Engineering Mathematics Kanodia

Eighth Edition

GATE Engineering Mathematics

RK Kanodia Ashish Murolia

NODIA & COMPANY

GATE Electronics & Communication Vol 2, 8e Engineering Mathematics RK Kanodia and Ashish Murolia

Copyright © By NODIA & COMPANY

Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither NODIA & COMPANY nor its author guarantee the accuracy or completeness of any information herein, and NODIA & COMPANY nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that NODIA & COMPANY and its author are supplying information but are not attempting to render engineering or other professional services. MRP 490.00

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To Our Parents

Preface to the Series For almost a decade, we have been receiving tremendous responses from GATE aspirants for our earlier books: GATE Multiple Choice Questions, GATE Guide, and the GATE Cloud series. Our first book, GATE Multiple Choice Questions (MCQ), was a compilation of objective questions and solutions for all subjects of GATE Electronics & Communication Engineering in one book. The idea behind the book was that Gate aspirants who had just completed or about to finish their last semester to achieve his or her B.E/B.Tech need only to practice answering questions to crack GATE. The solutions in the book were presented in such a manner that a student needs to know fundamental concepts to understand them. We assumed that students have learned enough of the fundamentals by his or her graduation. The book was a great success, but still there were a large ratio of aspirants who needed more preparatory materials beyond just problems and solutions. This large ratio mainly included average students. Later, we perceived that many aspirants couldn’t develop a good problem solving approach in their B.E/B.Tech. Some of them lacked the fundamentals of a subject and had difficulty understanding simple solutions. Now, we have an idea to enhance our content and present two separate books for each subject: one for theory, which contains brief theory, problem solving methods, fundamental concepts, and points-to-remember. The second book is about problems, including a vast collection of problems with descriptive and step-by-step solutions that can be understood by an average student. This was the origin of GATE Guide (the theory book) and GATE Cloud (the problem bank) series: two books for each subject. GATE Guide and GATE Cloud were published in three subjects only. Thereafter we received an immense number of emails from our readers looking for a complete study package for all subjects and a book that combines both GATE Guide and GATE Cloud. This encouraged us to present GATE Study Package (a set of 10 books: one for each subject) for GATE Electronic and Communication Engineering. Each book in this package is adequate for the purpose of qualifying GATE for an average student. Each book contains brief theory, fundamental concepts, problem solving methodology, summary of formulae, and a solved question bank. The question bank has three exercises for each chapter: 1) Theoretical MCQs, 2) Numerical MCQs, and 3) Numerical Type Questions (based on the new GATE pattern). Solutions are presented in a descriptive and step-by-step manner, which are easy to understand for all aspirants. We believe that each book of GATE Study Package helps a student learn fundamental concepts and develop problem solving skills for a subject, which are key essentials to crack GATE. Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall appreciate and greatly acknowledge all constructive comments, criticisms, and suggestions from the users of this book. You may write to us at rajkumar. [email protected] and [email protected].

Acknowledgements We would like to express our sincere thanks to all the co-authors, editors, and reviewers for their efforts in making this project successful. We would also like to thank Team NODIA for providing professional support for this project through all phases of its development. At last, we express our gratitude to God and our Family for providing moral support and motivation. We wish you good luck ! R. K. Kanodia Ashish Murolia

SYLLABUS Engineering Mathematics (EC, EE, and IN Branch ) Linear Algebra: Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors. Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series. Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems. Differential equations: First order equation (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s and Euler’s equations, Initial and boundary value problems, Partial Differential Equations and variable separable method. Complex variables: Analytic functions, Cauchy’s integral theorem and integral formula, Taylor’s and Laurent’ series, Residue theorem, solution integrals. Probability and Statistics: Sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Discrete and continuous distributions, Poisson, Normal and Binomial distribution, Correlation and regression analysis. Numerical Methods: Solutions of non-linear algebraic equations, single and multi-step methods for differential equations. Transform Theory: Fourier transform, Laplace transform, Z-transform.

Engineering Mathematics (ME, CE and PI Branch ) Linear Algebra: Matrix algebra, Systems of linear equations, Eigen values and eigen vectors. Calculus: Functions of single variable, Limit, continuity and differentiability, Mean value theorems, Evaluation of definite and improper integrals, Partial derivatives, Total derivative, Maxima and minima, Gradient, Divergence and Curl, Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gauss and Green’s theorems. Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Cauchy’s and Euler’s equations, Initial and boundary value problems, Laplace transforms, Solutions of one dimensional heat and wave equations and Laplace equation. Complex variables: Analytic functions, Cauchy’s integral theorem, Taylor and Laurent series. Probability and Statistics: Definitions of probability and sampling theorems, Conditional probability, Mean, median, mode and standard deviation, Random variables, Poisson,Normal and Binomial distributions. Numerical Methods: Numerical solutions of linear and non-linear algebraic equations Integration by trapezoidal and Simpson’s rule, single and multi-step methods for differential equations.

