Mathematics

By jitendra thakor

Session

Matrices and Determinants - 2

Session Objectives

Determinant of a Square Matrix

Minors and Cofactors

Properties of Determinants

Applications of Determinants

Area of a TriangleSolution of System of Linear Equations (Cramer’s Rule)

Class Exercise

DeterminantsIf is a square matrix of order 1,

then |A| = | a11 | = a11

ijA = a

If is a square matrix of order 2, then11 12

21 22

a aA =

a a

|A| = = a11a22 – a21a12

a a

a a

1 1 1 2

2 1 2 2

Example

4 - 3Evaluate the determinant :

2 5

( )4 - 3Solution : = 4×5 - 2× -3 = 20 + 6 = 26

2 5

Solution

If A = is a square matrix of order 3, then11 12 13

21 22 23

31 32 33

a a aa a aa a a

[Expanding along first row]

11 12 1322 23 21 23 21 22

21 22 23 11 12 1332 33 31 33 31 32

31 32 33

a a aa a a a a a

| A |= a a a = a - a + aa a a a a a

a a a

( ) ( ) ( )11 22 33 32 23 12 21 33 31 23 13 21 32 31 22= a a a - a a - a a a - a a + a a a - a a

( ) ( )11 22 33 12 31 23 13 21 32 11 23 32 12 21 33 13 31 22a a a a a a a a a a a a a a a a a a= + + − + +

Example2 3 - 5

Evaluate the determinant : 7 1 - 2-3 4 1

( )2 3 - 5

1 - 2 7 - 2 7 17 1 - 2 = 2 - 3 + -5

4 1 -3 1 -3 4-3 4 1

( ) ( ) ( )= 2 1+ 8 - 3 7 - 6 - 5 28 + 3

= 18 - 3 - 155= -140

[Expanding along first row]

Solution :

Minors

-1 4If A = , then

2 3

21 21 22 22M = Minor of a = 4, M = Minor of a = -1

11 11 12 12M = Minor of a = 3, M = Minor of a = 2

Minors4 7 8

If A = -9 0 0 , then2 3 4

M11 = Minor of a11 = determinant of the order 2 × 2 square sub-matrix is obtained by leaving first row and first column of A

0 0= =0

3 4

Similarly, M23 = Minor of a23

4 7= =12-14=-2

2 3

M32 = Minor of a32 etc.4 8

= =0+72=72-9 0

Cofactors

( ) i+ jij ij ijC = Cofactor of a in A = -1 M ,

ij ijwhere M is minor of a in A

Cofactors (Con.)

C11 = Cofactor of a11 = (–1)1 + 1 M11 = (–1)1 +1 0 0

=03 4

C23 = Cofactor of a23 = (–1)2 + 3 M23 = − =4 7

22 3

C32 = Cofactor of a32 = (–1)3 + 2M32 = etc.4 8

- =-72-9 0

4 7 8A = -9 0 0

2 3 4

Value of Determinant in Terms of Minors and Cofactors

11 12 13

21 22 23

31 32 33

a a aIf A = a a a , then

a a a

( )3 3

i jij ij ij ij

j 1 j 1

A 1 a M a C+

= == − =∑ ∑

i1 i1 i2 i2 i3 i3= a C +a C +a C , for i=1 or i= 2 or i= 3

Properties of Determinants1. The value of a determinant remains unchanged, if its

rows and columns are interchanged.

1 1 1 1 2 3

2 2 2 1 2 3

3 3 3 1 2 3

a b c a a aa b c = b b ba b c c c c

i e A A=. . '

2. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant is changed by minus sign.

[ ]1 1 1 2 2 2

2 2 2 1 1 1 2 1

3 3 3 3 3 3

a b c a b ca b c = - a b c R Ra b c a b c

Applying ↔

Properties (Con.)

3. If all the elements of a row (or column) is multiplied by a non-zero number k, then the value of the new determinant is k times the value of the original determinant.

