4, the value of z is

= 9 9

cos sin5 5

i

+

= cos 2 sin 25 5

i

− + −

= cos sin5 5

i

− + −

= cos sin5 5

i

−

Thus the 5 roots of the equation z5 + 1 = 0 are

3 3

–1, cos sin , cos sin .5 5 5 5

i i

Example 8. Find the roots of the equation z6 + z3 + 1 = 0

Solution. On solving the equation a quadratic in z3, we find

z3 = 1 3 1 3

2 2 2i

− = −

= 2

cos sin3 3

i

= 2 2

cos 2 sin 23 3

r i r

+ +

29

= 2

cos (3 1) sin (3 1)3 3

r i r

+ +

z =

1

32 (3 1) (3 1)cos sin

3 3

r ri

+ +

= 2 (3 1) (3 1)

cos sin , 0,1, 2.9 9

r ri r

+ + =

On giving r the values, 0, 1, 2, the six roots of the given equation can be written as

2 2 8 8 14 14

cos sin , cos sin , cos sin9 9 9 9 9 9

i i i

+

Now 14 14 4 4

cos sin cos 2 sin9 9 9 9

i i

= − −

= 4 4 4 4

cos sin cos sin9 9 9

i ip

− + − =

The roots are 2 2 4 4 8 8

cos sin , cos sin , cos sin9 9 9 9 9 9

i i i

Example 9. Solve the equation z5 + z4 + z3 + z2 + z + 1 = 0 multiply through by z – 1 we get z6 – 1 =

0.

Solution. z5 + z4 + z3 + z2 + z + 1 = 0

Now z6 – 1 = 0

z6 = 1 = cos 2k + i sin 2k, where k is an integer

z = cos 2 2

sin6 6

k ki

+

= cos sin , where 0,1, 2, 3, 4, 5.3 3

k ki k

+ =

z = 2 2 2

cos 0 sin 0, cos sin ; cos sin sin ; cos sin3 3 3 3 3

i i i i i

+ + + + +

4 4 5 5

cos sin ; cos sin ;3 3 3 3

i i

+ +

= 1 3 1 3 1 3 1 3

1, , , 1, ,2 2 2 2 2 2 2 2

i i i i+ − + − − − −

= 1 3

1,2 2

i +

The root 1 corresponds to the factor (z – 1)

The remaining roots 1 3

–1,2 2

i are the roots of the given equation

30

EXERCISE

1. Find all the values of

(i) (1 + i)1/3

(ii) 1 4(–1 3)i+

(iii) (–1)1/6

(iv) (1 + i)2/3

2. Solve the equation:

(i) z7 = 1

(ii) z7 + z = 0

3. Find the nth roots of unity and show that from a series in G.P. whose sum is zero.

Also prove that the sum of their pth power always vanishes unless p be a multiple of n, p being

an integer and that then the sum is n.

4. Solve the equation : z7 + z4 + z3 + 1 = 0.

[Hint : z7 + z4 + z3 + 1 = (z4 + 1) (z3 + 1) = 0]

5. Use De Moivre’s Theorem to solve : z10 + 11z5 + 1 = 0.

[Hint : Put z5 = y]

6. Solve the equation : z4 – z3 + z2 – z + 1 = 0.

[Hint : Multiply the equation by (z + 1)]

3.3 Expansion of sin n and cos n

De Moivre’s Theorem may be applied to express trigonometric functions of multiple angles in terms

of the trigonometric function of the angles. The theorem can be employed to express powers of sines and

cosines of angle in terms of trigonometric functions of multiple angles.

First we shall learn to express cos n, sin n, tan n in terms of cos , sin , tan , etc.

For example, by De Moivre’s Theorem

cos 2 + i sin 2 = (cos + i sin )2 = cos2 + 2 i sin cos + i2 sin2

= (cos2 – sin2 ) + 2i cos sin )

Equating the real and imaginary parts, we get

cos 2 = cos2 – sin2 ,

sin 2 = 2 cos sin

Generally

(cos n + i sin n) = (cos + i sin )n, n being a positive integer.

By the Binomial Theorem for a positive integral index, we have

(cos + i sin )n = 1 2 2( 1)

cos cos ( sin ) cos ( sin )2

n n nn nn i i− −−

+ +

3 3( 1) ( 2)

cos ( sin ) ...... ( sin )3

n nn ni i−+ −

+ +

But i2 = –1, i4 = 1, i3 = – 1 etc.

Hence we can write

31

cos n + i sin n = 2( 1)cos cos sin

2

n nn n − − −

4 4( 1) ( 2) ( 3)cos sin ......

4

nn n n n − − − −+ +

1 3 3( 1)( 2)

cos sin cos sin3

n nn n ni n − − − −

+ − +

By equating the real and imaginary parts, we get

cos n = 2 2( 1)cos cos sin

2

n nn n −− =

4 4( –1) ( 2) ( 3)

cos sin4

nn n n n −− −+ +

sin n = 1 3 3( 1)( 2)cos sin cos sin

3

n nn n nn − −− −

− +

By using the binomial coefficients, we can also express these results as follows

cos n = 2 2 4 42 4cos cos sin cos sin ......n n n n nC C− − − + +

= 2 42 4cos [1 tan tan ......]n n nC C − + + ...(i)

sin n = 1 3 31 3cos sin cos sin ......n n n nC C− − + +

= 31 3cos [ tan tan ......]n nn C C − + ...(ii)

Divinding (ii) by (i), we have

tan n = 3 5

1 3 5

2 42 4

tan tan tan

1 tan tan

n n n

n n

C C C

C C q

− + −

− + −

We have thus obtained the values of cos n, sin n, tan n in terms of cos , sin , tan .

Example 10. Obtain the values of cos 5, sin 5, and tan 5.

Solution. We can either apply the formulae or proceed as below:

cos 5 + i sin 5 = (cos + i sin )5

= cos5 + 5 cos4 (i sin ) + 10 cos3 (i2 sin2 )

+ 10 cos2 (i3 sin3 ) + 5 cos (i4 sin4 )

+ i5 sin5

Thus equating real and imaginary parts,

cos 5 = cos5 – 10 cos3 sin2 + 5 cos sin4

= cos5 – 10 cos3 (1 cos2 ) + 5 cos (1 – cos2 )2

= 16 cos5 – 20 cos3 + 5 cos

(Note that cos 5 has been expressed in powers of cos only).

Also tan 5 = 4 2 3 5

3 2 4

sin 5 5 cos sin 10 cos sin sin

cos 5 cos 5 10 cos sin 5 cos sin

− + =

− + +

32

= 3 5

2 5

5 tan 10 tan tan

1 10 tan 5 tan

− +

− +

(Note that tan 5 has been expressed in powers of cos only.

Note. We can apply a similar method to obtain

sin ( + + + ......), cos ( + + ) + ......), tan ( + + + .......)

and thus obtain expression for the sine, cosine and tangent of the sum of any numbers of angles in terms

of trigonometric ratio of individual angles.

We have

cos ( + + + ......) + i sin ( + + + ......)

= (cos + i sin ) (cos + i sin ) (cos + i sin ) .......

Since

cos + i sin = cos (1 + tan )

cos + i sin = cos (1 + i tan )

cos + i sin = cos (1 + i tan ) etc.

We have

cos ( + + + ......) + i sin ( + + + ......)

= cos cos cos cos ...... [(1 + i tan ) (1 + i tan )

(1 + i tan ) ......

= cos cos cos ...... [1 + i(tan + tan + ......)

+ i2(tan tan + tan tan + ......]

+ i3 (tan tan tan + ...... ) + ......]

= cos cos cos ...... [1 + iS1 – S

2 – iS

3 + ......]

where S1 = tan + tan cos ......

= tan ,

S2 = tan tan + tan tan + ......

= tan tan

S3 = tan tan tan + .....

= tan tan tan ,

i.e., S1 indicates the sum of the tangents of the angles, S

2 indicates the sum of the products of the

tangents of the angles taken two at a time and S3 indicates the sum of the products taken three at a time

and so on.

Equating real and imaginary parts, we have

cos ( + + ......) = cos cos cos ...... [1 – S2 + S

4 – S

6 + ......]

sin ( + + ......) = cos cos cos ...... [S1 – S

3 + S

5 ......]

Also, by division, we get

tan ( + + ) = 1 3 5

2 4 6

......

1 ......

S S S

S S S

− +

− + −

33

3.4 Expansion of sin and cos in Powers of

We know that for a positive integer n and an angle ,

cos n = 2 2( 1)

cos cos sin2

n nn n −− − −

4 4 4( 1) ( 2) ( 3)cos sin sin ......

4

nn n n n −− − −+

Now put n = , then n =

and

cos =

22( ) sin

cos cos2

n n − − −

42( ) ( 2 ) ( 3 ) sin

cos4

− − − +

Now if is mode indefinitely small, remaining constant and consequently n becoming indefinitely

large, cos and sin

both tend to unity, and hence in the limit when a → 0 we have

cos = 2 4 7

1 ......2 4 6

− + − +

Similarly by writing the value of sin n and proceeding as before, we can obtain.

sin = 2 4 7

......3 5 7

− + − +

The method we have applied here is only an indication to get the values of cos and sin and does

not give a rigorous proof.

We shall now obtain a series for tan in terms of by using the series for cos and sin .

Thus tan = sin

cos

=

3 5

2 4

3 5

12 4

− + −

− + −

=

13 5 2 4

16 120 2 24

− − + − − − −

Expanding the second bracket by the Binomial theorem, we have

tan = 3 5 2 4 2 2

16 120 2 24 2 24

− + − + − + − +

= 3 5 2 4 4

16 120 2 24 4

q − + − + − +

34

= 3 5 2

451

6 120 2 24q

− + − + +

= 3 3 5 5

55

2 6 24 12 120

+ − + − +

= 3

52

3 15

+ + +

Thus tan = 3

52

3 15

+ + +

If powers of more than five are neglected, i.e., for small values of , we have

sin = 3

,6

−

cos = 2 4

1 ,2 24

− +

tan = 3

,3

+

and if power more than 2 are to be neglected then

sin = , cos = 2

1 , tan2

− =

It is understood that the angle is expressed in radians.

Example 11. Find the value of when

cos = 9

100

Solution. Here cos is nearly 1 and is small, Hence we can take

cos = 12

−

i.e., 2

=

99 11 cos 1

100 100− = − =

= 2 1.41

.141radians10 10

= =

or = 180

0.141 degrees

= 0.8 degrees approx.

Example 12. Find the value of sin 3º correct to theree places of decimals.

Solution. We have 3º = 3

radians180

Now 3

sin180

=

313 1 3

...180 3 180

− +

35

= 3

nearly =180 60

sin 3º = 3.14159

0.0523 approx.60 60

= =

3.5 Expression of the Products of the Form cosm sinn Terms of sines or cosines of

Multiples of .

De Moivre’s Theorem can be employed to express a product of the form cosm sinm where m and

n are positive integers in terms of sines or consines of multiples or .

Let z = cos + i sin

then 1

z =

1cos sin

cos sini

i= −

+

Hence 1

zz

+ = 2 cos

1

zz

− = 2 i sin

Also from De Moivre’s Theorem, if r is an integer.

(i) zr = cos r = i sin r

(ii) 1

cos sinr

r i rz

= −

and, therefore, by adding and subtraction (i) and (ii), we get

2 cos r = 1r

rz

z+

and 2i sin r = 1r

rz

z−

We know that

(2 cos )m (2i sin )n = 1 1

m m

z zz z

+ −

The right hand side of this equation is expanded of z and terms with indices equal but opposite in

sign are grouped together. The resulting sum is made up, if n is even, of term of the form 1r

rz

z+ and if n

is odd, of term of the form 1

.r

rz

z−

Also we can express these terms 1 1

andr r

r rz z

z z

+ −

as 2 cos r and 2i sin r respectively.

The following examples will illustrate the method.

Example 13. Express sin6 in terms of cosines of multiples of .

Solution. Let z = cos + i sin , so that

36

1

zz

+ = 1

2 cos , zz

− = 2 i sin ,

1r

rz

z+ =

12 cos , 2 sinr

rr z i r

z − =

r being any positive integer.

Now (2i sin )6 =

66 4 6 2 6 0 6 2 6 4 6 4

1 2 3 4 5 6

1z z C z C z C z C z C z C z

z

− − − − = − + + + − +

64 i6 sin6 = 6 4 2 2 4 66 15 20 15 6z z z z z z− − −− + − + − +

sin6 = 6 4 2

6 4 2

1 1 1 16 15 20

64z z z

z z z

− + − + + + −

= 1

[2 cos 6 6.2 cos 4 15.2 cos 2 2 ]64

− − + −

= 1

[cos 6 6 cos 4 15 cos 2 20]32

− − + −

Example 14. Express sin4 cos2 in cosines of multiples of .

Solution. Let z = 1

cos sin then cos sin .i iz

+ = −

so that 1

zz

+ = 2 cos and 1

2 sinz iz

− =

Also 1r

rz

z+ =

12 cos and 2 sinr

rr z i r

z − =

2 (i sin )4 (2 cos )2 =

4 21 1

z zz z

− +

=

2 2 21 1 1

z z zz z z

− − +

=

2 221 1

z zz z

− −

= 2 4

2 4

1 12 2z z

z z

− + − +

=

24 2

6 4 2

1 1 12 4z z z

z z z

+ − + + + +

i.e., 64 sin4 cos2 = 2 cos 6 – 4 cos 4 + 2 cos 2 + 4.

Hence sin4 cos2 = 1

(cos 6 2 cos 4 cos 2 2)32

− + +

37

EXERCISE

1. Express sin 5 in terms of sin only.

2. Express cos 6 in terms of cos only.

3. Express tan 7 in terms of tan only.

4. Find the value of sin 1º approximately.

5. Find the value of , where sin 1013

.1014

=

6. Show that 16 sin5 = sin 5 – 5 sin 3 + 10 sin .

7. Express sin7 cos2 in terms of sines of multiples of .

8. Express cos6 in terms of cosines of multiples of .

38

LESSON 4

APPLICATIONS OF DE MOIVRE’S THEOREM

In this lesson, we shall study the application of De Moivre’s theorem to find the sum of certain types

of trigonometric series. The series may be finite or infinite. The general method is known as the C + iS

method. If we are given as series of cosines such as C = cos + cos 2 + ... cos n, we consider a similar

series of sines i.e. S = sin + sin 2 + ... sin n.

The new series C + iS so formed can be summed up by different methods. Equating real and

imaginary parts we obtain the values of C and S.

The following formulae are mainly used in the summation of series.

(i) 2 1 11 , 1

1

nn x

x x x xx

− −+ + + =

−

(ii) 21 2(1 ) 1n n n n

nx C x nC x C x+ = + + +

(iii) Euler’s Formula: If z is complex variable, then the complex exponential function ez is defined

by 2 3

12! 3!

z z ze z= + + + +

Replacing z by i we have

ei = 2 3 4 5( ) ( ) ( ) ( )

12 3 4 5

i i i ii

+ + + + + +

= 2 3 4 5

1 ...2 3 4 5

i i i

+ − − + + +

= 2 4 3

1 ...2 4 3

i

− + + − +

= cos + i sin

Similarly e–i = cos – i sin

ei + e–i = 2 cos

and ei – e–i = 2 i sin

Also (ei)n = eni = cos n + i sin n

where n is an integer.

We shall illustrate the method C + iS in the following solved examples.

Example 1. Find the sum to n terms of the series.

1 + x cos + x2 cos 2 + ... + xn – 1 cos (n – 1)

Let C = 1 + x cos + x2 cos 2 + ... + xn –1 sin (n – 1)

S = x sin + x2 sin 2 + ... + xn – 1 sin (n – 1)

C + iS = 1 + x(cos + i sin ) + x2(cos 2 + i sin 2)

+ ... + xn – 1 [cos(n – 1) + i sin (n – 1)] + i sin (n – 1) )]

= 1 + xei + x2e2i + ... + xn – 1 e(n – 1)i

39

= 1

,1

n ni

i

x e

xe

−

− summing the GP to n terms

= (1 ) (1 )

(1 ) ( )

n ni i

i i

x e xe

xe i xe

−

−

− −

− −

[Multiply the numerator and denominator by 1 – xe–i]

= 1 ( 1)

2

1

1 ( )

n ni i n n i

i i

x e xe x e

x e e x

− + −

− − +

− + +

= 1

2

1 [cos sin ] (cos sin ) [cos ( 1) sin ( 1) ]

1 2 cos

n nx n i n x i x n i n

x x

+− + − − + − + −

− +

(Using Euler’s Formulae)

Equating the real part we get

C = 1

2

1 cos cos cos( 1)

1 2 cos

n nx n x x n

x x

+− − + −

− +

Example 2. Find the sum to n term of the series

sin 2 + sin 4 + sin 6 + ...

Let C = cos 2 + cos 4 + cos 6 + ... + cos 2n

S = sin 2 + sin 4 + sin 6 + ... + sin 2n

C + iS = e2i + e4i + ... e2ni

= 2 2 2( 1)[1 ]i i n ie e e − + +

= 2 1[1 ]nx x x x −+ + + where x = e2i

C + iS = (1 )

1

nx x

x

−

−

= 2 2

2

[1 ]

1

i ni

i

e e

e

−

−

= 2 2 2

2 2

(1 ) (1 )

(1 ) (1 )

i ni i

i i

e e e

e e

−

−

− −

− −

(Multiplying numerator and denominator by the conjugate of the denominator)

= 2 2 2 2( 1)

2 2

[1

1 ( ) 1

i ni i n i

i i

e e e e

e e

− −

−

− − +

− + +

= 2 2( 1) 21

1 2 cos 2 1

i n i nie e e + − − +

− +

= (cos 2 sin 2 ) [cos 2( 1) sin 2 ( 1) ] 1 [cos 2 sin 2 ]

2(1 cos 2 )

i n i n n i n + − + + + − + +

−

Equating the imaginary parts on both sides, we get

40

S = sin 2 sin 2( 1) sin 2

2(1 cos 2 )

n n − + +

−

= 2

2 sin cos [sin 2( 1) sin 2 )]

4 sin

n n − + −

= 2

2 sin cos 2 cos (2 1) sin

4 sin

n − +

= 2

2 sin [cos cos (2 1) sin ]

4 sin

n − +

= ( )2 sin 1 sin

2 sin

n n+

= sin ( 1) sin

sin

n n+

where sin 0 i.e., n

Example 3. Find the sum of the series

1 2cos cos ( ) cos ( 2 ) ... cos ( )n n nnC C C n + + + + + +

Let C = 1 2cos cos ( ) cos ( 2 ) ... cos ( )n n nnC C C n + + + + + +

and S = 1sin sin ( ) ... sin ( )n nnC C n + + + + +

Here C + iS = ( ) ( 2 ) ( )1 2 ...i n i n i n i n

ne C e C e C e + + + + + + +

= 21 2[1 ...i n i n i n ni

ne C e C e C e + + + +

= [1 ]i i ne e +

= [1 cos sin ]i ne i + +

= 22 cos 2 sin cos

2 2 2

nie i

+

= 2 cos cos2 2 2

n nie i

+

= 2 cos [cos sin ] cos sin2 2 2

nn n

i i

+ +

= 2 cos cos sin2 2 2

nn n

i q

+ + +

Equating the real part on both sides, we get

C = 2 cos cos2 2

nn

+

Sometimes the sum to n terms of a trigonometric series can be obtained by using simple

trigonometric formulae viz.

2 sin A sin B = cos (A – B) – cos (A + B)

2 cos A sin B = sin (A – B) – sin (A + B)

41

These formulae can be useful when the nth term of a series can be expressed at the difference of two

consines of sines. The following example will illustrate this method.

Example 4. Find the sum to n terms of the series.

sin + sin 3 + sin 5 + ...

Let S = sin + sin 3 + sin 5 + ... + sin (2n – 1)

Multiplying both sides by 2 sin ( 0) we get,

2 sin · S = 2 sin2 + 2 sin · sin 3 + 2 sin · sin 5 + ... + 2 sin sin (2n – 1)

= [1 – cos 2] + [cos 2 – cos 4] + [cos 4 – cos 6] ...

+ [cos 2(n –1) – cos 2 n)

= 1 – cos2 n

= 2 sin2 n

S = 22 sinsin cosec

2 sin

nn

=

provided sin 0 i.e., n

Example 5. Find the sum of the infinite series.

sin 3 sin 5

sin3 5

− + +

Let S = sin 3 sin 5

sin3 5

− + +

C = cos 3 cos 5

cos3 5

− + +

and we have

C + iS = 3 5

3 5

i ii e e

e

− + −

= 3 5

3 5

x xx − + − where x = ei

= sin x

= sin [cos + i sin ]

= sin (cos ) · cos (i sin ) + cos(cos ) · sin (i sin )

= sin (cos ) · cos h (sin ) + i cos(cos ) · sin h (sin )

Equating imaginary parts on both sides, we get

S = cos (cos ) · sin h (sin ) [ cos i = cos h, sin iq = i sin h]

from the definition of Hyperbolic functions)

Example 6. Sum of the series

1 1.3 1.3.51 cos cos 2 cos 3

2 2.4 2.4.6− + − +

Let C = 1 1.3

1 cos cos 22 2.4

− + +

42

S = 1 1.3

sin sin 22 2.4

− + +

Here C + iS = 2 31 1.3 1.3.5

1 ...2 2.4 2.4.6

i i ie e e − + − +

=

1

2(1 )ie−

+

=

1

2[1 cos sin ]i−

+ +

=

1

222 cos 2 sin cos2 2 2

i

−

+

=

1

22 cos cos sin

2 2 2i

−

+

=

1

22 cos cos sin

2 4 1i

−

−

Equating real parts on both sides we get,

C =

cos4

2 cos2

Example 7. Find the sum to infinitely of the series

2

1 1sin sin 2 sin 3 ...

2 2 + + +

Let C = 2

1 1cos cos 2 cos 3 ...

2 2 + + +

S = 2

1 1sin sin 2 sin 3 ...

2 2 + + +

C + iS = 2

2

1 1...

2 2

i i ie e e + + +

= 2

2

1 11 ...

2 2

i i ie e e + + +

= 1

11

2

i

i

e

e

−

[Summing the G.P. to ]

=

11

2

1 11 1

2 2

i i

i i

e e

e e

−

− −

−

− −

43

=

1

21 1

1 ( )2 4

i

i i

e

e e

−

−

− + +

=

1cos sin

2

cos4

i + −

−

= 4(cos sin ) 2

5 4 cos

i + −

−

Equating the imaginary parts on both sides, we get

S = 4 sin

5 4 cos

−

Example 8. Find the sum to of the sides

2 2 3cos cos cos cos 2 cos cos 3 + + +

Let C = 2 2 3cos cos cos cos 2 cos cos 3 + + +

S = 2 3cos sin cos sin 2 cos sin 3 + + +

Here C + iS = 2 2 3 3cos cos cos ...i i ie e e + + +

= 2cos [1 cos (cos ) ]i i ie e e + + +

= 1

cos1 cos

i

ie

e

−

(Summing the G.P. to )

= ( ) ( )

cos [1 cos ]

1 cos 1 cos

i i

i i

e e

e e

−

−

− −

= ( )

2

2

cos cos

1 cos cos

i

i i

e

e e

−

−

− + +

= 2

2 2

cos [cos sin ] cos

1 2 cos cos

i + −

− +

= 2 2

2

cos sin cos cos

1 cos

i + −

−

= 2

sin cos

sin

i

It follows that C = 0

provided sin2 0

i.e., k where k is an integer

i.e., The sum of the given infinite series is zero, provided is not a multiple for .

44

EXERCISES

Find the sum to n terms of the following series :

1. cos + cos 2 + cos 3 + ...

2. sin · cos 2 + sin 2 · cos 3 + sin 3 · cos 4 + ...

3. sin + sin( + ) + sin ( + 2) + ...

4. 2 cos 2 + 22 cos 4 + 23 cos 6 + ...

Find the sum to infinity of the following series :

5. 1 1 .3

1 cos 2 cos 4 cos 6 ...2 2.4 2.4.6

+ − + +

6. 1 1

sin sin 2 sin 3 ...2 3

− + +

7. sin · sin + sin2 · sin 2 + sin3 · sin 3 + ...

8. 2 3sin sin

1 sin cos cos 2 cos 3 ...2 3

+ + +

9. 2 3cos cos

1 cos cos cos 2 cos 3 ...2 3

− + − +

10. sin( 2 ) sin ( 4 )

sin ...3 4

+ + − + +

45

LESSON 5

THEORY OF EQUATIONS I

5.1 In this lesson we shall deal with the theory of equations. We presume that you are all familiar with

the theory of quadratic equations. In the present course of study of theory of equations, we shall confine

ourselves to cubic and biquadratic equations.

5.2 Polynomial and Equations

We know that a function p(x) of the form

p(x) = 1 20 1 2 0... , 0,n n n

na x a x a x a a− −+ + +

where the coefficients a0, a

1, a

2, ... a

n are all complex numbers (a

0 0) is called a polynomial of

degree n, n being a positive integer. For example, is a polynomial of degree 3. The expansion p(x) = 0 is

called an nth degree equation, that is,

p(x) = 1 20 1 2 0... 0, 0n n n

na x a x a x a a− −+ + + = ...(1)

The fundamental theorem of Algebra states that every equation of the form (1), has at least one

complex number as a root. The proof of this result is beyond the scope of the present course of study and

hence is omitted.

In the discussion thereafter, it will be presumed that the coefficients like a0a

1 ... a

n are all complex

numbers. Now discuss the following theorem which is known as Factor theorem. We already know that

if is a root of an equation, say, a0x2 + a

1x + a

2 = 0, a

0 0 then x – is a factor of the corresponding

polynomial a0x2 + a

1x + a

2. For example, 3 is a root of the equation x2 – x – 6 = 0, then it is easy to verify

that x – 3 is a factor of the polynomial x2 – x – 6. We generalize this idea into the following theorem :

5.3 Factor Theorem

A number is a root of the equation

10 1 0n n

na x a x a− + + = ...(1)

if and only if (x – ) is a factor of the polynomial

p(x) = 10 1

n nna x a x a−+ + + ...(2)

where a0 0, a

1, a

2 ..., a

n are complex numbers and n is a positive integer.

Proof. In the first part, it is given that is a root of the equation (1) and we have to prove that

(x – ) is a factor of (2).

We divide the polynomial p(x) of degree n by a polynomial (x – ) of degree one. Then, the quotient

will be a polynomial of degree (n – 1). If the remainder is zero, then obviously (x – ) is a factor of p(x).

If the remainder is not zero, then the remainder is a constant and let it be R. Also let the quotient be

denoted by

1 2 30 1 2 2 1

n n nn na x b x b x b x b− − −

− −+ + + + +

Then we have

1 1 20 1 1 0 1 2 1( ) [ ]n n n n

n n n na x a x a x a x a x b x b x b R− − −− − −+ + + + − + + + +

46

Putting x = on both sides, we get

10 1 1

n nn na a a−

− + + + + 0 + R = R

In other words,

R = 10 1 1

n nn na a a−

− + + + + ...(3)

Now again, since is a root of the equation (1), therefore by putting x = in (1), we get

10 1 1

n nn na a −

− + + + + = 0 ...(4)

From (3) and (4), it follows that R = 0 so that (x – ) is a factor of p(x).

In the second part, it is given that (x – ) is a factor of the polynomial p(x) given by (2), we have to

show that is a root of the equation (1).

Since (x – ) is a factor of the polynomial p(x), therefore when we divide p(x) by (x – ), we must

get R = 0, and consequently from (3), we get

10 1 1 0n n

n na a a a−− + + + + =

which implies that a satisfies the equation

10 1 1 0n n

n nx a x a x a−− + + + + =

i.e., is a root of the equation (1).

This completes the proof of the theorem.

Now we discuss the theory of cubic equations.

5.4 Cubic Equation

The general cubic equation is of the form

a0x3 + a

1x2 + a

2x + a

3 = 0, a

0 0 ...(5)

where a0, a

1, a

2, a

3 are given complex numbers.

By the fundamental theorem of algebra, this equation (5) has atleast one root. Let this root be

denoted by ‘’. Then by Factor Theorem (x – ) is a factor of the polynomial a0x3 + a

1x2 + a

2x + a

3.

Divide the polynomial by x – and let the quotient be denoted by a0x2 + b

1x + b

2.

Then we have

a0x3 + a

1x2 + a

2 x + a

3 (x – ) (

0x2 + b

1x + b

2) ...(6)

Now consider the quadratic equation

a0x2 + b

1x + b

2 = 0 ...(7)

Again by the fundamental theorem of algebra equation (6) must have at least one root, say, and

accordingly (x – ) must be a factor of the polynomial 20 1 2 0.a x b x b+ + = Dividing a

0x2 + b

1x + b

2 by (x

– ), we get the quotient, say, (a0x + c

1) so that we have

20 1 2a x b x b+ + = 0 1( ) ( )x a x c− + ...(8)

Also a0 x + c

1 = 1

0 00

( )c

a x a xa

+ = −

...(9)

47

= 10

0

, 0c

aa

−

Combining (6), (8) and (9), we get

3 20 1 2 3a x a x a x a+ + + a

0 (x – ) (x – ) (x – ) ...(10)

Substituting x = , , successively on both sides of (9), we observe that each of the values, , ,

satisfies the equation (5). Thus , , are the root of the equation (5). Also it may be noted that no

number different from , , can be roots of the equation (5).

For, if possible let , a number different from each of , , be a root of this equation (5). Then by

substituting x = on both sides of (10), we find that while the left hand side is zero by virtue of the

supposition that is a root but the right hand side is non-zero and thereby we arrive at a contradiction.

Hence, , , are the only roots of the equation (5). Thus we have proved that a cubic equation has three

and only three roots. These roots may be all distinct (unequal), may be repeated (equal) or may be that

only two are repeated.

From the above discussion, it is clear that if , , are the three roots of the a cubic equation, then

the corresponding factors are (x – ), (x – ), (x – ) and hence the equation is given by

(x – ) (x – ) (x – ) = 0

or 3 2( ) ( ) 0.x x x a− + + + + + − =

We discuss the following examples :

Example 1. Find the equation whose roots are 1, 2, 3.

Solution. The required equation is given by

(x – 1) (x – 2) (x – 3) = 0.

or x3 – 6x2 + 11x – 6 = 0

[By putting = 1, = 2, = 3 in equation (11) above]

Example 2. Find the equation whose roots are 3, 3, –2.

Solution. (x – 3) (x – 3) (x + 2) = 0.

or x3 – 4x2 –3x + 18 = 0

5.5 Reactions between the Roots and Co-efficient of a Cubic Equation

Let , , be the roots of a cubic equation

3 20 1 2 3 0a x a x a x a+ + + =

Then we have the identify 3 2

0 1 2 3 0 ( ) ( ) ( )a x a x a x a a x x x+ + + − − −

which implies 3 2

0 1 2 3a x a x a x a+ + + 3 20 [ ( ) ( ) ]a x x x− + + + + + −

Equating the co-efficients powers of x, i.e., x3, x2, x and the constant terms, we get

a0 = a

0

a

1 =

–a

0 ( + + )

a2 = a

0 ( + + )ss

3 = 0 0( )a a− = −

48

These give

+ + = 1

0

a

a−

+ + = 2

0

a

a

= 3

0

a

a−

In other words, we can write these relations as

Sum of the roots = 2

3

coefficient of–

coefficient of

x

x

Sum of the products of roots taken in pairs = 3

coefficient of

coefficient of

x

x

= 3

– constant term

coefficient of x

Thus we note that though we are not able to find the values of , , separately, yet we have been

able to express the three conditions combinations namely + + , + + , in terms of

coefficients 0 0,

1,

2,

3.