********** .

CONTENTS CHAPTER 1

MATRIX ALGEBRA

1.1

INTRODUCTION

1

1.2

MULTIPLICATION OF MATRICES

1.3

TRANSPOSE OF A MATRIX

2

1.4

DETERMINANT OF A MATRIX

2

1.5

RANK OF MATRIX

2

1.6

ADJOINT OF A MATRIX

3

1.7

INVERSE OF A MATRIX

3

1

1.7.1

Elementary Transformations

1.7.2

Inverse of Matrix by Elementary Transformations

1.8

ECHELON FORM

5

1.9

NORMAL FORM

5

EXERCISE

6

SOLUTIONS

20

CHAPTER 2

4 4

SYSTEMS OF LINEAR EQUATIONS

2.1

INTRODUCTION

2.2

VECTOR

39

39

2.2.1

Equality of Vectors

39

2.2.2

Null Vector or Zero Vector

2.2.3

A Vector as a Linear Combination of a Set of Vectors

2.2.4

Linear Dependence and Independence of Vectors

39

2.3

SYSTEM OF LINEAR EQUATIONS

2.4

SOLUTION OF A SYSTEM OF LINEAR EQUATIONS

EXERCISE

42

SOLUTIONS

51

CHAPTER 3

40

40 40

EIGENVALUES AND EIGENVECTORS

3.1

INTRODUCTION

3.2

EIGENVALUES AND EIGEN VECTOR

3.3

DETERMINATION OF EIGENVALUES AND EIGENVECTORS

3.4

CAYLEY-HAMILTON THEOREM

3.4.1

40

65 65 66

66

Computation of the Inverse Using Cayley-Hamilton Theorem

67

3.5

REDUCTION OF A MATRIX TO DIAGONAL FORM

3.6

SIMILARITY OF MATRICES

EXERCISE

69

SOLUTIONS

80

CHAPTER 4

INTRODUCTION

4.2

LIMIT OF A FUNCTION

99 99

4.2.1

Left Handed Limit

99

4.2.2

Right Handed Limit

99

4.2.3

Existence of Limit at Point

4.2.4

L’ hospital’s Rule 100

CONTINUITY OF A FUNCTION

4.3.1 4.4

68

LIMIT, CONTINUITY AND DIFFERENTIABILITY

4.1

4.3

EXERCISE

102

SOLUTIONS

115

100

101

MAXIMA AND MINIMA

5.1

INTRODUCTION

139

5.2

MONOTONOCITY

139

5.3

MAXIMA AND MINIMA

EXERCISE

140

SOLUTIONS

147

CHAPTER 6

100

Continuity in an interval 100

DIFFERENTIABILITY

CHAPTER 5

139

MEAN VALUE THEOREM

6.1

INTRODUCTION

6.2

ROLLE’S THEOREM

6.3

LAGRANGE’S MEAN VALUE THEOREM 163

6.4

CAUCHY’S MEAN VALUES THEOREM

EXERCISE

164

SOLUTIONS

168

CHAPTER 7

163 163

163

PARTIAL DERIVATIVES

7.1

INTRODUCTION

7.2

PARTIAL DERIVATIVES

7.2.1

67

175 175

Partial Derivatives of Higher Orders

7.3

TOTAL DIFFERENTIATION

7.4

CHANGE OF VARIABLES

176 176

175

7.5

DIFFERENTIATION OF IMPLICIT FUNCTION

7.6

EULER’S THEOREM

EXERCISE

177

SOLUTIONS

182

CHAPTER 8

176

176

DEFINITE INTEGRAL

8.1

INTRODUCTION

8.2

DEFINITE INTEGRAL

8.3

IMPORTANT FORMULA FOR DEFINITE INTEGRAL

8.4

DOUBLE INTEGRAL

EXERCISE

193

SOLUTIONS

202

CHAPTER 9

191 191 192

192

DIRECTIONAL DERIVATIVES

9.1

INTRODUCTION

223

9.2

DIFFERENTIAL ELEMENTS IN COORDINATE SYSTEMS

9.3

DIFFERENTIAL CALCULUS

9.4

GRADIENT OF A SCALAR 224

9.5

DIVERGENCE OF A VECTOR

9.6

CURL OF A VECTOR

9.7

CHARACTERIZATION OF A VECTOR FIELD

9.8

LAPLACIAN OPERATOR

225

9.9

INTEGRAL THEOREMS

226

Divergence theorem

9.9.2

Stoke’s Theorem 226

9.9.3

Green’s Theorem 226

9.9.4

Helmholtz’s Theorem 227

SOLUTIONS

234

CHAPTER 10

223

224

225

9.9.1

EXERCISE

223

225

226

226

FIRST ORDER DIFFERENTIAL EQUATIONS

10.1

INTRODUCTION

247

10.2

DIFFERENTIAL EQUATION

247

10.2.1

Ordinary Differential Equation 247

10.2.2

Order of a Differential Equation

248

10.2.3

Degree of a Differential Equation

248

10.3

DIFFERENTIAL EQUATION OF FIRST ORDER AND FIRST DEGREE 248

10.4

SOLUTION OF A DIFFERENTIAL EQUATION

10.5

VARIABLES SEPARABLE FORM

249

249

10.5.1 10.6

HOMOGENEOUS EQUATIONS

10.6.1 10.7

Equations Reducible to Variable Separable Form 250

Equations Reducible to Homogeneous Form

LINEAR DIFFERENTIAL EQUATION

10.7.1

250

252

Equations Reducible to Linear Form

10.8

BERNOULLI’S EQUATION 253

10.9

EXACT DIFFERENTIAL EQUATION

251

253

254

10.9.1

Necessary and Sufficient Condition for Exactness

10.9.2

Solution of an Exact Differential Equation

10.9.3

Equations Reducible to Exact Form: Integrating Factors

10.9.4

Integrating Factors Obtained by Inspection

EXERCISE

257

SOLUTIONS

266

CHAPTER 11

254 255

HIGHER ORDER DIFFERENTIAL EQUATIONS

11.1

INTRODUCTION

11.2

LINEAR DIFFERENTIAL EQUATION

283 283

11.2.1

Operator

11.2.2

General Solution of Linear Differential Equation

283

11.3

DETERMINATION OF COMPLEMENTARY FUNCTION

11.4

PARTICULAR INTEGRAL

11.4.1

Determination of Particular Integral

11.6

EULER EQUATION 288 289

SOLUTIONS

300

287

INITIAL AND BOUNDARY VALUE PROBLEMS

12.1

INTRODUCTION

12.2

INITIAL VALUE PROBLEMS

317

12.3

BOUNDARY-VALUE PROBLEM

317

EXERCISE

319

SOLUTIONS

325

CHAPTER 13

284

285

HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION

EXERCISE

283

285

11.5

CHAPTER 12

254

317

PARTIAL DIFFERENTIAL EQUATION

13.1

INTRODUCTION

337

13.2

PARTIAL DIFFERENTIAL EQUATION

337

13.2.1

Partial Derivatives of First Order

337

13.2.2

Partial Derivatives of Higher Order

338

255

13.3

HOMOGENEOUS FUNCTIONS

13.4

EULER’S THEOREM

13.5

COMPOSITE FUNCTIONS

13.6

ERRORS AND APPROXIMATIONS

EXERCISE

342

SOLUTIONS

347

339

339 340 341

CHAPTER 14

ANALYTIC FUNCTIONS

14.1

INTRODUCTION

357

14.2

BASIC TERMINOLOGIES IN COMPLEX FUNCTION

14.3

FUNCTIONS OF COMPLEX VARIABLE

358

14.4

LIMIT OF A COMPLEX FUNCTION

358

14.5

CONTINUITY OF A COMPLEX FUNCTION

14.6

DIFFERENTIABILITY OF A COMPLEX FUNCTION

14.7

14.9

359

Cauchy-Riemann Equation: Necessary Condition for Differentiability of a Complex Function 360

14.6.2

Sufficient Condition for Differentiability of a Complex Function

ANALYTIC FUNCTION

362

Required Condition for a Function to be Analytic

HARMONIC FUNCTION

Methods for Determining Harmonic Conjugate

14.8.2

Milne-Thomson Method

364

14.8.3

Exact Differential Method

366

SINGULAR POINTS 367

SOLUTIONS

380

CHAPTER 15

CAUCHY’S INTEGRAL THEOREM

INTRODUCTION

15.2

LINE INTEGRAL OF A COMPLEX FUNCTION

405

15.2.1

405

15.4

Evaluation of the Line Integrals 406

Cauchy’s Theorem for Multiple Connected Region

CAUCHY’S INTEGRAL FORMULA

410

SOLUTIONS

420

CHAPTER 16

408

Cauchy’s Integral Formula for Derivatives

EXERCISE

16.1

405

CAUCHY’S THEOREM

15.4.1

363

366

15.1

15.3.1

362

363

14.8.1

EXERCISE

15.3

359

14.6.1

14.7.1 14.8

357

TAYLOR’S AND LAURENT’ SERIES

INTRODUCTION

439

409

407

361

16.2

TAYLOR’S SERIES

439

16.3

MACLAURIN’S SERIES

440

16.4

LAURENT’S SERIES

441

16.5

RESIDUES

443

16.5.1

The Residue Theorem

16.5.2

Evaluation of Definite Integral

EXERCISE

444

SOLUTIONS

453

CHAPTER 17

INTRODUCTION

469

17.2

SAMPLE SPACE

469

17.3

EVENT

17.5

17.6

469

17.3.1

Algebra of Events

470

17.3.2

Types of Events

470

DEFINITION OF PROBABILITY

471

17.4.1

Classical Definition

471

17.4.2

Statistical Definition

472

17.4.3

Axiomatic Definition

472

PROPERTIES OF PROBABILITY

Addition Theorem for Probability

17.5.2

Conditional Probability

17.5.3

Multiplication Theorem for Probability

17.5.4

Odds for an Event

SOLUTIONS

491

CHAPTER 18

473 473

473

RANDOM VARIABLE

18.1

INTRODUCTION

18.2

RANDOM VARIABLE

515 515

18.2.1

Discrete Random Variable

516

18.2.2

Continuous Random Variable

516

EXPECTED VALUE

18.3.1

18.5

472

BAYE’S THEOREM 474 476

18.4

472

17.5.1

EXERCISE

18.3

443

PROBABILITY

17.1

17.4

443

517

Expectation Theorems

517

MOMENTS OF RANDOM VARIABLES AND VARIANCE

18.4.1

Moments about the Origin

18.4.2

Central Moments

18.4.3

Variance

BINOMIAL DISTRIBUTION

518 518 519

518

518

18.6

18.7

18.5.1

Mean of the Binomial distribution

18.5.2

Variance of the Binomial distribution 519

18.5.3

Fitting of Binomial Distribution

POISSON DISTRIBUTION

519 520

521

18.6.1

Mean of Poisson Distribution

521

18.6.2

Variance of Poisson Distribution

521

18.6.3

Fitting of Poisson Distribution

522

NORMAL DISTRIBUTION

18.7.1

Mean and Variance of Normal Distribution

EXERCISE

526

SOLUTIONS

533

CHAPTER 19

522

STATISTICS

19.1

INTRODUCTION

19.2

MEAN

543

19.3

MEDIAN

544

19.4

MODE

545

19.5

MEAN DEVIATION

19.6

VARIANCE AND STANDARD DEVIATION

EXERCISE

547

SOLUTIONS

550

CHAPTER 20

523

543

545 546

CORRELATION AND REGRESSION ANALYSIS

20.1

INTRODUCTION

555

20.2

CORRELATION

555

20.3

MEASURE OF CORRELATION

555

20.3.1

Scatter or Dot Diagrams

20.3.2

Karl Pearson’s Coefficient of Correlation

556

20.3.3

Computation of Correlation Coefficient

557

20.4

RANK CORRELATION

20.5

REGRESSION

555

558

559

20.5.1

Lines of Regression

20.5.2

Angle between Two Lines of Regression

EXERCISE

562

SOLUTIONS

565

CHAPTER 21

559 560

SOLUTIONS OF NON-LINEAR ALGEBRAIC EQUATIONS

21.1

INTRODUCTION

569

21.2

SUCCESSIVE BISECTION METHOD

569

21.3

FALSE POSITION METHOD (REGULA-FALSI METHOD)

569

21.4

NEWTON - RAPHSON METHOD (TANGENT METHOD)

