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2 Applied Mathematics - I

APPLIEDMATHEMATICS - I(As per the Revised Syllabus of Mumbai University for

F.Y.B.Sc. (IT), Semester I)

Dr.(Mrs.) ABHILASHA S. MAGARM.Sc. (MATHS), B.Ed., Dip. C.P., Ph.D.

Department of Mathematics & Statistics,Annasaheb Vartak College of Arts,

K.M. College of Commerce &E.S. Andrades College of Science,Vasai Road (W), Thane – 401202.

MUMBAI NEW DELHI NAGPUR BENGALURU HYDERABAD CHENNAI PUNE LUCKNOW AHMEDABAD ERNAKULAM BHUBANESWAR INDORE KOLKATA GUWAHATI

Matrices 3

© AuthorsNo part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by anymeans, electronic, mechanical, photocopying, recording and/or otherwise without the prior written permission of thepublisher.

First Edition : 2015

Published by : Mrs. Meena Pandey for Himalaya Publishing House Pvt. Ltd.,“Ramdoot”, Dr. Bhalerao Marg, Girgaon, Mumbai - 400 004.Phone: 022-23860170/23863863, Fax: 022-23877178E-mail: [email protected]; Website: www.himpub.com

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Dedicated to my Parents

Mr. & Mrs. JAGTAP&

All My Respected Teachers“Who has always been my greatest source of

Inspiration and Courage”

Matrices 5

PREFACE

It gives me an immense pleasure to present the book on “Applied Mathematics - I” tostudents and teachers of F.Y.B.Sc. (IT) course of Mumbai University. This book is preparedexactly as per syllabus prescribed by the University.

I have made humble attempts, to fill up the gap and help the students and teacherscommunity giving them a suitable textbook catering to their special need with sufficient numberof solved problems on every topic.

Inspired by hard work of our college Principal Dr. Keshav N. Ghorude, Vice Principal,Dr. (Mrs.) Deepa Katre and (Mrs.) Shushama (HOD) B.Sc. (IT).

Special thanks to my all staff members and friends. I thankful to my husbandDr. (Mr.) Shashikant R. Magar and my loving son’s Shaurya and Samarath.

A big thanks to Himalaya Publishing House Pvt. Ltd. for brining out this book in time.

Author


SYLLABUS

Unit – I: Matrices (8 Lectures)Minors and Cofactors, Adjoint of a square matrix, Inverse of a Matrix, Rank of a Matrix,Solution of Homogeneous and Non-homogeneous linear equation using Matrix method.

Unit – II: Eigen Values and Eigen Vectors: (8 Lectures)Vectors, linear combination of vectors, Inner product of two vectors, Characteristic equation,Eigen Vector, Cayley-Hamilton Theorem, Similarity of Matrices, Derogatory and Non-derogatory matrices, Complex Matrices: Hermitian, Skew-Hermitian and Unitary matrices andtheir properties.

Unit – III: Vector Calculus: (8 Lectures)Vector Differentiation: Vector Operator Del, Gradient, and Geometrical Meaning of gradient,Divergence and Curl.

Unit –IV: Differential Equations: (8 Lectures)Differential Equations of 1st order and 1st degree and applications.

Unit – V: Linear Differential Equations: (8 Lectures)Linear Differential equations with constant coefficient, Differential equations of higher orderand applications.

Unit – VI: (8 Lectures)Successive differentiation, Mean value theorems, Partial differentiation, Euler’s Theorem,Approximation and errors, Maxima and Minima.

Term Work:(1) Assignments: Should contain at least 2 assignments covering the Syllabus.(2) Class Tests: One. Also Known as Unit Test or in-Semester Examinations.(3) Tutorial: Minimum Three tutorials covering the syllabus.

List of Problems:(1) Problem solving based on Matrices.(2) Problem solving based on Eigen Values and Eigen Vectors.(3) Problem solving based on Vector Analysis.(4) Problem solving based on Differential Equations.(5) Problem solving based on Linear Differential Equations.(6) Problem solving based on Successive Differentiation.(7) Problem solving based on Mean Value theorems.(8) Problem solving based on Partial differentiation.(9) Problem solving based on Euler’s Theorem.(10) Problem solving based on Approximation and Errors.(11) Problem solving based on Maxima and Minima.