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

ka kb kc a b ca b c = k a b ca b c a b c

which also implies

1 1 1 1 1 1

2 2 2 2 2 2

3 3 3 3 3 3

a b c ma mb mc1

a b c = a b cm

a b c a b c

Properties (Con.)4. If each element of any row (or column) consists of

two or more terms, then the determinant can be expressed as the sum of two or more determinants.

1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3

a +x b c a b c x b ca +y b c = a b c + y b ca +z b c a b c z b c

5. The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column).

[ ]1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 1 1 2 3

3 3 3 3 3 3 3 3

a b c a +mb - nc b ca b c = a +mb - nc b c C C + mC -nCa b c a +mb - nc b c

Applying →

Properties (Con.)6. If any two rows (or columns) of a determinant are

identical, then its value is zero.

2 2 2

3 3 3

0 0 0a b c =0a b c

7. If each element of a row (or column) of a determinant is zero, then its value is zero.

1 1 1

2 2 2

1 1 1

a b ca b c =0a b c

Properties (Con.)

( )a 0 0

8 Let A = 0 b 0 be a diagonal matrix, then0 0 c

a 0 0= 0 b 0

0 0 cA abc=

Row(Column) OperationsFollowing are the notations to evaluate a determinant:

Similar notations can be used to denote column operations by replacing R with C.

(i) Ri to denote ith row

(ii) Ri Rj to denote the interchange of ith and jth rows.

(iii) Ri Ri + λRj to denote the addition of λ times the elements of jth row to the corresponding elements of ith row.

(iv) λRi to denote the multiplication of all elements of ith row by λ.

↔

↔

Evaluation of DeterminantsIf a determinant becomes zero on putting is the factor of the determinant.( )x = , then x -α α

2

3

x 5 2

For example, ifΔ = x 9 4 , then at x =2

x 16 8

, because C1 and C2 are identical at x = 2

Hence, (x – 2) is a factor of determinant .

∆ = 0

∆

Sign System for Expansion of Determinant

Sign System for order 2 and order 3 are given by

+ – ++ –

, – + –– +

+ – +

( )42 1 6 6×7 1 6

i 28 7 4 = 4×7 7 414 3 2 2×7 3 2

[ ]1

6 1 6=7 4 7 4 Taking out 7 common from C

2 3 2

Example-1

6 -3 22 -1 2

-10 5 2

42 1 628 7 414 3 2

Find the value of the following determinants

(i) (ii)

Solution :

1 3= 7×0 C and C are identical

= 0

Q

Example –1 (ii)6 -3 22 -1 2

-10 5 2

(ii)

( )( )

( )

− × − −= − × − −

× −

3 2 3 2

1 2 1 2

5 2 5 2

− −= − − − −

= − × =

1

1 2

3 3 2

( 2) 1 1 2 Taking out 2 common from C

5 5 2

( 2) 0 C and C are identical

0

Q

Evaluate the determinant1 a b+c1 b c+a1 c a+b

Solution :

[ ]3 2 3

1 a b+c 1 a a+b+c1 b c+a = 1 b a+b+c Applying c c +c1 c a+b 1 c a+b+c

→

( ) ( ) 3

1 a 1= a+b+c 1 b 1 Taking a+b+c common from C

1 c 1

Example - 2

( ) 1 3= a+b + c ×0 C and C are identical

= 0

Q

2 2 2

a b cWe have a b c

bc ca ab

[ ]21 1 2 2 2 3

(a-b) b-c c= (a-b)(a+b) (b-c)(b+c) c Applying C C -C and C C -C

-c(a-b) -a(b-c) ab→ →

( ) ( )2

1 2

1 1 cTaking a-b and b-c common

=(a-b)(b-c) a+b b+c cfrom C and C respectively

-c -a ab

Example - 3

bc

2 2 2

a b ca b c

ca ab

Evaluate the determinant:

Solution:

[ ]21 1 2

0 1 c=(a-b)(b-c) -(c-a) b+c c Applying c c -c

-(c-a) -a ab→

2

0 1 c=-(a-b)(b-c)(c-a) 1 b+c c

1 -a ab

[ ]22 2 3

0 1 c= -(a-b)(b-c)(c-a) 0 a+b+c c -ab Applying R R -R

1 -a ab→

Now expanding along C1 , we get(a-b) (b-c) (c-a) [- (c2 – ab – ac – bc – c2)]= (a-b) (b-c) (c-a) (ab + bc + ac)

Solution Cont.