570

EXERCISE 21

571

SOLUTIONS 21

579

CHAPTER 22

INTEGRATION BY TRAPEZOIDAL AND SIMPSON’S RULE

22.1

INTRODUCTION

597

22.2

NUMERICAL DIFFERENTIATION

597

22.2.1

Numerical Differentiation Using Newton’s Forward Formula

597

22.2.2

Numerical Differentiation Using Newton’s Backward Formula

598

22.2.3

Numerical Differentiation Using Central Difference Formula

599

22.3

MAXIMA AND MINIMA OF A TABULATED FUNCTION

22.4

NUMERICAL INTEGRATION

Newton-Cote’s Quadrature Formula

22.4.2

Trapezoidal Rule

22.4.3

Simpson’s One-third Rule

601

22.4.4

Simpson’s Three-Eighth Rule

602

604

SOLUTIONS

608

CHAPTER 23

SINGLE AND MULTI STEP METHODS FOR DIFFERENTIAL EQUATIONS

INTRODUCTION

23.2

PICARD’S METHOD

617

23.3

EULER’S METHOD

618

23.3.1

600

601

23.1

23.4

600

22.4.1

EXERCISE

599

617

Modified Euler’s Method

RUNGE-KUTTA METHODS

618 619

23.4.1

Runge-Kutta First Order Method

619

23.4.2

Runge-Kutta Second Order Method

619

23.4.3

Runge-Kutta Third Order Method

619

23.4.4

Runge-Kutta Fourth Order Method

620

23.5

MILNE’S PREDICTOR AND CORRECTOR METHOD

23.6

TAYLOR’S SERIES METHOD

EXERCISE

623

SOLUTIONS

628

621

***********

620


EC, EE, ME, CE, PI, IN

Sample Chapter of Engineering

Mathematics

CHAPTER 1 Page 1 Chap 1

Matrix Algebra

MATRIX ALGEBRA

1.1

INTRODUCTION

This chapter, concerned with the matrix algebra, includes the following topics: • Multiplication of matrix • Transpose of matrix • Determinant of matrix • Rank of matrix • Adjoint of matrix • Inverse of matrix: elementary transformation, determination of inverse using elementary transformation • Echelon form and normal form of matrix; procedure for reduction of normal form. 1.2

in . o c . a i d o n . w w w

MULTIPLICATION OF MATRICES

If A and B be any two matrices, then their product AB will be defined only when number of columns in A is equal to the number of rows in B . If A = 6aij@m # n and B = 6bjk @n # p then their product, AB = C = 6cik@m # p will be matrix of order m # p , where cik =

n

/ aij bjk j=1

PROPERTIES OF MATRIX MULTIPLICATION

If A, B and C are three matrices such that their product is defined, then 1. Generally not commutative; AB ! BA 2. Associative law; (AB) C = A (BC) 3. Distributive law; A (B + C) = AB + AC 4. Cancellation law is not applicable, i.e. if AB = AC , it does not mean that B = C . 5. If AB = 0 , it does not mean that A = 0 or B = 0 . 6. (AB)T = (BA)T

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Sample Chapter of Engineering Page 2 Chap 1

1.3

Mathematics

TRANSPOSE OF A MATRIX

The matrix obtained form a given matrix A by changing its rows into columns or columns int rows is called Transpose of matrix A and is denoted by AT . From the definition it is obvious that if order of A is m # n , then order of AT is n # m .

Matrix Algebra

PROPERTIES OF TRANSPOSE OF MATRIX

Consider the two matrices A and B 1. (AT )T = A 2. (A ! B)T = AT ! BT 3. (AB)T = BT AT 4. (kA)T = k (A)T 5. (A1 A2 A 3 ...An - 1 An)T = A Tn A Tn - 1 ...A T3 A T2 A T1

1.4

DETERMINANT OF A MATRIX

i. n

The determinant of square matrix A is defined as a11 a12 a13 A = a21 a22 a23 a31 a32 a33

o .c

a i d

PROPERTIES OF DETERMINANT OF MATRIX

o n

Consider the two matrices A and B . 1. A exists + A is a square matrix 2. AB = A B

. w w 3.

w

AT = A

4. kA = kn A , if A is a square matrix of order n . 5. If A and B are square matrices of same order then AB = BA . 6. If A is a skew symmetric matrix of odd order then A = 0 . 7. If A = diag (a1, a2,...., an) then A = a1, a2 ...an . A n = An , n ! N . 9. If A = 0 , then matrix is called singular. 8.

Singular Matrix A square matrix A is said to be singular if A = 0 and non-singular if A ! 0. 1.5

RANK OF MATRIX

The number, r with the following two properties is called the rank of the matrix 1. There is at least one non-zero minor of order r . 2. Every minor of order (r + 1) is zero. This definition of the rank does uniquely fix the same for, as a consequence of the condition (2), every minor of order (r + 2), being the sum of multiples of minors of order (r + 1), will be zero. In fact, every minor of order greater Buy Online: shop.nodia.co.in

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Mathematics

than r will be zero as a consequence of the condition (2). The given following two simple results follow immediately from the definition 1. There exists a non-zero minor or order r & the rank is $ r . 2. All minors of order (r + 1) are zero & the rank is # r . In each of the two cases above, we assume r to satisfy only one of two properties (1) or (2) of the rank. The rank of matrix A represented by symbol r (A) .

Page 3 Chap 1 Matrix Algebra

Nullity of a Matrix If A is a square matrix of a order n, then n - r (A) is called the nullity of the matrix A and is denoted by N (A). Thus a non-singular square matrix of order n has rank equal to n and the nullity of such a matrix is equal to zero. 1.6

ADJOINT OF A MATRIX

If every element of a square matrix A be replaced by its cofactor in A , then the transpose of the matrix so obtained is called the Adjoint of matrix A and it is denoted by adj A. Thus, if A = 6aij @ be a square matrix and Fij be the cofactor of aij in A , then adj A + 6Fij @T .


PROPERTIES OF ADJOINT MATRIX

If A, B are square matrices of order n ad In is corresponding unit matrix, then 1. A (adj A) = A In = (adj A)A 2. adj A = A n - 1 3. adj (adj A) = A n - 2 A 4.

adj (adj A) = A (n - 1)

2

5. adj (AT ) = (adj A)T 6. adj (AB) = (adj B) (adj A)

7. adj (Am) = (adj A) m, m ! N 8. adj (kA) = kn - 1 (adj A), k ! R

1.7

INVERSE OF A MATRIX

If A and B are two matrices such that AB = I = BA, then B is called the inverse of A and it is denoted by A-1 . Thus, A-1 = B + AB = I = BA To find inverse matrix of a given matrix A we use following formula adj A A-1 = A Thus A-1 exists if A ! 0 and matrix A is called invertible.


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Page 4 Chap 1

PROPERTIES OF INVERSE MATRIX

Matrix Algebra

Let A and B are two invertible matrices of the same order, then 1. (AT ) -1 = (A-1)T 2. (AB) -1 = B-1 A-1 3. (Ak ) -1 = (A-1) k , k ! N 4. adj (A-1) = (adj A) -1 5. A-1 = 1 = A -1 A 6. If A = diag (a1, a2,..., an), then A-1 = diag (a 1-1, a 2-1, ..., a n-1) 7. AB = AC & B = C , if A ! 0

1.7.1

Elementary Transformations Any one of the following operations on a matrix is called an elementary transformation (or E -operation).

1. Interchange of two rows or two columns

i. n

(1) The interchange of i th and j th rows is denoted by Ri ) R j (2) The interchange of i th and j th columns is denoted by Ci ) C j .

o .c

2. Multiplication of (each element ) a row or column by a k.

a i d th

(1) The multiplication of i row by k is denoted by Ri " kRi (2) The multiplication of i th column by k is denoted by Ci " kCi

o n

3. Addition of k times the elements of a row (or column) to the corresponding elements of another row (or column), k ! 0

. w w

(1) The addition of k times the j th row to the i th row is denoted by Ri " Ri + kR j . (2) The addition of k times the j th column to the i th column is denoted by Ci " Ci + kC j . If a matrix B is obtained from a matrix A by one or more E -operations, then B is said to be equivalent to A. They can be written as A + B .

w 1.7.2

Inverse of Matrix by Elementary Transformations The elementary row transformations which reduces a square matrix A to the unit matrix, when applied to the unit matrix, gives the inverse matrix A-1 . Let A be a non-singular square matrix. Then, A = IA Apply suitable E -row operations to A on the left hand side so that A is reduced to I . Simultaneously, apply the same E -row operations to the pre-factor I on right hand side. Let I reduce to B, so that I = BA. Postmultiplying by A-1 , we get or or

IA-1 = BAA-1 A-1 = B (AA-1) = BI = B B = A-1


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Sample Chapter of Engineering 1.8

Mathematics

ECHELON FORM

A matrix is said to be of Echelon form if, 1. Every row of matrix A which has all its entries 0 occurs below every row which has a non-zero entry. 2. The first non-zero entry in each non-zero is equal to one. 3. The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row.


Rank of a matrix in the Echelon form The rank of a matrix in the echelon form is equal to the number of non-zero rows of the given matrix. For example, R0 2 6 1V W S is 2 the rank of the matrix A = S0 0 1 2W SS0 0 0 0WW X 3#4 T 1.9

NORMAL FORM


By a finite number of elementary transformations, every non-zero matrix A of order m # n and rank r (> 0) can be reduced to one of the following forms. Ir O Ir > O O H, > O H, 8Ir OB, 8IrB Ir denotes identity matrix of order r . Each one of these four forms is called Normal Form or Canonical Form or Orthogonal Form. Procedure for Reduction of Normal Form

Let A = 6aij@ be any matrix of order m # n . Then, we can reduce it to the normal form of the matrix A by subjecting it to a number of elementary transformation using following methodology. METHODOLOGY: REDUCTION OF NORMAL FORM

1. We first interchange a pair of rows (or columns), if necessary, to obtain a non-zero element in the first row and first column of the matrix A. 2. Divide the first row by this non-zero element, if it is not 1. 3. We subtract appropriate multiples of the elements of the first row from other rows so as to obtain zeroes in the remainder of the first column. 4. We subtract appropriate multiple of the elements of the first column from other columns so as to obtain zeroes in the remainder of the first row. 5. We repeat the above four steps starting with the element in the second row and the second column. 6. Continue this process down the leading diagonal until the end of the diagonal is reached or until all the remaining elements in the matrix are zero.