Matrices 7

CONTENTSUNIT 1: MATRICES 1 - 271.1 Definition1.2 Types of Matrix1.3 Properties of Addition Matrix1.4 Addition of Matrices1.5 Subtraction of Matrices1.6 Multiplication of Matrices1.7 Transpose of a Matrix.1.8 Orthogonal Matrix.1.9 Minors and Cofactors of a Matrix.1.10 Adjoint of a Suare Matrix1.11 Inverse of a Matrix1.12 Rank of Matrix1.13 Normal Form or Canonical Form1.14 Paq Normal Form1.15 Linear Algebraic Equations1.16 Types of Problems1.17 Non-homogeneous Equations

UNIT 2: EIGEN VALUES AND EIGEN VECTORS 28 - 602.1 Vectors2.2 Vector Space2.3 Linear Combination of Vectors2.4 Inner Product of Two Vectors2.5 Concept of Eigen Values and Eigen Vectors2.6 Method of Finding Eigen Values2.7 Type I2.8 Type II2.9 Type III2.10 Cayley-Hamiltion Theorem2.11 Transpose of a Matrix2.12 Symmetric Matrix2.13 Skew Symmetric Matrix2.14 Hermitian Matrix2.15 Skew Hermitian Matrix2.16 Unitary Matrix2.17 Orthogonal Matrix2.18 Diagonalizable Matrix2.19 Minimal Polynomial

UNIT 3: VECTOR CALCULUS 61- 733.1 Introduction3.2 Scalar


3.3 Vector3.4 Unit Vector3.5 Differentiation of Vectors3.6 Gradient of a Scalar Function3.7 Del Applied to Vector Point Function

UNIT 4 : DIFFERENTIAL EQUATION I 74 - 1034.1 Introduction4.2 Formation of Differential Equation4.3 General Solution (GS)4.4 Particular Solution4.5 Solution of Differential Equations4.6 Method of Separation of Variable4.7 Equation Reducible to Variable Separable Form4.8 Homogeneous Differential Equation4.9 Non-homogeneous Linear Differential Equation4.10 Exact Differential Equation4.11 Equations Reducible to Exact Equations4.12 Linear Differential Equation4.13 Bernoulli Differential Equation4.14 Application of Differential Equation

UNIT 5: LINEAR DIFFERENTIAL EQUATION 104 - 1185.1 Linear Differential Equations5.2 General Solution (GS) of Differential Equation5.3 Operator D5.4 Rules for Finding the Complementary Function5.5 Method of Finding Particular Integral (PI)5.6 Homogeneous Linear Differential Equation

UNIT 6: 119 - 1766.1 Successive Differentiation6.2 Higher Order Derivatives6.3 Nth Derivatives6.4 Standard Results6.5 Leibnitz Theorem6.6 Mean Value Theorem6.7 Rolle’s Mean Value Theorem6.8 Lagrange’s Mean Value Theorem6.9 Cauchy’s Mean Value Theorem6.10 Partial Differentiation6.11 Homogeneous Functions6.12 Euler’s Theorem6.13 Change F Variables

Matrices 9

6.14 Application of Partial Differentiation6.15 Maxima and Minima of Function of Two Variables


1.UNIT Matrices

1.1 DEFINITIONA set of mn numbers (real or complex numbers) arranged in the form of a rectangular arrayhaving m rows and n columns is called on n × m matrix.Different notations used for matrix are [ ], ( ), || || we are using notation [ ].Example:

1. 3column of No.1row of No.321 3 1

2. 2column of No.2row of No.