Without expanding the determinant,

prove that 3

3x+y 2x x4x+3y 3x 3x =x5x+6y 4x 6x

3x+y 2x x 3x 2x x y 2x xL.H.S= 4x+3y 3x 3x = 4x 3x 3x + 3y 3x 3x

5x+6y 4x 6x 5x 4x 6x 6y 4x 6x

3 2

3 2 1 1 2 1=x 4 3 3 +x y 3 3 3

5 4 6 6 4 6

Example-4

Solution :

[ ]3 21 2

3 2 1=x 4 3 3 +x y×0 C and C are identical in II determinant

5 4 6Q

Solution Cont.

[ ]31 1 2

1 2 1=x 1 3 3 ApplyingC C -C

1 4 6→

[ ]32 2 1 3 3 2

1 2 1=x 0 1 2 ApplyingR R -R and R R -R

0 1 3→ →

[ ]31

3

= x ×(3-2) Expanding along C

=x =R.H.S.

3

3 2 1=x 4 3 3

5 4 6

Prove that : = 0 , where ω is cube root of unity.

3 5

3 4

5 5

1ω ωω 1 ωω ω 1

3 5 3 3 2

3 4 3 3

5 5 3 2 3 2

1ω ω 1 ω ω .ωL.H.S=ω 1 ω = ω 1 ω .ω

ω ω 1 ω .ω ω .ω 1

[ ]

2

3

2 2

1 2

1 1ω= 1 1ω ω =1

ω ω 1

=0=R.H.S. C and C are identical

Q

Q

Example -5

Solution :

Example-6

2

x+a b ca x+b c =x (x+a+b+c)a b x+C

Prove that :

[ ]1 1 2 3

x+a b c x+a+b+c b cL.H.S= a x+b c = x+a+b+c x+b c

a b x+C x+a+b+c b x+c

Applying C C +C +C→

Solution :

( )

( ) 1

1 b c= x+a+b+c 1 x+b c

1 b x+c

Taking x+a+b+c commonfrom C

Solution cont.

[ ]2 2 1 3 3 1

1 b c

=(x+a+b+c) 0 x 00 0 x

Applying R R -R and R R -R→ →

Expanding along C1 , we get

(x + a + b + c) [1(x2)] = x2 (x + a + b + c)

= R.H.S

[ ]1 1 2 3

2(a+b+c) 2(a+b+c) 2(a+b+c)= c+a a+b b+c Applying R R +R +R

a+b b+c c+a→

1 1 1=2(a+b+c) c+a a+b b+c

a+b b+c c+a

Example -7

Solution :

Using properties of determinants, prove that

2 2 2

b+c c+a a+bc+a a+b b+c =2(a+b+c)(ab+bc+ca-a -b -c ).a+b b+c c+a

b+c c+a a+bL.H.S= c+a a+b b+c

a+b b+c c+a

[ ]1 1 2 2 2 3

0 0 1=2(a+b+c) (c-b) (a-c) b+c Applying C C -C and C C -C

(a-c) (b-a) c+a→ →

Now expanding along R1 , we get

22(a+b+c) (c-b)(b-a)-(a-c)

2 2 2=2(a+b+c) bc-b -ac+ab-(a +c -2ac)

Solution Cont.

2 2 2=2(a+b+c) ab+bc+ac-a -b -c

=R.H.S

Using properties of determinants prove that

2

x+4 2x 2x2x x+4 2x =(5x+4)(4-x)2x 2x x+4

Example - 8

1 2x 2x=(5x+4)1 x+4 2x

1 2x x+4

Solution :

[ ]1 1 2 3

x+4 2x 2x 5x+4 2x 2xL.H.S= 2x x+4 2x =5x+4 x+4 2x Applying C C +C +C

2x 2x x+4 5x+4 2x x+4→

Solution Cont.