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Mathematics

EXERCISE 1


Match the items in column I and II.

QUE 1.1

Column I

Column II

P.

Singular matrix

1.

Determinant is not defined

Q.

Non-square matrix

2.

Determinant is always one

R.

Real symmetric

3.

Determinant is zero

S.

Orthogonal matrix

4.

Eigenvalues are always real

5.

Eigenvalues are not defined

i. n

(A) P-3, Q-1, R-4, S-2 (B) P-2, Q-3, R-4, S-1 (C) P-3, Q-2, R-5, S-4 (D) P-3, Q-4, R-2, S-1

o .c

a i d

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QUE 1.2

ARIHANT/286/26

Cayley-Hamilton Theorem states that a square matrix satisfies its own characteristic equation. Consider a matrix -3 2 A=> - 2 0H If A be a non-zero square matrix of orders n , then (A) the matrix A + Al is anti-symmetric, but the matrix A - Al is symmetric (B) the matrix A + Al is symmetric, but the matrix A - Al is antisymmetric (C) Both A + Al and A - Al are symmetric (D) Both A + Al and A - Al are anti-symmetric

. w w

w

QUE 1.3 ARIHANT/285/3

If A and B are two odd order skew-symmetric matrices such that AB = BA , then what is the matrix AB ? (A) An orthogonal matrix (B) A skew-symmetric matrix (C) A symmetric matrix (D) An identity matrix

QUE 1.4

If A and B are matrices of order 4 # 4 such that A = 5B and A = a B , then a is_______.


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Mathematics

If the rank of a ^5 # 6h matrix A is 4, then which one of the following ARIHANT/292/117 statements is correct? (A) A will have four linearly independent rows and four linearly independent columns (B) A will have four linearly independent rows and five linearly independent columns (C) AAT will be invertible (D) AT A will be invertible

QUE 1.5

QUE 1.6

(C)

n

% (- 1) aii

(B)

i=1 n

/ (- 1) aii

(D)

i=1

n

% aii i=1 n

/ aii

i=1


QUE 1.7

If A ! Rn # n, det A ! 0 , then A is (A) non singular and the rows and columns of A are linearly independent. (B) non singular and the rows A are linearly dependent. (C) non singular and the A has one zero rows. (D) singular


Square matrix A of order n over R has rank n . Which one of the following statement is not correct? (A) AT has rank n (B) A has n linearly independent columns (C) A is non-singular (D) A is singular


Matrix Algebra

If An # n is a triangular matrix then det A is (A)

QUE 1.9

Page 7 Chap 1

V R S5 3 2 W Determinant of the matrix S1 2 6 W is_____ SS3 5 10 WW X T The value of the determinant a h g h b f g f c (A) abc + 2fgh - af 2 - bg2 - ch2 (B) ab + a + c + d (C) abc + ab - bc - cg (D) a + b + c


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Sample Chapter of Engineering Page 8 Chap 1 Matrix Algebra

Mathematics

67 19 21 The value of the determinant 39 13 14 is______ 81 24 26


QUE 1.12

1 If 0 2

3 2 5 - 6 = 26 , then the determinant of the matrix 7 8

QUE 1.13

R 0 1 S S- 1 1 The determinant of the matrix S 0 0 S S 1 -2 T

R S2 S0 SS1 T

V 7 8W 5 - 6 W is____ 3 2 WW X

2 VW 3W is______ 1W W 1W X

0 1 0 0

Let A be an m # n matrix and B an n # m matrix. It is given that determinant Îm + AB h = determinant În + BAh, where Ik is the k # k identity matrix. Using the above property, the determinant of the matrix given below is______ R V S2 1 1 1W S1 2 1 1W S1 1 2 1W S W S1 1 1 2W T X

QUE 1.14

i. n

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a i d

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. w w

3 1 - 2i H, then Let A = > 1 - 2i 2


3 1 - 2i H (1) A = > 1 + 2i 2

2 1 + 2i H (2) A * = > 1 - 2i 2

(3) A * = A

(4) A is hermitian matrix

w

Which of above statement is/are correct ? (A) 1 and 3 (B) 1, 2 and 3 (C) 1 and 4 (D) All are correct


QUE 1.17

For which value of l will the matrix given below become singular? R V S 8 l 0W S 4 0 2W SS12 6 0WW T X R V S 0 1 - 2W If S- 1 0 3 W is a singular, then l is______ SS 2 - 2 l WW T X


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Sample Chapter of Engineering QUE 1.18

Multiplication of matrices E and Rcos q - sin q S E = S sin q cos q SS 0 0 T What is the matrix F ? Rcos q - sin q 0V S W (A) S sin q cos q 0W SS 0 0 1WW RT cos q sin q 0XV S W (C) S- sin q cos q 0W SS 0 0 1WW T X

F is G . matrices E and G are R1 0 0V 0VW W S 0W and G = S0 1 0W SS0 0 1WW 1WW X X T


R cos q cos q 0V S W (B) S- cos q sin q 0W SS 0 0 1WW TR VX S sin q - cos q 0W (D) Scos q sin q 0W SS 0 0 1WW T X

QUE 1.19

1 2 3 4 Rank of matrix = is - 2 0 5 7G

QUE 1.20

V R S1 1 1 W The rank of the matrix S1 - 1 0 W is______ SS1 1 1 WW X T


Mathematics


Given,

(1) A = 0 (3) rank Âh = 2

R V S1 2 3W A = S1 4 2W SS2 6 5WW T X

(2) A = Y 0 (4) rank Âh = 3

Which of above statement is/are correct ? (A) 1, 3 and 4 (B) 1 and 3 (C) 1, 2 and 4 (D) 2 and 4


R V S2 1 - 1W Given, A = S0 3 - 2W is SS2 4 - 3WW X (1) A = 0 T (3) rank Âh = 2

(3) A = Y 0 (4) rank Âh = 5


V R S4 2 1 3W QUE 1.23 Given matrix 6A@ = S6 3 4 7W, the rank of the matrix is_____ SS2 1 0 1WW ARIHANT/292/111 X T Buy Online: shop.nodia.co.in

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Mathematics

QUE 1.24

R2 - 1 3V W S The rank of the matrix A = S4 7 lW is 2. The value of l must be SS1 4 5WW X T

QUE 1.25

R1 S S2 The rank of matrix S 3 S S6 T

QUE 1.26

Given,

Matrix Algebra

ARIHANT/306/10

2 4 2 8

3 3 1 7

1 0 A= 1 0

a c a c

0 VW 2W is______ 3W W 5W X

b d b d

0 1 0 1

(1) A = 0 (3) rank Âh = 2

(2) Two rows are identical (4) rank Âh = 3

i. n


o .c

a i d

o n

Two matrices A and B are given below : p q p2 + q2 pr + qs ; B=> A => H H r s pr + qs r2 + s2

QUE 1.27

. w w

If the rank of matrix A is N , then the rank of matrix B is (A) N/2 (B) N - 1 (C) N (D) 2N

w QUE 1.28 ARIHANT/286/27

If x, y, z are in AP with common difference d and the rank of the matrix R V S4 5 x W S5 6 y W is 2, then the value of d and k are SS6 k z WW T d =X x/2 ; k is an arbitrary number (A) (B) d an arbitrary number; k = 7 (C) d = k ; k = 5 (D) d = x/2 ; k = 6

QUE 1.29

The rank of a 3 # 3 matrix C = AB , found by multiplying a non-zero column matrix A of size 3 # 1 and a non-zero row matrix B of size 1 # 3 , is (A) 0 (B) 1 (C) 2 (D) 3


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QUE 1.30

QUE 1.31

Rx y 1V S 1 1 W If A = Sx2 y2 1W and the point (x1, y1),( y2, y2),( x3, y2) are collinear, then the SSx y 1WW 3 3 X T rank of matrix A is (A) less than 3 (B) 3 (C) 1 (D) 0

QUE 1.33

QUE 1.34

Matrix Algebra

(B) 1 (D) n

Let P be a matrix of order m # n , and Q be a matrix of order n # p, n ! p . If r (P) = n and r (Q) = p , then rank r (PQ) is (A) n (B) p (C) np (D) n + p


x = 8x1 x2 g xnBT is an n-tuple nonzero vector. The n # n matrix V = xxT (A) has rank zero (B) has rank 1 (C) is orthogonal (D) has rank n

If x, y, z in A.P. with common difference d and the rank of the matrix V R S4 5 x W S5 6 y W is 2, then the values of d and k are respectively SS6 k z WW X T (A) x and 7 (B) 7, and x 4 4 (C) x and 5 7

QUE 1.35

Page 11 Chap 1

Let A = [aij ], 1 # i, j # n with n $ 3 and aij = i.j Then the rank of A is (A) 0 (C) n - 1