4422

2 2


321

1 3

1.2 TYPES OF MATRIX(1) Row Matrix: A matrix contains one row and n No. of column is called Row Matrix.

Example

1. 3column of No.1row of No. 321


(2) Column Matrix: A matrix contains m No. of rows and one column is called ColumnMatrix .Example


321

1 3


5

(3) Zero Matrix of Null Matrix: A matrix whose elements are all zero is called null matrix.Denoted by “O”.

e.g. 1. 0000

2. 000 3.

000

Matrices 11

(4) Unit matix or Identity matrix: A square matrix each of whose diagonal elements are 1and Non-diagonal Elements are zero called unit matrix Denoted by “In×m”

e.g. 1. 1001

2 2

, 2.

3 3100010001

(5) Square Matrix: A matrix whose no. of rows in equal to no. of columns i.e. m = n iscalled square matrix.

e.g. 1. 2 2

dcba , 2.

3 3

ihgfedcba

, 3 1 1 1

(6) Scalar Matrix: A diagonal matrix containing all diagonal elements are equal is calledscalar matrix.

e.g. 1. 3003

2 2

, 2.

3 3400040004

(7) Symmetric matrix: A square matrix in which aij = aji is called symmetric matrix i.e.A = A.e.g.

1. A =

3663 and A =

3663

A = A

2. A =

cfgfbhgha

and A =

cfgfbhgha

A = A

Symmetric matrix.(8) Skew symmetric: A matrix A called skew symmetric matrix if aij = –aji for all i and j.

Therefore, every diagonal element of a skew symmetric matrix should be zero.e.g.

1. A =

04

40 , A =

0440

which is – A =

04

40

– A = A

2. A =

013102320

, A =

013102320

– A =

013102320


– A = A

1.3 PROPERTIES OF ADDITION MATRIX(1) Commutative: A + B = B + A(2) Associative: A + (B + C) = (A + B) + C(3) Additive entity = A + 0 = 0 + A = A(4) Existence of Addition Inverse: A + (–A) = 0 = (–A) + A

1.4 ADDITION OF MATRICESFor Addition of matrix, necessary condition is that the order of given matrices must be same.

e.g. A =2 223

41

, B = 12

322 2

A + B = 2341

+ 12

32

= 12233421

=

A + B = 3513

Both matrix A and B having same order 2 × 2.

1.5 SUBTRACTION OF MATRICESFor the subtraction any matrix, order must be same.

e.g. A = 7965

B = 76

42

A – B = 7965

–

7642 = 7769

)4(625

A – B = 03103

1.6 MULTIPLICATION OF MATRICESFor multiplication of matrix is that the number of elements in the row of matrix A must be equalNo. of column of matrix B, the matrix need not always be square matrix.

nmA , nmB

then

[A × B] m of first matrix equal m × n of second matrix

e.g. A =3 2254

331

, B =

2 3115102

Now AB =

254331

115102

=

125504121524135301131321

Matrices 13

=

22502583150332 =

2715188

AB =

2715188

Exercise 1

If A =

2243 , B =

32

06

Find(i) 3A + B (ii) 2A + 2B (iii) 5A (iv) A + 2B (v) 2A – 3B (vi) A – B

Solution:

1. Given A =

2243 , B =

32

06

3A + B = 3

2243 +

32

06 =

66129 +

32

06

3A + B =

362601269

3A + B =

341215

2. 2A + 2B = 2

2243 + 2

32

06

=

4486 +

64

012

=

64440826 =

2088

3. 5A = 5

2243 =

10102015

4. A + 2B =

2243 + 2

32

06

=

2243 +

64

012

=

424204123

=

22

415

5. 2A – 3B = 2

2243 – 3

32

06

=

4486 –

96

018


=

946408186 =

1310812

6. A – B =

2243 –

32

06

=

32220463 =

5443

Exercise 2: Find AB and BA

A =

000100010

, B =

010001000

Solution:

AB =

000100010

010001000

=

000000100000001000000100110000011000000100100100001100

=

000010001

BA =

010001000

000100010

=

001100100110000100001001110011000001001000000110000000

=

100010000

1.7 TRANSPOSE OF A MATRIXThe inter-change of rows and column is called transpose of given matrix, denoted by A or A,

e.g. A =

987654321

A or A =

963852741

but their diagonal elements remain in same position, unchanged.