[ ]2 2 1 3 3 2

1 2x 2x=(5x+4) 0 -(x-4) 0 ApplyingR R -R and R R -R

0 x-4 -(x-4)→ →

Now expanding along C1 , we get

2(5x+4) 1(x-4) -0

2=(5x+4)(4-x)

=R.H.S

Example -9Using properties of determinants, prove that

x+9 x xx x+9 x =243 (x+3)x x x+9

x+9 x xL.H.S= x x+9 x

x x x+9

[ ]1 1 2 3

3x+9 x x= 3x+9 x+9 x Applying C C +C +C

3x+9 x x+9→

Solution :

[ ]1=3(x+3) 81 Expanding along C

=243(x+3)=R.H.S.

×

1 x x=(3x+9)1 x+9 x

1 x x+9

Solution Cont.

( ) 2 2 1 3 3 2

1 x x=3 x+3 0 9 0 Applying R R -R and R R -R

0 -9 9

→ →

Example -10

Solution :

2 2 2 2 2

2 2 2 2 21 1 3

2 2 2 2 2

(b+c) a bc b +c a bcL.H.S.= (c+a) b ca = c +a b ca ApplyingC C -2C

(a+b) c ab a +b c ab

→

[ ]2 2 2 2

2 2 2 21 1 2

2 2 2 2

a +b +c a bca +b +c b ca Applying C C +Ca +b +c c ab

= →

2

2 2 2 2

2

1 a bc=(a +b +c )1 b ca

1 c ab

2 2

2 2 2 2 2

2 2

(b+c) a bc(c+a) b ca =(a +b +c )(a-b)(b-c)(c-a)(a+b+c)(a+b) c ab

Show that

Solution Cont.

[ ]2

2 2 22 2 1 3 3 2

1 a bc=(a +b +c ) 0 (b-a)(b+a) c(a-b) Applying R R -R and R R -R

0 (c-b)(c+b) a(b-c)→ →

[ ]2 2 2 2 21=(a +b +c )(a-b)(b-c)(-ab-a +bc+c ) Expanding along C

2 2 2=(a +b +c )(a-b)(b-c)(c-a)(a+b+c)=R.H.S.

2

2 2 2

1 a bc=(a +b +c )(a-b)(b-c) 0 -(b+a) c

0 -(b+c) a

( ) ( ) ( )2 2 2=(a +b +c )(a-b)(b-c) b c-a + c-a c+a

Applications of Determinants (Area of a

Triangle)

The area of a triangle whose vertices are

is given by the expression

1 1 2 2 3 3(x , y ), (x , y ) and (x , y )

1 1

2 2

3 3

x y 11

Δ= x y 12

x y 1

1 2 3 2 3 1 3 1 2

1= [x (y - y ) + x (y - y ) + x (y - y )]

2

ExampleFind the area of a triangle whose vertices are (-1, 8), (-2, -3) and (3, 2).

Solution :

1 1

2 2

3 3

x y 1 -1 8 11 1

Area of triangle= x y 1 = -2 -3 12 2

x y 1 3 2 1

[ ]1= -1(-3-2)-8(-2-3)+1(-4+9)

2

[ ]1= 5+40+5 =25 sq.units

2

Condition of Collinearity of Three Points

If are three points,

then A, B, C are collinear

1 1 2 2 3 3A (x , y ), B (x , y ) and C (x , y )

1 1 1 1

2 2 2 2

3 3 3 3

Area of triangle ABC=0

x y 1 x y 11

x y 1 =0 x y 1 =02

x y 1 x y 1

⇔

⇔ ⇔

If the points (x, -2) , (5, 2), (8, 8) are collinear, find x , using determinants.

Example

Solution :

x -2 15 2 1 =08 8 1

∴

( ) ( ) ( ) ( )x 2-8 - -2 5-8 +1 40-16 =0⇒

-6x-6+24=0⇒

6x=18 x=3⇒ ⇒

Since the given points are collinear.