QUE 1.32

Mathematics

(D) 5, and x 7

If the rank of a (5 # 6) matrix Q is 4, then which one of the following statement is correct ? (A) Q will have four linearly independent rows and four linearly independent columns (B) Q will have four linearly independent rows and five linearly independent columns T (C) QQ will be invertible (D) QT Q will be invertible


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QUE 1.36

Matrix Algebra

QUE 1.37

Mathematics

1 -4 The adjoint matrix of = is 0 2G 4 2 (A) = G 0 1

1 0 (B) = 4 2G

2 4 (C) = G 1 0

2 4 (D) = 0 1G

x y If A = > H, then adj(adj A) is equal to z b b -z (A) = -y x G (C)

QUE 1.38

QUE 1.39

b z (B) = y xG

1 = b yG xb - yz - y x

If A is a R S4 0 (A) S0 4 SS0 0 RT S1 0 (C) S0 1 SS0 0 T

(D) None of these

i. n

3 # 3 matrix and A = 2 then A (adj A) is equal to V V R 0W S2 0 0 W (B) S0 2 0 W 0W W SS0 0 2 WW 4W VX XV TR1 0W S2 0 0 W (D) S0 12 0 W 0W W SS0 0 1 WW 1W 2 X T X

o .c

a i d

. w w

o n

If A is a 2 # 2 non-singular square matrix, then adj(adj A) is (A) A2 (B) A -1 (C) A (D) None of the above

w

Common Data For Q. 40 to 42 If A is a 3 - rowed square matrix such that A = 3 .

QUE 1.40

QUE 1.41

QUE 1.42

The adj(adj A) is equal to (A) 3A (C) 27A

(B) 9A (D) 81A

The value of adj (ajd A) is equal to (A) 3 (C) 27

(B) 9 (D) 81

The value of adj (adj A2) is equal to (A) 3 4 (C) 316

(B) 38 (D) 332


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Sample Chapter of Engineering QUE 1.43

The rank of an n row square matrix A is (n - 1), then (A) adjA ! 0 (B) adjA = 0 (C) adjA = In

QUE 1.44

QUE 1.45

QUE 1.46

QUE 1.47

QUE 1.48

(D) adjA = In - 1


R- 1 - 2 - 2V S W The adjoint of matrix A = S 2 1 - 2W is equal to SS 2 - 2 1WW T X (B) 3A (A) A (C) 3AT (D) A t

The matrix, that has an inverse is 3 1 (A) = G 6 2 6 2 (C) = G 9 3

5 2 (B) = G 2 1 8 2 (D) = G 4 1


-1 2 The inverse of the matrix A = > is 3 - 5H 5 2 5 -3 (A) = G (B) = 3 1 3 1G -5 -2 5 3 (C) = (D) = G G -3 -1 2 1

1 2 The inverse of the 2 # 2 matrix = G is 5 7 -7 2 (A) 1 = G 3 5 -1

7 2 (B) 1 = G 3 5 1

7 -2 (C) 1 = G 3 -5 1

-7 -2 (D) 1 = G 3 -5 -1

3 4 If B is an invertible matrix whose inverse in the matrix = G, then B is 5 6 1 6 -4 3 4 (A) = (B) > 1 H 5 6 - 5 6G -3 2 (C) = 5 - 3 G 2 2

QUE 1.49

Mathematics

1

(D) > 13 5

1 4 1 6

H

A B Matrix M = > is an orthogonal matrix. The value of B is C 0H (A) 1 (B) 1 2 2 (C) 1

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QUE 1.50

Matrix Algebra

QUE 1.51

R1 2V S W If A = S2 1W then A-1 is SS1 1WW R TV X S2 3 W (A) S3 1 W SS2 7 WW RT VX S1 4 W (C) S3 2 W SS2 5 WW T X

QUE 1.53

V R S 1 - 2W (B) S- 2 1 W SS 1 2 WW X T (D) Undefined

The inverse of

R1 0 0V W S matrix A = S5 2 0W is equal SS3 1 2WW XR T V 0W S 2 0 1 (B) S- 5 1 0W 2S W S- 1 - 1 2W X TR V 0 0W S 4 1 (D) S- 10 2 0W 4S W S-1 -1 2W X T

R S 2 0 1 (A) S- 5 2 4S S- 1 - 1 R T 0 S 1 (C) S- 10 2 SS - 1 - 1 T

QUE 1.52

Mathematics

i. n

a i d

Ra b c V S W If det A = 7, where A = Sd e f W then det(2A) -1 is_____ SSg b c WW T X

o n

. w w

R1 0 - 1V W S If R = S2 1 - 1W, the top of R-1 is SS2 3 2WW X T (A) [5, 6, 4] (C) [2, 0, - 1]

(B) [5, - 3, 1] (D) [2, - 1, 12 ]

-1 2 Let B be an invertible matrix and inverse of 7B is = , the matrix B 4 - 7G is 1 2 7 2 (A) > 4 71 H (B) = G 4 1 7 7 1 (C) > 2 7

QUE 1.55

V 0W 0W 2 WW XV 0W 0W 2 WW X

o .c

w QUE 1.54

to

4 7 1 7

R S0 S0 If A = S S0 S1 T

7 4 (D) = 2 1G

H

1 2 3 4

0 0 2 1

V 0W 1W , then det (A-1) is equal to_____ 0WW 0W X


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QUE 1.56

QUE 1.57

Mathematics

R V S1 1 1 1W S1 1 - 1 - 1W T -1 Given an orthogonal matrix A = S W, 6AA @ is 1 1 0 0 S W S0 0 1 - 1W T R1 V R1 X V 0 0 0 S4 W S2 0 0 0 W S0 1 0 0 W S0 1 0 0 W (A) S 4 1 W (B) S 2 1 W S0 0 2 0 W S0 0 2 0 W 1 S0 0 0 2 W S0 0 0 12 W TR 1 XV RT1 0 0 0 V X 0 0 0 S W 4 W S S0 14 0 0 W S0 1 0 0 W W (C) S (D) S 0 0 14 0 W 0 0 1 0W S W S S0 0 0 1 W S0 0 0 14 W X T T X Given an orthogonal matrix R1 1 1 1 V W S S1 1 - 1 - 1W A =S 1 - 1 0 0W W S S0 0 1 1 W X T -1 6AAT @ is R1 V S4 0 0 0 W S0 14 0 0 W W (A) S 1 S0 0 2 0 W S0 0 0 12 W RT1 0 0 0 V X W S S0 1 0 0 W (C) S 0 0 1 0W W S S0 0 0 1 W X T



QUE 1.58

V 0 0 0W 1 W 2 0 0 W 1 0 2 0W 0 0 12 W V 0 0 0XW 1 W 4 0 0 W 0 14 0 W 0 0 14 W X

A is m # n full rank matrix with m > n and I is identity matrix. Let matrix Al = (AT A) -1 AT , Then, which one of the following statement is FALSE ? (A) AAlA = A (B) (AAl) 2 (C) AlA = I (D) AAlA = Al

3

QUE 1.59

R1 S2 S0 (B) S S0 S0 TR 1 S4 S0 (D) S S0 S0 T

5 For a matrix 6M @ = > x

4 5 3 5 T

H, the transpose of the matrix is equal to the

inverse of the matrix, 6M @ = 6M @ . The value of x is given by

(A) - 4 5 (C) 3 5

QUE 1.60

-1

(B) - 3 5 (D) 4 5

1 0 2x 0 If A = > and A-1 > then the value of x is_____ H x x - 1 2H


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QUE 1.61

Matrix Algebra

QUE 1.62

QUE 1.63

Mathematics

2 -1 2 The value of = is 3 2 G=1 G 8 (A) = G 3

3 (B) = G 8

(C) 8- 3, - 8B

(D) [3, 8]

1 2 If A = > 3 -1 1 3 (A) = - 1 4G 5 1 (C) = 1 26G

0 , then AAT is 4H 1 0 1 (B) = - 1 2 3G (D) Undefined

-2 1 -1 7 H=> H, then the matrix A is equal to If A > 3 5 - 1 20 1 2 2 1 (B) = G (A) = G 3 5 5 3 5 3 -5 3 (C) = G (D) = 2 1 2 1G

i. n

o .c

QUE 1.64

QUE 1.65

a i d

1 2 - 0.1 a Let, A = > and A-1 = > 2 H. Then (a + b) =_____ H 0 3 0 b

o n

. w w

1 2 - 0.1 a Let A = > and A-1 = > 2 H, Then (a + b) is H 0 3 0 b (B) 3 (A) 7 20 20 (C) 19 (D) 11 20 20

w QUE 1.66

2 6 3 x If A = > H and B = > H, then in order that AB = 0 , the values of x and 3 9 y 2 y will be respectively (A) - 6 and - 1 (C) 6 and - 3

QUE 1.67

(B) 6 and 1 (D) 5 and 14

R1V S W 1 1 0 If A = > and B = S0W, the product of A and B is H 1 0 1 SS1WW 1 1 0 T X (B) = G (A) = G 0 0 1 1 (C) = G 2