5.6 Symmetric Functions

Consider the following expressions :

(i) 2 2 2a b+ +

(ii) 2 2 2( ) ( ) ( ) + + + + +

(iii) 2 + 2 + 2 + 2 + 2 + 2

Each of the above is a function of , , with the property that if any two of , , are interchanged

the function remain unaltered. Such functions are called symmetric functions. More precisely a function f

(1,

2, ...

n) of n variables is said to be symmetric function of

1,

2, ...

n if it remains unaltered by

interchanging any two of

1,

2 ...

n.

Thus, for example

+ + , + + , are symmetric functions of , , .

In the case of a cubic equation whose roots are , , the functions

+ + , + + , are called basic symmetric functions.

Sigma Notation for Symmetric Functions

It is often convenient to describe a symmetric function by indicating only one or more terms in such

a manner that the other terms are obtainable from the given term on replacing the roots therein by other

roots in all possible ways. Also then we put down the sign (sigma) before the given term.

Thus for a cubic equation

= ( + + )

49

= ( + + )

2 = 2 + 2 + 2 + 2g + 2 + 2

2 = 2 + 2 + 2

3 = 3 + 3 + 3 and so on.

It is interesting to know that every symmetric function of , , can be expressed in terms of the

symmetric function + + , + + , and whose values can be written down from the cubic

equation whose roots are , , . We illustrate this fact by the following examples.

Example 3. If , , be the roots of the equation

x3 + px2 + qx + r = 0

find the value of

(i) ( + )2

(ii) ( + ) ( + ) ( + )

Solution. From the given equation, we have

( + + ) = 2

3

co-efficient

1co-efficient

x pp

x

−= − = − s

( + + ) = 3

co-efficient of

1co-efficient of x

x qp= =

= 3

constant term

1co-efficient of

rr

x− = − = −

(i) 2( ) − = 2 2 2 2 2( 2 ) 2 2 2 + − = + − = −

= 2 2 22[( ) 2 ] 2 2( ) 6 2 6a p q − − = − = −

(ii) We shall now express the symmetric function, ( + ) ( + ) ( + ) in terms of the basic

symmetric functions.

We have, by actual multiplication

( ) ( ) ( ) + + + = 2( ) ( ]a + + + +

= 2 + (2 + 2 + 2 + 2 + 2 + 2)

Also we have ( + + ) ( + + )

= (2 + 2 + 2 + 2 + 2 + 2) + 3

that (2 + 2 + 2 + + 2 + )

= ( + + ) ( + + ) –3

Thus we have ( + ) ( + ) ( + )

= 2 + {( + + ) ( + + ) –3}

= ( + + ) ( + + ) –

= – pq + r.

Another method: We have

( + ) ( + ) ( + )

= ( + + – ) ( + + – ) ( + + – )

= (–p – ) (–p – ) (–p – ), [ ( + + = – p)]

50

= – (p + ) (p + ) (p + )

= – [p3 + p2 ( + + ) + p) ( + + ) + ]

= 2–[ ( ) ( )] .p p p pq r pq r + − + − = − +

Example 4. If , , be the roots of the equation x3 + px2 + qx + r = 0. Find the value of

(i) 2 2

(ii)

+

Solution. We have + + = – p

+ + = q

= – r

Now

(i) ( + + )2 = 22 + 22 + 22 + 22 + 22 + 22

= (22 + 22 + 22) + 2 ( + + )

(22 + 22 + 22) = ( + + )2 – 2 ( + + )

= q2 – 2 (–r) (–p)

= 2 – 2rp.

(ii)

+

= 2 2 +

= 2 2 2 2 2 2 + + +

+ +

= 2 2 2 2 2 2( ) ( ) ( ) + + + + +

= 2

= 3 −

= ( ) ( ) 3( )p q r

r

− − −

−

= 3pq r

r

−

Example 5. If , , be the root of the cubic equation x3 + px2 qx + r = 0. Find the value of 3 3 3( ). + +

Solution. We have

( + + ) (2 + 2 + 2)

= (3 + 3 + 3) + (2 + 2 + 2 + 2 + 2 + 2)

2 = 3 + 2

51

3 = 2 – 2 ...(1)

But as shown in Example 3.

( + 2 + 2 + + 2 + 2)

= ( + + ) ( + + ) – 3

= – pq + 3r

Again

2 + 2 + 2 = ( + + )2 – 2( + + ) = p2 – 2q.

Also

+ + = – p

Hence from (1), we get

3 = (3 + 3 + 3) = – p (p2 – 2q) – (–pq + 3r)

= –p3 + 3pq – 3r.

Another method :

As , , are the roots of the equation x3 + px2 + qx + r = 0, we have

3 + p2 + q + r = 0.

3 + p2 + q + r = 0.

3 + p2 + q + r = 0.

Adding these three equations we obtain

(3 + 3 + 3) + p (2 + 2 + 2) + ( + + ) + 3r = 0 ...(1)

Now + + = – p

2 + 2 + 2 = ( + + )2 – ( + + )

= p2 – 2q

Substituting in (1), we obtain

(3 + 3 + 3) + p(p2 – 2q) – qp + 3r = 0

3 + 3 + 3 = – p3 + 3pq – 3r.

Example 6. , , be the roots of the equation

x3 + qx + r = 0,

then prove that

5 5 5

5

+ + =

3 3 3 2 2 2

3 2

b b + + + +

Solution. Since , , are the roots of equation x3 + qx + r = 0

3 + q + r = 0 ...(i)

3 + qb + r = 0 ...(ii)

3 + q + r = 0 ...(iii)

Adding (i), (ii), (iii), we get

3 3q r + + = 0

Also

= 0, = q, = –r

52

3 = –q – 3r = – 3r ...(iv)

2 = ()2 – 2 = 0 – 2q = – 2q ...(v)

Multiply (i) by 2, (ii) by 2 and (iii) by 2 and adding we get

5 + q 3 +sss r 2 = 0

5 = – q 3 – r 2 = –q(–3r) – r (–2q)

= 3qr + 2qr = 5qr by (iv) and (v)

51

5 = 3 21 1

3 2qr

=

Hence the result.

Example 7. Find the equation whose roots , , satisfy the relations :

+ + = 3

2 + 2 + 2 = 5

3 + 3 + 3 = 7

Also find the value of 4 + 4 + 4.

Solution. Let the required equation be

ax3 + bx2 + cx + d = 0

so that + + = b

a−

+ + = c

a

= d

a−

We, therefore have 3 = + + = b

a− ...(ii)

and 5 = 2 2 2 ( ) 2 ( )a b + + = + + − + +

= 2

9c

a−

Hence, from (ii), 3 and from ( ) 2.b c

iiia a

− =

Since , , , satisfy (i) we have

3 2a b c d + + + = 0

3 2a b c d + + + = 0

3 2a b c d + + + = 0

Adding these, we obtain

a(3 + 3 + 3) + b(2 + 2 + 2) + c ( + + ) + 3d = 0

i.e., 7a + 5b + 3c + 3d = 0

53

Dividing by ‘a’ we obtain

7 5 3 3b c d

a a a+ + + = 0

Substituting the values of b

a and

c

a obtained above, we obtain

7 15 6 3d

a− + + = 0

i.e., d

a =

2

3

Thus the required equation (i) on the diving by a is

3 2b c dx x x

a a a+ + + = 0

i.e., 3 2 23 2

3x x x− + + = 0

i.e., 3 23 9 6 2x x x− + + = 0

Since , , are the roots of this equation, they must satisfy the equation.

33 – 92 + 6 + 2 = 0

33 – 92 + 6 + 2 = 0

33 – 92 + 6 + 2 = 0

Multiplying the above equations by , , respectively and adding we get,

34 – 93 + 62 + 2 = 0

34 = 93 – 62 – 2

34 = 1

[9.7 6.5 6]3

− −

= 21 – 10 – 2

= 9

5.7 Biquadratic Equations

We consider the general biquadratic equation, i.e., the equation of degree four viz.,

4 3 20 1 2 3 4 0a x a x a x a x a+ + + + =

where a0 0, a

1, a

2, a

3, a

4 are complex coefficients.

Let ‘’ be a root of this equation so that (x – ) is a factor of the polynomial.

4 3 20 1 2 3 4a x a x a x a x a+ + + +

Then we have a certain

4 3 20 1 2 3 4a x a x a x a x a+ + + + 3 2

0 1 2 3( ) ( )x a a x b x b x b− + + + ...(1)

where 3 2

0 1 2 3a x b x b x b+ + + is the quoedient.

54

As proved previously 3 20 1 2 3 0a x b x b x b+ + + = being a cubic equation, has 3 roots. Let these roots

be , and .

Then, we have

3 20 1 2 3a x b x b x b+ + + 0 ( ) ( ) ( )a x x x− − − ...(2)

From (1) and (2) we obtain

4 3 20 1 2 3 4a x a x a x a x a+ + + + 0 ( ) ( ) ( ) ( )a x x x x ax− − − − ...(3)

Substituting x = , , , successively on the two dies of (3), we see that each of the values , , ,

satisfies the equation 4 3 20 1 2 3 4a x a x a x a x a+ + + + = 0.

Again, if substitute for x, a value different from each of , , , on both sides of (3), we see that

since R.H.S. is not zero, L.H.S. is also not zero. Thus no value other than , , , is a root of the

equation.

We have thus shown that a biquadratic equation has four and four roots.

The roots , , , may not be all different and we have the following different possibilities

(i) = Two roots equal.

(ii) = Roots equal in pairs.

(iii) = = Three roots equal.

(iv) = = = All the four roots equal.

(v) No two roots equal.

The factors of a 4 3 20 1 2 3 4a x a x a x a x a+ + + + are

(a) (x – )2 (x – ) (x – ) for the case (i)

(b) (x – )2 (x – )2 for the case (ii)

(c) (x – )3 (x – ) for the case (iii)

(d) (x – )4 for the case (iv)

(e) (x – ) (x – ) (x – ) (x – ) for the case (v)

5.8 Relation between the Roots and Coefficients of a Biquadratic Equation

4 3 20 1 2 3 4a x a x a x a x a+ + + + = 0

We have

4 3 20 1 2 3 4a x a x a x a x a+ + + + 0 ( ) ( ) ( ) ( )a x x x x ax− − − −

4 30[ ( )a x x− + + +

2( )x+ + + + + +

( ) ]x− + + + + +

Equating the coefficients of x3, x2, x and the constant term, we have

a1 = 0– ( )a + + +

a2 = 0 ( )a + + + + +

a3 = 0– ( )a + + +

a4 = 0a

55

These give

( + + + ) = 3

1

40

co-efficient of

co-efficient of

a x

a x− = −

)( + + + + + = 2

2

40

co-efficient of

co-efficient

a x

a x=

( + ) ( + ) + + = 2

0

a

a

( + + + = 3

40

co-efficient of

co-efficient

a x

a x− = −

( ) ( ) + + + = 3

0

a

a−

= 4

40

constant term

co-efficient of

a

a x=

Any function of , , , which does not change its value of inter-changing any two of them is a

symmetric function.

It may be easily seen that

+ + + ( )

+ + + + + ( )

+ + + ( )

are four symmetric functions known as basic symmetric functions of the roots of the biquadratic.

Every symmetric function of , , , can be expressed in terms of the above symmetric functions.

EXERCISE

1. If , , , are the roots of the equation

x3 – 5x2 + x + 12 = 0

Calculate the value of 2.

2. If , , , be the roots of the equation x3 + qx + r = 0, prove that

2 53( ) ( ) = 35( ) ( ).

[Hint : Since, , , , are the roots of the given cubic equation]

3 + q + r = 0

3 + q + r = 0

3 + q + r = 0

To find 4 multiply equations (1) , , , respectively and add.

To find 5 multiply (1) 2, 2, 3, respectively and add.

3. Find the sum of the cubes of the roots of the equation

x3 – 6x2 + 11x – 6 = 0

56

4. If , , , be the roots of the equation

x3 + 5x2 – 6x + 3 = 0

Find the values of 3−

3 3

3

3 3 3 3 3 3

1 1 1−

+ + =

Hint :

Again 3 3 2 2 23( ) [ ( ) ] − = −

5. If , , , be the roots of the cubic x3 + qx + r = 0 find the values of

(i) 1

b c

+

(ii) 2 2b c

b c

+

+

(iii) (b + c)2

6. If , , , are the roots of the equation

x3 – px2 + qx – r = 0.

Find the value of

(i) 2

1

(ii) 2

(iii) ( ) ( ) ( ) + + +

57

LESSON 6

THEORY OF EQUATIONS II

In this lesson, we shall discuss the methods of transforming a given equation into another equation

or to solve the given equation under certain given conditions.

6.1 Transformation of Equations

We shall discuss methods of finding an equation whose roots are related to those of a given

equation. The example below, will indicate and illustrate and methods which may be employed for the

purpose.

We know that the cubic equation, with roots , and is given by

(x – ) (x – ) (x – ) = 0

i.e., x3 – x2 ( + + ) + x ( + + ) – = 0

i.e., x3 – x2 + x – = 0

Also the biquadratic equation with roots, , , , is

(x – ) (x – ) (x – ) (x – ) = 0

i.e., 4 3 2x x x x− + − + = 0

Examples

Example 1. Find the equation whose roots are the negatives of the roots of the equation

5x3 – 3x2 + 7x + 2 = 0 ...(1)

Solution. If , , , are the roots of the given equation, then the roots of the new equation are –, –

, –.

We write y = –x so that when x takes values, , , then y takes the values –, –, – respectively,

Write y = –x x = – y ...(2)

Eliminating x between (1) and (2)

We get, 5(–y)3 – 3 (–y)2 + 7(–y) + 2 = 0

or –5y3 – 3y2 – 7y + 2 = 0

If 5y3 + 3y2 + 7y – 2 = 0 is the required equation.

Example 2. Find the equation whose roots are the roots of

x4 – 5x3 + 7x2 – 17x + 11 = 0 ...(1)

each diminished by 5.

Solution. If , , , are the roots of the given equation we have to find the equation whose roots

are – 4, – 4, – 4, and – 4.

Write y = x – 4 x = y + 4

Eliminating x between (1) and (2) we get

(y + 4)4 – 5 (y + 4)3 + 7 (y + 4)2 – 17 (y + 4) + 11 = 0

i.e., y4 + 16y3 + 96y2 + 256y + 256 – 5(y3 + 12y2 + 48y + 64)

+7(y2 + 8y + 16) – 17 (y + 4) + 11 = 0

58

Simplifying we get,

y4 + 11y3 + 43y2 + 55y – 9 = 0

as the required equation.

Example 3. Find an equation whose roots are three times those of the equation

2x3 – 5x2 + 7 = 0

Solution. If , , are the roots of given equation, then we write equation whose roots are 3, 3,

3.

We write y = 3x

x = .3

y

Substituting x = .3

y in the given equation, we gets

3 2

2 5 73 3

y y − +

= 0

2y3 – 15y2 + 189 = 0

which is the required equation.

Example 4. Find the equation whose roots are the squares of the roots of the equation.

x3 + qx + r = 0

Solution. We shall give two method for the solution.

First Method:

Let , , , be the roots of the given equation so that we require the equation with roots 2, 2. 2 we

write

y = x2 ...(1)

so that when x takes up the vales , , , then y takes up the values 2, 2, 2.

Aslo if x3 +qx + r = 0 ...(2)

then x necessarily takes up the values, , , and these only.

Thus the required equations will be obtained by eliminating x between (1) and (2) with y indicating

the unknown.

Thus we have

y y q y r+ + = 0

or ( 9)y y + = – r

Squaring, we get

y (y + q)2 = r2

i.e., y3 + 2y2 + q + yq2 + r2 = 0

which is the required equation.

Second Method:

Indicating the unknown by y, the equation with roots 2, 22, is

(y – 2) (y – )2 (y – )2 = 0

59

i.e., y3 – y2 (2 + 2 + 2) + y (22 + 22 + 22) – 222 = 0 ...(3)

We have now to compute the three symmetric functions.

viz., (2 + 2 + 2), (22 + 22 + 22) and 222

From the given equation, we have

( + + ) = 0, ( + + ) = , = – r

Hence (2 + 2 + 2) = ( + + )2 – 2( + + ) = – 2q

and (22 + 2 + 2) = [( + + )2 – 2 ( + + )] = q2.

Also 222 = r2

Making substitutions in (3), we obtain the required equation.

Example 5. If , , are the roots of the cubics x3 + px2 + qx + r = 0, find the equation whose roots

are 1 1 1

, and where , , 0. − − −

Solution. 1

−

= 1 1 1 1

andr r

r r

− − − − + = =

− [ = –r]

Similarly 1

−

= 1r

r

+

1

−

= 1r

r

+

We write 1r

y xr

+ =

so that as x takes , , then y takes up the values

1 −

,

1 −

,

1 −

respectively. Eliminating x between the given equation and 1r

y xr

+ =

we get

3 2

1 1 1

ry ry ryp q r

r r r

+ + +

+ + + = 0

i.e., 3 3 2 2 2 3( 1) ( 1) ( 1) 0r y pr r y qry r r r+ + + + + + =

2 3 2 2 3( 1) ( 1) ( 1) 0r y pr r y q r y r+ + + + + + =

Example 6. If , , are the roots of x3 – px2 + r = 0, form the equation whose roots are

, , and . + + +

Solution. , , are the roots of

x3 – px2 + r = 0

+ + = p, + + = 0, = – r ...(1)

We have +

=

( ) p + + − − =

Similarly +

= ,

p p− + − =

60

We write y = p x

x

− ...(2)

so that when x takes the values , , ; y takes the values , andp p p− − −

respectively.

The required equation is obtained by eliminating x between (1) and (2).

From (2) xy = p – x x(y + 1) = p 1

px

y=

+

Putting the value of x in (1) we have

2

3 2(1 ) (1 )

p pp r

y y

− ++ +

= 0

or p3 – p3 (1 + y) + r (1 + y)3 = 0

i.e., ry3 + 3ry2 + (3r – p3) + y + r = 0

is the required equation.

Example 7. Find the equation whose roots are less by 2 less than the roots of the equation

x4 + 2x3 – 3x2 – 2x + 2 = 0 ...(i)

Solution. If a, b, c, d be the roots of the given equation

We require the equation with roots a – 2, b – 2, c – 2, d – 2,

We write y = x – 2 i.e., x = y + 2

and eliminate x between (ii) and the given equation (i).

The eliminant is

(y + 2)4 + 2(y + 2)3 – 3(y + 2)2 – 2(y + 2) + 2 = 0

i.e., (y4 + 8y3 + 24y2 + 32y + 16) + 2(y3 + 6y2 + 12y + 8)

– 3 (y2 + 4y + 4) – 2 (y + 2) + 2 = 0

i.e., y4 + 10y3 + 33y2 + 42y + 18 = 0

Example 8. Transform the equation

ax3 + by2 + cx + d = 0 ...(i)

into one, from which the second term is missing.

Solution. Let , , be the roots of the given equation.

We shall find another equation whose roots are

+ h, + h, + h

where h is a number to be so chosen that the new equation is free of the second term

We write y = x + h x = y – h ...(ii)

so that the new equation will be obtained on eliminating x between (i) and (iii)

The eliminant is

a(y – h)3 + (y – h)2 + c (y – h) + d = 0

i.e., ay3 + y2 (b – 3ah) + y(3ah2 – 2bh + c) + (–ah3 + bh2 – ch + d) = 0 ...(iii)

61

We choose h so that – 3ah = 0, i.e., b/3a

This choice of h will make the coefficient of y2 zero, i.e., the second term missing.

Substituting the value of h (iii), we obtain

2 2 3 3

3

2 3 2

23 0

3 39 27 9

b b ab b bcay a c y d

a aa a a

−+ − + + + − + =

i.e., 2 2 3

3

2

3 27 9 20

3 27

ac b a d abc bay y

a a

− − ++ + =

i.e., 3 3 2 2 327 9 (3 ) (27 9 2 ) 0y a ac b y a d abc b+ − + − + =

which is the required equation.

EXERCISE

1. From an equation whose roots are negatives of the roots of the equation

x3 + 6x2 + 8x – 9 = 0.

2. Form an equation whose roots are twice the roots of

2x4 + 3x2 + 7x – 9 = 0.

3. Find the equation whose roots are the squares of the roots of the equation

x3 – x2 – 8x + 6 = 0.

4. Find the equation whose roots are the reciprocals of the roots of the equation

ax3 + bx2 + cx + d = 0

5. If a, b, c are the roots of x3 + qx + r = 0, for the equation whose roots are

b2c2, c2a2, a2b2.

6. Find the equation whose roots are the roots of the equation

x4 – 3x3 + 7x2 + 5x – 2 = 0

each diminished by 2.

7. Transform the equation

x3 – 6x2 + 4x – 7 = 0

into one, from which the second term is missing.

6.2 Roots Related by Some Conditions

Sometimes we are required to find conditions under which the roots of given equation may be

related in a given manner. We are also sometimes required to solve an equation it is given that the roots of

the equation are related in some given manner. The examples given below will illustrate the procedure to

be adopted in solving the two types of problems referred to here.

Example 1. Solve the equation

x3 – 12x2 + 39x – 28 = 0

whose roots are given to be in arithmetical progression.

Solution. We put down a general set of three number in A.P. We could take this set simply as

a, a + d, a + 2d

But it would, however, be found more convenient to take this set as

a – d, a, a + d.

62

Let us suppose the roots of the given equation are

a – d, a, a + d.

So that we have,

(a – d) + a + (a + d) = sum of the roots = 12.

(a – d) a + a (a + d) + (a – d) (a + d) = sum of the product of the roots takes in pairs = 39.

(a – d) × a × (a + d) = product of roots = 28

Thus we

3a = 12 ...(i)

3a2 – d2 = 39 ...(ii)

a(a2 – d2) = 28 ...(iii)

From (i), we get a = 4

Substituting a = 4 in (ii), we get

d2 = 9; i.e., d = ±3

We may easily see that a = 4, d = 3 as also a = 4, d = –3 satisfy (iii).

Taking a = 4, d = + 3 and putting these values in a – d, d, a + d, we see that the required roots are 1,

4, 7.

Again taking a = 4, d = – 3, we shall see that the roots are 7, 4, 1 which are the same as those

obtained above except for the order.

Thus we have proved that the roots of the given cubic are 1, 4, 7.

Example 2. Solve

8x3 – 14x2 + 7x – 1 = 0

give that its roots are in geometrical progression.

Solution. We suppose that its roots are a/r, a, ar.

We have a

a arr

+ + = sum of the roots 14 7

8 4=

( ) ( )a a

a a ar arr r

+ + +

= sum of the products of the roots taken in pairs =

7

8

( )a

a arr

= 1

product of roots =8

Thus we have

(1 )a r r

r

+ + =

7

4 ...(i)

2 (1 )a r r

r

+ + =

7

8 ...(ii)

and a3 = 1

8 ...(iii)

63

Dividing (i) by (ii), we get

1

a = 2 so that a =

1

2

Substituting this value of a in (i), we get

21 r r

r

+ + =

7

2

or 22 5 2r r− + = 0

or (2r – 1) (r – 2) = 0

The value 1

2 of a satisfied (iii).

Taking a = 1

2, r = 2, we see that the roots

1 1, , are , ,1.

4 2

aa ar

r

Taking a = 1 1

,2 2

r = we obtain the same roots but in the reverse order.

Thus 1 1

1, , ,2 4

are the required roots.

Examples 3. Solve the equation 3 23 11 12 4 0,x x x+ + + = the roots being in H.P.

(Harmonoic Progression)

Solution. Let the roots of the given equation be , , then

, , = 11

3−

+ + = 4

= 4

.3

−

Since , , are in H.P.

2

=

1 1+

2

+ =

( + ) = 2

+ + = 3

From (ii) and (iv), we get

3 = 4

= 4

3

Substituting = 4

3 in (iii), we get

64

4

3 =

4

3−

= –1.

Thus –1 is the roots of the given equation and (x + 1) is the factor of

3x3 + 11x2 + 12x + 4.

The given equation may be written as

3x2 (x + 1) + 8x(x + 1) + 4(x + 1) = 0

(x + 1) (3x2 + 8x + 4) = 0

Now the other two roots of the given equation are obtained by the solving the equation

3x2 + 8x + 4 = 0

x = 8 64 48 – 8 4 2

, 26 6 3

− − −= = −

Hence the required roots are –1, – 2, –2/3.

Note : If , , , are the roots of biquadratic equation

3

0 1 2 3 4 0a x a x a x a x a + + + + =

Then + + + = 1

0

,a

a−

( + ) ( + ) + + = 2

0

,a

a

+

( + ) + ( + ) = 3 4

0 0

and .a a

a a

− =

Example 4. Solve the equation

4 3 28 14 8 15x x x x− + + − = 0

the roots being in A.P.

Solution. Let the required roots be a – d, a, a + d, a + 2d.

Then (a – d) + a + (a + d) + (a + 2d) = 8 4a + 2d = 8 d = 4 – 2a.

(a – d + a) (a + d + a + 2d) + a(a – d) + (a + d) (a + 2d) = 14

Substituting d = 4 – 2a in this equation, we get

(2a – 4 + 2a) (2a + 12 – 6a) + a (a – 4 + 2a) + (a + 4 – 2a) (a + 8 – 4a) = 14

(4a – 4) (12 – 4a) + a(3a – 4) + (4 – a) (8 – 3a) = 14

48a – 48 – 16a2 + 16a + 3a2 – 4a + 32 – 8a + 12a + 3a2 = 14

– 10a2 + 40a – 30 = 0

a2 – 4a + 3 = 0

(a – 1) (a – 3) = 0 a = 1, 3

Now d = 4 – 2a = 2 when a = 1

And d = 4 – 6 = – 2 when a = 3.

a = 1, d = 2 and a = 3, d = –2.

65

Now when a = 1, d = 2 ; Roots are –1, 1, 3, 5.

And when a = 3, d = – 2 ; Roots are 5, 3, 1, –1. This is the same roots but in different order.

Required roots are –1, 1, 3, 5.

Example 4. Solve the equation

x4 + 15x3 + 70x2 + 120x + 64 = 0

the roots being in G.P.

Solution. We know that if four number be in G.P. then the product of the first and fourth is equal to

the product of the second and third. Thus, if the roots are in G.P. then the product of the two of the roots

must be equal to the product of the other two.

Let the roots be , , , then

= ...(i)

+ + + = –15 ...(ii)

( + ) ( + ) + + = 70 ...(iii)

( + ) + ( + ) = – 120 ...(iv)

= 64 ...(iv)

From (i) and (v), we get

()2 = 64 = 8

= 8 =

Substituting these values in (iii), we get

( + ) ( + ) = 70 – 16

= 54 ...(vi)

From (ii) ( + ) ( + ) = –15 ...(vii)

From (vi) and (vii), we find that ( + ) and ( + ) are the roots of the equation

t2 + 15t + 54 = 0

t2 + 9t + 6t + 54 = 0

(t + 9) (t + 6) = 0

t = – 6, –9

Thus ( + ) = – 6 and + = – 9.

Also, = 8 and = 8

Thus , are the roots of equation

y2 + 6y + 8 = 0

and , are the roots of equation

y2 + 9y + 8 = 0 ...(viii)

Solving (viii), we get

y = – 2, – 4 ...(ix)

and solving (ix), we get

y = –1, – 8

66

Thus roots of given equations are –1, – 2, – 4, – 8.

Example 5. The equation

x4 – 2x3 + 4x2 + 6x – 21 = 0

has two real roots whose sum is zero. Solve the equation.

Solution. Let the roots be , , , , where + = 0

We have, + + + = sum of the roots = 2

( + + + + + ) = sum of the products of the roots taken in pairs = 4.

( + + + ) = sum of the product of the roots taken in traids = –6.

= product of the roots = –21.

We now make use of the fact that + = 0

Now + + + + + = = ( + ) + ( + ) +

= ( + ) + ( + ) ( + )

= ( + ) + [ + =0]

Also + + + = ( + ) + ( + )

= ( + ), [ + = 0]

Thus we have

+ = 2 ...(i)

+ = 4 ...(ii)

( + ) = – 6 ...(iii)

= – 21 ...(iv)

From (i) and (iii), = – 3

Also + = 0 , i.e., = –

2 = 3 or = 3

2 = , we have = 3−

Putting = –3 in (iii) and (iv), we obtain

+ = 2

= 7

These give, on elimination ,

(2 – ) = 7

i.e., 2 – 2 + 7 = 0

= 2 4 28

1 62

i −

=

It will be seen that other choices of the values and give us the same of roots.

The roots are therefore, 3, 3,1 6 ,1 6 .i i− + −

67

Example 6. Solve the equation 4 3 16 4 48 0x x x x+ − − + =

Give that the product of the two roots of the equation is 6.

Solution. Let , , , be the roots of the given equation.

Then + + + = – 1, ...(1)

( + ) + ( + ) + + = – 16 ...(2)

( + ) + ( + ) = 4 ...(3)

= 48 ...(4)

It is given that = 6. Then by (4), = 8.

By (2), ( + ) ( + ) + 6 + 8 = – 16

( + ) ( + ) = – 30 ...(5)

From (1) and (5) if follows that + and + are the roots of

y2 + y – 30 = 0

(y + 6) (y – 5) = 0

= – 6, 5

+ = 5,

+ = – 6.

( – )2 = ( + )2 – 4 = 25 – 24 = 1

– = ±1

Also ( – )2 = ( + )2 – 4 = 36 – 4 × 8 = 4

– = ± 2

Let us take – = 1 and – = 2

Also + = 5 and + = – 6

2 = 6, 2 = –4 = 3, = – 2

and = – 1= 3 – 1= 2, = – 2 = – 4

Required roots are 3, 2, –2, – 4.

Example 7. Solve the equation 4 35 10 4 0x x x− + + =

given that the product of two of the roots of the equation is equal to the product of the other two.

Solution. Let , , , the roots of the given equation,

Then = ...(1)

( + ) + ( + ) = 5 ...(2)

( + ) + ( + ) + + = 10 ...(3)

() ( + ) + ( + ) = 0 ...(4)

= 4 ...(5)

Substituting = in (5), we get

22 = 4

= 2 =

68

Substituting in (3), we get

( + ) + ( + ) + 2 + 2 = 10

( + ) ( + ) = 6

Also ( + ) ( + ) = 5 From (2)

+ and + are the roots of equation

y2 – 5y + 6 = 0

(y – 3) (y – 2) = 0

y = 3, 2

Then a + = 3 and y + = 2

( – )2 = ( + )2 – 4

= 9 – 8 = 1

– = ±1

Also ( – )2 = ( + )2 – 4

= 4 – 8 = – 4 = 4i2

– = ± 2i

Let – = 1 and – = 2i

Also + = 3 and + = 2

= 2, = 1, = 1 + i, = 1 – i

Required roots are 2, 1, 1 ± i.

Examples 8. Find the condition that the roots of the given equation

px2 + qx2 + rx + s = 0

may be in arithmetical progression.

Solution. Let a – d, a, a + d be the roots of the given equation.

We have (a – d) + a + (a + d) = q

p

i.e., 3a = q

p− ...(i)

(a – d) a + a (a + d) + (a + d) (a – d) = r

p

i.e., 3a2 – d2 = r

p ...(ii)

a (a + d) (a – d) = s

p−

a(a2 – d2) = s

p− ...(iii)

From (i) and (ii), d2 = 23r

ap

−

= 2 2

2 2

3

3 3

q r q pr

pp p

−− =

69

Substituting the values of a and d2 in (iii), we obtain

2 2

2 2

3

3 9 3

q q q pr

p p p

−− −

= s

p−

2 2

3( 3 9 )

27

qq q pr

p− + =

s

p−

q(9pr – 2q2) = 27p2s

which is the condition to be necessarily satisfied if the roots of the given equation are in A.P.