1.8 ORTHOGONAL MATRIXFor a orthogonal matrix if A.A = I, then matrix A is called orthogonal matrix.

Matrices 15

Exercise: 3 Check given matrix are orthogonal

A =

1000θcosθsin0θsinθcos

Solution:Given,

A =


A =


For orthogonal matrix, A.A = I



=

1000000000000θcos θsin0θsinθcosθcosθsin000θcosθsinθsinθcos0θsinθcos

22

22

=

100010001

= I

Hence, A.A = I given matrix is orthogonalExercise 4 : Express the matrix A as the sum of a symmetric and a skew-symmetric matrix. When

A =

705631324

Solution:Given:

A =

705631324

and A =

763031514

A + A =

468663838


A – A =

062601210

A =21 (A + A) +

21 (A – A)

=21

468663838

+21

062601210

EXERCISE

1. If A =

9174 , B =

2781 , then

Find (i) 3A + 2B (ii) 2A – 4B

2. If A =

1786 , B =

4334

Find (i) A + 4B (ii) 2A + B (iii) A – 2B

3. If A =

1231 and B =

312101 , then verify (AB) = BA

4. Express the matrix A as the sum of symmetric and skew symmetric, when

A =

505432171

5. Given that A =

3211 show that A2 – 4A + 5I = 0

6. Show that given matrix is orthogonal, when

A =

21

61

31

062

31

21

61

31

1.9 MINOR AND CO-FACTORS OF A MATRIXThe minor of a square matrix is the determinant obtained by deleting the row and column whichintersect in that element.The co-factor of any element is it’s minor with (–1)i + j sign

1.10 ADJOINT OF A SQUARE MATRIXThe Adjoint of a matrix A is the transpose of the matrix formed by the co-factor of matrix A.It is denoted by Adj A.

Matrices 17

Matrix of co-factor=

333231

232221

131211

aaaaaaaaa

Adj A =

332313

322212

312111

aaaaaaaaa

1.11 INVERSE OF A MATRIXIf A be any non-singular square matrix and non-singular matrix B such that AB = I, then B iscalled Inverse of A such that denoted by A–1.Since, AB = A B = I = 0

A 0 and B 0

Hence, we can find A–1 by using Adjoint Method.

A–1 =A

AAdj.

Exercise 5: Find Inverse of the matrix by adjoint method

A =

120031221

Solution:Given,

A =

120031221

Step I: We have show A 0, then only A–1 exits.

A = 1 (3 – 0) – 2 (–1 – 0) – 2(2 – 0)

= 3 + 2 – 4 = 5 – 4 = 1 A = 1 0,

Hence, A–1 Exits.Step II: Co-factors of A

a11 = + 1203

= 3

a12 = – 1001 = –(–1) = 1

a13 = + 2031

= 2

a21 = – 1222

= – (2 – 4) = – (–2) = 2


a22 = + 1021 = 1

a23 = – 2021

= – (–2 – 0) = 2

a31 = + 0322 = 0 – (–6) = 6

a32 = – 0121

= – (0 –2) = 2

a33 = + 3121

= 3 + 2 = 5

Co-factor of A=

526212213

Step III: Adjoint of A =

522211623

Adjoint of A = (Co-factor of A)

A–1 =A

AAdj. =11

522211623

A–1 =

522211623

Exercise 6 : Find inverse of A, by using Adjoint Method.

A =

312321111

Solution:Given:

A =

312321111

Step I: A = 1(6 – 3) – 1(3 + 6) + (–1 – 4)

= 3 – 9 – 5 = –11 A = – 11 0.

Hence, A–1

Step II: Co-factor of A

a11 = + 3132

= 3

Matrices 19

a12 = – 3231 = – 9

a13 = + 1221

= – 5

a21 = – 3111

= – 4

a22 = + 3211 = 1

a23 = – 1211

= 3

a31 = + 3211

= – 5

a32 = – 3111

= 4

a33 = 2111 = 1

Co-factor of A =

145314593

Step III:

Adjoint of A =

135419543

Now, A–1 =A

AofAdj

Hence, A–1 =111

135419543

EXERCISE

1. Find co-factor of given matrix

1. A =

120031221

II. A =

765542321

III. A =

012113231

Ans: – I.