1 0 (D) = G 0 2


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QUE 1.68

R 1 S 1 1 2 If A = > and B = S 2 H 2 1 0 SS- 1 1 4 T (A) = 4 4G 1 4 (C) = 4 1G

QUE 1.69


QUE 1.71

QUE 1.72

Mathematics

2VW 0W, then (AB)T is 1WW 1 4 X (B) = G 1 4


1 1 (D) = 4 4G

R 2 - 1V W S 1 -2 -5 If A = S 1 0W and B = > then AB is 3 4 0H SS- 3 4WW X V V R T R 0 - 10 W S 0 S- 1 - 8 - 10 W (A) S- 1 - 2 (B) S- 1 - 2 - 5 W 5W SS 9 22 15 WW SS 0 21 - 15 WW VX VX RT TR S- 1 - 8 - 10 W S0 - 8 - 10 W (C) S 1 - 2 - 5 W (D) S1 - 2 - 5 W SS 9 22 15 WW SS9 21 15 WW X X T T


R V S0 1 0 0W S0 0 1 0W W, then the rank of XT X , where XT denotes the transpose If X = S S0 0 0 1W S0 0 0 0W T X of X , is______

Consider the matrices X(4 # 3), Y(4 # 3) and P(2 # 3) . The order of [P (XT Y)T PT ]T will be (A) (2 # 2) (B) (3 # 3) (C) (4 # 3) (D) (3 # 4)

cos a sin a If A a = > , then consider the following statements : - sin cos aH 1. A a .A b = A ab 2. A a .A b = A(a + b) n n cos na sin na cos a sin a 3. (A a) n = > 4. (A a) n = > n n H - sin na cos naH - sin a cos a Which of the above statements are true ? (A) 1 and 2 (B) 2 and 3 (C) 2 and 4 (D) 3 and 4

QUE 1.73

0 - tan a2 cos a - sin a2 If A = > then (I - A) > is equal to H a tan 2 0 sin a cos a H (A) I - 2A (C) I + 2A

(B) I - A (D) I + A Buy Online: shop.nodia.co.in

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QUE 1.74

Matrix Algebra

ARIHANT/306/13

Mathematics

cos q sin q H Let A = > - sin q cos q 1 0 (1) AAT = > H 0 1

(2) AAT = 1

(3) A is orthogonal matrix

(4) A is not a orthogonal matrix


QUE 1.75

If the product of matrices cos2 q cos q sin q A == G and cos q sin q sin2 q cos2 f cos f sin f B == G cos f sin f sin2 f is a null matrix, then q and f differ by (A) an even multiple p2 (B) an even multiple p (C) an odd multiple of p2 (D) an odd multiple of p

i. n

o .c

QUE 1.76

a i d

o n

. w w

1 1 For a given 2 # 2 matrix A, it is observed that A > H =-> H and -1 -1 1 1 A > H =- 2 > H. The matrix A is -2 -2

w

2 1 -1 0 1 1 (A) A = > - 1 - 1H> 0 - 2H>- 1 - 2H 1 1 1 0 2 1 (B) A = > - 1 - 2H>0 2H>- 1 - 1H 1 1 -1 0 2 1 (C) A = > - 1 - 2H> 0 - 2H>- 1 - 1H 0 -2 (D) = 1 - 3G

QUE 1.77

3 -4 If A = > , then for every positive integer n, An is equal to 1 - 1H 1 + 2n 4n (A) = n 1 + 2n G

1 - 2n - 4n (B) = n 1 + 2n G

1 - 2n 4n (C) = n 1 + 2n G

1 + 2n - 4n (D) = n 1 - 2n G


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QUE 1.78

QUE 1.79

Mathematics

R V S3 W For which values of the constants b and c is the vector Sb W a linear R VR V R V SSc WW S1 W S2 W S- 1W T X Combination of S3 W, S6 W and S- 3 W SS2 WW SS4 WW SS- 2 WW T XT X T X (A) 9, 6 (B) 6, 9 (C) 6, 6 (D) 9, 9


V R Sa b c W The values of non zero numbers a, b, c, d, e, f, g, h such that the matrix Sd k e W SSf g h WW is invertible for all real numbers k . X T (A) finite sol n (B) infinite sol n (C) 0 (D) none

in . o c . a i d o n . w w w ***********


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Mathematics

SOLUTIONS 1


SOL 1.1

Correct option is (A). (P) Singular Matrix " Determinant is zero A = 0 (Q) Non-square matrix " An m # n matrix for which m ! n , is called nonsquare matrix. Its determinant is not defined (R) Real Symmetric Matrix " Eigen values are always real. (S) Orthogonal Matrix " A square matrix A is said to be orthogonal if AAT = I Its determinant is always one.

SOL 1.2

Correct option is (B). Here, if A be a non-zero square matrix of order n , then the matrix A + Al is symmetric, but A - Al will be anti-symmetric.

SOL 1.3

Correct option is (C). If A and B are both order skew-symmetric matrices, then A =- AT and B =- BT Also, given that AB = BA

i. n

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a i d

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. w w

w

...(1)

[from Eq. (1)] = ^- BT h^- AT h T T T = B A = ÂB h AB = ÂB hT i.e., AB is a symmetric matrix

SOL 1.4

Correct answer is 625. If k is a constant and A is a square matrix of order n # n then kA = kn A . A = 5B & A = 5B = 5 4 B = 625 B or a = 625

SOL 1.5

Correct option is (A). If rank of ^5 # 6h matrix is 4, then surely it must have exactly 4 linearly independent rows as well as 4 linearly independent columns. Hence, Rank = Row rank = Column rank

SOL 1.6

Correct option is (B). From linear algebra for An # n triangular matrix det A is equal to the product of the diagonal entries of A.


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Sample Chapter of Engineering SOL 1.7

Mathematics

Correct option is (B). If det A ! 0, then An # n is non-singular, but if An # n is non-singular, then no row can be expressed as a linear combination of any other. Otherwise det A=0

SOL 1.8

Correct option is (D). Since, if A is a square matrix of rank n , then it cannot be a singular.

SOL 1.9

Correct answer is - 28 .


5 3 2 1 2 6 = 5(20 - 30) - 3(10 - 18) + 2(5 - 6) 3 5 10 =- 50 + 24 - 2 =- 28

SOL 1.10

SOL 1.11


Correct option a h g

is (A). h g a f h f h b b f =a -h +g f c g c g f f c

= a ^bc - f 2h - h ^hc - fg h + g ^hf - gb h = abc = af 2 - h2 c + hfg + ghf - g2 b = abc + 2fgh - af 2 - bg2 - ch2

Correct answer is - 43 .

Determinant = 67

13 14 39 14 39 13 - 19 + 21 24 26 81 26 81 24

= 134 + 2280 - 2457 =- 43

SOL 1.12

Correct answer is 26. By interchanging any row or column, the value of determinant will remain same. For the given matrix, the first and third row are interchanged, thus the value remains the same.

SOL 1.13

Correct answer is - 1.

R 0 1 0 2V S W S- 1 1 1 3 W We have A =S 0 0 0 1W S W S 1 - 2 0 1W T X Expanding cofactor of a34 0 1 0 A = - 1 -1 1 1 1 -2 0 =- [0 - 1 (0 - 1) + 0] =- 1 Buy Online: shop.nodia.co.in

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SOL 1.14

Mathematics

Correct answer is 5. Consider the given matrix be

Matrix Algebra

Im + AB where m = 4 so, we obtain AB

A

Hence, we get

BA = 81 1

Therefore,

R V S2 1 1 1W S1 2 1 1W =S W S1 1 2 1W S1 1 1 2W T X R V R V S2 1 1 1W S1 0 0 0W S1 2 1 1W S0 1 0 0W =S W-S W S1 1 2 1W S0 0 1 0W S1 1 1 2W S0 0 0 1W X RT VX TR V S1 1 1 1W S1W S1 1 1 1W S1W =S W = S W 61 1 1 1@ S1 1 1 1W S1W S1 1 1 1W S1W TR V X T X S1W S1W = S W, B = 61 1 1 1@ S1W S1W T RX V S1W S1W 1 1BS W = 64@ S1W S1W T X

i. n

o .c

a i d

From the given property, Det Îm + AB h = Det În + BAh R V S2 1 1 1W S1 2 1 1W Det S W = Det "61@ + 64@, S1 1 2 1W S1 1 1 2W T X = Det 65@ = 5

o n

. w w

w

NOTE : Determinant of identity matrix is always 1.

SOL 1.15

Correct option is (D). A = conjugate of A 3 1 + 2i H => 1 + 2i 2 and A* = Â hT = transpose of A 3 1 + 2i H => 1 - 2i 2 Since, A* = A Hence, A is hermitian matrix.