Note : You may verify that the equation in Ex. 1 does satisfy this condition.

Example 9. Find the condition for the equation to have a pair of equal roots.

Solution. Let the roots be , , .

We have 2 + = 0 ...(i)

2 + 2 = 0 ...(ii)

2 = – q ...(iii)

Now (i) and (ii) give

2 + 2(–2) = p i.e., 2 = 3

p−

= –2 23

p− −

Substituting in (iii), the values of 2 and we have

23 3

p p −− −

= – q

Squaring we obtain

4

27

p− = q2

i.e., 3 24 27p q+ = 0

which is the required condition.

EXERCISES

1. Solve the equation

3 26 3 10x x x− + + = 0

whose roots are in A.P.

2. Solve the equation

3 23 26 52 24x x x− + − = 0 the roots being in G.P.

3. Solve the equation

3 228 39 12 1x x x− + − = 0

the roots being in Harmonical Progression.

70

4. Solve the equation

3 24 16 9 36x x x− − − = 0

where the sum of the two of the roots being equal to zero.

5. Solve the equation

3 22 22 24x x x− − − = 0

two of its roots being in the ratio 3 : 4.

6. Solve the equation

3 22 7 6x x x+ − − = 0

given that the difference of two roots is 3.

7. Solve the equation

3 25 2 24x x x− − + = 0

given that the product of two of its roots is 12.

8. Solve the equation

4 3 23 40 130 20 27x x x x− + − + = 0

given that product of two of its roots is equal to the product of the other two.

9. Prove that the necessary condition for the roots of the equation

3 20 1 2 32 3 3a x a x a x a+ + + = 0

to be in G.P. is 20 3a a = 3

1 3a a

10. Prove that the necessary condition for the root of the equation

3 2x px qx r− + − = 0

to be in H.P. is 3 2 227 9 2r r pq rq− + = 0

11. Prove that the necessary condition for the sum of two roots of the equation

3 2x px qx rx s − + − + = 0

to be equal to the sum of the other two is

3 4 8p pq r− + = 0

12. Solve the equation

22 15 35 30 8x x x − + − + = 0

given that the product of two roots is equal to the product of the other two.

13. Solve the equation

4 3 28 21 20 5x x x x− + − + = 0

given that the sum of two roots is equal to the sum of the other two.

14. Solve the equation

4 22 15 35 30 8x x x x− + − + = 0

given that the product of the two roots is equal to the product of the other two.

71

6.3 NOW WE DISCUSS CERTAIN IMPORTANT THEOREMS REGARDING THE COMPLEX ROOTS AND SURD ROOTS OF AN EQUATION

Theorem 1. In an equation with real coefficient complex roots occur in conjugate pairs.

Proof. Let f(x) = 0 ... (1)

be an equation, with real co-efficients. Let a + ib be a root of (1). Then we have to prove that a – ib is also

a root of (1).

Since a + ib is a root of (1), therefore, the factor of f(x) corresponding to the root a + ib is

[x – (a + ib)]

Now consider the product

[x – (a + ib)] [x – (a – ib)]

or [(x – a) – ib] [(x – a) + ib]

or [(x – a)2 + b2]

We divide f(x) by the second degree polynomial 2 2[( ) ].x a b− + The remainder will be a

polynomial of degree one. Also since the dividend f(x) and the divisor (x – a)2 + b2 are polynomials with

real coefficient the quotient and remainder will also be polynomials with real coefficient. Let the quotient

be denoted by f(x) and the remainder by rx + rx where r and r are real numbers. We have numbers, thus

f(x) = 2 2[( ) ] ( ) ( )x a b x rx r− + + + ...(i)

We put x = a + ib on the two dies and obtain

0 = 0 + (a + ib) r + r

or 0 = (ra + r) + irb

Equating the real and imaginary party to zero, we obtain

ra + r = 0 ...(ii)

rb = 0 ...(iii)

Now b 0 for if b = 0, a + ib would be real so we can suppose that b 0.

Therefore from (iii), r = 0.

Finally from (ii), r = 0

Substituting r = 0, r = 0 in (i), we obtain

f(x) 2 2[( ) ] ( )x a b x− +

[ ( )][ ( ) ( )]x a ib x a ib x− + − −

Putting x = a – ib, on the two sides, we see that the R.H.S is zero and hence the L.H.S. is zero.

Thus a – ib is a root of f(x) = 0

Hence the result.

Theorem 2. In an equation with rational coefficients, surd roots occur in pairs, i.e., if ,p q+

where p and q are rational number and q is not the square of a rational number, is a root of an equation

with rational coefficients then p – q is also a root of the same equation f(x) = 0 with rational numbers.

Proof. We have [ ( )][ ( )]x p q x p q− − −

= [( ) ] ][( ) )]x p q x p q− − − +

= [(x – p)2 – q]

72

We divide f(x) by [(x – p)2 – q]. Since the coefficients in the divident f(x) and the divisor

[(x – p]2 – q)] are all the rational, the co-efficients in the quotient and the remainder will also be

ratational. Moreover the remainder will be of the first degree.

We write

f(x) = [(x – p)2 – q] (x) + rx + r. ...(i)

where r and r are rational numbers.

Putting x = p q+ on the two dies of (i), we obtain

0 = ( )r p q r+ +

= ( )rp r r q+ +

Equating the surd part and the non-surd part secondary to zero, we have

r q = 0 and rp + r = 0

From these, we deduce r = 0, r = 0 ( q 0)

Thus f(x) [(x – p)2 – q] (x)

[ ( ) ( ( ) ( )]x p q x p q x− + − −

so that ( )x p q− − is a factor of f(x) and as such ( )p q− is a root of f(x) = 0.

Hence the result.

Example 1. From an integer with lowest degree whose roots are 2, 3 4 .i+

Solution. Since in an equation with real co-efficients surd roots occur in pairs and complete roots

occur in conjugate pairs.

Roots of required equation are 2, 3 4 .i

Hence required equation is

( 2) ( 2) [ (3 4 ) [ (3 4 )x x x i x i− + − + − − = 0

(x2 – 2) [(x – 3)2] – (4i)2 = 0

(x2 – 2) (x2 – 6x + 9 + 16) = 0

(x2 – 2) (x2 – 6x + 25) = 0

x4 – 2x2 – 6x3 + 12x + 25x2 – 50 = 0

x4 – 6x3 + 23x2 + 12x – 50 = 0

Example 2. Solve the equation

4 3 23 10 4 6 0x x x x− + − − =

some root being 1 3

2

+ −

Solution. The coefficients of the given equation are all real.

Now 1 3 1 3

2 2 2i

+ −= + is a root and accordingly the conjugate complex of this root viz.

1 3

2 2i−

is also a root.

73

Now 1 3 1 3

2 2 2 2

i ix x

− + − −

= 1 3 1 3

2 2 2 2

i ix x

− − − +

=

2 221 3 1 3

( 1)2 2 2 4

ix x x x

− − = − + = − +

Thus (x2 – x + 1) is a factor of

3x4 + 10x3 + 4x2 – x – 6

Dividing (i) by (x2 – x + 1), we may see that

3x4 + 10x3 + 4x2 – x – 6 (x2 – x + 1) (3x2 – 7x – 6)

The roots of 3x2 – 7x – 6 = 0 are 7 49 72 7 11

, . ., 3,6 6 3

i e +

= −

Hence the required roots are 2 1 3

3, , .3 2 2

i −

74

Exercise-1

1. Simplify the following:

(i) (2 – 5i) + (–3 + 4i) + (8 + 3i)

(ii) 5

3 2

i

i

+

−

2. Evaluate the following:

(i) (–7 –2i) (–1 – 5i)

(ii) 2

2

(2 4 )

(3 )

i

i

+

+

3. Solve the following equations:

(i) [(x + 2y), (2x – y – 6)] = (3, 2)

(ii) 2 3

( ) ( )1

ix y i x y

i

− +− + + =

+

4. Express the following complex numbers in the polar form.

(i) –1 3i+

(ii) –2 12i−

5. Find the trigonometric representation of

(i) sin – i cos .

(ii) 1 + cos + i sin .

6. Show that | cos + i sin | = 1.

7. Express the complex number 5

2

i

i

+

+ in the form x + iy where and x and y are real.

Find its modulus and amplitude.

Exercise-2

1. Simplify 8 9

9 8

(cos sin ) (cos 2 sin 2 )

(cos 2 sin ) (cos sin )

i i

i i i

+ −

+ −

2.

10 10

6

cos sin cos sin15 15 15 15

cos sin15 15

i i

i

+ + −

+

3. Use De Moivre’s theorem to simplify 4

5

(cos sin )

(sin cos )

i

i

+

+

4. Prove that (sin cos ) cos sin2 2

nx i x n x n x

− = − − −

5. Obtain the value of (3 + 4i)3 with the help of De Moivre’s theorem.

6. In n is a positive integer, show that 1( 3 1) ( 3 ) 2 cos .

6

n n n ni +

+ + − =

7. Prove that

75

[(cos cos ) (sin sin )] [(cos cos ) (sin sin )]n ni i + + + + + − +

= 1 ( )2 cos cos

2 2

n n n q− − +

[ (cos cos ) (sin sin )]i + + + Hint :

= 2 cos cos 2 sin cos2 2 2 2

i + − + −

+

Exercise-3

1. If 1 1

2 cos , 2 cos show thatx yx y

= + = +

1m n

m nx y

x y+ = 2 cos (m + n)

and m n

n m

x y

y x+ = 2 cos (m – n),

where m and n are integer. What happens if m and n are rational numbers?

2. If x = cos + i sin , y = cos + i sin and z = cos + i sin , such that x + y + z = 0 then prove

that 1 1 1

0x y z

+ + =

and 1

xyzxyz

+ = 2cos (pa + qb + rc)

3. Let the complex numbers x, y, z be given respectively by cos a + i sin a, cos b + i sin b and cos c +

i sin c, then prove that for any integers p, q, r

1p q r

p q rx y z

x y z+ = 2 cos (pq + qb + rc)

4. 1 1 2 2Let ; , ...... n nx iy x iy x iy+ + + be any n complex number and A + Bi be some other complex

number, such that

1 1 2 2( ) ( ) ...... ( )n nx iy x iy x iy A iB+ + + = +

then show that

(i) 1 1 1 11 2

1 2

tan tan tan tann

n

yy y B

x x x A

− − − −+ + + =

(iii) 2 2 2 2 2 2 2 21 1 2 2( ) ( ) ( )n nx y x y x y A B+ + + + = +

5. Expand (1 + i)n in two different ways and find the sum of the two series.

2 4 6 1 3 5(1– ) and ( )n n n n n nC C C C C C+ − + − + )

both the series being finite.

[Hint : First expand (1 + i)n by Binomial theorem and then expand it by the help of De Moivre’s Theorem.]

76

Exercise-4

1. Find the values of

(i)

1

3(1 )i+

(ii)

1

4(–1 3)i+

(iii)

1

6( 1)−

(iv) 2 / 3(1 )i+

2. Solve the equations (i) z7 = 1 (ii) z7 + z = 0.

3. Find the nth roots of unity and show that they form a series in G.P. whose sum is zero. Also prove

that the sum of their pth power always vanishes unless p be a multiple of n, p being an integer and

that then the sum is n.

4. Solve the equation 7 1z z z + + + = 0.

[Hint : 7 4 3 41 ( 1) ( 1) 0]z z z z z+ + + = + − + =

5. Use De Moivre’s theorem to solve

10 511 1z z+ + = 0

[Hint : Put z5 = y]

6. Solve the equation

4 3 2 1z z z z− + − − = 0

[Hint : Mutiply the equation by (z – 1)]

7. Solve the equation z12 – 1 = 0 and find which of its roots satisfy the equation

4 1z z+ + = 0.

Exercise-5

1. Express sin 5 in terms of sin only.

2. Express cos 6 in terms of cos only.

3. Express tan 7 in terms of tan only.

4. Find the value of sin 1º approximately.

5. Find the value of , where sin 1013

.1014

=

6. Show that sin 5 = 16 sin 5 – 5 sin 3 + 10 sin .

7. Express sin7 cos2 in terms of sines of multiples of .

8. Express cos9 in terms of cosines of multiples of .

Exercise-6

1. cos + cos 2 + cos 3 + ......

2. sin · cos 2 + cos 2 · sin 3 · cos 4 + ......

3. sin + sin ( + ) + sin ( + 2) + ......

77

4. 2 cos 2 + 22 cos 4 + 23 cos 6 + ......

Find the sum to infinity of the following series.

5. 1 1 1.3

1 cos 2 cos 4 cos 6 ......2 2.4 2.4.6

+ − + +

6. 1 1

sin sin 2 sin 3 .....2 3

− + +

7. sin · sin + sin2 · sin 2 + sin3 · sin 3 + ......

8. 2 3sin sin

1 sin cos cos 2 cos 3 ......2 3

+ + + +

9. 2 3cos cos

1 cos cos cos cos 32 3

− + + +

10. sin ( 2 ) sin ( 4 )

sin2 4

+ + − + +

Exercise-7

1. , , are the roots of the equation

3 25 12x x x− + + = 0

Calculate the value of 2.

2. If , , are the roots of the equation x3 + qx + r = 0, prove that

2 53( )( ) = 3 45( )( ).

[Hint: Since , , are the roots] of the given cubic equation

3 + q + r = 0

3 + q + r = 0

3 + q + r = 0

To find 4 multiply equations (1) by , , respectively and add.

To find 5 multiply equation by (2, 2, 2 respectively and add.

3. Find the sum of the cubes of the roots of the equation

x3 – 6x2 – 11x + 6 = 0

4. If , , be the roots of the equation

x3 +5x2 – 6x + 3 = 0

Find the values of –3.

3 33

3 3 3 3 2 2

3 3 2 2 2

1 1 1

Again – 3( ) = [ ]

−

+ + =

−

Hint :

78

5. If a, b, c, be the roots of the cubic x3 + qx + r = 0

(i) 1

,b c

+

(ii) 2 2

,b c

b c

+

+

(iii) (b + c)2

6. If , , , be the roots of the equation

4 3 2x px qx rx s+ + + + = 0

Find the value of

(i) 2

(ii) 1

.

7. If , , are the roots of the equation

3 2x px qx r− + − = 0

Find the value of

(i) 2

1,

(ii) 2

(iii) ( + ) ( + ) ( + )

Exercise-8

1. From an equation whose roots are negatives of the roots of the equation

3 26 8 9x x x− + − = 0

2. From an equation whose roots are twice the roots of the equation

4 22 3 7 9x x x+ + − = 0

3. Find the equation whose roots are the squares of the roots of the equation

x3 – x2 – 8x – 6 = 0

4. Find the equation whose roots are the reciprocals of the roots of the equation

ax3 + bx2 + cx + d = 0.

Exercise-9

1. If , , are the roots of the equation

3 2x px qx r+ + + = 0

form the equation whose roots are

1 1

, ,a

− − −

79

2. If a, b, c are the roots of 3 0,x qx r+ + = from the equation whose roots are

2 2 2 2, , .b c c a a b

3. Find the equation whose roots are the roots of the equation

x4 – 3x2 + 7x2 + 5x – 2 = 0

each diminished by 2.

4. Transform the equation

3 26 4 7x x x− + − = 0

into one, from which the second term is missing.

5. If , , are the roots of the equation

3 2 0, where 0x px qx r r− + − =

Find the equation whose roots are 1 1 1

, , + + +

Exercise-10

1. Solve the equation

4 3 23 40 130 20 27x x x x− + − + = 0

given that product of two of its roots is equal to the product of the other two.

2. Prove that the necessary condition for the roots of the equation

3 20 1 23 3a x a x a x a+ + + = 0

to be in G.P. is

a0 + a

23 = a

13a

3

3. Prove that the necessary condition for the roots of the equation

x3 – px2 + qx – r = 0

to be in H.P. is

2 327 9 2r pqr q− + = 0.

4. Prove that the necessary condition for the sum of two roots of the equation

x4 – px3 + qx2 – rx + s = 0

to equal to the sum of the other two is,

3 4 8p pq r− + = 0.

Exercise-11

1. Solve the equation

3 26 3 10x x x− + + = 0

whose roots are in A.P.

2. Solve the equation

3 23 26 52 24x x x− + − = 0

the roots being in G.P.

80

3. Solve the equations

28x3 – 39x2 + 12x – 1 = 0

the roots being in Harmonical Progression.

4. Solve the equation

4x3 + 16x2 – 9x – 36 = 0

5. Solve the equation

2x3 – x2 – 22x – 24 = 0

two of its roots being in the ratio 3 : 4

6. Solve the equation

2x3 + x2 – 7x – 6 = 0

given that the difference of two roots is 3.

7. Solve the equation

x3 – 5x2 – 2x + 24 = 0

given that the product of two if roots is 12.

81

UNIT 2

LESSON 1

MATRICES

(BASIC CONCEPTS)

1. INTRODUCTION

You are already familier with addition and multiplication of matrices. We shall now talk about some

important types of matrices such as symmetric and skew-symmetric matrices, hermitian and skew-

hermitian matrices etc., elementary operations on a matrix inverse of a matrix, rank of a matrix, and

characteristic equation of a matrix. In the end we shall apply some of these concepts to solutions of

systems of lineare equations. However, before we do so, we shall briefly recapitalate the main facts about

addition and multiplication of matrices.

2. DEFINTION OF A MATRIX

Let S be any set. A set of mn elements arranged in a rectangular array of m rows and n column as

11 12 1

21 22 2

1 2

n

n

m m mn

a a a

a a a

a a a

is called an m × n (“m by n”) matrix over S. A matrix may be represented by the symbols || aij ||, [a

ij], [a

ij]

or by a single letter such as A. The aij’s in a matrix are called the element of the matrix. The indices is and

j of an element indicate respectively the row and the column in which the elements aij is located.

Since we shall be dealing only with matrices over the set of complex number therefore, we shall use

the word “matrix” so as to mean “matrix over C” throughtout, unless we state to the contrary.

Thes 1 × n matrics are called row vectors and the m × l matrices are called column vectors. The m ×

n matrix whose elements are 0 is called the null matrix (or zero matrix) of the type m × n. It is usually

denoted by Om × n

or simply by O if there is no possibility of confusion.

If the number of rows and the number of columns of a matrix are equal (say each equal to n) the

matrix is said to be a square matrix of order n or an n-row square matrix. The elements a11

a22

, ... amn

of a

square matrix A are said to constitute the main doagonal of A. A square matrix in which all the off

diagonal elements are zero is called a diagonal matrix. Thus an n-rowed square matrix [aij] is a diagonal

matrix if aij = 0 whenever i + j. An n-rowed diagonal matrix is often written as

dia. [a11

, a22

, ..., amn

]

A diagonal matrix in whic all the diagronal elements are equal is called a scalar matrix. In other

words, an n-rowed square matrix [aij] is a scalar matrix if for some number k.

aij = , when ,

0, when .

=

k i j

i j

A scalar matrix in which each diagonal element is unity, is called a unit matrix. Thus, an n-rowed

square matrix [aij] is called a unit matrix if

82

aij = 1, whenever ,

0, whenever .

=

i j

i j

The n-rowed unit matrix is usually denoted by In (or simply by I if there is no possibility of

confustion).

The matrix of elements which remain after deleting any number of rows and columns of a matrix A

is called a sub matrix of A.

Illustrations :

1.

0 0 0 0

0 0 0 0

0 0 0 0

is the 3 × 4 null matrix.

2.

3 1 2

5 4 7

1 2 8

−

is a 3-rowed square matrix. 3, 4, 8 constitute the main diagonal of this matrix.

3.

1 0 0

0 7 0

0 0 2

−

is a 3-rowed diagonal matrix.

4.

1 0 0

0 7 0

0 0 2

−

is a 3-rowed scalar matrix.

5.

1 0 0

0 1 0

0 0 1

is the 3-rowed unit matrix. We denote it by I3.

6. The matrix 3 4

6 5 −

is submatrix of

1 8 7

2 3 4

1 6 5

−

−

because it can be obtained from the latter by

deleting the first row and the first columne.

3. EQUALITY OF MATRICES

Two matrices A = [aij] and B = [b

ij] are said to be equal if (i) they are comparable, i.e., the number

of rows in B is the same as the number or rows A, and the number of columns in B is the same as the

number of columns in A; (ii) aij = b

ij = for every pair of subscripts i and j. Thus, for example, the matrices

3 7

8 9

and 1 5 4

3 6 2

are not comparable; the matrices 1 2 3

3 1 0

−

and 4 3 6

1 8 9

are comparable but

not equal; the matrices 2 4 7

6 3 1 −

and 4 4 7

2.3 9 1

− are equal.

From the definition of equality of matrices, it can be easily verified that if A, B, and C be any

matrices, then

(i) A = A (reflexivity)

(ii) A = B B = A (symmetry)

(iii) if A = B and B = C, the A = (transitivity)

83

The above statements (i)—(iii) can be summed up by saying that the relation of equality in the set of

all matrices is an equivlance relation.

4. ADDITION OF MATRICES

If A = [aij], and B = [b

ij] be two matrices of the same type, say m × n, their sum is the m × n matrix C

= [cij], where c

ij = a

ij + b

ij for every pair of subscripts i and j. In other words,

If A =

11 12 1 11 12 1

21 22 2 21 22 2

1 2 1 2

=

n n

n n

m m mn m m mn

a a a b b b

a a a b b bB

a a a b b b

then A + B = 11 11 12 12 1 1

21 21 22 22 2 2

1 1 2 2

+ + + + + +

+ + +

n n

n n

m m m m mn mn

a b a b a a

a b a b a b

a a a b a b

Illustrations. If A = 2 1 3 4 2 5

and4 2 1 3 0 6

− = −

B

then A + B = 2 4 1 2 3 5 2 3 8

4 3 2 0 1 6 7 2 5

− + + + = + + − +

Properties of matrix addition

Addtion of matrices, has the following properties:

(i) Addition of matrices is associative. That is, if A, B, and C be matrices of the same type, then A

+ (B + C) = (A + B) + C

(ii) Addition of matrices is commutative. That is, if A and B be matrics of the same type, then A +

B = B + A.

(iii) Property of zero matrix. If A be an m × n matrix and 0 denotes the m × n zero matrix, then A +

0 = 0 + A = A.

(iv) Negative of a matrix. If a be an m × n matrix, there exists an m × n matrix B, called the

negative of the matrix A, such that A + B = B + A = 0.

5. MULTPLICATION OF A MATRIX BY A SOLAR

If A = [aij] be an m × n matrix, and k be any complex number, then kA is defined to be the m × b

matrix whose (i, j)th the elements is k aij. The matrix kA is called the scalar multiple of A by k. The

following properties of scalar multiplication are worthnothing :

(i) If A and B are comparable matrices, and k is any complex number, then

k(A + B) = kA + kB.

(ii) If A be any matrix, and k and l be any two complex numbers, then

(k + 1) A = kA + lA.

(iii) If A be any matrix, and k and l be any two complex numbers, then

k(lA) = (kl) A.

(iv) For every matrix A,

1 A = A

84

6. MULTIPLICATION OF MATRICES

Definition 1. Let A = [aij] and B = [b

ij] be m × n and n × p matrices respectively. The m × p matrix

[cij], where

cij = a

ij, b

ij + a

i2 + b

2j + a

in b

nj =

1

,=

n

ij kj

k

a b

is called the product of the matrices A and B and is denoted by AB.

The above definition expresses two facts:

(i) We can talk of the product AB if two matrices of and only if the number of columns of A is

equal to the number of rows of B. In the case this condition is satisfied, we say that A and B are

conformable to multiplication.

(ii) If A and B are conformable to multiplication, the (i, j)th element of the matrix AB is obtained by

multiplying the elements of the ith row of A by the corresponding elements of the jth column of

B and adding the products. The sum so obtained is the desired (ij)th elements of AB.

Properties of matrix multiplication

The following are same of the important properties of matrix mulplication :

(i) Matrix multiplication is associative. That is, if A, B, and C be of suitable sizes for the products

A(BC) and (A + B) C to exist, then A(BC) = (AB) C.

(ii) Matrix multiplication is not commutative. That is, given two matrices A and B, AB = BA is not

always true. It is important to note here that for pair of matrices A and B several different

possibilities arise.

(a) Neither of the products AB and BA exits;

(b) only one of the products AB and BA exist and the other one does not exist;

(c) both AB as well as BA exist but they are of different type;

(d) both AB as well as BA exist and are of the same type, but are not equal

(e) AB = BA.

All the above possibilities do exist for certain pairs of matrices.

The important thing to note is that the phase ‘matrix multiplication is not cummutative’ means

that AB is not always equal to BA. It does not exclude the possibility of AB and BA being

equal in some cases.

(iii) Multiplication of matrices is distributive with respect to addition, i.e.,

A(B + C) = AB + BC

and (B + C)D = BD + CD

where A, B, C and D are of the suitable sizes for the above relations to be meaningful.

(iv) Multiplication by the unit matrix. If A be any m × n matrix then Im

A = A = A In

7. POSITIVE INTEGRAL POWERS OF SOURCES MATRIX

If A be an n-rowed square matrix, and n be a positive integer, then An is defined by setting A1 = A,

Ak + 1 = AkA. By the principle of finite induction this defines An for all positive integer An.

If A be an n-rowed square matrix and p and q be positive integers, it can be easily shown that

Ap · Aq = Ap + q, (Ap)q = Apq.

85

8. TRANSPOSE OF A MATRIX

Consider the matrices

A =

3 63 1 4

, 1 06 0 7

4 7

− =

−

B .

The matrix A is a 2 × 3 matrix, and the matrix B is a 3 × 2 matrix. Also, the first column of B is the

same as the first row of A, and the second columne of B is the same as the second row of A. In other

words, B is the matrix obtained from A by writing the row of A as columns. We say that the matrix B is

the transpose of A.

Defintion 2. If A = [aij] be an m × n matrix, then the n × m matrix B = [b

ij], such that b

ij = a

ji is

called the transpose of A and is deonted by At.

From the above defintion we find that

(i) the transpose of an m × n matrix is an n × m matrix ;

(ii) the (i, j)th element of At is the (j, i)th element of A.

Example 1. Let A = 3 1 2 1 3 5

and .0 4 7 6 2 1

− − =

B

Compute At, (At)t, Bt, (A + B)t, At + Bt, (3A)t, 3At.

Solution. At =

3 0

1 4

2 7

−

(At)t = 3 1 2

0 4 7

−

Bt =

1 6

3 2 .

5 1

−

A + B =

2 6 2 62 2 7

, ( ) 2 6 , 2 6 .6 6 8

7 8 7 8

+ = + =

t t tA B A B

3A =

9 0 9 09 3 6

, (3 ) 3 12 , 3 3 12 .0 12 21

6 21 6 21

− = − = −

t tA A

Remark. In the above example we find that ( ) , ( ) ,= + = +t t t t tA A A B A B and (3A)t = 3At. These

results are only special cases of the following theorem :

Theorem 1. If At and Bt are transposes of A and B respectively, then

(i) (At)t = A

(ii) C A + Bt = At + Bt, if A and B are comparable.

(iii) kAt = kAt, k being any complex number

Proof. (i) Let A = [aij] be an m × n matrix. Then At is an n × m matrix and (At)t is an m × n matrix.

The matrices (At)t and A are, therefore, comparable.

Also, (i, j)th element of (At)t

= (j, i)th element of At

= (i, j)th element of A

Since the matrices (At)t and A are comparable and their (i, j)th elements are equal, threfore, (At)t = A.

86

(ii) Let A = [aij] and B = [b

ij] be m × n matrices. Since A and B are both m × n matrices, therefore A

+ B exists and is an m × n matrix. Consequently (A + B)t is an n × m matrix, therefore

Again, At and Bt are both n × m matrices, so that At + Bt also exists and is an n × m matrix.

The matrices (A + B)t and At + Bt are both of the type n × m, and are therefore comparable.

Also, (i, j)th element (A + B)t = (j, i)th element of A + B

= aji + b

ji

= (j, i)th element of At + (i, j)th element Bt

= (j, i)th element of (A + B)t

Thus the matrices (A, B)t and At + Bt are comparable, and their corresponding elements are equal.

Hence (A + B)t + At + Bt .

(iii) Let A = [aij] be an m × n matrix. kA is an m × n matrix and therefore (kA)t is an n × m matrix.

Also At being an n × m matrix, kAt is an n × m matrix. The matrices (kA)t and kAt are both of type n × m,

and are, therefore, comparable. Also (i, j)th element of (kA)t = (j, i)th element of kA

= kaji

=

k[(i, j)th elements of At]

Since the matrices (kA)t and kAt are comparable and their (i, j)th elements are equal, therefore

(kAt) = kAt.

Remark. If At = B, the Bt = (At)t = A, i.e., if B is the transpose of A, then A is the transpose of B.

Example 2. If

1 3 43 1 2

and 2 1 1 ,1 0 3

0 4 2

− = = − − − −

A B

Compute (AB)t and BtAt.

Solution. AB =

1 3 43 1 2

2 1 11 0 3

0 4 2

− − − − −

= 5 0 17

,1 15 2

−

so that (AB)t =

5 1

0 15

17 2

−

Also, Bt =

1 2 0 3 1

3 1 4 , 1 0 .

4 1 2 2 3

− − = −

− −

tA

Therefore Bt At =

5 1

0 15

17 2

−

.

Remark. In the above example (AB)t = BtAt. This is of course, only a particular case of the general

result which we state and prove in the following theorem.

Theorem 2. If A and B be matrices conformable to multiplication, then

(AB)t = BtAt

87

Proof. Let A = [aij] and B[b

ij] be m × n and n × p matrices particularly.

The At = [cij], where c

ij = a

ji, is an n × m matrix.

Bt = [cij], where d

ij = b

ji, is an p × n matrix. The matrices (AB)t and BtAt are

both of type of p × m, and are therefore comparable.

Also (i, j)th elements of (AB)t

= (j, i)th element of (AB)t

= 1=

n

jk ki

k

a b

= 1=

n

jk ki

k

a b

= 1=

n

ij kj

k

d c

= (i, j)th element of BtAt

Since the matrices (AB)t and BtAt are of the same type, and their (ij)th elements are equal, therefore,

(AB)t = BtAt

9. SYMMETRIC AND SKEW-SYMMETRIC MATRICES

Consider the matrices

A =

3 0 2 0 1 3

0 4 1 , 1 0 4

2 1 5 3 4 0

− = − −

− −

B

In matrix A, (1, 2)th element is equal to (2, 1)th element, (1, 3)th element is equal to (3, 1)th element,

and (2, 3)th element is equal to (3, 2)th element. Because of these properties we say that the matrix A is

symmetric.

In matrix B, (2, 1)th element is the negative of (1, 2)th element, (3, 1)th element is the negative of the

(1, 3)th element, (3, 2)th element is the negative of the (2, 3)th element, and (1, 1)th element, (2, 2)th

element, and (3, 3)th element are own negatives, (i.e., they are all zero). Because of these properties we

say that the matrix B is skew-symmetric.

Symmetric and skew-symmetric matrices play useful (an important) roles in the theory of matrices.

Definition 3. A square matrix A = [aij] is said to be symmetric if a

ij = a

ji for all i and j.

Illustrations 1. The matrices

4 1 2 1 0 0

1 3 7 and 0 1 0

2 7 0 0 1

− −

i a h g

i h b f

i g f c

are all symmetric.