312321113

II.

012133231

III.

653542221


2. Find inverse of given matrices by using Adjoint Method.

I. A =

312321111

Ans: A–1 =111

135419543

II. Find A–1 A =

113321210

Ans: A–1 =21

135268111

3. Find A–1

A =

1000αcosαsin0αsinαcos

A–1 =

1000αcosαsin0αsinαcos

1.12 RANK OF MATRIXLet, A be a m × n Matrix. Then, A is said to be a matrix of rank r, if1. There exists atleast one Non-zero minor of A of order r, and2. Each minor of A of order greater than r is zero i.e. each minor of order (r + 1) or higher is

zero or vanishes, we generally denoted rank of matrix by (A).Let, A be a square matrix of order m × n. Then, rank of Matrix A is

1. (A) = n if A 0 i.e. A is non-singular

(A) < n if A = 0 i.e. A is singular

Exercise 7: Find Rank of A =

7421

Solution:Given, A is square matrix

A = 7 – 8 = –1 0

So, rank A = order of 2rank A = 2 (A) = 2

Exercise 8: Find the rank of matrix A.

A =

601245012

Matrices 21

Solution: Given, Matrix 3 × 3 andA = 76 0

Therefore, Rank A = 3, (A) = 3

Exercise 9: Find rank of A =

604132

Solution:Since, A is of 2 × 3i.e. Rank A 2rank A Min (2, 3)

Now,

A = 0432

= 2 × 0 – 3 × 4= –12 0

Rank A = Rank A1 = 2(A) = 2

1.13 NORMAL FORM OR CANONICAL FORM

Let, A be a matrix of m × n order with rank r. The matrix obtain in the form of

000rI

after

applying a finite chain of E-operations in referred as normal form of Matrix A.Here, Ir is the rowed Identity matrix other normal forms are

rI or 0 :rI

OR

000rI

0rI

Remark:To reduce the matrix to normal form, apply row as well as column transformations.1. First try to convert left and Top element I and other first column element zero.2. Try to convert diagonal element 1 nondiagonal element zero.3. Continue the process till matrix reduces to normal form.

Exercise 10: Find the rank of matrix by reducing it to normal.

Where, A =

576243212341

Solution: We have

A =

576243212341

By Applying, R2 R2 – R1


R3 R3 – 2R1

We get,

112020202341

By Applying, c2 c2 + 4c1, c3 c3 – 3c1, c4 c4 – 2c1

112020200001

Now, Applying, c2 21 c2 we get

111020100001

Applying, c2 c2 – c3 we get

110020100001

Applying, c4 c4 – 2c2 , – 1 × c2

110000100001

Applying, c4 c4 – c3

010000100001

[I3 : 0]. Hence, (A) = 3

Exercise 11: Reduce the following matix to normal form and find its rank.

A =

15124167931016248316

Solution: Given,

A =

15124167931016248316

Applying, c1 c2

Matrices 23

15121647910316428361

by c2 c2 – 6c1, c3 c3 – 3c1, c4 c4 – 8c3

170841708317082

0001

Apply

81 c2 and

171 c4

1014101310120001

R4 R4 – R2, R4 R4 – 2R1, R3 R3 – R2

R3 R3 – R1

0000000010120001

c1 c1 – 2c2, c4 c4 – c2

0000000000100001

[I2 : 0] is in the Normal form (A) = 2

Exercise 12: Reduce the matrix A =

820103420121

to canonical form (Normal) and find its rank.