SOL 1.16

Correct answer is 4. For singularity of matrix, Buy Online: shop.nodia.co.in

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Mathematics

8 l 0 4 0 2 =0 12 6 0


8 ^0 - 12h - l ^0 - 2 # 12h = 0 & l = 4

SOL 1.17

Correct answer is - 2 . Matrix A is singular if A = 0

R V S 0 1 - 2W S- 1 0 3 W = 0 SS 2 - 2 l WW 1 -2 0 3X 1 -T2 - (- 1) +2 +0 =0 -2 l -2 l 0 3

or

(l - 4) + 2 (3) = 0 l =- 2

or or

SOL 1.18


Correct option is (C). Given where G = I = Identity matrix EF = G Rcos q - sin q 0V R1 0 0V S W S W cos q 0W # F = S0 1 0W S sin q SS 0 SS0 0 1WW 0 1WW T X T X We know that the multiplication of a matrix and its inverse be a identity matrix AA-1 = I So, we can say that F is the inverse matrix of E 6adj.E @ F = E -1 = E Rcos q - (sin q) 0VT R cos q sin q 0V S W S W adj E = S sin q cos q 0W = S- sin q cos q 0W SS 0 SS 0 1WW 0 0 1WW T X T X E = 6cos q # (cos q - 0)@ - 8^- sin qh # ^sin q - 0hB + 0

Hence,

SOL 1.19

= cos2 q + sin2 q = 1 R cos q adj . E 6 @ S F = = S- sin q E SS 0 T

sin q cos q 0

0VW 0W 1WW X

Correct answer is 2. 1 2 3 4 A == - 2 0 5 7G It is a 2 # 4 matrix, thus r (A) # 2 The second order minor 1 2 =4!0 -2 0 We have

Hence,

r (A) = 2


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SOL 1.20

Mathematics

Correct answer is 2. We have

V R V R S1 1 1 W S1 1 1 W A = S1 - 1 0 W + S1 - 1 0 W SS1 1 1 WW SS0 0 0 WW X T r (A) X< 3T Since one full row is zero, 1 1 Now =- 2 ! 0 , thus r (A) = 2 1 -1

Matrix Algebra

SOL 1.21

R3 - R1

Correct option is (B).

R V S1 2 3W Here, A = S1 4 2W SS2 6 5WW R V T X S1 2 3W Performing operation R 31 ^- 1h, we get A + S1 4 2W SS1 4 2WW X R VT S1 2 3W By operation R 32 ^- 1h, we get A + S1 4 2W SS0 0 0WW T X A =0

i. n

and

SOL 1.22

1 2 =0 1 4 Y Rank Âh = 2

o .c

a i d

o n

. w w

Correct option is (B). Here, A = 2 ^- 9 + 8h + 2 ^- 2 + 3h = 0

w

But

Hence,

2 1 =0 0 3 Y

Rank Âh = 2

SOL 1.23

Correct answer is 2. Consider 3 # 3 minors, maximum possible rank is 3. Now we can obtain 4 2 1 2 1 3 4 1 3 4 2 3 6 3 4 = 0 , 3 4 7 = 0 , 6 4 7 = 0 and 6 3 7 = 0 2 1 0 1 0 1 2 0 1 2 1 1 Since, all 3 # 3 minors are zero. Now, we consider 2 # 2 minors 4 2 2 1 = 0, = 8-3 = 5 = Y 0 6 3 3 4 Hence, rank = 2

SOL 1.24

Correct answer is 13. Since r (A) = 2 < order of matrix


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Mathematics

2 -1 3 Thus A = 4 7 l =0 1 4 5 or (235 - 4l) + 1 (20 - l) + 3 (16 - 7) = 0 or 70 - 8l + 20 - l + 27 = 0 = 0 or, l = 13

SOL 1.25


Correct answer is 3. It is 4 # 4 matrix, So its rank r (A) # 4 1 2 3 0 2 4 3 2 We have A = 3 2 1 3 6 8 7 5 1 2 = 3 0

2 4 2 0

3 3 1 0

0 2 3 0

applying R4 - (R1 + R2 + R3) " R4

in . o c . a i d o n . w w w 1 2 3 0 0 -3 = 0 -4 -8 0 0 0

0 2 3 0

applying

R2 - 2R1 " R2 R3 - 3R1 " R3

The only fourth order minor is zero. 1 2 3 Since the third order minor 0 - 4 - 8 = (1)( - 4)( - 3) = 12 ! 0 0 0 -3 Therefore its rank is r (A) = 3

SOL 1.26

Correct option is (C). Here, A = 0 All minors of order 3 are zero, since two rows are identical. The second minor Hence,

SOL 1.27

1 0 =0 0 1 Y

Rank Âh = 2

Correct option is (C). Given the two matrices, p A => r and

q sH

p2 + q2 pr + qs B => H pr + qs r2 + s2

To determine the rank of matrix A, we obtain its equivalent matrix using the operation, a2i ! a2i - a21 a1i as a11


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Mathematics

p q A = >0 s - r q H p


s- r q = 0 p

If

or ps - rq = 0 then rank of matrix A is 1, otherwise the rank is 2. Now, we have the matrix p2 + q2 pr + qs B => H pr + qs r2 + s1 To determine the rank of matrix B , we obtain its equivalent matrix using the operation, a2i ! a2i - a21 a1i as a11 R 2 V 2 pr + qs Sp + q W 2 S W + pr qs ^ h B = ^r2 + s2h - 2 S 0 2 W p +q W S T X ^pr + qs h2 2 2 2 If = ^ps - rq h = 0 ^r + s h - 2 p + q2 or ps - rq = 0 then rank of matrix B is 1, otherwise the rank is 2. Thus, from the above results, we conclude that If ps - rq = 0 , then rank of matrix A and B is 1. If ps - rq ! 0 , then rank of A and B is 2. i.e. the rank of two matrices is always same. If rank of A is N then rank of B also N .

i. n

o .c

a i d

o n

SOL 1.28

. w w

Correct option is (B). It is given that x, y, z are in A.P. with common difference d x = x , y = x + d , z = x + 2d 4 5 x 4 5 x 4 5 x Let A = 5 6 y = 5 6 x+d = 1 1 d 6 k z 6 k x + 2d 1 k-6 d Applying R2 - R1 = R2 and R 3 - R2 = R 3 4 5 x = 1 1 d R3 = R3 - R2 0 k-7 0

w

A = 0 & ^k - 7h^4d - x h = 0 d = x , k = 7. 4

SOL 1.29

Correct option is (B). Let Let

Ra V S 1W A = Sb1W, B = 6a2 b2 c2@ SSc WW 1 T X C = AB


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Mathematics

Ra a a b a c V Ra V 1 2W 1 2 S1 2 S 1W = Sb1W # 8a2 b2 c2B = Sb1 a2 b1 b2 b1 c2W SSc a c b c c WW SSc WW 1 2 1 2 1 1 2 X are also T Xis zero and all Tthe 2 # 2 minors The 3 # 3 minor of this matrix zero. So the rank of this matrix is 1, i.e.


r 6C @ = 1

SOL 1.30

Correct option is (A). Since all point are collinear, x1 y1 1 Thus x2 y2 1 = 0 x3 y3 1 Therefore

r (A) < 3


SOL 1.31

Correct option is (B). Let n =3 V R S1 2 3 W Then A = S2 4 6 W SS3 6 9 WW X T 1 2 3 1 2 3 R3 - 3R1 " R3 A and applying = 2 4 6 = 0 0 0 R2 - 2R1 " R2 3 6 9 0 0 0 Thus rank if n = 3 then r (A) = 1 so possible answer is (B).

SOL 1.32

Correct option is (B). If P is a matrix of order m # n and r (P) = n then n # m In the normal form of P only n rows are non-zero Now Q is a matrix of order n # p and r (PQ) = p then p # n but p ! n but p ! n so p < n . In the normal form of Q only p rows are non-zero. Thus is the normal form of PQ only p rows are non zero. r (PQ) = p

SOL 1.33

Correct option is (D).

So rank of V is n .

x = 8x1 x2 g xnBT V = xxT R V R V Sx1W Sx1W Sx2W Sx2W =S W S W ShW ShW SxnW SxnW T X T X


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SOL 1.34

Matrix Algebra

Mathematics

Correct option is (A). Given that x, y, z are in A.P. with common differences d . Thus y = x + d, z = x + 2d 4 5 x 4 5 x 4 5 x A = 1 1 d = 5 6 x+d = 1 1 d 1 k-6 d 6 k x + 2d 1 k-6 d

Now

Apply R2 - R1 = R2 and R 3 - R2 = R 3 4 5 x applying R3 - R2 " R3 = 1 1 d 0 k-7 0 Thus A = (k - 7)( 4d - x) Since r (A) = 2 < order of matrix, Thus A = 0 or we get (k - 7)( 4d - x) = 0 or d = x ,k = 7 4

SOL 1.35

SOL 1.36

i. n

o .c

Correct option is (A). Rank of a matrix is no. of linearly independent rows and columns of the matrix. Here Rank r (Q) = 4 So Q will have 4 linearly independent rows and flour independent columns.

a i d

o n

. w w

Correct option is (D). We have

w

1 -4 A == 0 2G

C11 = 2, C12 = 0, C21 =- (- 4) = 4 , and C22 = 1 2 0 C == 4 1G adj A = C T 2 4 == 0 1G

SOL 1.37

Correct option is (D). A=

x y z b

b -y adj A = = -z x G x y adj(adj A) = = z bG

SOL 1.38

Correct option is (B). Since, A(adjA) = A In


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Sample Chapter of Engineering We have

Mathematics

V R V R S1 0 0 W S2 0 0 W A(adjA) = 2 S0 1 0 W = S0 2 0 W SS0 0 1 WW SS0 0 2 WW X T X T

SOL 1.39

Correct option is (B). We know that adj(adj(A) = A n - 2 A Here n = 2 so we get adj(adjA) = A 2 - 2 A = A 0 A = IA = A

SOL 1.40

Correct option is (A). We know that adj(adj A) = A n - 2 A Putting n = 3 and A = 3 . so we get adj(adj A) = A 3 - 2 A


in . o c . a i d o n . w w w = A A = 3A

SOL 1.41

Correct option is (D).