2. The matrices

0 1 0 3 6

0 3 and 3 0 4

1 3 0 6 4 0

+ − − − − −

− − − +

i i i

i

i i

are both skew-symmetric

88

3. The matrices

1 1 0 1 2

2 6 and 1 0

1 6 3 2 0

+ − −

− − −

i i

i i i

i i i

are neither symmetric nor skew-symmetric.

In the following theorem we state and prove some basic facts about symmetric and skew-symmetric

matrices.

Theorem 3.

(i) A necessary and sufficient condition for a matrix A to be symmetric is that At = A.

(ii) A necessary and sufficient condition for a matrix A to be skew-symmetric is that At = –A.

(iii) The diagonal elements of a skew-symmetric matrix are all zero.

Proof. (i) Necessity. Let A = [aij] be a symmetric matrix. Since A is symmetric, it must be square

matrix, say of order n, At is then also of order n, so that At and A are comparable. Also, (i, j)th element of

At = aji = a

ij = (i, j)th elements of A.

Therefore At = A.

Sufficiently. Let A = [aij] be an m × n matrix such that At = A. Since A is an m × n matrix., therefore

At is an m × n matrix. Since At and A are equal matrices, they are comparable, so that n = m, and

consequently A is a square matrix. Also, as given, (i, j)th element of At = (i, j)th, which gives aji = a

ij.

Since A a square matrix such that aij = a

ij for all i and j, therefore A is symmetric.

(ii) Necessity. Let A = [aij] be a skew-symmetric matrix. Since A is skew-symmetric, it must be a

square matrix, say of order n. At is then also of order n, so that At and A are comparable. Also, (i, j)th

element of At = aji = –a

ij = (i, j)th element of –A threfore At = A.

Sufficiently. Let A = [aij] b an m × n matrix such that At = –A. Since A is m × n matrix, therefore At is

an n × m matrix and –A is an m × n matrix. Since the matries At and –A are equal, they are comparable, so

that n = m, and consequently A is a square matrix. Also (i, j)th element of At = –[(i, j)th element of A],

which gives aji = –A

ij.

We shall now state and prove a theorem which assures us that every square matrix can be expressed

uniquely as a sum of a symmetric and skew-symmetric matrix.

Theorem 4. Every square matrix can be expressed uniquely as the sum of a symmetric and a skew-

symmetric matrix.

Proof. Let A be an n-rowed square matrix.

Let A = X + Y,

where X is an n-rowed symmetric, and Y is an n-rowed skew-symmetric matrix. Taking the

transpose of both sides of (i), we have

At = ( )+ = = = −t t tX Y X Y X Y ... (2)

since X is symmetric and Y is skew-symmetric so that Xt = X and Yt = –Y

From (1) and (2), we get

X = 1

( ),2

+ tA A ... (3)

Y = 1

( )2

− tA A ... (4)

89

We have shown that if A is expressible as the sum of a symmetric matrix X and a skew-symmetric

matrix Y, then X and Y must be given by (3) and (4). This establishes the uniqueness part. To demonstrate

the existence of a symmetric matrix, X and a skew-symmetric matrix Y such that A = X + Y, we have only

to see that if we write.

X = 1 1

( ), ( ),2 2

+ = −t tA A Y A A

then Xt = 1

( ) ,2

+

ttA A

= 1

( ) ,2

+

ttA A

= 1

( ),2

+tA A

= X,

Yt = 1

( )2

−

ttA A

= 1

( )2

−

ttA A

= 1

( )2

−tA A

= –Y,

so that X is an n × n symmetric matrix and Y is an n × n skew-symmetric matrix. Furthermore, X + Y

= A, which completes the proof.

Example 3. Express the matrix

A =

2 6 5

3 1 4

9 1 7

−

as the sume of symmetric and a skew-symmetric matrix.

Solution. Let

2 6 5

3 1 4

9 1 7

−

= X + Y,

where X is a 3-rowed symmetric matrix and Y is a 3-rowed skew-symmetric matrix.

Taking transposes of both sides (1), and using the facts that (X + Y)t, = , we have

2 3 6

6 1 1

5 4 7

−

= X – Y.

From (1) and (3) we find that

2X =

2 3 6

6 1 1

5 4 7

−

2Y =

0 3 4

3 0 5 ,

4 5 0

− −

−

90

so that

X =

9 32 7 0 2

2 29 3 3 5

1 , 02 2 2 2

3 57 7 2 0

2 2

−

= −

−

Y

Verification. Since Xt = X, Yt = Y, therefore that matrix X is symmetric and the matrix Y is skew-

symmetric. Also, by actual addition we find that X + Y = A.

10. HERMITIAN AND SKEW-HERMITIAN MATRICE

Let A = [aij] be an m × n matrix. The m × n matrix B = [b

ij] such that b

ij = is called the conjugate of

A and is denoted by A. The transpose of A is called the tranposed conjugate or transjugate of A and is

enoted byA.

For example

If A = 3 2 4

,7 3 4

+ − −

i i

i

then = 3 2 4

,7 3 4

− − − +

i i

i

A = 3 2 7

4 3 4

− − − +

i

i i

It can be easily the seen that :

(i) if A be any matrix, then (A) = A.

(ii) if and B be two matrices conformable to addition, then

(A + B) = A + B

(iii) if A be any matrix and k be any complex number, then

( )kA = .kA

(iv) if A and B be two matrix conformable to multiplication, then

(AB) = BA.

We have the proofs of (i), (ii) to the reader and prove only (iii).

Let A = [aij] be an m × n matrix and let B = [b

ij] be an n × p matrix.

The matrices (AB) B, A are respectively of types p × m, p × n, and n × m, so that BA is a p × m

matrix.

Since (AB) and BA are both the type p × m, therefore they are comparable. Also (ij)th element of

(AB) = (j, i)th element of 1 1

( )= =

= =

n n

jk ki jk ki

k k

AB a b a b

Also, (i, j)th element of BA = th th

1

[( ) element of ( ) ]

=

n

k

ik B kj element of A

= th th

1

[( ) element of ( ) ]

=

n

k

ki B jk element of A

91

= 1=

n

ki jk

k

b a

= 1

,=

n

kl ki

k

a b

which is the same as the (i, j)th element of (AB).

Since (i) (AB) and BA are matrices of the same type, and (ii) (ij)th elements of (A, B) and BA

are equal for all i and j, therefore (AB) = BA.

Defintion 4. A square matrix A = [aij] is said to be Hermitian if for all i and j A square matrix A

[aij] is said to be skew-hermitina if for all i and j.

Illustration 1. The matrices

2 1 3 4

, are Hermitian.3 3 4 4

− + − −

i i

i i

2. The matrices

2 1 3 1 5 7

,1 3 5 5 7 3

+ − − + − −

i i i

i i i iare skew-hermitian.

3. The matrix 4

4 1

+ −

i i

i is neither Hermitian nor skew-hermitian.

In the following theorem we prove the important facts about hermitian and skew-hermitina matrices.

Theorem 5. Let A be an n-rowed square matrix.

(i) A necessary and sufficient condition for A to be hermitina is that A = A.

(ii) A necessary and sufficient condition for A to be skew-hermitian is that A = –A.

(iii) The diagonal elements of a hermitian matrix are all real, and the diagonal elements of a skew

hermitian matrix are either pure imagineries of zero.

Proof. (i) First let us assume that A is hermitian. Since A is an n-rowed square matrix, therefore A

( ( ) )= tA is also an n-rowed square matrix. Consequently A and A are matrices of the same type. Also, (i,

j)th element of A (j, i)th element of th( , )= = =ji ijA a a i j element of A. Hence . =A A Conversely, let us

suppose that th( , ) = A A i j element of . = jiA a Since A = A, therefore , . ., ( )= =ji ij ij jia a i e a a and

consequently A is hermitian.

(ii) First lets assume that A is skew-hermitian. Since A is an n-rowed square matrix, therefore, A

and –A are both n-rowed square matrices, and as such they are comparable. Also (i, j)th element of A = (j,

i)th element of A = ,jia and th( , )i j element of –A = –aij. Since A is skew-hermitian, therefore .= −ji ija a

Hence (i, j)th element of A = (i, j)th element of –A.

Since A and –A are comparable, and their (i, j)th elements are equal for all i and j, therefore A = –

A.

Conversely, let us suppose that = –A then (i, j)th elements of A and –A must be equal for all i and

j. This gives .= −ji ija a Taking conjugates of both sides, we have , . ., – ,= −ji ij ij jia a i e a a so that A is

skew-hermitian.

92

(iii) Let A be a hermitain matrix. then =ij jia a for all i and j. For a diagonal elements j = i, which

gives =ij jia a for all i, i.e., aij thus real. Thus the diagonal elements of A are all real.

If A is skew-hermitian, = −ij jia a for all i, j. For j = i this gives .= −ij jia a If aij = p + iq, then a

ji = p

– iq and so we have (p – iq) = –(p + iq) i.e., 2p = 0 or p = 0. The real part of aij is, therefore, zero and

consequently aji is either pure imaginary (if q 0).

We shall now state and prove a decomposition theorem which tells us that every square matrix can

be uniquely decomposed as the sum of the hermitian and a skew-hermitian matrix.

Theorem 6. Every square matrix can be express uniquely in the form X + iY, where X and Y are

hermitian matrices.

Proof. Let A be an n-rowed squrae matrix.

Let A = .= −ij jia a

= ( ) +X iY

= −X iY

= ,−X iY

Since X = X and Y = Y.

From (1) and (2), we have

X = 1 1

( ), ( )2 2

+ = −A A Y A Ai

We have thus shown that if A is expressivle as X + iY, where X and Y, are hermitian, the XZ and Y

must be given by (3). Thus establishes the uniqueness part.

To demonstrate the existence hermitain matrices X and Y such that A = X + iY, we have only to see

that if we write

X = 1 1

( ), ( )2 2

+ = −A A Y A Ai

then X = 1

( )2

+

A A

= 1

( )2

+A A

= 1

( ), since .2

+ = +A A A A

= X,

Y = 1

( )2

−

A Ai

= 1

( )2

− −A Ai

= 1

( ), since .2

− − =A A A Ai

= Y,

93

so that X and Y are hermitian.

Also, X + Y = A, which completes the proof.

Exampe 4. Express the matrix.

A =

1 7 1

1 2

3 1 4 2

− +

− +

i

i i

i

in the form X + iY, where X and Y are hermitian matrices.

Solution. We know that every square matrix A can be uniquely expressed in the form X + iY, whre

X and Y are hermitian matrices given by

X = 1 1

( ), ( )2 2

+ = −A A Y A Ai

Here A =

1 1 3

7 1

1 2 4 2

− − −

+ −

i

i

i i

Therefore X =

1 11 4 1

2 21 1 3

( ) 4 02 2 2

1 31 4

2 2

− − −

+ = +

− +

i i

A A i

i

Y =

1 10 3 2

2 21 1 1

( ) 3 12 2 2

1 12 2

2 2

− − −

− = + −

− +

i i

A A i ii

i i

Verification. Since X = X and Y = Y, the matrices X and Y are hermitian. Also X + iY = A.

Exercise

1. For each of the following matrices A, verify that At = A :

1 2 1 1 6 7

2 0 4 , 6 3 8 ,

1 4 3 7 8 5

− − − −

− − − −

a h g

h b f

g f c

2. For each of the following matrices A, verify that At = –A.

0 1 0 3 1 0 3 2

0 2 , 3 0 4 , 3 0 4

1 2 0 1 4 0 2 4 0

+ − − − − −

− − − − −

i i i

i

i i

3. If A =

1 2 21

2 1 23 2 2 1

−

− −

, verify that A At = At = A = I3.

94

4. If A be any square matrix, verify that the matrix A + At is symmetric and the matrix A —At is

skew-symmetric.

5. If A be any square matrix, prove that A + A, AA, A A are the hermitian and A – A is

skew-hermitian.

6. Express the matrix

3 1 7

2 4 8

6 1 2

−

−

as X + Y where X is symmetric and Y is skew-symmetric.

7. Find hermitian matrices A and B such that

A + iB =

1 1 3

4 2 1

3 2 3

+ −

− +

i

i

i i

95

LESSON 2

ELEMENTARY OPERATIONS ON A MATRIX AND INVERSE OF A MATRIX

1. INTRODUCTION

In this lesson we shall study two important concepts—namely elementary operations on a matrix,

and inverse of a matrix. Both are devices for producing new matrices out of a given matrix. However,

there is one difference. Elementary operations can be applied to a matrix of any type but the process of

finding the inverse can be applied only to some special types of matrices. Furthermore, while it is possible

to produce as many new matrices as we like from a given matrix by means of elementary operations, the

operation of finding the inverse, if applicable, can produce only one matrix from a given matrix.

Both the types of operations that we are going to study have one things in common. Both can be

used for solving a system of linear equations. There is one important connection between the two types of

operations. Elementary operations can be used to find the inverse of a matrix—and that is one reason for

studying them together in this lesson.

2. ELEMENTARY OPERATIONS

Consider the matrices

2 3 4 5 0 6 6 9 12 2 3 4

, , , ,5 0 6 2 3 4 5 0 6 1 6 2

− − − − = = = = − − −

S A B C

2 6 4 2 3 16

, .5 0 6 5 0 6

− − − = = − −

E F

The matrices A, B, C, D, E and F are related to S in as much as :

A can be obtained from S by interchanging the first and second rows.

B can be obtained from S by multiplying the first row of S by 3.

C can be obtained from S by adding 2 times the first row to the second row.

D can be obtained from S by interchanging the first and third columns.

E can be obtained from S by multiplying the second column of S by –2.

F can be obtained from S by adding 4 times the second column of S to the third column.

The operations described above are only examples of operations known as elementary operations.

Definition 1. An elementary operation is an operation of the following types:

Type I. Interchange of two rows of columns

Type II. Multiplication of a row or a column by a non-zero number

Type III. Addition of a multiple of one row or column to another row or column.

An elementary operation is called a row operation or a column operation according as it applies to

rows or columns.

It can be easily seen that if a matrix B is obatined from a matrix A by an E-operation then the matrix

A can be obtained form B by an E-operation of the same type.

It is convenient (and usual) to use the following notation of E-operations :

96

We shall denote the interchange of ith and jth rows (columns) by R1 R

j (C

i C

j); multiplication

of ith row (column) by a number k 0 by Ri → kR

j (C

i → + kC

i); and addition of k times the jth row

(column) to the ith row (column) by Ri → R

i + k R

j(C

i → C

i + kC

j).

If a matrix B is obtained form a matrix A by a finite chain of E-operations, we say that A is

equivalent to B and write it is as A ~ B.

Elementary Matrices. A matrix obtained form a unit matrix by a single E-operation is called an

elemenatary matrix of an E-matrix. For example, the matrices.

0 1 0 1 0 0 1 0 0

1 0 0 , 0 3 0 , 0 1 0

0 0 1 0 0 1 3 0 1

are the E-matrices obtained from I3 by the E-operations R

1 R

2 1 2 2 2(or ), 3 ) →C C R R or (C

2

→ 3C2) and R

3 → R

3 + 3R

1 (or C

1 → C

1 + 3C

3) respectively.

Remark. It can be easily see that the operations Ri R

j and C

i C

j have the same effect on I

n; R

i

→ kRi and C

i → kC

i have the same effect on I

n; R

i → R

i + kR

j and C

j → C

j + kC

i have the same effect

on In. It an E-matrix is obtained from I

n by an E-operation , we say that it is the E-matrix corresponding

to the operation .

Effect of an elementary operation on the product of two matrices.

Before we consider the effect of an E-operation on the product of two matrices, let us consider the

following example.

Example 1. Let A =

3 1 2 1 2

1 0 4 and = 1 4 .

3 5 6 3 5

− − −

− −

B

Let C and D be the matrices obtained form A and AB respectively by the E-row operation R1 R

2

Compute C, D and CB and show that D = CB.

Solution.

AB =

3 1 2 1 2

1 0 4 1 4

3 5 6 3 5

− − −

− −

=

4 20

11 22

10 16

− −

−

,

C =

1 0 4

3 1 2

3 5 6

− −

−

,

D =

11 22

4 20 ,

10 16

− −

−

CB =

1 0 4 1 2

3 1 2 1 4

3 5 6 3 5

− − −

− −

97

=

11 22

4 20

10 16

− −

−

= D.

In the above example we find that the E-row operation R1 R

2 on the product AB is equivalent to

the same E-operation on the pre-factor A. in other words.

Whether we apply the E-row operation R1 R

2 to the matrix A and then post-multiply the resulting

matrix by B, or first multiply the matrices A and B, and then apply the E-row operation R1 R

2 to the

product we get the same result.

The above example suggests the following theorem :

Theorem 1. An elementary row operation on the product of two matrices is equivalent to the same

elementary row operation on the pre-factor.

Proof. Let A [aij] and B = [b

ij] be m × n and n × p matrices respectively. We shall show that if A* be

the matrix obtained form A by an E-row operation, and (AB)* be the matrix obtained from AB by the same

E-row operations, then A * B = (AB)*.

We shall consider the three types of E-rows operations one by one and prove the result in each case.

Type I. Consider the E-row operation Ri R

j

Let A * and (AB)* be the matrices obtained from A and B respectivelyss by the E-row operation Ri

Rj since A is of type m × n, therefore A* is also of type m × n, and consequently A * B is of type m ×

p. Also since AB is of type m × p, (AB)* is also of type m × p. The matrices A * B and (AB)* are, therefore

of the same type.

The matrices A* and A differ form each other in the ith and jth rows only, therefore, it follows that A

* B and AB differ form each other in the ith and jth row only. Also (AB)* and AB differ from each other in

the ith and jth row only.

Consequently, the matrices A* B and (AB)* can differ at the most in ith and jth rows only. It follows

that in order to prove the equality of A * B and (AB)* it is enough to show that the ith (as well as jth) row of

the matrices A * B and (AB)* are identical.

Now (i, k)th element of A* B

= 1=

n

jr rk

r

a b

= th( , ) elementofj k AB

= th( , ) element of ( )*;i k AB

showing that the ith rows of A * B and (AB)* are identical.

Similarly the jth rows of (AB)* and A * B are identical.

Hence A * B = (AB)*

Type II. Consider the E-row operations Ri kR

i (k 0)

Let A* and (AB)* be the matrices obtained from A and AB respectively by the E-row operation

Ri kR

i (k 0).

98

As in the case of the Type I, (i) the matrices A * B and (AB)* are of the same type (ii) A * B and

(AB)* can differ in the ith row only at the most.

= th th

1

{( , ) element of A* ( ) element of }=

n

p

i p p l B

= 1=

n

ip pl

p

ka b

= 1=

n

ip pl

p

k a b

= th{( , ) element of }.k i l AB

= th( , ) element of ( )*,i l AB

so that the ith rows of A * B and (AB)* are identical.

Hence A * B = (AB)*

Type III. Let A* and (AB)* be the matrices obtained from A to AB respectively by the E-row

operation Ri → R

i + kR

j. As in the case of Type I, (i) the matrices A * B and (AB)* are of the same type.

(ii) A * B and (AB)* can differ in the ith row at the most. Therefore in order to complete the proof it

is enough to show that the ith rows of A * B and (AB)* are identical.

Now (i, j)th element A * B

= th th

1

{( , ) element of A* ( ) element of }=

n

i

i r r l B

= 1

( )=

+n

ir jr rl

r

a ka b

= 1 1= =

+ n n

ir rl jr rl

r r

a b k a b

= (i, l)th element of AB + k times = (j, l)th element of AB

= (i, l)ss element of (AB)*,

showing that the ith rows of A * B and (AB)* are identical.

Hence A * B = (AB)*

From the above we find that if A* and (AB)* be the matrices obtained from A and AB respectively

by the same E-row operation of any one of the three types, then A * B = (AB)*.

A theorem similar to the above holds for E-column operations.

Theorem 2. An elementary column operation on the product of two matrices is equivalent to the

same column operation on the post-factor.

Proof. Given a matrix M and an E-column operation c, let us denote by c(M) the matrix obtained for

M by the E-column operation c. Also, let us denote by c* (M) the matrix obtained from M by the

corresponding E-row operation.

99

Let A and B be two matrices of types m × n and n × p respectively. Denoting by c(A) and c(AB) the

matrices obtained from A and AB respectively by the E-column operation c, we have

c(AB) = [ * ( ) ] ,t tc AB

= [ * ( ] ,t t tc B A

= [ * ( ) ]t t tc B A

= ( ) [ * ( )]t t t tA c B

= Ac(B).

This proves the theorem.

Remark. Instead of deducing the theroem 2 from theorem 1, we can prove it by imitating the proof

of theorem 1.

As direct consequences of the above theorems we can prove the following important theorem.

Theorem 3. Every elementary row (resp. column) operation on a matrix is equivalent to pre-

multiplication (resp. post-multiplication) by the elementary matrix corresponding to that theorem.

Proof. Let A = [aij] be an m × n matrix. With the same notation as in the proof of theorem 2, we

have

A* = (Im

A)* = I*m A,

from which we find that A* can be obtained from A by pre-multiplying A by the elementary matrix

Again, writing

A = AIn, we have

c(A) = c(Aln)

= Ac(ln),

From which we find that c(A) can be obtained from A by post-multiplying A by the elementary

matrix c(In). Hence the theorem.

Example 2. Reduce the matrix.

A =

–1 2 1 8

3 1 1 0

4 2 1 7

−

to triangular from by E-row operations.

Solution.

Step 1. By suitable E-row operations on A well shall reduce A to a matrix in which all elements in

the first column except the first element are zero. We can do this by the E-row operations, R2 → R

2 + 3R

1

and R3 → R

3 + 4R

1. These two operationa yield

A ~

1 2 1 8

0 7 2 24 , say

0 10 5 39

− − =

B

Step 2. By suitable E-row operations, we shall reduce B to a matrix in which all the elements in the

second column except the first two are zero. Since we would like that in this process the elements of the

100

first column remain unaltered, therefore we apply the E-row operations R3 → R

3 – 2

10to .

7R B This will

reduce the (3, 2)th element to zero and will not alter any element of the first column.

We thus get

B ~

1 2 1 8

0 7 2 24 , say.

55 330 0

7 7

− − =

C

The matrix C is a triangular matrix. We thus find that the E-row operations R2 → R

2 + 3R

1, R

3 → R

3

– 4R1, R

3 → R

3 – 2

10

7R reduce A to the triangular matrix C.

Remarks. 1. If a matrix has more than three rows, then we shall have several steps which would be

similar to step 2 of the above example.

2. If one of the elements in the first column is either 1 or — 1, it is convenient but not essential to

bring it ot the (1, 1)th place as the initial step in reducing a matrix to triangular form.

3. If every element in the first column of the given matrix happens to be zero then step 1 of the

above example is not required and we have to start with step 2.

4. If the (1, 1)th element is zero, but there is at least one element in the first column different from

zero, we apply a suitable E-row operation and bring this non-zero element to (1, 1)th place. For example,

in order to reduce the matrix

0 3 1

2 4 5

3 1 9

− − − −

to triangular form we apply the E-row operation R1 R

2 so as to obtain the matrix.

2 4 5

0 3 1

3 1 9

− − − −

in which the (1, 1)th element is different from zero. We can now effect triangular reduction of this

matrix in the same way as the in the above example.

5. In the illustration in remark 4 above, we could as well have applied the E-row operation R1 R

3

to the given matrix to obtain a matrix in which the (1, 1)th element is difficult from zero.

6. Triangular reduction of a matrix is not unique. In fact, if we apply the E-row operation R3 → 7R

3

on the matrix C in the above example, then

We get the matrix

1 2 1 8

0 7 2 24

0 0 55 33

− −

which is also a traingular matrix.

Theorem 4. Every matrix can be reduced to a triangular form by elementary row operations.

Solution. We shall prove the theorem by inductive on the number of rows.

101

Let A = [aij] be an m × n matrix.

The theorem is trivially true when m = 1, for every matrix having only one row is a triangular

matrix.

Let us assume that the theorem holds for all matrices having (m – 1) rows, i.e., every matrix having

(m – 1) rows can be reduced to triangular form by E-row operations. We shall show that A can be reduced

to triangular form by E-row operations.

Three different cases arise according as (i) a11

0, (ii) a11

= 0 but a11

0 for some i, (iii) a11

= 0 for

all i. We shall consider these cases onle by one.

Case (i). Let a11

0. The E-row operation R1 → a

11–1 R

1 reduces A to an m × n matrix B = [b

ij] in

which b11

= 1. The E-row operations Rf → R

f – b

f1 R

1 (f = 1, 2, ..., m) reduce B to an m + n matrix C in

which Cf1

= 0 whenever f > 1. The matrix C is of the form

12 12 1

22 23 2

2 3

1

0

0

n

n

m m mn

c c c

c c c

c c c

By our hypothesis, the (m – 1) rowed matrix

22 23 2

2 3

n

m m mn

c c c

c c c

can be reduced to triangular from by E-row operations. If we apply the corresponding E-row

operations to C, it will be reduced to triangular form.

Case (ii). Let a11

= 0, but ali 0 for some f such that 1 f m. By applying the E-row operation R

1

→ R1, to A we get a matrix D = [d

ij] in which d

11 = a

11 0.

We can now proceed as in case (i) and reduce D to triangular form by E-row operations.

Case (iii). If ail = 0 for all i such that l i m, (i.e., all the elements in the first column are zero),

then A is of the form

12 13 1

22 23 2

2 3

0

0

0

n

n

m m mn

a a a

a a a

a a a

By hypothesis we can reduce the (m – 1) - rowed matrix

22 23 2

2 3

n

m m mn

a a a

a a a

to triangular form by E-row operations. When we apply the corresponding E-row operations to A, it will

be reduced to triangular form.

From the above we find that in all the three cases the matrix A can be reduced to trinagular form by

E-row operations.

The proof is now complete by the principle of finite induction.

102

Exercise 1

1. Apply the elementary operation R2 R

3 to the matrix

1 1 3

4 2 6

5 8 9

− −

2. Apply the elementary operation C2 → 2C

2 to the matrixss

1 3 7 6

5 1 4 2

− − −

3. Write down the elementary matrix obtained by applying R3 → R

3 – 4R

1 to l

3.

4. Reduce the matrix

1 2 3 4

3 1 2 0

2 1 1 5

− −

−

to triangular form by applying E-row operations

5. Reduce the matrix

1 1 1

4 1 0

8 1 1

− −

to I3 by E-row operations.

6. Verify that the E-row equation R1 → R

1 – R

3 on the matrix AB is equivalent to the same E-row

operation on A, where

A =

1 1 2 2 3 5

3 1 4 , 1 2 6 .

0 1 3 3 1 1

− − − = −

B

3. INVERSE OF A MATRIX

Consider the matrices

A =

1 2 1 4 1 1

4 7 4 , 4 1 0 .

4 9 5 8 1 1

− − − − − =

− −

B

It can be easily seen that

AB = BA = I3.

Because of this relationship between A and B we say that the matrix B is an inverse of the matrix A.

In fact, we have the following definition.

Definition 2. If A = [aij] be an n-rowed square matrix, then a square matrix B = [b

ij] is said to be an

inverse of A if AB = BA = In.

Since the relation AB = BA remains unaltered when we interchange A and B, it follows that if a

matrix B is an inverse of a matrix, A then A must be the inverse of B. Furthermore, it is obvious from the

defintion of an inverse of a matrix that if a matrix A has an inverse, then it must be a square matrix. In

fact, if A be an m × n matrix and B be a p × q matrix such that AB and BA both exist, we must have p = n,

and g = m. If, AB and BA are to he comparable (which is a necessary condition for AB and BA to be

equal) we must also have m = n, i.e., A must be a square matrix.

A matrix a having an inverse is called an invertible matrix.

103

It is quite natural to ask the following questions regarding inverses of matrices :

Question 1. Does every square matrix have an inverse?

Question 2. In case the answer to question 1 is ‘No’, how can we test as two whether a given matrix

in invertible?

Questions 3. Can a matrix have more than one inverse?

Questions 4. In case the answer to question 2 is ‘No’ how can we proceeed to determine the inverse

of invertible matrix?

We shall try to answer to above quetions one by one.

Answer to question 1. Let us consider the matrix.

A = 0 0

.0 0

If be any 2-rowed square matrix, we find that

AB = BA = 0.

We thus find that there cannot be any matrix B for which AB and BA are both equal to I*. Therefore

A is not invertible. We, therefore, find that a given square matrix need not have an inverse.

Before taking up question 2, we shall try to answer question 3.

Answer to questions 3. As the following theorem shows, if a matrix possesses an inverse, it must

be unique.

Theorem 5. If B and VC be inverses of a square matrix, A then B = C.

Proof. Let B and C be inverses of a square matrix A. Since B is an inverse of A, therefore

AB = BA = I ... (1)

Again, since C is an inverse of A, therefore

AC = CA = I ... (2)

From (1) we find that

C(AB) = CI = I A... (3)

Also, from (2) we find that

(CA)B = IB = B ... (4)

Since C(AB) = (CA)B , it follows from (3) and (4), that

C = B

In view of the above theorem it is only proper to talk of the inverse of a square matrix rather than

taking of an inverse.

The inverse of an inversible matrix is denoted by A–1.

Answer to question 2. Suppose a square matrix A possesses an inverse B. Then

AB = BA = I.

Since the determinant of the product of two matrices is equal to the product of their determinantsm,

we find that

| AB | = | I |

i.e., | A | × | B |= 1

104

From the above relation we find that | A | 0. Thus a necessary condition for a square matrix to

have an inverse is that | A | 0. We shall now show that this conidtion is sufficient as well. In order to do

so, let us consider the matrix 1

adj. .| |

=C AA

By using the idendity

A(adj. A) = (adj. A) A = | A | I,

We find that AC = 1 1

( adj. ) | | ,| | | |

= =A A A I IA A

and CA = 1 1 1

adj. (adj. ) | | .| | | | | |

= = =

A A A A A I I

A A A

Since AC = CA = I,

it follows that C is the inverse of A.

From the above discussion we find that a square matrix possess an inverse if and only if

| A | 0.

This answers questgion 2. In order to test whether a square matrix possesses an inverse, we have

simply to calculate | A |. If | A | = 0 then A does not possesses an inverse but if | A | 0 then A possesses in

inverse.

Answer to question 4. White trying to answer quetions 2, we saw that if | A | 0, then the matrix

1adj.

| |A

A is the inverse of A. This provides one possible answer to question 4.

If a square matrix A possesses an inverse A, then in order to find the inverse we compute adj A, and

multiply it by the scalar1

.| |A

The resulting matrix is the desired inverse of A.

Example 3. Find the inver of the matrix

2 5 0

0 1 1

1 0 3

−

by first computing its adjoint.

Solution. The given matrix be denoted by A.

The co-factors of the elements of the first row of A are 3, – 1, and 1 respectively.

The co-factors of the elements of the second row of A are –15, 6, and –5 respectively.

The co-factors of the elements of the third row are 5, –2, and 2 respectively.

Therefore adj. A =

3 1 1

15 6 5

5 2 2

− − −

−

Also | A | = 2.3 + 5(–1) + 0.1 = 1

Therefore A–1 =

3 1 1

15 6 5

5 2 2

− − −

−

105

4. USE OF ELEMENTARY OPERATIONS TO COMPUTE THE INVERSE OF A MATRIX

We have already described one method for finding the inverse of an invertible matrix. We shall now

describe another method for the same, namely by applying elementary row operations.

Suppose we wish to find the inverse of an n × n matrix A. We consider the identity

IA = A.

We reduce the matrix A on the right hand side to triangular from by E-row operations, and apply the

same E-row operations to the pre-factor I on the left hand side. In this manner we get the identity

PA = Q,

where P and Q are some triangular matrices. As our next step, we apply E-row operations on Q and

reduce it to the unit matrix I. The same E-row operations are, of course, applied to P. We get the identity

BA = I.

where B is obtained from P by E-row operations. The matrix B is the desired inverse of A. We shall

illustrate the procedure by considering a few examples.