Solution: We have,

A =

820103420121

Applying, C2 C2 – 2C1, C3 C3 – C1

812105820001


Apply, R2 R2 + 2R1, R3 R3 – R1

812005800001

Apply, C2 81 C2

81410

05100001

by C3 C3 – 5C2

849

410

00100001

by R3 4R3

3291000100001

R3 R3 + R2

3290000100001

C3 ×91

3210000100001

by C4 C4 + 32C3

010000100001

[I3 : 0] = (A) = 3

1.14 PAQ NORMAL FORM

If A be m × n, matrix of rank r can be reduced to

000rI from by E-operations. Since,

E-operation are equivalent to fore or post multiplication by the corresponding elementarymatrices, therefore, E-matrix.

Matrices 25

PAQ =

000rI

Where, P = pk, pk – 1 … p1

Q = Q1, Q2 … Qj

Thus, we concluded that, for any matrix A of m × n order with rank r, there exist non-singularmatrices P and Q.

Such that PAQ =

000rI

To find P and Q1. We write A = Im A In, m = row’s, n = columns unit matrix of order m × n.2. We can Apply row operation to A on left and column operation to A on Right.3. When A to the left reduces to normal form we have A = PAQ

Exercise 13: Find Non-singular matrices P and such that PAQ is in the Normal form.

A =

110321211

Solution: Given,

A =

110321211

We can write A = I3 A I3

110321211

=

100010001

A

100010001

Apply, C2 C2 – C1, C3 C3 – 2C1 we get

110111001

=

100010001

A

100010211

Apply, R2 R2 – R1

110111001

=

100011001

A

100010211

Apply, C3 C3 – C2

010011001

=

100011001

A

100110111

Apply, R3 R3 + R2


000011001

=

111011001

A

100110111

Thus, we have required form, where

P =

111011001

and Q =

100110111

Exercise 14 : Find the Non-singular Matrix P and Q such that PAQ is in Normal form.

A =

113111111

Solution: Given, A =

113111111

We can write,A = I3 A I3

113111111

=

100010001

A

100010001

Applying, C2 C2 – C1, C3 C3 – C1 we get

223221001

=

100010001

A

100010111

by R2 R2 – R1, R3 R3 – 3R1

220220001

=

100011001

A

100010111

by

21 C2,

21 C3 we get,

110110001

=

100011001

A

2100

0210

21

211

Applying, C3 C3 – C2

010010001

=

100011001

A

21002

12

10

0211

Matrices 27

Applying, R3 – R2 we get

000010001

=

111001001

A

2100

21

210

0211

PAQ =

0002I = Normal form

P =

111011001

and Q =

2100

21

210

0211

EXERCISE

1. Reduce the following matrix to the Normal form find it’s rank.

(a)

431822306318

(b)

113111111

(c)

112271611716413113

(d)

101312111111

Ans: (a) (A) = 3 (b) (A) = 3 (c) (A) = 22. Find rank of the given below.

(a)

1345 (b)

210610145

(c)

121363

Ans: (a) (A) = 2 (b) (A) = 3 (c) (A) = 1

3. If A =

957442412

find P and Q. Two Non-singular matrices such that PAQ = I when I is unit

matrix.4. Find two non singular matrix P and Q so that PAQ is in Normal form

A =

19114124155123


Ans: P =

3/13/21510100

Q =

10000100

731

717

2110

79

79

2141

1.15 LINEAR ALGEBRAIC EQUATIONSConsider, a set of equation

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

In Matrix form,AX = D

333

222

111

cbacbacba

zyx

=

3

2

1

ddd

Now, augmented matrix join the matrix A and D

i.e. [A : D] =

3333

2222

1111

:::

dcbadcbadcba

NOTE:Homogeneous EquationsHomogeneous set of equation is always consistent. Since, it has always atleast one solution.1. Viz. x1 = x2 = x3 = 0 called trivial solution or unique solution or zero solution.

(A : 0) = (A) = No. of unknowns.2. If (A : 0) = (A) < No. of known the system may posses infinitely many solution.3. If A 0 then (A) = (A : 0) No. of unknown then system has zero solution.4. If A = 0 then the system has non zero solution.

1.16 Types of Problems1. No. of equation = No. of Unknown.2. No. of equation > No. of Unknown.3. No. of equation < No. of Unknown.