Solution of System of 2 Linear Equations (Cramer’s

Rule)

Let the system of linear equations be

( )2 2 2a x+b y = c ... ii

( )1 1 1a x+b y = c ... i

1 2D DThen x = , y = provided D 0,

D D≠

1 1 1 1 1 11 2

2 2 2 2 2 2

a b c b a cwhere D = , D = and D =

a b c b a c

Cramer’s Rule (Con.)

then the system is consistent and has infinitely many solutions.

( ) 1 22 If D = 0 and D =D = 0,

then the system is inconsistent and has no solution.

( )1 If D 0

Note :

,≠

then the system is consistent and has unique solution.

( ) 1 23 If D = 0 and one of D , D 0,≠

Example

2 -3D= =2+9=11 0

3 1≠

1

7 -3D = =7+15=22

5 1

2

2 7D = =10-21=-11

3 5

Solution :

1 2

D 0D D22 -11

By Cramer's Rule x= = =2 and y= = =-1D 11 D 11

≠

∴

Q

Using Cramer's rule , solve the following system of equations 2x-3y=7, 3x+y=5

Solution of System of 3 Linear Equations (Cramer’s

Rule)Let the system of linear equations be

( )2 2 2 2a x+b y+c z = d ... ii

( )1 1 1 1a x+b y+c z = d ... i

( )3 3 3 3a x+b y+c z = d ... iii

31 2 DD DThen x = , y = z = provided D 0,

D D D, ≠

1 1 1 1 1 1 1 1 1

2 2 2 1 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3

a b c d b c a d c

where D = a b c , D = d b c , D = a d c

a b c d b c a d c

1 1 1

3 2 2 2

3 3 3

a b d

and D = a b d

a b d

Cramer’s Rule (Con.)Note:

(1) If D ≠ 0, then the system is consistent and has a unique solution.

(2) If D=0 and D1 = D2 = D3 = 0, then the system has infinite solutions or no solution.

(3) If D = 0 and one of D1, D2, D3 ≠ 0, then the system is inconsistent and has no solution.

(4) If d1 = d2 = d3 = 0, then the system is called the system of homogeneous linear equations.

(i) If D ≠ 0, then the system has only trivial solution x = y = z = 0.

(ii) If D = 0, then the system has infinite solutions.

ExampleUsing Cramer's rule , solve the following system of equations5x - y+ 4z = 5 2x + 3y+ 5z = 25x - 2y + 6z = -1

Solution :

5 -1 4D= 2 3 5

5 -2 6

1

5 -1 4D = 2 3 5

-1 -2 6

= 5(18+10)+1(12+5)+4(-4 +3)= 140 +17 –4= 153

= 5(18+10) + 1(12-25)+4(-4 -15)= 140 –13 –76 =140 - 89= 51 0≠

3

5 -1 5D = 2 3 2

5 -2 -1

= 5(-3 +4)+1(-2 - 10)+5(-4-15)= 5 – 12 – 95 = 5 - 107= - 102

Solution Cont.

1 2

3

D 0D D153 102

By Cramer's Rule x= = =3, y= = =2D 51 D 51

D -102and z= = =-2

D 51

≠

∴

Q

2

5 5 4D = 2 2 5

5 -1 6

= 5(12 +5)+5(12 - 25)+ 4(-2 - 10)= 85 + 65 – 48 = 150 - 48= 102

ExampleSolve the following system of homogeneous linear equations:

x + y – z = 0, x – 2y + z = 0, 3x + 6y + -5z = 0

Solution:

( ) ( ) ( )1 1 - 1

We have D = 1 - 2 1 = 1 10 - 6 - 1 -5 - 3 - 1 6 + 63 6 - 5

= 4 + 8 - 12 = 0

The system has infinitely many solutions. ∴

Putting z = k, in first two equations, we get

x + y = k, x – 2y = -k

Solution (Con.)

1

k 1D -k - 2 -2k + k k

By Cramer's rule x = = = =D -2 - 1 31 1

1 - 2

∴

2

1 kD 1 - k -k - k 2k

y = = = =D -2 - 1 31 1

1 - 2

k 2kx = , y = , z = k , where k R

3 3∴ ∈

These values of x, y and z = k satisfy (iii) equation.

Thank you