Example 4. Find the inver of the matrix.

A =

3 1 1

15 6 5

5 2 2

− − −

−

Solution. Consider the identity

1 0 0 2 3 5

0 1 0 3 5 9

0 0 1 1 1 2

− − −

−

=

2 3 5

3 5 9

1 1 2

− − −

−

By performing the E-row operation R2 → R

2 + 1

3

2R on the matrix on right as well as on the pre-

factor on the left, we have

1 0 02 3 5

31 0 3 5 9

2 1 1 20 0 1

−

− − −

=

2 3 5

1 33

2 21 1 2

−

− − −

Performing R3 → R

3 – 1

1

2R on the matrix on the right as well as on the pre-factor on the left, we

have

1 0 02 3 5

31 0 3 5 9

2 1 1 21

0 02

−

− − −

−

=

2 3 5

1 30

2 21 1

02 2

−

− −

Performing R3 → R

3 – R

2 on the matrix on the right as well as on the pre-factor on the left, we have

1 0 02 3 5

31 0 3 5 9

2 1 1 22 1 1

−

− − − − −

=

2 3 5

1 30

2 20 0 1

−

− −

Performing R1 → R

1 – 5R

3, on the matrix on the right as well as on the pre-factor on the left, we

have

106

11 5 52 3 5

3 1 33 5 9

2 2 2 1 1 22 1 1

− −

− − − − − − −

=

2 3 0

10 0

20 0 1

− −

Performing R1 → R

1 + 6R

2 on the matrix on the right as well as on the pre-factor on the left, we have

2 2 42 3 5

3 1 33 5 9

2 2 2 1 1 22 1 1

−

− − − − − − −

=

2 0 0

10 0

20 0 1

− −

Performing 1 2 2 3 3

1, 2 , and

2→ − → −R R R R R on the matrix on the right as well as on the pre-factor

on the left, we have

1 1 2 2 3 5

3 1 3 3 5 9

2 1 1 1 1 2

− − − −

− −

=

1 0 0

0 1 0

0 0 1

From the above identity we find that the desired inverse is

B =

1 1 2

3 1 3

2 1 1

−

−

Verification. By actual multiplication we find that

AB = BA = I,

The above working an often be arranged in a more convenient and compact from as in the following

example.

Example 5. Find the invers of the matrix

4 1 4

3 0 4 .

3 1 3

− − −

− −

Solution. We shall first reduce the given matrix to I3 by E-row operations

4 1 4

3 0 4

3 1 3

− − −

− −

~ 2 2 1 3 3 1

4 1 4

3 3 30 1 ,

4 4 41

0 04

− −

− → − → −

−

R R R R R R

~ 3 3 2

4 1 4

3 10 1

4 31

0 03

− −

− → +

−

R R R

107

~ 1 1 3 2 2 3

4 1 0

30 0 12 , 3

41

0 03

−

→ − → − −

R R R R R R

~ 1 1 2

4 0 0

3 40 0

4 31

0 03

→ +

−

R R R

~ 1 1 2 2 3 3

1 0 01 4

0 1 0 , , 34 30 0 1

→ → → −

R R R R R R

The given matrix has been reduced to I3 by the E-row operation 2 2 1

3,

4→ −R R R R

3 → R

3 2

3,

4− R

3 3 2 1 1 3 2 2 3 1 1 2 1 1 2 2,

1 4 1 4, 12 , 3 , , ,

3 3 4 3→ + → − → − → + → →R R R R R R R R R R R R R R R R and R

3 → 3R

3.

We shall now perform these operations on l3 in this order. We then have

1 0 0

0 1 0

0 0 1

~ 2 2 1 3 1

1 0 0

3 3 31 0 ,

4 4 43

0 14

− → − −

−

R R R R R

~ 3 3 2

1 0 0

3 11 0

4 31

1 13

− → +

−

R R R

~ 1 1 3 2 2 3

13 4 12

90 3 [ 12 , 3 ]

71

1 13

− −

− → − → − −

R R R R R R

~ 1 1 2

16 4 10

9 40 3

4 31

1 13

− −

− → + −

R R R

~ 1 1 2 2 3 3

4 1 41 4

3 0 4 , , 34 33 1 3

− − − → → → − − −

R R R R R R

The last matrix is the desired invers.

The following theorem is sometimes useful for conputing inverses:

108

Theorem 6. If A and B are n-rowed invertible matrices, then At and AB are both invertible and

(a) (At)–1 = (A–1)t.

(b) (AB)–1 = B–1A–1.

Proof. (a). Since AA–1 = A–1A = In, therefore by the reversal law for transposes we have

(AA–1)t = (A–1A)t = Itn,

i.e., (A–1)t At = A(A–1)t = In.

By the definition of the inverse of a matrix it follows that the matrix At is invertible and its inverse is

the matrix (A–1)t.

(b) Since A and B are both n-rowed invertible matrices, therefore A–1 and B–1 exist and are both n-

rowed square matrices, and consequently B–1A–1 is also an n-rowed square matrix. Let us denote the

matrix B–1A–1 C. We wish to show that B–1A–1 is the inverse of the n-rowed square matrix AB. In order to

achieve our aim we have to verify that

(AB) C = C(AB) = In,

where C denotes the matrix B–1A–1.

Now (AB)C = 1 1( ) ( )− −AB B A

= –1 1( ) −A BB A

= AIn A–1

= AA–1

= In

Also, C(AB) = (B–1 A–1) AB

= B–1(A–1 A)B

= B–1In B

= B–1B

= In.

From the above relations we find that (AB) C = (AB) = In.

Hence it follows that the matrix AB is invertible and its inverse is the matrix, i.e.,

(AB)–1 = C = B–1A–1.

Remark. Result (b) of the above theorem is often referred to as the reversal law of inverses.

The following example illustrates how the reversal law for inverse can be used for computing the

inverses of the product of two invertible matrices.

Example 6. Obtain the invers of the matrices

A =

1 0 1 0 0

0 1 , 1 0

0 0 1 0 1

=

p

p B q

q

and hence that of the matrix

C =

1 0 1 0 0

0 1 , 1 0

0 0 1 0 1

=

p

p B q

q

109

Solution. In the very begining let us observe the following two facts which will be very useful :

(i) B is simply the matrix At with p replaced by q every where in At.

(ii) C = AB.

Step 1. Compute A–1

To compute A–1 we reduce A to I3, by E-row operations.

1 0

0 1

0 0 1

p

p ~ 2 2 3

1 0

0 1 0 [ ]

0 0 1

→ −

p

R R pR

~ 1 1 2

1 0 0

0 1 0 [ ]

0 0 1

→ −

R R pR

We find that A can be reduced to I3 by applying the E-row operations R

2 → R

2 – pR

3 and R

1 → R

1 –

pR2. We now apply these E-row operations to I

3 in this very order. We than get the matrix.

I3 ~ 2 2 3

1 0 0

0 1 [ ]

0 0 1

− → −

p R R pR

=

2

1 1 2

1

0 1 [ ]

0 0 1

− − → −

p p

p R R pR

Thus A–1 =

21

0 1

0 0 1

− −

p p

p .... (1)

From (i) we find that

1( )−tA = 1

2

1 0 0

( ) 1 0

1

−

= − −

tA p

p p

Replacing p and q in the above identity throughout and using (i), we have

B–1 = 2

1 0 0

1 0

1

− −

q

q q

... (2)

By applying the reversal law to (ii) we have

C–1 = (AB)–1 = B–1A–1

=

2

2

1 0 0 1

1 0 0 1

0 0 11

− − −

−

p p

q p

q q

=

2

2

2 2 2 2

1

1 .

1

−

− + − − − − + +

p p

q pq p q p

q pq q p q pq

110

Exercise 2

1. Find the invers of the matrix.

3 15 5

1 6 2

1 5 2

− − −

−

by calculating its adjoint.

2. Show that the matrix.

1 2 3

0 1 4

2 2 1

− −

−

possesses an inverse.

3. Given that

1 1 0

2 1 0 ,

0 0 2

= − −

A calculate A2 and A–1.

4. If A =

1 2 1 2 1 2

4 7 4 , 2 2 1 ,

4 9 5 1 2 2

− − − =

− −

B verify that (AB)–1 = B–1A–1

5. By applying elementary row operations find the inverse of the matrix

1 1 2

3 1 3 .

2 1 1

−

−

111

LESSON 3

RANK OF A MATRIX

1. INTRODUCTION

In this lesson we shall introduce the important concept of the rank of a matrix. We shall show that

the rank of matrix remains unaltered by elementary operations. This will give us a neat method for finding

the rank of matrix.

In the next lesson we shall use the concept of rank to determine whether a given system of linear

equation possesses a solution, and if so, then how many linearly independent solutions are there.

2. RANK OF MATRICES

Consider the matrix

A =

2 3 1 4

7 0 6 5 .

1 4 3 2

− − −

−

A is a 3 × 4 matrix. By delecting any one column of A we can obtain a 3 × 3 matrix. Let us recall

that such a matrix is called a sub-matrix of A. The determinant of such a matrix is called a minor of A.

Thus for example,

2 3 1 2 1 4 3 1 4

7 0 6 , 7 6 5 , 0 6 5

1 4 3 1 3 2 4 3 2

− − − − −

− −

− − −

are all minor of A. Since each of these determinants has 3 rows, these minors are called 3-rowed

minors.

By detecting any one row and any two columns of A we can obtain sub-matrices of the type 2 × 2.

The determinant of any such sub-matrices is called a 2-rowed minor of A. Thus

2 3 1 4 0 6

, ,7 0 6 5 4 3

− −

− −

are all 2-rowed minors of A.

In general, we have the following definitions:

Definition 1. Let A be an m × n matrix. The determinant of any s × s submatrix of A, obtained by

deleting m-s rows and n-s columns of A is called and s-rowed minor of A.

Let us now consider the matrix

A =

2 3 5

1 7 8 .

9 19 9

− −

−

The matrix A has only one 3-rowed minor, namely

2 3 5

1 7 8 ,

9 16 9

−

−

−

and this can be seen to be zero.

112

A has 9 two-rowed minors (why ?) These are not all zero. In fact,

2 3

1 7−

is a 2-rowed minor which has the value 2. (–7) – 3.1 = –17.

The fact that every 3-rowed minor of A is zero and there is at least one 2-rowed minor which is not

zero, is usally expressed bny saying that the rank of the matrix A is 2.

Let us now consider the matrix

I =

1 0 0

0 1 0 .

0 0 1

It has only one 3-rowed minor, namely

1 0 0

0 1 0 ,

0 0 1

and this minor has the value 1. Since I a 3-rowed non-zero minor, and there are not minors having more

than three rows, we say that I is of rank 3.

In general, we have the following definition :

Definition 2. (1) A number r is said to be the rank of mark of a matrix A if

(i) A possesses at least one r-rowed minor which is different from zero.

(ii) A does not possess any non-zero (r + 1)-rowed minor

(2) The rank of every zero matrix is, by definition zero.

The rank of a matrix A is usually denoted by p(A).

Illustrations 1. If A In, then p(A) = n.

2. If 0 0 0 0

, then ( ) 0.0 0 0 0

= =

A A

From the definition of the rank of a matrix we have the following useful criteria for determining the

rank of a matrix;

(i) If all the (r + 1)-rowed minors of a matrix vanish or if the matrix does not possess any (r + 1)-

rowed minor, then the rank of the matrix must be either less than or at the most equal to r.

(ii) It atleast one r-rowed minor of a matrix does not vanish, then the rank of the matrix must be

either grether than or at least equal to r.

It is important (and useful) to note that if every (r + 1)-rowed minor of a matrix is zero, then every

higher order minor is automatically zero (because we can expand every higher order minor ultimately in

terms of (r + 1)-rowed minors.

Example 1. Find the rank of the matrix.

A = 2 4 5 6

.3 8 7 1

− −

Solution. The given matrix has at least one non-zero minor of order 2, namely

2 4

3 8

−

113

(In fact this minor = 2.8 – (–4).3 = 28) Actually, A has 6 2-rowed minors all the of which are non-

zero.

Since A has a non-zero 2-rowed minor, of therefore (A) 2.

Again, since A is a 2 × 4 matrix, therefore it does not have any 3-rowed minors. Consequently (A)

2.

From the relations (A) 2 and (A) 2 we have (A) = 2.

Example 2. Find the rank of the matrix

A = 2 2 2

1 1 1

,

a b c

a b c

all of a, b, and c being real numbers.

Solution.

| A | = 2 2 2

1 1 1

a b c

a b c

= (b – c) (c – a) (a – b).

The following different cases arise:

Case I. a, b, and c are all different. If a, b, and c all the different from each other, then | A | 0.

Therefore A has a non-zero three-rowed minor, namely

2 2 2

1 1 1

,a b c

a b c

so that (A) 3.

Also, A has no higher order minors.

Therefore (A) 3.

Hence (A) = 3.

Case II. Two of the numbers a, b, and c are equal but a, b, and c are not all equal. For the sake of

definiteness assume that a = b c.

In this case | A | = 0 since | A | is the only 3-rowed minor and this minor vanishes, therefore

| A | 2. Also, A has atleast one non-zero 2-rowd minor, namely

1 1

b c

Therefore (A) 2.

Hence (A) = 2.

In the same way as above we find that (A) = 2 in the cases b = c a or c = a b.

Case III. a = b = c,

114

Since a = b = c, therefore | A | = 2 2 2

1 1 1

a a a

a a a

= 0 so that the only 3-rowed minor is zero. Also all

the 2-rowed minors of A vanish. In fact all three columns of A are identical in this case and therefore all

the nine 2-rowed minors of A vanish. Therefore, (A) 1.

Also, A has a non-zero of 1-rowed monor. (In fact these are at least three 1-rowed non-zero minors,

namely those which are obtained by deleting the 2nd and 3rd rows and any tow columns of A). Therefore

(A) 1.

Hence (A) = 1.

3. RANK AND ELEMENTARY OPERATIONS

In the two examples discussed above, we has obtained the ranks of the matrices being considered by

a straight forward consideration of all the minors. This may not be so easy in all the cases. It we have to

find the rank of A 4 × 4 matrix, we may have to consider a fourth order determinants, 16 third order

determinants, and perhaps some second order determinants.

By using elementary operations we can conveniently handle the problem of determining the rank of

a matrix. The following theorem gives the relation between elementary operations on a matrix and its

rank.

Theorem 1. The rank of a matrix remains unaltered by elementary operations.

Proof. We shall prove the theorem for elementary row operations only. The proof for elementary

column operations is similar.

Let A = [aij] be an m × n matrix of rank r.

Case I. Rp R

q.

Let B be the matrix obtained from A by the E-row operation Rp R

q and let the rank of B be s.

In order to verify that (B) = r, we shall show that all (r + 1)-rowed minors of B vanish and atleast

one r-rowed minor does not vanish.

Let B* be any (r + 1) - rowed squae sub-matrix of B. Since B has been obtained from A by the

operation Rp R

q, therefore every row of B is also a row of A (and every row of A is a row of B). This

implies that * *| | | |= B A for some (r + 1) -rowed submatrix A* or A. But every (r + 1)-rowed minor of A

is zero (because (A) = r). Therefore | B* | = 0, and consequently every (r + 1)-rowed minor of B is zero.

This implies that (B) r, i.e., s r.

Since B has been obtained from A by interchaning the pth and qth rows, therefore we can get A from

B by applying Rp R

q to B. By interchanging the roles of A and B in the above proof it follows that r s.

The relations s r and r s together yield r = s.

Case II. Rp → cRp (C 0)

Let us denote by B the matrix obtained from A by the opeation Rp → cR

p, and let the rank of B be s.

Let B* be any (r + 1) - rowed sub-matrix of B and let A* be the correspondingly placed sub-matrix of A.

Let us compare | B* | and | A* |. Two possiblities arise :

(i) pth row of B is one of the row struck off to obtain B* from B.

In this case B* is identical with A* and | B* | = | A* |.

115

(ii) pth row of B is not among the rows struck off to obtain B* from B. In this case | B |* =

c | A* |, for B* and A* agree with each other in all rows except one row, and every element of this

paritcular row of B* is c times the corresponding element of A*.

Since A is of rank r, therefore every (r + 1) -rowed minor of A is zero. In particular | A* | = 0 and in

each of the two cases (i) and (ii) above, we have | B* | = 0.

Since every (r + 1) - rowed minor of B is zero, therefore P(B) , i.e., s r. Also, since A can be

obtained from B by the E- operation Rp → C–1R

p, therefore by interchanging the roots of A and B it

follows that r s. The relations s r and r s together yield r = s.

Case III. Rp → R

p + kR

q (k 0).

Let us denote by B the matrix obtained from A by the E-operation Rp → kR

q, and the rank of B be s.

Let B* be any (r + 1) - rowed sub-matrix of B and let A* be the corresponding sub-matrix of A.

Three different possibities arise :

(i) Neither the pth row nor qth row of B has been struck off while obtaining B*. In this case A* and

B* differ in only one row (the rows which correspond to pth row of A) and have all the other rows

identical. We have | B *| = | A* |. Since every (r + 1) - rowed minor of A is zero, threfore is particular | A*

| = 0, and consequently | B* | = 0.

(ii) The pth row of B has not been struck off while obtaining B* from B but the qth row has been

struck off.

Let us denote by C* the (r + 1) - rowed matrix obtained from A* by replacing apj

by aqj

. By the

property of determinants we have

| B* | = | A* | + k| C* |.

Observe that C* can be obtained from A by first performing the E-operation Rp → R

q and then

striking off those very rows and columns of the new matrix thus obtained as are strucke off for obtaining

B* from B. This implies that | C* | = ±1 times some (r + 1) - rowed minor of A. Since (A) = r,s therefore

every (r + 1) - rowed minor of A is zero, and so | A* | = 0, and | C* | = 0. From (1) it follows that | B* | =

0.

(iii) The pth row of B is one of those rows which have been struck off while obtaining B*. Since A

and B differ only in the pth row, therefore in this case B* = A* an so | B* | = | A* |. Since A is of rank r,

therefore | A* | = 0 and consequently | B* | = 0. Since | B* | = 0, in all the cases (i) - (iii), therefore every

(r + 1) - rowed minor of B is zero and so (B) r, i.e., s r. Also since A can be obtained from B by the

E-operation Rp → R

p – kR

q, therefore by interchanging the roles of A and B it follows that r s. Hence r

= s.

We have thus shown that the rank of a matrix remains unaltered by E-row operations, and so

remarked in the beginning this complest the proof.

Example 3. Reduce the matrix

3 1 1 3

1 4 2 7

2 1 3 0

1 2 3 0

− − − − − − −

to triangular form by E-row operations and hence determine its rank.

Solution. Let us denote the given matrix by A. We shall first of all interchange R1 and R

4 so that we

get –1 as the (1, 1) the entry. This will ultimately simplify the entire working.

116

A =

3 1 1 3

1 4 2 7

2 1 3 0

1 2 3 0

− − − − − − −

~ 1 4

1 2 3 0

1 4 2 7, by

2 1 3 0

3 1 1 3

− − − − − −

−

R R

~ 2 2 1 3 3 1 4 4 1

1 2 3 0

0 2 5 7, by , 2 , 3

0 3 9 0

0 7 10 2

− − − − −

→ − → + → + −

−

R R R R R R R R R

~ 3 3 2 4 4 2

1 2 3 0

0 2 5 73 733 21

, by ,0 02 22 2

55 550 0

2 2

− − − − − → − → −

R R R R R R

~ 4 4 3

1 2 3 0

0 2 5 75

, by33 210 0 3

2 20 0 0 10

− − − − − → −

R R R

If we denote the above matrix by B, then we see that B = 330 0, so that rank of B = 4.

But A and B have the same rank.

Therefore A is of rank 4.

Example 4. Find the rank of the matrix

2 3 4 5 6

3 4 5 6 7

.4 5 6 7 8

9 10 11 12 13

14 15 16 17 18

Solution. Let us denote the given matrix by A. Observe that if we subtract the lement of the first row

from the corresponding elements of the second row, then the second row will consist of ones only. In fact

by 2 2 1 3 1 4 4 1 5 5 1, , , ,→ − → − → − → −R R R R R R R R R R R R and then 3 3

1,

2→R R 4 4

1,

7→R R

5 12

1,

12→R R we find that

A ~

2 3 4 5 6

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

117

~ 2 2 1 3 3 1 4 4 1 5 5 1

2 1 2 3 4

1 0 0 0 0

, by C , , , ]1 0 0 0 0

1 0 0 0 0

1 0 0 0 0

→ − → − → − → −

C C C C C C C C C C C

~ 3 3 2 4 2 5 5 2

2 1 0 0 0

1 0 0 0 0

[by 2 , 3 , 4 ]1 0 0 0 0

1 0 0 0 0

1 0 0 0 0

→ − − → −

C C C C C C C C

Denoting the last matrix by B, we find that every 3-rowed minor of B is zero, and B has atleast one

non-zero two-rowed minor, namely 2 1

.1 0

There (B) = 2.

But (A) = (B), so that (A) = 2.

Remark. Observe that in the above example we did not proceed in the normal way for triangular

reduction.

Example 5. If x, y, z are all unequal, show by using elementary operations or otherwise that the

matrix

A =

0

0

0

0

− − + − − +

− − + + + +

x y x z y z

y x y z x z

z x z y x y

y z z z x y

is of rank 2.

Solution.

A =

0

0

0

0

− − + − − +

− − + + + +

x y x z y z

y x y z x z

z x z y x y

y z z x x y

~ 1 1 2

0[by ]

0

0

− − − − − − +

→ − − − +

+ + +

x y x y x y y x

y x y z x zR R R

z x z y x y

y z z x x y

~ 1 1

1 1 1 1

10, by R

0

0

− − − +

→ − − + − + − +

y x y z x zR

z x z y x y x yy z x y x y

~ 2 2 1 3 3 1 4 4 1

1 0 0 0

, [by C , , ]

− − − +

→ − → − → + − − − +

+ − − +

y x x y x z y zC C C C C C C C

z x x y x z y z

y z x y x z y z

118

~ 2 2 1 3

1 0 0 0

0, [by ( ) ,

0

0

− − +

→ − − → − − +

− − +

x y x z y zR R y z R R

x y x z y z

x y x z y z

Two different cases arise according as y + z 0 or y + z = 0.

Case I. y + z 0.

A ~ 2 2 3 3 4 4

1 0 0 0

1 1 10 1 1 1, by , ,

0 1 1 1

0 1 1 1

→ → → − − +

C C C C C Cx y x z y z

~ 3 3 2 4 4 2

1 0 0 0

0 1 0 0, [by , ]

0 1 0 0

0 1 0 0

→ − → −

C C C C C C

~ 3 3 2 4 4 2

1 0 0 0

0 1 0 0, [by , ~ ]

0 0 0 0

0 0 0 0

→ − →

R R R R R R

Since the matrix

1 0 0 0

0 1 0 0

0 0 0 0

0 0 0 0

is of rank 2, therefore it follows that (A) = 2

Case II. y + z = 0

A ~

1 0 0 0

0 0

0 0

0 0

− −

− − − −

x y x z

x y x z

x y x z

~ 2 2 3 3

1 0 0 0

1 10 1 1 0, by ,

0 1 1 0

0 1 1 0

→ → − −

C C C Cx y x z

~ 3 3 2 4 4 2

1 0 0 0

0 1 1 0, [by , ]

0 0 0 0

0 0 0 0

→ − → −

R R R R R R

~ 3 3 2

1 0 0 0

0 1 0 0, [by ]

0 0 0 0

0 0 0 0

→ −

C C C

which is of rank 2.

Hence (A) = 2 in this case also.

119

Thus (A) = 2

Example 6. Find the rank of the matrix

0

0,

0

0

− −

− − − −

r q x

r p y

q p z

x y z

where px + qy + rz = 0, and all of p, q, and r are positive real numbers.s

Solution. Let the given matrix be denoted by A.

A =

0

0,

0

0

− −

− − − −

r q x

r p y

q p z

x y z

~ 4 4

0

0, [by , since 0]

0

2 0

− −

→ −

− −

r q px

r p pyC pC p

q p py

q p

~ 4 4 3

0

0 0, [by ]

0

− + −

→ − −

− − −

r q px qy

r pC C yC

q p pz

x y z yz

~ 4 4 2

0

0 0, [by ]

0 0

0

− + + −

→ + −

− − −

r q px qy rz

r pC C zC

q p

x y z

~

0 0

0 0, since 0.

0 0

0

− −

+ + = −

− − −

r q

r ppx qy rz

q p

x y z

~ 3 3

0 0

0 0, [by ]

0 0

0

− −

→ −

− − −

r qr

r prC rC

q p

x y rz

~ 3 2 2

0 0 0

0 0, [ ]

0

0

−

→ + − −

− − − −

r

r prC C qC

q p pq

x y qy rz

~ 3 3 2

0 0 0

0 0 0, [ ]

0 0

( ) 0

−

→ + −

− − − + +

r

rby C C qC

q p

x y px qy rz

120

~

0 0 0

0 0 0, since = 0.

0 0

0 0

−

+ + −

− −

r

rpx qy rz

q p

x y

= B(say)

Now every 3-rowed minor of the matrix B vanishes, and there is atleast one non-zero 2-rowed minor

namely0

0−

r

r, therefore the rank of B is 2. Since A ~ B, therefore (A) = (B) = 2.

Example 7. If the number a, b, and c are all different from each other, and the numbers, x, y,z are

also all different from each other slow that the matrix

A =

1

1

1

a x ax

b y by

c z cz

is of rank 3.

Solution.

A =

1

1

1

a x ax

b y by

c z cz

~ 2 2 1 3 3 1

1

0 , [by , ]

0

− − − → − → −

− − −

a x ax

b a y x by ax R R R R R R

c a z x cz ax

~ 2 2 1 3 3 1 4 4 1

1 0 0

0 , [by , , ]

0

− − − → − → − → −

− − −

a

b a y x by ax C C aC C C xC C C axC

c a z x cz ax

~ 4 4 2

1 0 0 0

0 ( ) , [by ]

0 ( )

− − − → −

− − −

b a y x b y x C C xC

c a z x c z x

~ 4 4 3

1 0 0 0

0 0 , [by ]

0 ( ) ( )

− − → −

− − − −

b a y x C C bC

c a z x c b z x

= B(say).

Let us first of all consider the rank of B. since B is a 3 × 4 matrix, (B) 3. Also, since B has a 3-

rowed non-zero minor, namely

1 0 0

0 0 ,

0 ( )( )

−

− − −

y x

z x c b z x

therefore r(B) 3.

From the inequalities (B) 3 and (B) 3 it follows that (B) = 3.

Also, since the rank of a matrix remains unaltered by elementary operations, therefore (A) = (B) =

3.

Hence the rank of the given matrix is 3.

121

Exercise

1. Find the rank of the matrix

3 1 2 6

.6 2 4 9

− − − −

2. Find the rank of the matrix

1 2 3

0 3 4 .

0 0 0

−

3. Find the rank of the matrix.

3 3 3

1 1 1

.

a b c

a b c

4. Reduce the matrix

1 1 2 3

4 1 0 2.

0 3 1 4

0 1 0 2

− −

to triangular form and hence find its rank.

5. Find the rank of the matrix.

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

1 2 3 4

2 3 4 5.

3 4 5 6

4 5 6 7

122

LESSON 4

SYSTEMS OF LINEAR EQUATIONS

1. INTRODUCTION

In this lesson we shall apply our knowledge of matrices to obtain solutions of systems of linear

equations.

Consider a system of m linear equations.

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

,

,

+ + + =

+ + + = + + + =

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

... (1)

in n unknows x1, x

2, ..., x

n.

We can write the above system of equation as

1 1

2 211 12 1

21 22 2

1 2

. .

. .

. .

. .

=

n

n

m m mn

n n

x b

x ba a a

a a a

a a a

x b

... (2)

If we write

A =

1

211 12 1

21 22 2

1 2

.

, .

.

.

=

n

n

m m mn

n

x

xa a a

a a aX

a a a

x

and B =

1

2

.

.

.

n

b

b

b

, so that A is an m × n

matrix, X is an n × 1 matrix, and B is an m × 1 matrix, then we can write (2) as

AX = B

123

The matrix A is called the co-efficient matrix, and [A, B] is called the augmental matrix, where [A,

B] stands for the m × (n + 1) matrix obtained by adjoining one column, namely (b1, b

2) ... b

m)t to the m ×

n matrix A.

We shall confine out study to the following aspects :

I. Application of the inverse of a matrix to study a system of n equations AX = B in n-unknowns. We

shall show that if the matrix A possesses an inverse, then the system has the unique solution which is

given by X = A–1B. As a corollary of this result we shall show that if the matrix A is invertible, then the

system of equations AX = 0 does not possess any non-zero solution.

II. We shall use elementary row operations to put AX = B in a simplified form from which the

solution of the system can be written easily.

III. We shall apply the concept of rank of a matrix to discuss the existence and uniqueness of

solutions of AX = B. We shall show that if A be m × n matrix of rank r, then (i) the general solution of the

system of homogeneous equations AX = 0 contains n–r arbitrary constants, and (ii) the system of non-

homogeneous equations AX = B posses a solution if and only if the matrix [A, B] is also of rank r. In case

this condition is satisfied, the solution contains n–r arbitrary constant.

2. USE OF INVERSE OF MATRIX TO SOLVE A SYSTEM OF LINEAR EQUATIONS

Consider a system of n linear equations

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

,

,

+ + + = + + + =

+ + + =

n n

n n

n n nn n n

a x a x a x b

a x a x a x b

a x a x a x b

... (1)

in n unknowns x1, x

2, ... x

n.

We can write the above system of equation as

1 1

2 211 12 1

21 22 2

1 2

. .

. .

. .

. .

=

n

n

n n nn

n n

x b

x ba a a

a a a

a a a

x b

... (2)

If we write

A =

1

211 12 1

21 22 2

1 2

.

, ,.

.

.

=

n

n

n n nn

n

x

xa a a

a a aX

a a a

x

and B =

1

2

.,

.

.

n

b

b

b

124

so that A is an n × n matrix, X is an n × 1 matrix, and B is an n × 1 matrix then we can write (2) as

AX = B ... (3)

If A is a non-singular matrix, so that A–1 exists, we pre-multiply (3) throughout by A–1 , as that

A–1 (AX) = A–1 B

(A–1A)X = A–1B

IX = A–1B

= A–1B. ... (4)

From (4) we find that if the system of equations (3) has a solution, it is given by (4). By substituting

X = A–1B in (3) we find that

AX = A(A–1B) = (AA–1) B = IB = B, showing that X = A–1B is indeed a solution. Thus we have the

following result.

Corrollary. If A is an n-rowed non-singular matrix, the system of homogeneous linear equations.

AX = 0 has X = 0 as its only solution.

Example 1. Show that the system of equations

4x + 3y + 3z = 1

–x – z = 2

–4x – 4y – 3z = 3

has a unique solution, and obtain it by first computing the inverse of the co-efficient matrix.

Solution. The given system of equations can be written as

4 3 3

1 0 1

4 4 3

− −

− − −

x

y

x

=

1

2 ,

3

or as AX = B.

where A =

4 3 3 1

1 0 1 , , 2 .

4 4 3 3

− − = =

− − −

x

X y B

z

| A | =

4 3 3

1 0 1

4 4 3

− −

− − −

= 0 1 1 1 1 0

3 34 3 4 3 4 4

− − − −− +

− − − − − −

= 4(–4) – 3. (–1) + 3.4

= –1.

Since | A | 0, the matrix is invertible. Since the co-efficient matrix is invertible, therefore the given

system of equations possesses a unique solution.