Type-I (No. of equation = No. of Unknown)Exercise: 15 Solve

` 3x – 2y + z = 02x + 2y – z = 03x + y + 6z = 0

Solution: In the matrix form,AX = 0

Matrices 29

613122123

zyx

=

000

Here,

A =6 1 3 1 2 2 1 2 3

= 75 0

= 3 (12 + 1) + 2 (12 + 3) + 1 (2 – 6)= 3 13 + 2 15 + 1 (–4)= 49 + 30 – 4 = 45 + 30 = 75

system is consistent and has Zero solution viz x = y = z = 0.Exercise: 16 Solve

2x – y + 3z = 03x + 2y + z = 0x –4y + 5z = 0

Solution: In Matrix form

541123312

zyx

=

000

A = 0 i.e. system has non-zero solution

[A] =

5 411 2 33 12

R1 R3 R1

3 121 2 35 41

R2 R2 – 3R1, R3 R3 – 2R1

77 0141405 41

141 R2, 7

1 R3

11 011 05 41

R3 R3 – R2

0 0 011 05 41

(A : 0) = (A) = 2 < No. of unknownLet z = t, y – z = 0, x – 4y + 5z = 0, y = – t.


Type-II (No. of equation > No. of unknown)Exercise: 17 Solve

2x1 + 2x2 + 2x3 = 02x1 + 3x2 + x3 = 04x1 + x2 + x3 = 0x1 + x2 – 2x3 = 0

Solution: Consider first three equation

114132222

zyx

=

000

Now,

A =1 1 41 3 22 2 2

= – 12 0

= 2 (3 – 1) – 2 (2 – 4) + 2 (2 – 12)= 2 2 – 2 –2 + 2 –10 = 4 + 4 – 20 = –12

A = –12 0 system has zero solution.

Viz x1 = x2 = x3 = 0Type-III: No. of equations < No. of UnknownExercise: 18 Solve,

x1 + 2x2 + 3x3 + x4 = 0x1 + x2 – x3 – x4 = 03x1 – x2 + 2x3 + 3x4 = 0

Solution: In Matrix form,AX = 0

321311111321

4

3

2

1

xxxx

=

000

Now,

321311111321

R2 R2 – R1, R3 R3 – 3R1

077024101321

Matrices 31

71 R3, (–1) R2

011024101321

R3 R3 – R2

2 30 02 4 1 01 3 2 1

Now, (A) = (A0) = 3 < No. of Unknown’s, so the system has non-zero solution.Let, x4 = t

by R3, –3x3 –2x4 0 x3 =32t

by R2, x2 + 4x3 + 2x4 = 0 x2 = t314

by R1, x1 + 2x2 + 3x3 + x4 = 0 x1 =3t

1.17 NON-HOMOGENEOUS EQUATIONSIf in a system of equation atleast one constant term is Non-zero, then the system is said to beNon-homogeneous.Note:(a) We reduce [A:D], to Echelon form and thereby find the rank of A and [A:D](b) If (A) (A:D), then the system is in consistent, i.e. it has no solution.(c) If (A) = (A:D), then the system is consistent and if1. (A) = (AD) = No. of unknown’s It has unique solution, we can find out directly2. If (AD) = (A) = r Number of unknown’s, then the system is consistent has infinitely

many solution.Type - I = No. of equations = No. of unknownsExercise: 18 Discuss the consistency of

x + y + z = 05x – y + z = – 62x – y – 2z = – 1

Solution: The above given system of equations can be write as.AX = D

212115111

zyx

=

16

0


The augmented Matrix

[A:D] =

1: 2126: 1150: 111

by Applying, R2 R2 – 5R1, R3 R3 – 2R1

1: 0306: 6600: 111

by Applying, R2 61 R2

1: 0300: 1100: 111

Apply, R1 R1 – R2, R3 R3 + 3R2

2: 3001: 1101: 001

Apply, R3 R3

31

2/3: 1001: 1101: 001

Apply, R2 R2 + R3

2/3: 1001/3: 010

1: 001

(A:D) = (A) = 3Also, (A) = 3Then the system of equation is consistent and has unique solution.