We shall now find A–1.

Now adj. A =

4 3 3

1 0 1

4 4 3

− − −

125

A–1 = 1

adj.| |

AA

=

4 3 3

1 0 1

4 4 3

− −

− − −

X = A–1B

=

4 3 3 1

1 0 1 2 ,

4 4 3 3

− −

− − −

=

19

4 .

21

−

−

Therefore x = 19, y = –4, z = –21 is the desired solution.

Verification. When x = 19, y = –4, z = –21,

4x + 3y + 3z = 4.19 + 3(–4) + 3 (–21) = 1,

–x –z = –19 + 21 = 2

–4x – 4y –3z = –4 (19) –4 (–4) –3 (–21) = 3.

Example 2. Show, by considering the co-efficient matrix, that the system of equations

x + 2y – z = 0,

–4x –7y + 4z = 0,

–4x – 9y + 5z = 0,

does not possess a non-zero soltion.

Solution. The co-efficient matrix of the given system of equation of given by

A =

1 2 1

4 7 4

4 9 5

− − −

− −

| A | = 7 4 4 4 4 7

29 5 4 5 4 9

− − − −− −

− − − −

= 1 – 2(–4) –8

= 1.

Since | A | 0, therefore A is invertible. Since A–1 exists, the given system of equations does not

possess a non-zero solution.

3. USE OF ELEMENTARY ROW OPERATION TO SOLVE A SYSTEM OF LINEAR EQUATONS

Consider the system of m linear equations

126

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

,

,

+ + + = + + + =

+ + + =

n n

n n

m m mn n m

a x a x a x b

a x a x a x b

a x a x a x b

s ...(1)

By writing

A =

11 12 1

21 22 2

1 2

n

n

m m mn

a a a

a a a

a a a

X =

1

2

.,

.

.

n

x

x

x

and B =

1

2

.

.

.

n

b

b

b

we can write (1) as

AX = B,

where A = [aij] is an m × n matrix, X = 1 2( , , )t

nx x x is an n × 1 matrix, and B = 1 2( , , )tnb b b is an m ×

1 matrix.

The system of equations (1) remains unaltered by the following operations :

(i) Interchange of any two equations;

(ii) Multiplication of both sides of an equation by non-zero number k;

(iii) addition of k times an equation to another equation.

The above equations are equivalent to the following elementary row operations on the matrices A

and B.

(i) interchange of two rows.

(ii) multiplying a row of A and the corresponding row of B by a non-zero number k;

(iii) adding k times an equation to another equation.

Let us recall that :

(a) A matrix can be reduced to triangular form by E-row operations.

(b) An E-row operation on the product of two matrices is equivalent to the same E-row operation

on the pre-factor.

In view of the all the facts stated above we find that in order to solve a given system of linear

equation we may proceed as follows :

(1) Write the given system in matrix notation as AX = B.

(2) Reduce A to a triangular matrix say A* by E-row operations.

(3) Keep on applying the E-row operations in (2) above to the matrix B in the same order as in (2)

above.

(4) Suppose the result of (2) and (3) gives us the equation A * X = B* where A* is a triangular

matrix. By writing A * X = B* as a system of linear equations, we shall arrive at a system of

127

equations which can be solved easily. Of course, it can be happen that the system does not

possess a solution. This will happen if one of the equations in A * B = B* is of the form

0 · x1 + 0 · x

2 + ... + 0 · x

n = b,

for some non-zero number b.

We shall illustrate the method with the help of examples.

Example 3. Solve the system of equations

2x – 5y + 7z = 6,

x – 3y + 4z = 3,

3x – 8y + 11z = 11.

Solution. Let A =

2 5 7 6

1 3 4 , 3

3 8 11 11

− − =

−

B

and assume that there exists a matrix X = ,

x

y

z

such that AX = B

Now

2 5 7

1 3 4

3 8 11

− −

−

x

y

z

=

6

3

11

1 3 4

2 5 7

3 8 11

− −

−

x

y

z

= 1 2

3

6 , by

11

R R

1 3 4

0 1 1

0 1 1

− −

−

x

y

z

= 2 2 1 3 3 1

3

0 , by 2 , 3

2

→ − → −

R R R R R R

1 3 4

0 1 1

0 0 0

− −

x

y

z

= 3 3 2

3

0 , by

2

→ −

R R R

x – 3y + 4z = 3,

y – z = 0,

0 = 2.

Since the conclusion 0 = 2 is false, therefore our assumption that for some X, AX = B, is false.

Consequently there is no X for which AX = B, that is, the given system of equations has no solution.

Remark. A system of equations having no solution is said to be inconsistent.

Example 4. Solve the system of linear equations

x – 2y + 3z = 6,

3x + y – 4z = –7,

5x – 3y + 2z = 5.

128

Solution. Let A =

1 2 3 6

3 1 4 , 7

5 3 2 5

− − = −

−

B , and assume that there exists a matrix X =

x

y

z

, such

that AX = B.

Then

1 2 3

3 1 4

5 3 2

− −

−

x

y

z

=

6

7

5

−

1 2 3

0 7 13

0 7 13

− −

x

y

z

= 2 2 1 3 3 1

6

25 , by 3 , 5 .

25

− → − → −

−

R R R R R R

1 2 3

0 7 13

0 7 0

− −

x

y

z

= 3 3 2

6

25 , by

0

− → −

R R R

2 3 6,

7 13 25

− + =

− = −

x y z

y z

Form the second of above equations, we have y = 13 25

.7 7

−z Substituting this value of y in the first

equation, we have x = 2y – 3z + 6 = 13 25 5 8

3 6 .7 7 7 7

− − + = −

z z z

Therefore we have

x = 5 8

7 7−z

y = 13 25

,7 7

−z

z = z,

i.e., x = 5 8 13 25

, , ,7 7 7 7

− = − =k y k z k where k is any real number. By actual substitution we find

that these values satisfy the given equations.

Hence all the solutions of the given system of equations are

x = 5 8 13 25

, , ,7 7 7 7

− = − =k y k z k

where k is any real number

Example 5. Solve the following system of linear equations :

2x – 3y + z = 0,

x + 2y – 3z = 0,

4x – y –2z = 0.

Solution. Let

2 3 1

1 2 3 ,

4 1 2

− −

− −

and assume that there exists X = ,

x

y

z

such that AX = O, where O

stand for the zero-matrix of type 3 × 1.

129

Then

2 3 1

1 2 3

4 1 2

− −

− −

x

y

z

= O

1 2 3

2 3 1

4 1 2

− −

− −

x

y

z

= O, by R1 R

2

1 2 3

0 7 7

0 9 10

− −

−

x

y

z

= O, by R2 R

2 – 2R

1, R

3 → R

3 – 4R

1

1 2 3

0 7 7

0 9 10

− −

−

x

y

z

= O, R2 →

1 2 3

0 7 7

0 9 10

− −

−

x

y

z

= O, R3 → R

3 + 9R

2

x + 2y – 3z = 0

y – z = 0

z = 0

x = 0, y = 0, z = 0.

Thus we find that if the given system possesses a solution, it must be given by

x = 0, y = 0, z = 0.

Also, by actual substitution it follows that x = y = z = 0 is a solution.

Hence x = y = z = 0 = is the only solution of the given system.

Example 6. Solve the following system of equations :

x + 3y + 4y + 7w = 0,

2x + 4y + 5z + 8w = 0,

3x + y + 3z + 3w = 0,

Solution. Let A =

1 3 4 7

2 4 5 8

3 1 2 3

and assume that there exists a matrix X =

x

y

z

w

such that AX = O,

where O stands for the zero-matrix of type 4 × 1.

Then

1 3 4 7

2 4 5 8

3 1 2 3

x

y

z

w

= O

1 3 4 7

0 2 3 6

0 8 10 18

− − − − − −

x

y

z

w

= O, by R2 → R

2 – 2R

1, R

3 → R

3 – 3R

1

130

1 3 4 7

0 2 3 6

0 0 2 6

− − −

x

y

z

w

= 0, R3 → R

3 – 4R

2

x + 3y – + 4z + 7w = 0

–2y – 3z – 6w = 0

2z + 6w = 0

z = 3 1

3 , ,2 2

− = =w y w x w

x = 1 3

, , 3 , ,2 2

= = − =k y k z k w k

where k is any real number. Thus we find that if the given system possesses a solution it must be

given by

x = 1 3

, , 3 , ,2 2

= = − =k y k z k w k where k is any real number.

By actual substitution we find that (1) is indeed a solution of the given system.

Hence all the solutions of the given system of equations are given by x = 1 3

, ,2 2

=k y k

3 , ,= − =z k w k where k is any real number.

Example 7. Solve the following system of equations :

x – 3y + 2z = 0,

7x – 21y + 14z = 0,

–3x + 9y – 6z = 0.

Solution. Let A =

1 3 2

7 21 17

3 9 6

− −

− −

and assume that there exists a matrix X =

x

y

z

, such that AX = O,

where O is the zero-metrix of type 3 × 1.

Then

1 3 2

7 21 14

3 9 6

− − =

− −

x

y O

z

1 3 2

0 0 0

0 0 0

−

x

y

z

= 0, by R2 → R

2 – 7R

2, R

3 → R

3 + R

1

x – 3y + 2z = 0

x = 3y – 2z, y = y, z = z

x = 3k1 – 2k

2, y = k

1, z = k

2, ... (1)

where k1, k

2 are any real numbers.

Thus we find that if there is a solution of the given system, it must be (1).

Again, by actually substituting x = 3k1 – 2k

2, y = k

1, z = k

2 in the given system of equations, we find

that

131

x – 3y + 2z = (3k1 – 2k

2) – k

1 + 2k

2 = 0,

7x – 21y + 14z = 7(3k1 – 2k

2) – 21k

1 + 12 k

2 = 0,

–3x + 9y – 6z = – 3(3k1 – 2k

2) + 9k

1 – 6k

2 = 0,

so that (1) is indeed a solution. Therefore we conclude that

x = 3k1 – 2k

2, y = k

1, 3 = k

2, where k

1, k

2 are real numbers, are all the solutions of the given system of

equations.

4. APPLICATION OF RANK OF A MATRIX TO A SYSTEM OF LINEAR EQUATIONS

Let us consider a system of m linear equations in n variables:

AX = B,

where A is an m × n matrix, B is an m × 1 matrix, and X is an n × 1 matrix. It can be shown that the above

system of equtions is consistent if and only if the matrices A and [A, B] have the same rank. Further more,

if the common rank be r, the general solution contains n–r arbitrary constants.

We shall not prove the about result, but will contnet ourselves with the followig illustrative

examples.

Example 8. Show, without actually solving, that the following system of equations is inconsistent.

x – y + z = – 1,

2x + y – 4z = –1,

6x – 7y + 8z = 7.

Solution. The given system of equations can be written as

AX = B

where A =

11 1 1

2 1 4 , , 1 .

6 7 8 7

−− − = =

−

x

X y B

z

Let us find the ranks of the matrices A and [A, B]

A =

1 1 1

2 1 4

6 7 8

− −

−

~ 2 2 1 3 3 1

1 1 1

0 3 6 , by R 2 , 6

0 1 2

− − → − → −

−

R R R R R

~ 2 2

1 1 11

0 1 2 , by30 1 2

− − →

− −

R R

~ 3 3 2

1 1 1

0 1 2 , by

0 0 0

− − → +

R R R

Since

1 1 1

0 1 2 , is of rank 2,

0 0 0

− −

132

therefore (A) = 2

Also, [A, B] =

1 1 1 1

2 1 4 1

6 7 8 7

− − − −

−

~ 2 2 1 3 3 1

1 1 1 1

0 3 6 1 , by R 2 , 6

0 1 2 13

− − − → − → −

−

R R R R R

~ 3 3 2

1 1 1 11

0 3 6 1 , by340

0 0 03

− − − → +

R R R

~ 3 3

1 1 1 13

0 3 6 1 , by400 0 0 1

− − − →

R R

~ 1 1 3 2 2 3

1 1 1 0

0 3 6 1 , by ,

0 0 0 1

− − → + → −

R R R R R R

~ 2 2 1 3 3 1

1 0 0 0

0 3 6 1 , by ,

0 0 0 1

− → + → −

C C C C C C

~ 2 2 3 3

1 0 0 01 1

0 1 1 0 , by ,3 60 0 0 1

− → →

C C C C

~ 3 3 2

1 0 0 0

0 1 0 0 , by

0 0 0 1

→ −

C C C

showing that the rank of arugmented matrix is 3.

Since the ranks of the co-efficient matrix and augmented matrix are not equal, the given system of

equations is inconsistent.

3x + 4y – 6 z + w = 7

x – 2y + 3z + 2w = –1

x – 3y + 4z – w = –2

5x – y + z – 2w = 4.

Solution.

A B

? =

3 4 6 1 : 7

1 2 3 2 : 1

1 3 4 1 : 2

5 1 1 2 : 4

− − − −

− − − − −

133

? ~ 1 2

1 2 3 2 : 1

3 4 6 1 : 7, by

1 3 4 1 : 2

5 1 1 2 : 4

− − − −

− − −

− −

R R

? ~ 2 2 1 3 3 1 4 4 1

1 2 3 2 : 1

0 10 15 7 : 10, by 3 , , 5

0 1 1 1 : 1

0 9 14 8 : 9

− − − −

→ − → − → − − −

−

R R R R R R R R R

? ~ 2

3 3 2 4 4

1 2 3 2 : 1

0 10 15 7 : 10

1 91 17, ,0 0 : 0

10 102 101 17

0 0 : 02 10

− − − − → + → −−

−

R R R R R R

? ~ 4 4 3

1 2 3 2 : 1

0 10 15 7 : 10

, by .1 170 0 : 0

2 100 0 0 0 : 0

− − − − → − −

R R R

From the above we find that the ranks of A and [A, B] are both equal to 3. Since the co-efficient

matrix and augment both have the same rank, therefore the given system of equations is consistent.

Remark. Observe that in the above example we have computed the ranks of A and [A, B]

simultaneously. This has meant considerable simplification in workings.

Exericise

1. Does the following system of equations possess a unique common solution?

x + 2y + 3z = 6,

2x + 4y + z = 7,

3x + 2y + 9z = 10.

It so, find it.

2. Show that x = y = z = 0 is the only common solution of the following system of equations:

2x + 3y + 4z = 0,

x – 2y – 3z = 0,

5x + y – 8z = 0.

3. Solve the system of equations :

x + y + z = 3,

3x – 5y + 2z = 8,

5x – 3y + 4z = 14.

4. Solve the followng system of equations :

x – 3y + 2z = 0,

7x – 21y + 14z = 0,

–3x + 9y – 6z = 0.

134

5. Which of the following system of equations are consistent ?

(a) x – 4y + 7z = 8,

3x + 8y – 2z = 6,

7x – 8y + 26z = 31.

(b) x – y + 2z = 4,

3x + y + 4z = 6,

x + y + z = 1.

135

LESSON 5

THE CHARACTERISTIC EQUATION OF A MATRIX

1. INTRODUCTION

In this lesson we shall study the important concept of characteristic equation of a square matrix.

The characterstic equation of a square matrix has many applications. It can be used to compute

power of a square matrix, and inverse of a non-singular matrix. It is also useful in simplifying matrix

polynomials.

We shall begin with some definitions.

Definition. Let A = [Aij] b an n × n matrix. The matrix A – I, where is a scalar and I is the n-

rowed unit matrix, ims called the characteristic matrix of A. The determinant |A – | is called the

characteristic polynomial of A. The equation |A – I| = 0 is called the characteristic equation of A and its

roots are called the characteristic roots (or latent roots or eigen values) of A.

Remark. It can be easily seen that the diagonad elements of A are of the first degree in and the

off-diagonal elements are independent of . |A – I| is, therefore a polynomial of degree in , and the co-

efficient of n is (–1)n. It follows that |A – I| = 0 is of degreen n is . By the fundamental theorem of

algebra, every polynomial equation of degree n has exactly n roots. Therefore it follows that every square

matrix of order n has exactly n characteristic roots.

Example 1. Find the characteristic roots of the matrix.

1 1 4

0 3 7

0 0 5

−

=

A

Solution.

1 1 4

| | 0 3 7 (1 )(3 )(5 )

0 0 5

− −

− = − = − − −

−

A I

The roots of |A – l| = 0 are, therefore, 1, 3, and 5.

Hence the characteristic roots of A are 1, 3, and 5.

Example 2. Show that the matrices A and A have the same characteristic roots.

Solution. Le A = [aij] be an n n matrix. The characteristic equation of A is

|A – I| = 0 (i)

Also the characteristic equation of At is

|At – I| = 0,

which is the same as

|(A – I)|t = 0 (ii)

Since the value of a determinant remains unaltered by taking its transpose, therefore

|A – I| = |(A – I)t|

Consequently

|A – I| = 0, |(A – I)|t = 0 (iii)

136

i.e., |A – I| = 0 and |(A – I)t| = 0 are the same equation. Since |A – I| = 0 is the characteristic

equation of A, and |(A – I)t| = 0 is the characteristic equation of At, it follows that A and At have the same

characteristic roots.

Example 3. Show that the characteristic roots of a triangular matrix are just the diagonal elements

of the matrix.

Solution. Let

11 12 1

22 2

33 3

0

0 0

0 0 0

=

n

n

n

nn

a a a

a a

a a

A

a

be an n × n triangular matrix.

The characteristic equation of A is

11 12 1

22 20

. .0

. .

. .

0 0 1

−

−

=

−

n

n

nn

a a a

a a

a

L

L

i.e., (a11

– ) (a22

– ) … (ann

– ) = 0. (i)

The roots of (i) are given by

= a11

, a22

, a33

, …, and ann which are just the diagonal elements of A.

Hence the charactristic roots of A are just the diagonal elements of A.

Example 4. Show that if 1

2, …

n be the charactertistic roots. of A. then k

1, k

2, …, k

n are the

characteristic roots of kA.

Solution. Let A = [aij] be an n-rowed square matrix. The characteristic equation of A is

11 12 1

21 22 2

1 2

. .0

. .

. .

−

−

= −

n

n

n n nn

a a a

a a a

a a a

L

L

Let the roots of (i) be 1,

2, …,

n. to obtain the equation whose roots are k

1, k

2, … k

n, we have

to substitute * = k in (i). The resulting equation in * will have roots k1, k

2, …, k

n,

By the transformation * = k, equation (i) is transformed to

137

*

11 12 1

*

21 22 2

*

1 2

. . .

. . .

. . .

. . .

−

−

−

n

n

n nn

a a ak

a a ak

a an ak

K

K

= 0

*11 12 1

*1 2

. . .1

. . .

. . .

−

−

n

n

n n mn

ka ka ka

k

ka ka ka

K

K

= 0

* *11 12

22

*1 2

. .

. .

. .

− −

−

−

nn

n n nn

ka ka ka

ka

ka ka ka

K

L

K

= 0

*− kA I = 0 ...(ii)

Since (ii) is the characteristic equation of kA, therefore it follows that the characteristics roots of (ii)

are k times characteristics roots of (i), i.e., the characteristics roots of kA are k1, k

2, ... k

n.

Example 5. If A and P be square matrices of the same type and if P be invertible, show that the

matrices A and P–1 AP have the same characteristic roots.

Solution. The characteristics matrix of P–1 AP is P–1 AP – I.

Now P–1 AP – I = 1 1( ( ) ) ,− −− P A P I P P

= P–1 (A – I) P, ... (i)

since P (I) P–1 = (I) P P–1 = (I) I = I.

The characteristic equation of P–1 AP is | P–1 AP – I) | = 0. Using (i) we find that this equation is

the same is

| P–1 (A – I) P | = 0,

| P–1 | | A – I | | P | = 0, since the determination of the product of two or more matrices

is equal to the product of the determines of the matrices.

| P–1 | | P | | A – I | = 0, since | P–1 |, | A – I |, | P | commute

| P–1 P | | A – I | = 0, since the determinant of a product is equal to the product of

determinants

138

| I | | A – I |= 0, since P–1 P = 1

| A – I | = 0, since | I | = 1

Since |P–1 A P – I | = 0 |A – I | = 0, if follows that the characteristic roots of P–1AP are the

same as the characteristic roots of A.

Exercise

1. Find the characteristics roots of each of the following matrices:

(i)

1 2 3

0 4 2

0 0 7

−

(ii)

1 1 1

1 1 0 .

1 0 1

− − −

−

2. Show that the matrices.

, ,

o g f o f h o h g

g o h f o g h o f

f h o h g o g f o

have the same characteristic equation.

3. Show that the matrices

, ,

a b c b c a c a b

b c a c a b a b c

c a b a b c b c a

4. Show that the characteristics roots of A* are the conjugates of the characteristics roots of A.

5. Show that 0 is a characteristics root of a matrix if and only if the matrix is singular.

6. If A and B are n-rowed square matrices and if A be invertible, show that the matrices A–1B and B

A–1 have the same characteristic roots.

2. CAYLEY-HAMILTON THEOREM

We shall now establish the relation between a square matrix and its characteristics equation. The

relation, which is known after it discovers Arthur Cayley and W.R. Hamilton is known as Cayley-

Hamilton Theorem. It was established by Hamilton in 1953 for a special class of matrices and was stated

(without proof) by Caylay in 1858.

Theorem. Let | A – I | = (–1)n = 1 21 2{ }− − + + + +n n n

na a a be the characteristics polynomial

of an n + n matrix A. Then the matrix equation 1 21 2

− −+ + + + =n n nnX a X a X a I O is satisfied by X =

A.

Proof. The elements of A – I are at most of the first degree in . (In fact if we write the matrix in

full, we find that the diagonal elements are of the first degree in , and all the off-diagonal elements are

independent of ). The elements of adj (A – I) are (n – 1) × (n – 1) determinants, whose elements are at

most of the first degree in . Therefore adj. (A – I) is an n × n matrix whose elements are atmost of

degree n – 1 in . The matrix adj (A – I) can, therefore, be written as a matrix polynomial of degree n – 1

in .

Let adj (A – I) = B0n–1 + B

1n–2 + ... + B

n–1, where B

0, B

1, ... B

n–1 are n × n matrices whose

elements are functions of aij’s (the elements of A).

Since (A – I) adj (A – I) = | A – I | I,.

Therefore(A – I) (B0n–1 + B

1n–2 + ... + B

n–1) = (–1)n

139

Comparing the co-efficients of like powre of on both sides, we hve

– IB0 = (–1)n I,

AB0 – IB

1 = (–1)n a

1I,

AB1 – IB

2 = (–1)n a

2I,

....................................

ABn – 1

= (–1)n an

I.

Premultiplying the above equations by An, An–1 ... I, respectively and adding the corresponding sides

of the resulting equation, we have

O = (–1)n [An + a1An–1 + ... + a

nI].

Thus An + a1An–1 + ... + a

2An –2 + ... + a

nI = O, and this proves the theorem.

Remark. Calyey-Hamilton theorem is often expressed by saying that every square matrix satisfies

its characteristics equation.

We shall now illustrate the Cayley-Hamilton therorem with the help of an example.

Example 6. Verify that the matrix A =

1 0 2

6 1 2

1 2 0

−

satisfies its characteristics equation.

Soluiton. For the given matrix A,

| A – I | =

1 0 2

6 1 2

1 2

− −

− −

= (1 – ) [{– (1 – ) + 4} + 2 {– 12 – (1 – )}

= (1 – ) (2 – + 4) + 2 ( – 13)

= – 3 + 22 – 3 – 22

The characteristics equation of A is therefore

– 3 + 22 – 3 – 22 = 0

or – 3 + 22 + 3 + 22 = 0 ...(i)

Now A2 =

6 0 2 6 0 2

6 1 2 6 1 2

1 2 0 1 0

− −

=

3 4 2

14 3 14 ,

11 2 2

− −

− − −

A3 =

3 4 2 1 0 2

14 3 14 6 1 2

11 2 2 1 2 0

− −

− − − −

=

19 8 2

10 31 22

25 2 26

− − − −

− −

Substituting A for in the left-hand side of (i) we have

140

A3 – 2A2 + 3A + 22 I

=

19 8 2 3 4 2 1 0 2 1 0 0

10 31 22 2 14 3 14 3 6 1 2 22 0 1 0

25 2 26 11 2 2 1 2 0 0 0 1

− − − − − − − + +

− − − − − −

=

0 0 0

0 0 0 0.

0 0 0

=

Thus we have

A3 – A2 + 3A + 22I = 0, showing that the matrix A satisfies its characteristic equation.

3. APPLICATION OF CAYLEY-HAMILTON THEOREM TO COMPUTE POWERS AND INVERSE OF A GIVEN SQUARE MATRIX

Cayley-Hamilton theorem can be used to compute powers of a square matrix and invers of a non-

singular square matrix.

Let the characteristic equation of an n-rowed square matrix A be

n + a1n – 1 + a

2n – 2 + ... + a

n = 0. ...(i)

By Cayley-Hamilton theorem the matrix A satisfies (i).

Therefore we have form (i),

Am + a1An – 1 + a

2An – 2 + ... + a

nAm – n = 0. ...(ii)

Substituting m = n + 1, n + 2, ... n + m in (iii) we have the relations

An + 1 + a1A

n + a

2An – 1 + ... + a

nA = 0. ...(iii)

An + 2 + a1An + 1 + a

2An + ... + a

nA2 = 0.

.....................................................................................

An + k + a1An + k – 1 + a

2An + k – 2 + ... + a

nAk = 0.

By substituting the values of A, A2, ... An in the first of the above equations, we get the value of An +

1. Similarly by substituting the values A2, A3, ... An + 1 in the second of the above equations we can get the

value of An + 2, and so on.

If the matrix A is non-singular, then an 0.

But multiplying (ii) throughout by 11 −

n

Aa

we have

1 2 1110− − −

+ + + =

n n

n n

aA A A

a a

or A–1 = 11 211 −− −

− + − + −

nn n

n n n

aaA A I

a a a ...(iv)

By substituting the values of I, A, A2, ... An – 1 in the right-hand side of (iv) and simplifying we can

get the value of A–1.

141

Example 7. Find the characteristic equation of the matrix

A =

1 2 3

2 3 4 ,

1 0 1

−

and hence compute its cube.

Solution. The characteristic equation of the matrix A is

1 2 3

? 2 3 4 0.

1 0 1

−

− =

− −

l

(1 – ) (3 – ) (–1 – ) – {2 (–1 – ) – 4} + 3 (–3 + ) = 0

3 – 32 + 8 = 0 ...(i)

By Cayley-Hamilton theorem by substituting A for in (i), we have

? A3 –3A2 – 8A = 0,

so that A3 = 3A2 + 8A. ...(ii)

Now A2 =

1 2 3 1 2 3

2 3 4 2 3 4

1 0 1 1 0 1

− −

? =

8 8 8

12 13 14

0 2 4

...(iii)

Substituting the values of A2 and A in (ii) we have

A3 =

8 8 8 1 2 3

3 12 13 14 8 2 3 4

0 2 4 1 0 1

+

−

=

32 40 48

52 63 74 .

8 6 4

Example 8. Find the characteristics equation of the matrix

A =

1 0 2

0 1 2

1 2 0

and hence compute is inverse.

Solution. The characteristic equation of the matrix A is

1 0 2

0 1 2 0

1 2

−

− =

−

(1 – ) {–(1 – ) – 4} + 2 (–1 + ) = 0

3 – 22 – 5 + 6 = 0. ...(i)

By Cayley-Hamilton theorem, A satisfies (i) so that we must have

A3 – 2A2 – 5A + 6I = 0. ...(ii)

142

Multiplying (ii) throughout by A–1 we have

A2 – 2A – 5I + 6A–1 = 0,

so that

6A–1 = –A2 + 2A + 5I

=

21 0 2 1 0 2 1 0 0

0 1 2 2 0 1 2 5 0 1 0

1 2 0 1 2 0 0 0 1

− + +

=

3 4 2 1 0 2 1 0 0

2 5 2 2 0 1 2 5 0 1 0

1 2 6 1 2 0 0 0 1

− + +

=

4 4 2

2 2 2

1 2 1

− −

−

whence A–1 =

4 4 21

2 2 2 ,6 1 2 1

− −

−

=

2 2 1

3 3 31 1 1

3 3 31 1 1

6 3 6

−

−

−

Verification. By actual multiplication it can be easily seen that AA–1 = I.

4. APPLICATION OF CAYLEY-HAMILTON THEOREM : SIMPLIFICATION OF MATRIX POLYNOMIALS

By using Cayley-Hamilton theorem of an n-rowed square matrix. A, a polynomial in A of degree

greater than n can be reduced to a polynomial in A of degree less than n, i.e., a polynomial of degree

greater than two in a 2 × 2 matrix A can be reduced to a linear polynomial in A, a polynomial of degree

greater than three in a 3 × 3 matrix A can be reduced to a quadratic polynomial in A, and so on. The

following example will illustrate the method.

Example 9. If A =3 1

,1 2

−

express 2A5 – 3A4 + A2 – 4I as a linear polunomial in A.

Solution. The characteristic equation of A is

3 1

1 2

−

− − = 0,

i.e., (3 – ) (2 – ) + 1 = 0,

i.e., 2 – 5 + 7 = 0.

The given polynomial is f (A), where

f () = 25 – 34 + 2 – 4.

By division alogrithm for polynomials we can write

f() = (2 – 5 + 7) (23 + 72 + 21 + 57) + 138 – 403

143

f(A) = (A2 – 5A + 7I) (2A3 + 7A2 + 21A + 57) + (138A – 403 I) ...(i)

Since every square matrix satisfies its characteristic equation, therefore

A2 – 5A + 7I = 0.

Substituting from (ii) in (i) it follows that f(A) = 138A – 403 I.

Exercise

1. Verify that the matrix

A =

1 2 1

1 0 3

2 1 1

−

−

satisfies its characteristic equation.

2. Obtain the characteristic equation of the matrix

A =

1 0 2

0 2 1

2 0 6

and hence calculate its inverse.

3. If A = 1 2

, express1 3

−

A6 – 4A5 + 8A4 – 12A3 + 14A2 as a linear polynomial in A.

4. Evaluate the matrix polynomial A5 – 27A3 + 65A2 as a 3 × 3 matrix, where

A =

0 0 1

3 1 0

2 1 4

−

5. If A =

1 0 0

1 0 1

0 1 0

show that A3 – A = A2 – I.

Deduce that for every integer n 3, An – An – 2 = A2 – I. Hence otherwise determine A50

144

Exercise-1

1. Show that the set

C2 = {(x

1, x

2) : x

1 C, x

2 C}

is a vector space over C with respect to co-ordinate wise addition and scalar multiplication.

2. Show that the set of all 2 × 2 matrices over C is a vector space over C with respect ot matrix

addition and multiplication of a matrix by a scalar.

3. Let V = {(a1, a

2, a

3, a

4) : a

1, a

2, a

3, a

4 are integers}

Is V a vector space over R with respect to co-ordinatewise addition and scalar multiplication ?

Justity your answer.

4. Let V = {(x1, x

2, x

3,) : x

1, x

2, x

3, are complex numbers, and x

1, x

2, = 0}

It V a vector space of over C with respect to co-oridnatewise addition and scaler multiplication ?

Justify your answer.

5. Show that the set of all matrices of the form ,

x o

o y where y C is a vector space over C with

respect to matrix addition and multiplication of a matrix by a scalar.

6. Show that the set of all matrices of the form , −

a b

b a where a, b C is a vector space over C.

With respect to matrix addition and multiplication of a matrix by a scalar.