100010001

zyx

=

3/23/11

x = – 1, y = 1/3, z = – 2/3Ex: 19 Test the consistency of the given system of linear equations

x + y + z = 6x + 2y + 3z = 14x + 4y + 7z = 30

Solution: We can writeAX = D

Matrices 33

741321111

zyx

=

3014

6

[A:D] =

30: 74114: 3216: 111

Apply, R2 R2 – R1, R3 R3 – R1, we get

24: 6308: 3206: 111

Apply, R3 R3 – 3R2

0: 0008: 2106: 111

Rank [A:D] = Rank A = 2 (A) = 2The given system of equation is consistent since r < n i.e. Rank is Less then No. of

Unknown, so given system will have an infinite number of solution.Now, number of variable to be assigned arbitrary values n – r = 3 – 2 = 1.

Now, equation 1 can reduced asx + y + z = 6y + 2z = 8

Let, z = k y = 8 – 2zy = 8 – 2k

x + 8 – 2k + k = 6x = 6 – 8 + kx = k – 2

x = k – 2, y = 8 – 2k, z = kK is arbitrary constant

Exercise: 20 Discuss the consistency of2x + 3y – 4z = –2x – y + 3z = 43x + 2y – z = –5

Solution: We can write

123311432

zyx

=

542

(A:X) = D

(A:D) =

5: 1234: 3112: 432


Apply, R2 –21 R1 , R3 R3 –

23 R1

2: 5253

5: 5250

2: 432

Apply, R3 R3 – R2

7: 0005: 52

502: 432

Thus, (A:D) = 3, (A) = 2 (A:D) (A)

The system is not consistent and has no solutions.Type = II No. of equations > No. of UnknownExercise : 21 Solve the system of equation

3x + 3y + 2z = 1x + 2y = 410y + 3z = –22x – 3y – z = 5

Solution: We can write

1323100021233

zyx

=

52

41

AX = DNow,

[A:D] =

5: 1324: 0211: 233

R1 R2

5: 1322: 31001: 2334: 021

Apply, R2 R2 – 3R1, R4 R4 – 2R1

3: 1702: 3100

11: 2304: 021

Matrices 35

Apply, R3 R3 +3

10 R2 , R4 R4 –37 R2

368: 3

17003

116: 32900

11: 2304: 021

Apply,293 R3 and

317 R4

4: 1004: 100

11: 2304: 021

Now, Apply R4 R4 – R3

0: 0004: 100

11: 2304: 021

it is an Echelon form (A) = (A:D) = 3 = No. of Unknown system has unique and consistently.Now, byby R3 z = – 4by R2 – 3y + 2z = – 11 y = 1by R1 x + 2y = 4 x = 2

EXERCISE

1. Find Rank of the following by reducing to Echelon form

(i) A =

3211462163421143

(ii) A =

654543432

(iii) A =

51322310040211233

Ans: (i) –4 (ii) –2 (iii) –3.


2. Examine, whether the following equations are consistent and solve if they are consistent.(a) 2x + 6y + 11 = 0, 6x + 20y – 6z + 3 = 0, 6y – 18z + 1 = 0(b) x + y +z = 6, 2x + y + 3z = 13, 5x + 2y + z = 12

3. Investigate for consistency of the following equation and if possible find the solution.4x – 2y + 6z = 8, x + y – 3z = – 1, 15x + 3y + z = 21

4. Test for consistency and solve2x – 3y + 7z = 5, 3x + y – 3z = 13, 2x + 19y – 47z = 32

5. Test for consistency and solve5x + 3y + 7z = 4, 3x + 26y + 2z = 9, 7x + 2y + 10z = 5

Ans: x =117 , y =

113 , z = 0 is particular solution.

6. Prove that the following matrix is orthogonal

3/23/23/13/13/23/23/23/13/2

7. Is the matrix

913134132

orthogonal?

If not it be converted into an orthogonal matrix.

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