Exercise-2

1. For each of the following matrices A, verify that At = A:

1 2 1 1 6 7

2 0 4 , 6 3 8 ,

1 4 3 7 8 5

− − − −

− − − −

a h g

h b f

g f c

2. For each of the following matrices A, verify that At = – A:

0 1 0 3 1 0 3 2

0 2 , 3 0 4 , 3 0 4

1 2 0 1 4 0 2 4 0

+ − − − − −

− − − − −

i i i

i

i i

3. If A =

1 2 21

2 1 2 ,3 2 2 1

−

− −

verify that A At = At = A = I3.

4. If A be any square matrix, verify that he matrix A + At is symmetric and the matrix A – At is

skew-symmetric.

5. If A be any square matrix, prove that A + A, AA, AA are all hermitian and A – A is

skew-hermitian.

6. Express the matrix

3 1 7

2 4 8

6 1 2

−

−

as X + Y where X is symmetric and Y is skew-symmetric.

145

7. Find remain matrices A and B such that

A+ iB =

1 1 3

4 2 1

3 2 3

+ −

− +

i

i

i i

Exercise-3

1. Apply the elementary operation R2 R

3 to the matrix

1 1 3

4 2 6

5 8 9

− −

2. Apply the elementary operation C2 → 2C

2 to the matrix

1 3 7 6

5 1 4 2

− − −

3. Write down the elementary matrix obtained by applying R3 → R

3 – 4R

1 to I

3.

4. Reduce the matrix

1 2 3 4

3 1 2 0

2 1 1 5

− −

−

to triangular from by applying E-row operations

5. Reduce the matrix

1 1 1

4 1 0

8 1 1

− −

to I3 by E-row operations

6. Verify that the E-row operation R1 → R

1 – R

3 on the matrix AB is equivalent to the same E-row

operation on A, where

A =

1 1 2 2 3 5

3 1 4 , 1 2 6

0 1 3 3 1 1

− − − = −

B .

Exercise-4

1. Find the inverse of the matrix

3 15 5

1 6 2

1 5 2

− − −

−

by calculating its adjoint.

2. Show that the matrix

1 2 3

0 1 4

2 2 1

− −

−

possesses an inverse.

146

3. Given that A =

1 1 0

2 1 0 ,

0 0 2

− −

calculate A2 and A–1

4. If A =

1 2 1 2 1 2

4 7 4 , 2 2 1 ,

4 9 5 1 2 2

− − − =

− −

B verify that (AB)–1 = B–1A–1.

5. By applying elementary row operations find the inverse of the matrix

1 1 2

3 1 3

2 1 1

−

−

.

Exercise-5

1. Find the rank of the matrix

3 1 2 6

.6 2 4 9

− − − −

2. Find the rank of the matrix.

1 2 3

0 3 4 .

0 0 0

−

3. Find the rank of the matrix

3 3 3

1 1 1

.

a b c

a b c

4. Reduce the matrix

1 1 2 3

4 1 0 2.

0 3 1 4

0 1 0 2

− −

to triangular form and hence find rank.

5. Find the rank of the matrix

2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

1 2 3 4

2 3 4 5.

3 4 5 6

4 5 6 7

Exercise-6

1. Does the following system of equations possess a unique common solution?

x + 2y + 3z = 6,

2x + 4y + z = 7,

3x + 2y + 9z = 10.

It so, find it.

147

2. Show that x = y = z = 0 is the only common solution of the following system of equations:

2x + 3y + 4z = 0,

x – 2y – 3z = 0,

3x + y – 8z = 0.

3. Solve the system of equations:

x + y + z = 3,

3x – 5y + 2z = 8,

5x – 3y + 4z = 14.

4. Solve the following system of equations:

x – 3y + 2z = 0,

7x – 21y + 14z = 0,

–3x + 9y – 6z = 0.

5. Which of the following system of equations are consistent?

(a) x – 4y + 7z = 8,

3x + 8y – 2z = 6,

7x – 8y + 26z = 31.

(b) x – y + 2z = 4,

3x + y + 4z = 6,

x + y + z = 1.

Exercise-7

1. Find the characteristics roots of each of the following matrices:

(i)

1 2 3

0 4 2

0 0 7

−

(ii)

1 1 1

1 1 0 .

1 0 1

− − −

−

2. Show that the matrices.

, ,

o g f o f h o h g

g o h f o g h o f

f h o h g o g f o

have the same characteristics equation.

3. Show that the matrices

, ,

a b c b c a c a b

b c a c a b a b c

c a b a b c b c a

4. Show that the characteristics roots of A* are the conjugates of the characteristics roots of A.

5. Show that 0 is a characteristics root of a matrix if and only if the matrix is singular.

6. If A and B are n-rowed square matrices and if A be invertible, show that the matrices A–1B and B

A–1 have the same characteristics roots.

148

Exercise-8

1. Verify that the matrix

A =

1 2 1

1 0 3

2 1 1

−

−

satisfies its characteristic equation.

2. Obtain the characteristic equation of the matrix

A =

1 0 2

0 2 1

2 0 6

and hence calculate its inverse.

3. If A = 1 2

, express1 3

−

A6 – 4A5 + 8A4 – 12A3 + 14A2 as a linear polynomial in A.

4. Evaluate the matrix polynomial A5 – 27A3 + 65A2 as a 3 × 3 matrix, where

A =

0 0 1

3 1 0

2 1 4

−

5. If A =

1 0 0

1 0 1

0 1 0

show that A3 – A = A2 – I.

Deduce that for every integer n 3, An – An – 2 = A2 – I. Hence otherwise determine A50.

149

UNIT 3

LESSON 1

GROUPS

In this chapter we study the properties of an important algebraic structure called Groups.

An algebraic structure is any set which is endowed with one or more binary operations satisfying

some given axioms.

The motivation for the study of these structures stems from our familiarity with the number

system. In the set I, of all integers, we have the following properties of addition:

1. a + b I a, b I (Closure Axiom)

2. a + (b + c) = a + (b + c) a, b, c I (Associative Law)

3. a + 0 = a a I (Existence of identity)

4. a + (–a) = 0 a I (Existence of inverse)

Similarly in the sets of all non-zero rational, real or complex numbers, the operation of

multiplication has similar properties, with 1 playing the role of the identity and 1

𝑎 playing the role of the

inverse of any element a.

We would like to introduce a binary operation in any abstract set, satisfying the properties

mentioned above.

Definition and examples of Groups

Let G be an arbitary set in which there exists a binary operation () satisfying the following

axioms.

G1 a b G a, b G

G2 a (b c) = (a b) c a, b, c G

G3 e G such that a e = e a = a a G

G4 To every a G, a' G a a' = a' a = e

Then G together with this operation () is called a group. We say that (G, ) is a group. Note that a

group is actually a pair – the set together with the operation. With respect to the binary operation, we may

say that G is a group.

Obvious examples of groups are [I, +], [Q – {0},.], {R –[0],.} {C –[0],.} where I, Q, R and C are

the sets of all integers, rational numbers, real numbers and complex numbers respectively.

We give some other examples of groups.

Example 1. Let G be the set of all even integers i.e. G = {2 p, p Z} where Z is the set of

integers. It is easy to see that G forms a group with respect to addition. The closure and associative

150

axioms are obviously true. 'o' serves as the identity element and for any 2p G, –2p is the additive

inverse and as such all the group axioms hold. Therefore the set of all even integers form a group under

ordinary addition.

Example 2. Consider the set S of all fourth roots of unity.

i.e. S = [1, –1, i, –i] where i = √−1

Let the binary composition defined on S be ordinary multiplication obviously closure and

associative axioms hold in S. It is easy to see that 1 is the identity element. The inverse of 1 is 1 itself. –1

is the inverse of itself. The inverse of i is –i and that of –i is i [ i2 = –1]

S forms a group under ordinary multiplication.

Example 3. Let G be the set of all non-singular n × n matrics, with real or complex entries. Let the

group operation be the usual matrix multiplication. it is easy to see that all the axioms are satisfied. The

unit matrix is the identity and the inverse of any matrix is the inverse.

Example 4. Let (G, ) be the group given by the following:

e x y z

e e x y z

x x e z y

y y z e x

z z y x e

TABLE-1

Verify that all the group axioms are satisfied.

Example 5. Let (G, ) be the group given by the following table where e is the identity element.

e a b c

e e a b c

a a e c b

b b c a e

c c b e a

TABLE-2

It is easily verified that all the group axioms are satisfied.

Example 6. Consider the following Table 3:

e a b c d

e e a b c d

a a e c d b

b b d e a c

c c b d e a

d d c a b e

151

TABLE-3

Note that (b c) d = a d = b

and b (c d) = b a = d

so that the associative property is not satisfied. Thus the set {e, a, b, c, d} under the given operation is not

a group.

Example 7. Consider the set of all vectors in space. Let the binary operation be the usual scalar or

dot product. Then this set together with this operation is not a group. The dot product of two vectors,

being a scalar, the closure property G1 is not satisfied.

Example 8. Consider the set V of all vectors in space, together with the operation a b = a × b

(the cross product). This is not a group as a × (b × c) (a × b) × c i.e. Associative Axiom is not true.

Remark: Note that one of the properties of addition and multiplication in the number system,

namely the commutative property (a + b = b + a, ab = ba) has not beeh stipulated as an axiom in the

definition of a group.

If a group (G, *) satisfies the additional property a * b = b * a a, b G, then G is called an

Abelian or Commutative of a group.

[I, +], [Q – {0},], {R –[0],} {C –[0],} are all Abelian groups.

The groups of Ex. 3 is non Abelian as matrix multiplication is not communtative.

Remark: If the group G is not abelian, it is not true that a * e = e * a a, b G. However, note

that G3 requires that the identity should commute with all elements of G. The axiom G4 requires that

element a should commute with its inverse.

Remark: Note that G3 merely states that e G satisfying a * b = b * a a G. It does not

insist on the uniqueness. For all that we know, there may be more than one element satisfying the

requirement of G3. Similarly G4 does not insist on the uniqueness of the inverse element of any a G.

We now prove that the axioms G1 to G4 ensure the uniqueness of the identity and the inverse.

Theorem: The identity of a group is unique.

Prof: Suppose two elements e, G such that

e * a = a e = a a G

and a = a = a a G

Treating e as in identity, e =

and treating f as an identity, e = e

= e = e

152

So that the identity of a group is unique.

Theorem 2: The inverse of any element a G is unique.

Prof: Suppose ā and a' such that

ā a = a ā = e

and a' a = a a' = e

We have ā = ā e = ā (a a') = (ā a) a' = e' a' = a'

so that ā = a'

Thus the inverse of any element is unique.

The inverse of any element is generally written as a–1. If the group operation is +, we denote the

inverse of a by –a.

Remark: (a–1)–1 = a and (ab)–1 = b–1 a–1, for abb–1 a–1 = aea–1 = aa–1 = e

Definition: if the number of elements in a group G is finite, we say that G is a finite group.

Otherwise, G is said to be an infinite group.

The groups of examples, 2, 4, 5 are finite groups. The groups of example 1 and 3 are infinite

groups.

Definition: Powers of an element.

Let (G, ) be a group. Let x G.

We define x0 = e and xn = x x ...... x, n times, for any positive integer n.

Remark: In (I, +), x0 = 0 for any integer x.

and xn = x + x + ... + n times

= nx.

DEFINITION:

1. Cyclic groups

If every element x G can be expressed as an for some fixed a G and some +ve integer n, then

G is said to be cyclic group. 'a' is said to be a generator of the group.

The set of all integers under addition is a cycle group. 1 is a generator.

2. Subgroups

Let (G, ) be any group. A subset S of G is said to be a subgroup is the elements of S satisfy all the

axioms of a group under the same binary composition as in G.

Example 1: In (I, +), the set of all even integers forms a subgroup. The set of all multiples of any

given integer is also a subgroup.

Example 2: Consider the group of Table 2.

The subset {a, e} is a subgroup.

153

The subset {b, e} is not a subgroup. b b = a and hence the closure axiom is not satisfied.

For every group G, the singleton set {e} and the entire set G, are always subgroups. Thus every

group contains atleast two subgroups. These two subgroups are called trivial subgroups.

A subset S of G is said to be a proper subgroup if it is a non-trivial subgroup. i.e., S {e}, S G.

In order that a subset S of a group G, be a subgroup, it is necessary that all the four axioms

G1-G4 are satisfied in S. However, that associative law is automatically satisfied, as every element of S is

also an element of G. Thus it is only necessary to verify the other 3 axioms. However, the next theorem

shows that it suffices to verify only two axioms.

Theorem 3: Let (G, ) be group. Let H be a non-empty subset of G. Then H is a subgroup of

G if,

(i) a, b H a b H

(ii) a H a-1 H

Proof: If H is a subgroup, the conditions (i) and (ii) are axiomatically satisfied.

Conversely, let (i) and (ii) be satisfied. As the associative property is automatically satisfied, we

have to only verify the existence of the identity in H.

By (ii) a H a–1 H. Also a a–1 H by (i) i.e. e H.

i.e. identity e exists in H.

H is a subgroup of G.

Theorem 4: If H is a nonempty finite subset of a group G, and H is closed under multiplication

then H is a subgroup of G.

Proof: Let a H. Since H is closed under multiplication

a2, a3, a4 .............................. all belong to H.

Again H being a finite set, this is possible only if there exists integers m and n such that am = an,

where m > n > 0.

am–n = e, e H

Also m > n > 0 m – n 1 m – n – 1 0

am–n–1 H

Also a. am–n–1 = am–n = e

Similarly am–n–1 . a = a m–n = e,

am–n–1 is the inverse of a H

For every a H, a–1 = am–n–1 H

H is a subgroup G.

154

Let G be a given group and H and K be two subgroups of G. What can we say about the subset

H K. It is interesting to note that H K need not necessarily be a subgroup of G. To illustrate this let

us consider the set I of all integers which is a group under addition.

Let I2 = [2P | p I ] and I3 = [3q | q I ] be subsets of I obviously I2, I3 are subgroups of I under

addition composition. But I2 I3 is not a subgroup of I.

∵ We see that 2, 3 I2 I3. But 2 + 3 = 5 I2 I3, and as such closure axiom is not valid in

I2 I3 we conclude that the union of two subgroups of a given group G need not necessary be a subgroup

of G. However the intersection of two subgroups of a given is necessarily a subgroup as proved in the

ensuing Theorem.

Theorem 5: Let G be a group. Let H, K be 2 subgroups of G. Then S = H K is also a subgroup.

Proof: Let a, b, S

Then a, b H, a, b K

As H is a subgroup, ab H.

Similarly ab K

Hence ab H K = S

Similarly a S a H, a K

So that a–1 H, a–1 K as H and K are subgroups of G

Hence a–1 H K = S

Conditions (i) and (ii) of theorem 3 are satisfied.

Hence S is a subgroup

Note: We will write ab for a b in future.

Definition: Congruence modulo a subgroup.

Let G be a group and let H be a subgroup of G. Define a relation, called congruence modulo H,

written as, a b (mod H). [a is congruent to b modulo H]

if ab–1 H.

Theorem 6: The relation a b (mod H) is an equivalence relation in G.

Proof: We show that this relation satisfies the properties of reflexivity, symmetry and transitivity.

Reflexivity: a a (mod H)

for aa–1 = e H. (H, being a subgroup, e H)

Symmetry: Let a b (mod H)

Then ab–1 H. As H is a subgroup, (ab–1) –1 = ba–1 H.

i.e. b a (mod H).

155

Transitivity: Let a b (mod H) and b e (mod H).

i.e. ab–1 H and bc–1 H.

Because of the closure property in H, (ab–1) (bc–1) = ac–1 H

a c (mod H).

Hence the relation of congruence is an equivalence relation.

Theorem 7: A non-empty subset H of a group G is a subgroup of G if ab–1 H for all a, b H

Proof: Firstly we show that the condition is necessary. Let H be a subgroup of G and a, b H.

Since H itself is a group.

b H b–1 H

Again a H, b–1 H ab–1 H

Now we shall show that the condition is sufficient as well.

Let ab–1 H a, b H

Putting b = a in the above we get aa–1 = e H i.e. identity exists in H.

Taking a = e, we find that eb–1 = b–1 H b H

Let a, b H a H, b–1 H. a(b–1) H i.e. ab H

H is a subgroup of G.

SOLVED EXAMPLES

Example 1: If a group G is such that (ab)2 = a2 b2 a, b G, prove that G is abelian.

Solution: We have (ab)2 = a2b2

i.e. abab = aabb

Premultiply by a–1 and postmultiply by b–1, on both sides.

a–1 (ab ab)b–1 = a–1 (aabb)b–1 Using Association axiom and remembering

a-1 a = e, eb = b etc. ]

ba = ab a, b G

Hence G is abelian.

Example 2: Prove that in a group G, the equation

ax = b and xa = b

have unique solution

Solution: Consider the equation

ax = b.

156

It is easily verified that a–1 b is solution, for

a(a–1b) = (aa–1) b = eb = b

To prove uniquences, suppose x and y are solutions.

Then ax = ay = b

Premultiplying by a–1, we have

a–1(ax) = a–1 (ay) = a–1 b

(a–1 a) x = (a–1 a)y ex = ey

giving x = y

Example 3: Let Q be the set of non-zero rational numbers and let the group operation be the

usual multiplication. Is the subset of integers I, a subgroup of Q ?

Solution: Note that the subset I satisfies the three axioms G1, G2 and G3 required of a group

namely, the closure axiom, the associative property and the existence of the identity.

However, the inverse of any integer a (under multiplication) is 1

𝑎 which does not belong to I.

Hence I is not a subgroup of (Q,.)

Example 4. Let G be the group of all 2×2 non singular matrics with real entries.

Let

𝑆 = [(𝑎 𝑜𝑏 𝑐

) 𝑎, 𝑏, 𝑐, ∈ 𝑅, 𝑎𝑐 ≠ 0]

Prove that S is a subgroup of G, the group operation being the usual matrix multiplication.

Solution: (i) The closure axiom is satisfied, for if A, B S

𝐴 = (𝑎1 𝑜𝑏1 𝑐1

) 𝐵 = (𝑎2 𝑜𝑏2 𝑐2

)

𝐴𝐵 = ( 𝑎1 𝑎2𝑏1𝑎2+𝑐1𝑏2

𝑜 𝑐1𝑐2

)

a1c1 0, a2 c2 0 a1a2c1c2 0

hence AB S

(ii) The identity (1 00 1

) S

(iii) Consider any element A of S where A = (𝑎 0𝑏 𝑐

)

1

𝑎𝑐(

𝑐 0−𝑏 𝑎

) which is also an element of S is the inverse of A.

Note the characteristic property of elements of S is that the entry in the first row and second

column should be zero.

Hence S is a subgroup of G.

157

LESSON 2

SUBGROUP

Let G be a group and H, a subgroup of G. Let a G. The set Ha = (h a|h H) is called a right coset of H

in G. Similarly the set

a H = (h a|h H) is called a left coset of H in G

In an abelian group, as ha = ah, the distinction between left and right cosets disappears. However,

if G is non abelian, the coset Ha, will in general, be different from aH.

Example 1 : Consider the group (I, +) and the subgroup H, consisting of all multiples of 4. (As the

group operation is +, we will denote a coset by H + a, instead of Ha). The coset H + 1 is

H + 1 = (................, –3, 1, 5, 9, ................)

H + 2 = (................, –2, 2, 6, 10, ................)

Note that the cosets H + 1, H + 2, are merely subsets of I.

They are not subgroups.

H + 5 = (................, –3, 1, 5, 9, ................)

Note that the coset H + 1 = the coset H + 5.

We thus see that, in general, the cosets Ha and Hb may be equal even if a b.

Note also that if h H, Hh = H

Theorem 1 : Let G be a group and H a subgroup of G. If a, b G, then the cosets Ha and Hb are

equal if ab–1 H i.e. a b (mod H)

In other words Ha = Hb a b (mod H)

Proof : Firstly we shall show that Ha = Hb a b (mod H)

Ha = Hb h1 a = h2b for some h1, h2 H

a = h1–1 h2 b = h3 b (say) [ ∵ h1

–1 h2 H]

ab–1 = h3 bb–1 = h3 i.e. ab–1 H or a b (mod H)

Conversely if a b (mod H)

i.e. ab–1 H.

Let ab–1 = h2

Consider any element x Ha

x Ha x = h1 a for some h1 a H

xb–1 = h1 ab–1

158

= h1 (ab–1) = h1 h2 = h3 (say) [ab–1 = h2]

x = h3b x Hb

It follows that Ha Hb. Similarly, it can be shown that Hb Ha

Ha Hb, Hb Ha Ha = Hb

Ha = Hb a b (mod H)

Theorem 2 : Two right (or left) cosets of H are either identical or disjoint i.e. If Ha, Hb are any

two right cosets of H in a group G then Ha = Hb or Ha Hb = .

Proof : If Ha Hb = , there is nothing to prove. Let us assume that Ha Hb . We shall show

that Ha = Hb.

Let x Ha Hb

x Ha, x Hb

h1h2 H such that

x = h1a, x = h2b

so that h1 a = h2b

Premultiplying by h1–1, we have

h1–1h1a = h1

–1h2b

i.e. a = (h1–1h2) b = h3b where h3 H

a = h3b

ab–1 = h3 H

a b (mod H)

By theorem 1, Ha = Hb

Similarly it can be shown that if aH, bH are any two left cosets H of in a group G, then aH = bH or

aH bH = .

Theorem 3 : Let G be a group. Let He be a subgroup of G. Let Ha, Hb be two distinct cosets of H

in G. Then there exists a one-one correspondence between Ha and Hb.

Proof : Any element x of Ha is of the form ha, for some h H.

If x Ha, define a function f as follows.

f(x) = hb. where x = ha

f(x) Hb f is a function from Ha to Hb.

f is obivously onto, for f–1(hb) = ha, for any element hb of Hb.

f is one-one. For if h1 a and h2 a are two different elements of Ha, their images h1 b and h2 b are

different. i.e. there exists a one-one correspondence between Ha and Hb.

159

Remark : The above theorem implies that if G is a finite group, then the number of elements in

any two cosets is the same.

Definition : Order of a group.

If G is a finite group, the number of elements in the group is called the order of G, written as o(G).

Lagrange’s Theorem : Let G be a finite group and H any subgroup of G. Then the order of H

divides the order of G. i.e. o(H) divides o(G).

Proof : Let o(G) = n, o(H) = m

Let H = {h1, h2 ....................... hm}. Consider any element a G. Firstly we shall prove that

o(aH) = m a G.

aH = {ah1 ah2 .............a hm}.

It is easy to see that the elements of aH are all distinct.

ahi = a hj, i j

hi = hj [by the cancellation law in a group] which is a contradiction for H is a finite group of

order m o(aH) = m

Let all possible distinct left cosets of H in G be a1 H, a2 H, ................ ak H

where a1 a2 .................. ak G

We shall prove that

G = a1H a2H .................. ak H.

obviously a1H G, a2H G .............. ahH G for each coset of H in G is a subset of G.

a1H a2H a3H .................... akH G

Let g G, be any element of G.

Then g aiH for some i satisfying 1 i k

For otherwise g g H for where g H is a left coset of H distinct from a1H,

a2H, .......ak H which is a contradiction

G a1H a2H a3H ................... akH

G = a1H a2H a3H ................... akH

o(G) = 0(a1H) + 0(a2H) + ...........0(akH)

= k0(H) 0(H) divides 0(G)

Definition : Index of a subgroup.

Let G be a group and H, a subgroup of G. The number of distinct cosets of H in G is called the

index of H in G.

If G is a finite group, the index of a subgroup H = 0(G) / 0(H).

160

Note, however, that the notion of index is valid even if G is an infinite group.

For example in (I, +), if H is the subgroup consisting of multiples of 4, the distinct cosets of H in I

are H, H + 1, H + 2 and H + 3.

Hence the index of H in I is 4

Normal subgroups

While defining the concept of cosets, we have observed earlier that left cosets are in general,

different from right cosets. However, subgroups for which left cosets are identical with right cosets

deserve to be studied specially. A subgroup H of G which is such that every left coset is also right coset

will be called a Normal subgroup. However, we start the definition of normal subgroups in a slightly

different manner and ultimately prove that the two definitions are equivalent.

Definition : A subgroup N of a group G is said to be a normal subgroup of G if gng–1 N g

G and n N

Remark : If G is abelian, gng–1 = gg–1n = n N g G, n N

Hence every subgroup of an abelian group is a normal subgroup.

In an arbitrary group, if N is a normal subgroup, gng–1 is some element of N

Remark : It is not necessary that gng–1 is equal to n. However, if G is abelian then the element

gng–1 is the same as n.

We state some important consequences of our definition in the form of a series of theorems.

Definition : Let G be any group and N is a subgroup of G, we define

gNg–1 = {gng–1 : n N}

where g is any element of G.

Theorem 4 : N is a normal subgroup of a group G if gNg–1 = N g G

Proof : In order to show that gN–1 g = N, it is sufficient to prove that gN–1g N and N gNg–1.

From the definition of normal subgroup it is clear that

gng–1 N g G, n N gNg–1 N

We have only to show that N gNg–1. Now g G g–1 G and N being a normal subgroup of

G it follows that

g–1ng N where n is any element of N.

Let g–1ng = n1 where n1 N

g(g–1ng) = gn1

(gg–1) (ng) = gn1

eng = gn1 [where e is the identity in G]

(ng) g–1 = gn1g–1

161

n = gn1g–1

N gNg–1

Now gNg–1 N, N gNg–1 gNg–1 = N

Theorem 5 : N is normal in G if every left coset of N is also a right coset of N, in G.

Proof : Let N be normal in G.

By Theorem 5, gNg–1 = N g G

Consider any n N. From the definition of normal subgroup, gng–1 N for all g G

let gng–1 = n1 where n1 N

gng–1g = n1g

gn = n1g

gN = Ng

i.e. every left coset of N is also a right coset of N in G. Conversely if gN = Ng g G we can show that

N is a normal subgroup of G. Consider any n N. For a given g

gN = Ng There exists n1 N such that gn = n1g

gng–1 = n1gg–1 = n1

i.e. gng–1 N

N is a normal subgroup of G.

This theorem establishes the equivalence of the two definitions of a normal subgroup.

Theorem 6 : If G is a group and N, a subgroup of index 2 in G, then N is a normal subgroup of G.

Proof : Let g be any element of G. If g N then obviously gN = N = Ng

If g N, then gN N, Ng N

From the definition of cosets, it follows that

[𝑔𝑁 𝑁 =

𝑁𝑔 𝑁 = ] ..........(1)

Since the index of N in G is 2, we have

[𝑁 𝑔𝑁 = 𝐺

𝑁 𝑁𝑔 = 𝐺] ..........(2)

Using (1) we see that gN = Ng g G.

N is a normal subgroup of G.

Theorem 7 : The intersection of two normal subgroups of G is again a normal subgroup of G.

Proof : Let H, K be 2 normal subgroups of G.

162

Let N = H K

Let n N, g G

Then n H, n K

As H is normal, gng–1 H

As K is normal, gng–1 K

so that gng–1 H K = N

Hence N is normal in g.

Theorem 8 : Let G be any group and H, an arbitrary subgroup of G and N, a normal subgroup of

G. Then H N is a normal subgroup of H

Proof : Let x H N. Let h H.

h H h G [∵ H < G]

x H N x H, x N

The hxh–1 N as N is normal in G.

Moreover, hxh–1 H as h, x and h–1 are in H.

hxh–1 H N h H, x H N

Hence H N is a normal subgroup of H.

SOLVED EXAMPLES

Example 1 : Let G be the group of integers (I, +). Let S be the subgroup containing multiples of 3.

List out the different cosets of S in I. What can you say about the cosets S + 1 and S + 7?

Solution : The different cosets of S in I are

S + 0 = S = { ........, –3, 0, 3, 6, .........}

S + 1 = { ........., –5, –2, 1, 4, 7, 10, ..........}

and S + 2 = { ........., –4, –1, 2, 5, 8, 11, ..........}

Note that I = S S + I S + 2.

The index of S in I is 3, as there are 3 distinct cosets.

Note also that 7 1 (mod S)

as 7 * 1–1 = 7 + (–1) = 6 S.

(Here * is + and a–1 = –a)

Hence S + 7 = S + 1

Example 2 : Let G be a group and let a G be a fixed element of G. Consider the set

S = [x g / xa = ax]

163

Then S is a subgroup of G.

Solution : In order to prove that S is a subgroup of G, it is just sufficient to verify that

x1, x2 S, x1x2 S

Also x S x–1 S

Let x1, x2 S. Then by definition of S.

𝑎𝑥1= 𝑥1𝑎

𝑎𝑥2= 𝑥2𝑎]

Now

a(x1x2) = (ax1)x2

= (x1a)x2

= x1 (ax2)

= x1 (x2a) = (x1x2)a

a(x1x2) = (x1x2)a x1x2 S

i.e. Closure Axiom is valid in S

Also x S xa = ax

x–1xa = x–1ax

[x G, x–1 exists and xx–1 = x–1 x = e

where e is the identity in G]

ea = (x–1a)x

ea x–1 = (x–1a)xx–1

ax–1 = x–1a

x–1 S

S is a subgroup of G.

EXERCISES

1. Show that each of the following sets is a group with respect to multiplication.

(a) {1}, (b) {1, –1}, (c) {1, , 2} where 3 = 1 (d) {1, i –1, –i} (i2 = –1).

2. Check if the following sets are groups.

(a) The set of all positive integers with respect to addition.

(b) The set of all rational numbers with respect to multiplication. present

(c) The set of all odd integers with respect to addition.

164

3. Let I be the set of integers. Define an operation (o) by a o b = a + b – 1. Prove that (I, o) is a group.

Identify the identity of I and the inverse of any element a I.

4. Let G be the group of all 2 × 2 non singular matrices with real entries, under the binary operation of

the usual matrix multiplication. Let S consist of all matrices of the form

(𝑎 0𝑏 1

), 𝑎 0. Check if S is a subgroup of G.

5. Let G = {0, 1, 2, 3, 4, 5} and let the binary operation be addition modul 6. Prove that G is a group.

Prove that the subset S = {0, 2, 4} is a subgroup of G. Write down the different cosets of S.

6. In the group of ex 5, consider the subset S1 = {0, 2, 3, 4}. Is S1 a subgroup of G?

165

LESSON 3

RING

Introduction : In the first two lessons we discussed Groups which are Algebraic structures only

one binary composition satisfying certain axioms. In the present lesson we introduce the notion of a ring

which is an algebraic structure having two binary compositions satisfying certain axioms. We give the

formal definition of a ring as follows:

Definition : A non-empty set R with two binary compositions * and o is to be ring with respect to

these compositions if the following axioms hold.

R–1 For all a, b, c R

a * (b * c) = (a * b) * c

R–2 There exists an element e R

such that a * e = e * a = a a R

R–3 a R, there exists b R

such that a * b = b * a = e

R–4 a, b R, a * b = b * a

R–5 a, b R, a o (b o c) = (a o b) oc

R–6 a, b c R

i) a o (b * c) = a o b * a o c

ii) (b * c) o a = b o a * c o a

It is a common practice to use the symbols + and reply for * and a and these compositions are

referred to as addition and multiplication respectively. You will do well to note that when we use the

symbols + and * for the compositions in a ring or refer to them as addition and multiplication, we do not

have to confuse them with addition and multiplication of numbers. They may be quite different from

ordinary addition and multiplication of numbers. They may stand for addition, multiplication of ordered

pairs, or vectors, or of matrices or they may be some other compositions. We choose to call them as

addition and multiplication for reasons to be seen later.

When we use the symbols + and * instead of * and o respectively we give a simple name to the

element e introduced in R-2. We call it the zero of the ring R and denote it by o of course the o of a ring

has nothing to do with the number zero of the number system. In view of our simplified notation as

introduced above, the definition of a Ring could be as follows:

Definition : A non-empty set R along with two binary composition + and * is said to be a Ring if

the following conditions hold:

R–1 For all a, b, c R [Associative Law]

a + (b + c) = (a + b) + c

166

R–2 o R such that a R

a + o = o + a = a [Existence of Zero]

R–3 For each a R, b R such that a + b = b + a = o

(Existence of additive inverse)

R–4 a, b R, a + b = b + a [Commutative Law of addition]

R–5 For all a, b, c R,

a(bc) = (ab)c [Associative Law of Multiplication]

R–6 For all a, b, c R

i) a(b + c) = ab + ac [Left Distributive Law]

ii) (b + c)a = ba + ca [Right Distributive Law]

The element b R in R–3 is called the additive inverse or the negative of ‘a’. We noticed that the

Axioms R–1 to R–4 are simply the Axioms for an Abelian group. We may state that a triple (R, +,.) is

said to be Ring if (R, +) is an Abelian group, (R,.) is a semi-group and multiplication is left as well as

right distributive with respect to addition.

SOME SPECIAL TYPES OF RINGS

1. Commutative Ring

A Ring R is said to be commutative if a, b R, ab = ba

A ring is said to be non-commutative if multiplication is not commutative. For a ring to be non-

commutative it is just sufficient to see if the ring possesses at least one pair of elements a, b for which ab

ba.

In a commutative Ring there is no distinction between left distributive and Right distributive laws

for each implies the other. In such a ring R, R–6 would read as follows. For all a, b, c R

a(b + c) = ab + ac (Distributive Law)

2. Ring with unity (or Identity)

R is said to be a ring with unity (or Identity) if there exists 1 R such that a.1 = 1.a = a a R

1 is called the Identity element of the Ring.

3. Ring without Zero divisors

A Ring (R, +,.) is said to be without zero divisors if the following condition holds.

For all a, b R, ab = o a = o or b = o

We say that a ring (R, +,.) has zero divisors if for some pair a, b of elements of the ring which are

both different from the zero of the ring, we have

ab = o

167

4. Integral Domain

A commutative ring with Identity and without zero divisors is called an Integral Domain.

SOME EXAMPLES OF RINGS

Example 1. The set Z of integers is a commutative ring with unit element under the usual addition

and multiplication composition. Similarly the set R of all real numbers and the set C of all complex

numbers are commutative rings with identity 1 under the usual addition and multiplication of real

numbers and complex numbers respectively.

Example 2. The set E of all even integers is a commutative ring without identity under the usual

addition and multiplication of numbers.

Note : The set F of all odd integers is not even group under addition, for the closure axiom is not

valid [ The sum of two odd integers is an even integer which is not a member of F]

Example 3. The set S of all matrices of the form

(𝑥 𝑜𝑦 𝑜) , 𝑥, 𝑦𝑄,

where Q is the set of rational numbers is a non-commutative ring without identity with respect to addition

and multiplication of matrices. Consider any two elements of S.

A = (𝑥1 𝑜𝑦1 𝑜) , 𝐵 = (

𝑥2 𝑜𝑦2 𝑜)

Then A + B = (𝑥1 + 𝑥2 𝑜𝑦1 + 𝑦2 𝑜)

[ x1 + x2 y1 + y2 Q for all x1 x2 y1 y2 Q]

Obviously Associative Law is true for matrix addition.

i.e. For All A, B, C S

(A + B) + C = A + (B + C)

The matrix (𝑜 𝑜𝑜 𝑜

) serves as the additive identity in S.

If A = (𝑥1 𝑜𝑦1 0), then 𝐵 = (

– 𝑥1 𝑜– 𝑦1 0

)

is the additive inverse of A.

Also A + B = B + A A, B S

for matrix addition is commutative

If A = (𝑥1 𝑜𝑦1 0) , 𝐵 = (

𝑥2 𝑜𝑦2 0)

then AB = (𝑥1𝑥20

𝑦1𝑥20) 𝑆

168

Also BA = (𝑥2𝑥10

𝑦2𝑥10)

AB BA

For all A, B, C S

A (BC) = (AB)C

as matrix multiplication is associative

Also [𝐴(𝐵 + 𝐶) = 𝐴𝐵 + 𝐴𝐶(𝐵 + 𝐶)𝐴 = 𝐵𝐴 + 𝐶𝐴

] 𝐴, 𝐵, 𝐶 𝑆

Since matrix multiplication distributes over addition so we conclude that S is a ring. The ring is non

commutative and does not have the identity element.

Example 4. The set S = {0, 1, 2, 3, 4} is a commutative ring with unity I under addition and

multiplication modulo 5.

Definition : We define the binary compositions n and ʘn in the set Z of integers as follows:

a n b = least non negative remainder obtained when a + b is divided by n.

a ʘn b = least non negative remainder obtained when a b is divided by n.

The compositors n and ʘn are known as addition and multiplication modulo n respectively. Now

consider the set S = {0, 1, 2, 3, 4}. It is easy to verify that

2 5 4, since the least non negative remainder.

when, 2 + 4 = 6 is divided by 5 is 1.

2 5 3 = 0, as the least non negative remainder when 2 + 3 = 5 is divided by 5 is zero.

Also 0 5 1 = 1, 2 5 0 = 2 etc.

0 serves as the identity element under the operation addition modulo 5.

2 5 3 = 0, 1 5 4 = 0 2 is the inverse of 3 and 4 is the inverse of 1 under addition modulo 5.

Also the inverse of 5 is 0 5 5 0 = 0

Also 2 ʘ5 3 = 1, 4 ʘ5 2 = 3 etc.

The set S = {0, 1, 2, 3, 4} is a ring under the operations of addition and multiplication modulo 5.

You may easily verify that the associative law is valid under the binary operations addition and

multiplication modulo 5. 1 serves as the identity element under multiplication modulo 5.

Example 5. Let C = {(a, b) : a, b R} where R is set of all real numbers i.e. C is the set of all

complex numbers. We define addition and multiplication is C as follows:

(a, b) + (c, d) = (a + c, b + d)

(a, b) . (c, d) = [ac – bd, ad + bc]

Then C is a commutative Ring with identity (1, 0) with respect to the operations of addition and

multiplication. Also (0, 0) is the zero of C and the additive inverse of (a, b) is (–a, –b).

169

The associative law of multiplication and one of the distributive laws are verified as follows:

(a, b) [(c, d) – (e, f)] = (a, b) . (ce – df, cf + de)

= [ace – adf – bcf – bde, acf + ade + bce – bdf]

Also {(a, b) . (c, d)} (e, f) = {ac – bd, ad + bc} . (e, f)

= {ace – bde – adf – bcf, acf – bdf + ade + bce}

and these are equal elements of C. Again, for one of the distributive laws, we get

(a, b) [(c, d) + (e, f)] = (a, b) . (c + e, d + f)

= (ac + ae – bd – bf, ad + af + bc + be)

Also (a, b) . (c, d) + (a, b) (e, f) = (ac – bd, ad + bc) + (ae – bf, af + be)

= (ac + ae – bd – bf, ad + af + bc + be)

The commutative law of multiplication follows easily as shown below:

(c, d) (a, b) = (ca – db, cb + da)

= (ac – bd, ad + bc)

= (a, b) . (c, d)

Example 6. Let R = (𝑎 𝑏𝑐 𝑑

) = a, b, c, d Z i.e. the set of all 2 × 2 matrices with integers as

elements. We define the usual addition and multiplication in R as follows:

(𝑎1 𝑏1

𝑐1 𝑑1) + (

𝑎2 𝑏2

𝑐2 𝑑2) = (

𝑎1 + 𝑎2 𝑏1 + 𝑏2

𝑐1 + 𝑐2 𝑑1 + 𝑑2)

(𝑎1 𝑏1

𝑐1 𝑑1) × (

𝑎2 𝑏2

𝑐2 𝑑2) = (

𝑎1𝑎2 + 𝑏1𝑐2 𝑎1𝑏2 + 𝑏1𝑑2

𝑐1𝑎2 + 𝑑1𝑐2 𝑐1𝑏2 + 𝑑1𝑑2)

Then R is a ring with respect to addition and multiplication as defined above. It is called the ring of

matrices of order 2 over integers. It is easy to see that R is a non-commutative ring with identity and with

zero divisors. The zero element and the identity of R are given by

O = (0 00 0

) and e = (1 00 1

) respectively

Note that (1 10 0

) (1 0

– 1 0) = (

0 00 0

)

We find that (1 10 0

) and (1 0

– 1 0) are divisors of zero in R

Theorem 1 : If R is a ring then for all a, b R

i) a.0 = 0.a = 0

ii) a(–b) = (–a)b = –ab

iii) (–a) (–b) = ab

If in addition R has an identity 1, then (–1) a = –a

170

(–1) (–1) = 1

Proof : If R has additive identity 0, then a + 0 = a a R

Also, in particular

0 + 0 = 0

a(0 + 0) = a.0

By distributive law a.0 + a.0 = a.0

= a.0 + 0

By Left Concellation Law in the group (R, +) we get

a.0 = 0

Similarly 0.a = 0

Also b + (–b) = 0 b R

a[b + (–b)] = a.0 = 0 [as proved above]

i.e. ab + a(–b) = 0 [By Distributive Law]

a(–b) = –ab

Similarly starting with

[a + (–a)] b = 0.b = 0 we can prove that (–a) b = –ab

To prove (iii) (–a) (–b) = ab

We have (–a) (–b) = (–a) c where c = –b

= –ac

= –[a (–b])

= –[–ab]

= ab

–(–x) = x is true in the additive group (R, +)

Definition : Integral Domain

Def : A commutative ring with identity and without zero divisions is called an Integral Domain.

If {R, +,.} is an integral Domain

i) R is commutative

i.e. a, b R, ab = ba

ii) ab = 0 a = 0 or b = 0

i.e. a 0, b 0 ab 0

171

i.e. R does not have zero divisors. For example, the set Z of integers is an integral Domain, as we

know that commutative law is true for multiplication of integers.

Also ab = 0 a = 0 or b = 0 where a, b Z

It is not necessary that every Ring be an integral Domain for example, you may verify that the set

Z6 = {0, 1, 2, 3, 4, 5} forms a Ring with respect to the binary operations of Addition modulo 6 and

multiplication modulo 6.

Also 2 Z6, 3 Z6 2 0

3 0]

But 2 ʘ6 3 = 0

Z6 is not an integral Domain.

Theorem 2. A finite integral Domain is a field.

Before proving the theorem, we shall define a field. A field is an Integral Domain in which every

non zero element has the multiplicative inverse. It is easy to see that the set of all rational numbers

𝑄 = [𝑝

𝑞/𝑝, 𝑞 𝑍, 𝑞 0] is a field. Every non-zero rational number

𝑝

𝑞 has the multiplicative inverse

𝑞

𝑝 as

𝑝

𝑞×

𝑞

𝑝= 1

Proof : Let R = {1, 2, 3, .......n

be a finite integral Domain

ai aj, i j .....(i)

Let a o be a non zero element of R [i.e. a = k for some k, 1 k n]

Consider the set aR.

aR = {a1, a2, a3, ..............an}

It is easy to see that the elements of aR are all distinct.

For if ai = aj i = j

i j i j

which contradicts (1)

By the closure property in R it follows that

R = aR [ ai R for all i R]

If 1 denotes the identity of R then for some k R

1 = adh dh is the inverse of a

R is a field.

172

Theorem 3. If p is a prime number then Jp the ring of integers modulo p is a field.

Proof : Let Jp = {0, 1, 2, 3, ............, p – 1}. It is easy to see that Jp is a ring under the operations of

addition and multiplication modulo p. As we have proved in Theorem 2 that every finite integral domain

is a field, it is just sufficient to show that Jp is an integral domain.

Let x, y Jp be such that

xy = 0 then x = 0 or y = 0

or p divides xy. If p divides xy, then p divides x or p divides y since p is prime.

x = 0 or y = 0 in Jp

xy = 0 in Jp x = 0 or y = 0 in Jp

It follows that Jp is a finite integral domain and therefore a field.

SUBRINGS : Let R be a ring. Any nonempty subset S of R is said to be subring if it forms a ring

under the binary compositions of R. Every ring R has two trivial subrings vis {0} and the ring R itself.

Example 1. The set of all multiples of three

viz S {3, 6, 9, ........ 0, –3, –6, –9} is a subring of the ring of integers.

You may easily verify that S satisfies all the Axioms required for a ring.

Example 2. The set of all 2 × 2 matrices of the form (𝑜 𝑥𝑜 𝑦) where x, y R, the set of all real

numbers, is a subring of the ring of all 2 × 2 matrices

We shall prove a theorem which gives sufficient conditions for a non empty subset S of a ring R to

be a subring of R.

Theorem 4. A non empty subset S of a ring is a subring of R, if and only if

(i) x – y S

(ii) xy S

x, y S

Proof : Firstly we shall show that the conditions are necessary. If S be a subring of R we have x, y

S, xy S [By Def. of a ring]

Also {S, +} is an abelian group

y S –y S x – y S For all x, y S.

The conditions are also sufficient if x – y S, x, y S it follows that S is an Abelian group with

respect to addition. Also (S, +) is subgroup of (R, +).

Let x, y, z S Then xy S by (ii) obviously x(yz) = (xy)z

x(y + z) = xy + xz

and (y + z)x = yx + zx, since S is a subset R and associative and distributive laws are true in R

S is a subring of R.

173

Theorem 5. The intersection of two subrings of a ring is a also a subring i.e. If S1, S2 are two

subrings of a griven ring R, then S1 S2 is also a subring in R.

Proof : Let x, y S1 S2

x S1, x S2, y S1, y S2

x – y S1, x – y S2 and xy S1, xy S2

as S1, S2 are subrings of R {By Theorem 4}

x – y S1 S2

and xy S1 S2 x, y, S1 S2

S1 S2 is a subring of R.

Note : The union of two sub-rings need not necessarily be a subring as is evident from the

following example.

Example 3. Consider the set of integers Z. Obviously Z is a ring. It is easy to see that

R2 = [2p / p Z]

R3 = [3p / p Z]

are subrings of Z. Also R2 R3 = [2, 3, 4, 6, ........0, –2, –3, –4, –6, ........]

2, 3 R2 R3

But 2 + 3 = 5 R2 R3

R2 R3 is not even a group with respect to addition in Z

R2 R3 is not subring of Z although R2, R3 are subrings of Z.

174

LESSON 4

VECTOR SPACES

1.1 INTRODUCTION

You are already familier with several algebric structures such as groups, rings, integral domains and

fields. In this lesson we shall tell you about another equally important algebric structures, namely, a

vector space.

Let V be a non-emply set and let F be a field. Let us agree to call elements of V vectors and

elements of F scalars.

A maping from V × V to V will be called addition in V and a mapping from F × V to V will be

called multiplication by a scalar multiplication, V is said to a vector space over F if addition and scalar

multiplication satisfy certain properties. Of course, these conditions are to be chosen in such a manner

that the resulting algebric structure is rich enough to be useful. Before presenting the definition of a vector

space, let us note that addition in V is denoted by the symbol ‘+’, and scalar multiplication is denoted by

juxtaposition, i.e., if x V, y V, and F, the x + y denoted the sum of x and y, and x denotes the

scalar multiople of x by .

Definition 1. A non-empty set V is said to be a vector space over a field F with respect to addition

and scalar multiplication if the following properties hold.

V 1 Addition in V is associative, i.e.,

x + (y + z) = (x + y) + z, for all x, y, z, V

V 2 There exists of natural element for addition in V, i.e., there exists an element 0 Î V such that

x + 0 = 0 + x = x, for all x V

V 3 Every element of V possesses a negative (or addition inverse), i.e., for each x V, there

exists an element y V such that

x + y = y + x = 0.

V 4 Addition in V is commulative, i.e., for all elements x, y V,

x + y = y + x

V 5 Associtiavity of scalar multiplication, i.e.,

i.e., (x) = ( ) x, for all , , F and x V

V 6 Property of 1. For all xV,

1x = x, where 1 is the multiplicative identity of F.

V 7 Distributivity properties for all , F and x, y V.

( + )x = x + x

(x + y) = x + y

Remarks 1. The first of the two distributivity properties stated in V 7 above is generally called

distributivity of scalar multiplication over addition in F, and the second of the two distributibity

properties is called distributivity of scalar multiplication over addition in V.

175

2. We generally refer to properties V 1 – V 7 above by saying that (V, +) is a vector space over F. If

the underlying field F is fixed, we simply say that (V, +,) is a vector space, and do not make an explicit

reference to F.

In case, the two vector space compositions are known, we denote a vectors space over a field’ F by

the symbol V(F). If there is no chance of confusion about the underlying field, then we simply talk of ‘the

vector space V’.

3. You might have observed that the axions V 1 to V 4 simply assert the V is an abelian group for

the composition ‘+’. In view of the we can re-state the definition of a vector space as follows:

2. DEFINNITION AND EXAMPLS OF A VECTOR SPACE

Defintion 2. A triple (V, +,) is said to be a vector space over a field F if (V, +) is an abelian group,

and the following properties are satisfied :

(x) = ()x, , F and x V

1x = x, x, V, where 1 is the multiplicative identity of F

( + )x = x + x, , F, and x, y V

(x + y) = x + y, F and x, y V

We shall now consider some examples of vector spaces.

Example 1. Let R be the set of number (R, +) is vector space over R. The addition is addition in R

and scalar multiplication is simply multiplication of real numbers.

It is easy to verify that all the vector space axioms are verified. In fact, V1-4 are satisfied because R

is an abelian group with respect to addition, V5 is nothing but the associative property of multipication,

V6 is the property of the multiplicative identity in (R, +,) and the properties listed in V7 are nothing but

the distributivity of multiplication over addition.

Example 2. (C, +, ) is a vector space over C

Example 3. (Q, +, ) is a vector space over Q.

Example 4. Let F be any field. F is a vector space over itself for the usual compositions of addition

and multipication (to be called scalar multiplication) in F.

Example 5. C is a vector space over R, and R is a vector space over Q.

Example 6. R is not a vector space over C. Observe that if C and x R, the x is not in R.

Therefore the multiplication composition in R fails to give rise to the scalar multiplication composition.

The examples considered above are in a way re-labelling of the field properties C, R or Q. We shall

now consider some examples of a different type.

Example 7. Let V be the set of all vectors in a plane. You know that addition of two vectors is a

vector, and that V is a group with respect to sum of vectors. Let us take addition of vectors as the first

compostion for the purpose of our example. Also, we know that if d be any vector and k be any real

numbers, then k d is a vector. Let us take R as the underlying field and multiplication of vector by a

scalar as the second vector space composition. It is easy to see that V is vector space over R for these two

compositons.

Example 8. Let R3 be the set

1 2 3 1 2, 3{( , , ) : , }x x x x x x R

and the addition and scalar multiplication R3 be defined as follows:

176

If x R3 and y R3 let

x + y = 1 1 2 2 3 3( , , ).+ + +x y x y x y

Also if x R3 and c R, let

cx = 1 2 3( , , ).cx cx cx

It can be seen that R3 is a vector space over R for the two compositions—addition and scalar

multiplication, as defined above.

We may note before passing on to the next example that the vector space being considered here is

nothing but the space of the vectors (in space) with addition and scalar multiplication as the composition.

This example is of special interest because it was in fact motivation for the present terminology of vector

spaces.

The next three examples are a little abstract in nature but are quite important.

Example 9. Let Rn be the set of ordered n-tuples of real numbers, so that a typical element of Rn is

We shall denote this element by x (printed as a bold-face better), and write

x = 1 2( , , ..., )nx x x

Let us take R to be the underlying field and define addition and scalar multiplication in Rn by setting

x + y = 1 1 2 2( , , ..., ),+ + +n nx y x y x y

where x = ( )1 2 1 2( , , ... ), , , ...=n nx x x y y y y

and x = ( )1 2, , ... , nx x x R

Let us first of all see that addition scalar multiplication as defined above are meaningful in the sense

that they define the two compositions that we need for making Rn vector space.

Since 1 2 3 1 2, , , and , , ...n nx x x x y y y are all real numbers, therefore 1 1 2 2, , ...,+ +n nx y x y x y are all

real numbers and therefore 1 1 2 2( , , ..., )+ + +n nx y x y x y is an ordered n-type of real numbers and

consequently it is in Rn. Again, since is a real number and 1 2( , , ..., )nx x x are also real numbers,

therefore 1 2, , ..., nx x x are also real numbers, and consequently 1 2( , , ..., ) nx x x is an n-type of real

numbers and so is Rn.

Having defined addition and scalar multiplication in Rn, let us see in some detail that all the

properties needed for Rn to be a vector space are actually satisfied.

1. Let x = 1 2 1 2 1 2( , , ..., ), ( , , ..., ), ( , , ..., )= =n n nx x x y y y y z z z z be any three elements of Rn.

Then (x + y) + z = 1 1 2 2 1 2( , , ..., ) ( , , , )+ + + +n n nx y x y x y z z z

= 1 1 1 2 2 2[( ) , ( ) , ..., ( ) ]+ + + + + +n n nx y z x y z x y z

= 1 1 2 2 2[ ( ), ( ), , ( )]1 + + + + + +n n nx y z x y z x y z

= 1 2 3 1 1 2 2( , , , , ) ( , , ... )+ + + +n n nx x x x y z y z y z

= x + (y + z).

2. Let o = (0, 0, ..., 0). so that o Rn and x + 0 = o + x = x for all x Rn

3. Let x = 1 2( , , ..., )nx x x be any element of Rn. If y = (–x1, –x

n, ..., –x

n) then y Rn and y is an

element of Rn such that

x + y = y + x = 0

4. If x = 1 2 1 2( , , , ), ( , , ..., )=n nx x x y x y y are any two elements of Rn then

177

x + y = 1 2 1 2( , , , ), ( , , ..., )+n nx x x y y y

= 1 1 2 2( , , , )+ + +n nx y x y x y

= 1 1 2 2( , , , )+ + +n ny x y x y x

= 1 2 1 2( , , ... ) ( , , ..., )+n ny y y x x x

= y + x.

5. If x = 1 2( , , ... ),nx x x be any element of Rn and p, q be any real numbers, then

(pq)x = 1 2( ) ( , , ... )npq x x x

= 1 2[( ) , ( ) , ..., ( ) ]npq x pq x pq x

= 1 2[ ( ), ( ), ... ( )]np qx p qx p qx

= 1 2( , , ... )np qx qx qx

= p(qx)

6. If x = (x1, ..., x

n) be any element of Rn and p, q be any real numbers, then

(p + q) x = 1 2[( ) , ( ) , ..., ( ) ]+ + + np q x p q x p q x

= 1 2 1 2( , , ..., ) ( , , ..., )+n npx px px qx qx px

= px + qx

7. If 1 2( , , ..., )= nx x x x and 1 2( , , ..., )= ny y y y be any two elements of Rn, and p be any real number,

then

p(x + y) = p(x1 + y

1, x

2 + y

2, ..., x

n + y

n)

= [p(x1 + y

1), p(x

2 + x

2), ..., p(x

n + y

n)]

= (px1, px

2, ..., px

n) + (py

1, py

2, ..., py

n)

= p(x1, x

2, ..., x

n) + p(y

1, y

2, ..., y

n)

= px + py.

8. If 1 2( , , ..., )= nx x x x be any element of Rn, then

1x = 1 21( , , ..., )nx x x

= 1 2(1 ,1 , ,1 ) nx x x

= 1 2( , , ... )nx x x

= x.

From 1-8 above we find that Rn is a vector space over R with co-ordinatewise addition and

co-ordinatewise scalar multiplication as the two vector space compositions.

The use of the word co-otherwise, is due to the fact that if 1 2( , , ..., )= nx x x x be any element of Rn

then 1 2, , ..., nx x x are called the co-ordinates of x, and while defining x + y, we add the corresponding

co-ordinates of x and y, and while defining cx we multiply the co-ordinates of x by c.

We may note that the space in example 9 is only a spacing case of the example 9 for n = 3.

Example 10. The set Cn of all ordered n-tuples of complex number is a vector space over C for

co-ordinatewise addition and co-ordinatewise scalar multiplication as the two vector space compositions.

178

Example 11. Let F be any field. The set Fn of all ordered n-tuples of elements of F is a vector space

over F with co-ordinatewise addition and co-ordinatewise scalar multiplication as the two vector space

compositions.

Example 12. Let Mmn

is a vector space over C with respect to matrices over C. Mmn

is a vector space

over C with respect to matrix addition and multiplication of a matrix by a scalar, for

1. The sum of two m × n matices with complex entries is an m × n matrix with complex entries.

2. Addition of matrices is associative.

3. The m × n zero-matrix is a natural element for addtion.

4. It A be an m × n matrix with complex entries, then –A is also an m × n with complex entries

such that (–A) + A + (–A) = 0

5. Addition of matrices is commutative.

6. If A Mmn

and c be any complex number, then –A is also m × n matrix with complex entries

and so cA Mmn

7. If p, q be any complex numbers and A, B be any two m × n matrices with complex entries, then

p(q A) = (pq) A,

(p + q) A = pA + qA,

p(A + B) = pA + pB,

1 A = A, for all A Mmn

Example 13. The set S of all matrices of the form −

a b

b a, where a, b are any complex numbers, is

a vector space over C with respect to matrix addition and multiplication by a scalar for the following

reasons:

1. If A, B, S, the A + B S.

For, if A = , − −

a b c d

b a d c

then A + B = , where , .( )

+ + = = + = + − + + −

a c b d p qp a c q b d

b d a c q p

2. Matrix addition is associative.

3. The matrix ,

=

o oO S

o o and A + O = O + A = A, for all A S.

4. If ,

= −

p qA S

q pthen the matrix ,

− − = −

p qB S

q p and is such that A + B = B + A = 0.

5. Addition f matrices is commulative.

6. If c be any complex number, and A = −

a b

b a S, then cA is the matix given by

cA = −

ca cb

cb ca.

It is obvious that cA S.

179

7. If p, q be any complex numbers, and A, B be any two matrices in S, say ,

= −

a bA

b a

= −

c dB

d c, then

(i) (pq)A = ( ) ( ) ( ) ( )

( ).( )( ) ( ) ( ) ( )

= = = − − −

a b pq a pq b p qA p qbpq p qA

b a pq b pq a p qb p qa

(ii) (p + q) A = pA + qA

(iii) p(A + B) = pA + pB

(iv) 1 A = A.

Example 14. The set of all matrices of the form ,

x y

z o where x, y, z, C, is a vector space over

C with respect to matrix addition and multiplication of a matrix by a scalar.

The verification of the vector space axioms is straight forward.

Example 15. The set s of all hermitian matrices of order n is a vector space over R with respect to

matrix addition and multiplication of matrix by a scalar.

To verify that (S, +, ) is a vector space over C, we proceed as follows:

1. Let A, B be two hermitian matrices of order n. Then A + B is a matrix of order n. It is hermitian

because (A + B) = A + B = A + B, since A = A, B = B.

2. Addition of matrices is associative.

3. The n-rowed, zero matrix O is a matrix is a hermitian matrix such that A + 0 = 0 + A = A

4. It A S, so that At = A, then (–A)t = –At = –A, so that –4A S, and A + (–A) = (–A) + A = 0.

5. Matrix addition is commutative

6. If C R, and A S, then (CA)t = CAt = CA, so that A S.

7. If p, q, R and and A B S, then

(i) (p, q)A = p(qA)

(ii) (p + q)A = pA + qA

(iii) p(A + B) = pA + pB

(iv) 1 A = A

In view of the above properties it follows that (S, +, ) is a vector space over R.

Example 16. The set S of all real symmetric matrices of order n is a vector space over R with

respect to matrix addition and multiplication of a matrix by a scalar.

In order to convince ourselves that (S, +, ) is a vector space over R, let us note the following:

1. A, B be n- rowed real symmetric matrices, then At = A, Bt = A. It follows that

(A + B)t = At + Bt = A + B

2. Addition of matrices is associative, so that for all A, B, C S,

A + (B + C) = (A + B) + C.

3. The n-rowed zero matrices o is a real symmetric matrix, and therefore it is in S.

Also, A + 0 = 0 + A = A

for all A in S.

4. If A S so that At = A, then (–A)t = –At = –A, so that—17 A S. Also, A + (–A) = (–A) + A = 0.

180

5. Matrix addition is commulatative.

6. If C R and A S, then (cA)t = cAt = cA, so that A S.

7. If p, q R, and A, B S, then

(i) (p, q) A = p(qA)

(ii) (p + q)A = pA + qA

(iii) P(A + B) = pA + pB

(iv) 1 A = A

3. SOME DIRECT CONSEQUENTS OF VECTOR SPACE AXIOMS

We shall now state and prove some elementary consequences of the vector space axioms. These will

help us in dealing with vector in a convenient way in many situations.

Theorem. Let V be a vector over a field F. Then for all a F and x V,

(i) 0 = 0,

(ii) 0x = 0,

(iii) (–) x = –ax

(iv) (–) (–x) = x

(v) x = 0 if either = 0 or x = 0.

Proof.

(i) 0 = (0 + 0), by the property of 0 V

= 0 + 0, by distributivity of scalar multiplication over addition in V

Since 0 + 0 = 0 = 0 + 0, by the property of 0 in V by cancellation law in (V, +), it

follows that 0 = 0

(ii) 0 x = (0 + 0) x, by the property of 0 in F

= 0x + 0x, by distributivity of scalar multiplication over addition in F.

Since 0x + 0x = 0x + 0, by, the property of 0 in V, therefore by cancellation law in (V, +),

it follows thats

0x = 0

(iii) 0 = a + (–) 0x = ( + (–))x

0 = x + (–) x, since ox = 0.

Now x, (–)x are two elements of V such that ax + (–)x = 0, therefore (–) x is the negative of.

x, i.e, (–) x = –(x).

(iv) Now –(0) = – (x + (–x))

0 = –x + ((–) (–x)), since –(0) = 0, by (i) above

(–) (–x) is the negative of –x in V

(–) (–x) = –(–x) –x,

because –x V and therefore negative of –x in V is simply x.

(v) Let us first suppose that ax = 0. If = 0, we are done. If 0, then –1 F, because F

and F is a field

Therefore 0 = –1 0

181

= –1 (x), because x = 0 by hypothesis

= (–1 )x, by associativity of scalar multiplication

= 1x

= x.

Thus x = 0 either = 0 or x = 0. Conversely, let us assume taht either = 0 or x = 0

In case = 0, by (ii) above x = 0x = 0.

In case x = 0, by (i) above x = 0 = 0.

Thus in both cases we find that x = 0.

It is now obvious that x = 0 if either x = 0 or = 0.

Exercise

1. Show that the set

C2 = {(x1, x

2) : x

1 C, x

2 C}

is a vector space over C with respect to co-ordinateswise addition and scalar multiplication.

2. Show that the set of all 2 × 2 matrices over C is a vector space over C with respect to matrix

addition and multiplication of a matrix by a scalar.

3. Let V = {(a1, a

2, a

3, a

4) : a

1, a

2, a

3, a

4 are integers}

Is V a vector space over R with respect to co-ordinatewise addition and scalar multiplication?

Justify your answer.

4. Let V = {(x1, x

2, x

3) : (x

1, x

2, x

3, are complex numbers, and x

1, x

2, = 0}

Is V a vector space over C with respect to co-ordinateswise addition and scalar multiplication?

Justify your answer.

5. Show that the set of all matrices of the form ,

x o

o y where y C is a vector space over C with

respect to matrix addition and multiplication of a matrix by a scalar.

6. Show that the set of all matrices of the form , −

a b

b a where a, b C is vector space over C

with respect to matrix addition an multiplication of matrix by a scalar.