2

We have adj (adjA) = A (n - 1) Putting n = 3 and A = 3 we get adj (adj A) = A 4 = 3 4 = 81

SOL 1.42

Correct option is (B). Let B = adjA2 then B is also a 3 # 3 matrix. adj (adjA2) = adj B = B 3 - 1 = B 2 = adjA2 = 8 A2

2

B = A2 4 = A 8 = 38

(3 - 1) 2

SOL 1.43

Correct option is (A). Since r (A) = n - 1, at least one (n - 1) rowed minor of A is non-zero, so at least one minor and therefore the corresponding co-factor is non-zero. So, adjA ! 0

SOL 1.44

Correct option is (C). If A = [aij ] n # n then detA = [Cij ]Tn # n where Cij is the cofactor of aij Also Cij = (- 1) i + j Mij , where Mij is the minor of aij , obtained by leaving the row and the column corresponding to aij and then taken the determinant of the remaining matrix. 1 -2 Now, M11 = minor of a11 i.e. (- 1) =- 3 -2 1 Buy Online: shop.nodia.co.in

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Mathematics

Similarly

Page 30 Chap 1

2 -2 = 6; 2 1 -2 -2 M21 = =- 6 ; -2 1 M12 =

Matrix Algebra

M23 =

-1 -2 = 6; 2 -2

M32 =

-1 -2 = 6; 2 -2

2 1 =- 6 2 -2 -1 -2 M22 = 3; 2 1 = M13 =

M31 = M33 =

-2 -2 =6 ; 1 -2 -1 -2 =3 2 1

= (- 1) 1 + 1 M11 =- 3; C12 = (- 1) 1 + 2 M12 =- 6; = (- 1) 1 + 3 M13 =- 6; C21 = (- 1) 2 + 1 M21 = 6; = (- 1) 2 + 2 M22 = 3; C23 = (- 1) 2 + 3 M23 =- 6; = (- 1) 3 + 1 M31 = 6; C32 = (- 1) 3 + 2 M32 =- 6; = (- 1) 3 + 3 M33 = 3 VT R R- 1 - 2 - 2VT S- 3 - 6 - 6 W S W adj A = S 6 3 - 6 W = 3 S 2 1 - 2W = 3AT SS 6 - 6 3 WW SS 2 - 2 1WW T X X T C11 C13 C22 C31 C33

i. n

Thus

SOL 1.45

o .c

a i d

Correct option is (B). If A is zero, A-1 does not exist and the matrix A is said to be singular. Except (B) all satisfy this condition. 5 2 A = = (5)( 1) - (2)( 2) = 1 2 1

o n

SOL 1.46

. w w

w

Correct option is (A). We know Here Also,

A-1 = 1 adjA A -1 2 A == =- 1 3 - 5G -5 -2 adj A = = - 3 - 1G A-1 =

SOL 1.47

1 =- 5 - 2G = =5 2G 3 1 -1 -3 -1

Correct option is (A). We know Here Also,

A-1 = 1 adj A A A =

1 2 =- 3 5 7

7 -2 adj A = = - 5 1G


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Sample Chapter of Engineering A-1 =

Mathematics

1 = 7 - 2G = 1 =- 7 2 G -3 -5 1 3 5 -1


SOL 1.48

Correct option is (C). Let We know

3 4 B-1 = = G = A and B = A-1 5 6 A-1 = 1 adj A A

Here

A =

Also,

adj A = A-1

SOL 1.49

3 4 =- 2 5 6

6 -4 -5 3 6 -4 = 1= G -2 -5 3 -3 2 == 5 -3G 2 2


Correct option is (C). For orthogonal matrix det M = 1 and M-1 = MT , A C MT = = B 0G = M-1 =

This implies or

1 > 0 - BH - BC - C A

B = -C - BC B = 1 or B = ! 1 B

SOL 1.50

Correct option is (D). Inverse matrix is defined for square matrix only.

SOL 1.51

Correct option is (D). We know

A-1 = 1 adj A A 1 0 0 A = 5 2 0 = 4 ! 0, 3 1 2 4 10 - 10 T RS 4 0 0VW adj A = 0 2 - 1 = S 10 2 0W SS- 1 - 1 2WW 0 0 2 T X R 4 0 0VW S A-1 = 1 S 10 2 0W 4S S - 1 - 1 2WW T X


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SOL 1.52

Mathematics

Correct answer is 0.01786. 1 = 1 2A 2n A Since A is a 3 # 3 matrix, therefore n = 3 and we have A = 7 . det (2A) -1 = 3 1 = 1 = 1 8 # 7 56 2 A det (2A) -1 =

Matrix Algebra

SOL 1.53

Correct option is (B). C11 = 2 - (- 3) = 5 C21 =- [0 - (- 3)] =- 3 C31 = [- (- 1)] = 1 R = (1) C11 + 2C21 + 2C31 = 5 - 6 + 2 = 1

SOL 1.54

Correct option is (A). Let

. w w

SOL 1.55

A-1 =

1 =- 7 - 2G -1 -4 -1

7 2 7B = A-1 = > H 4 1 1

B = >4 7

2 7 1 7

H

Correct answer is 0.5. We have

Now

SOL 1.56

-1 2 =- 1 4 -7

-7 -2 adj A = = - 4 - 1G

o n

Also,

or

a i d

A =

Here

w

o .c

A-1 = 1 adj A A

We know

or

i. n

-1 2 and 7B = A-1 (7B) -1 = A = > 4 - 7H

R S0 S0 A =S S0 S1 T

V 0W 1W 0WW 0W X 0 0 1 A =- 0 2 0 =- 1 (0 - 2) = 2 1 1 0

det (A-1) =

1 2 3 4

0 0 2 1

1 =1 det A 2

Correct option is (C). For orthogonal matrix we know that


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Mathematics

AAT = I [AAT ] -1 = I-1 = I

and


SOL 1.57

Correct option is (C). From orthogonal matrix [AAT ] = I Since the inverse of I is I , thus [AAT ] -1 = I-1 = I

SOL 1.58

Correct option is (D). Al = (AT A) -1 AT = A-1 (AT ) -1 AT = A-1 I Put Al = A-1 I in all option.

in . o c . a i d o n . w w w AAlA AA-1 A A (AAl) 2 (AA-1 I) 2 Î h2 AlA -1 A IA I AAlA AA-1 IA

option (A)

option (B)

option (C)

option (D)

SOL 1.59

=A =A =A =I =I =I =I =I =I = Al

=A= Y Al

(true)

(true)

(true) (false)

Correct option is (A). 3 5

4 5 3 5

M => x

Given :

H

[M]T = [M] -1

And

T -1 We know that when 6A@ = 6A@ then it is called orthogonal matrix. 6M @T = I 6M @

6M @T 6M @ = I

Substitute the values of M and M T , we get 3

x

5

5

> 45

3

5 3 H.> x

R 3 S b # 3 l + x2 5 S 5 S 4 # 3 + 3x Sb 5 5l 5 T

4 5 3 5

1 0

H = >0 1H

3 4 + 3 x VW # b5 1 0 5l 5 W = > 4 4 3 3 W 0 1H b 5 # 5 l + b 5 # 5 lW X


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Mathematics 9 25

+ x2

>12 + 3 x 25

Matrix Algebra

5

12 25

1 0 + 53 x => H 0 1H 1

Comparing both sides a12 element, 12 + 3 x = 0 " x =- 12 5 =- 4 25 5 25 # 3 5

SOL 1.60

Correct answer is 0.5. 2x 0 1 0 1 > x x H>- 1 2H = =0 2x 0 1 or = 0 2x G = =0

0 1G 0 1G

So, 2x = 1 & x = 1 2

SOL 1.61

SOL 1.62

Correct option is (B). 2 -1 2 (2) (2) + (- 1) (1) 3 >3 2H>1H = > (3) (2) + (2) (1) H = =8 G

i. n

o .c

Correct option is (C).

R1 3V W S 1 2 0 AAT = > S2 - 1W H 3 -1 4 S S0 4WW X T (1) (1) + (2) (2) + (0) (0) (1) (3) + (2) (- 1) + (0) (4) => (3) (1) + (- 1) (2) + 4 (0) (3) (3) + (- 1) (- 1) + (4) (4)H

a i d

o n

. w w

SOL 1.63

w

5 1 => 1 26H

Correct option is (B). - 1 7 - 2 1 -1 A => - 1 20H> 3 5H -1 7 1 > 5 - 1H => H b - 1 20 - 13 l - 3 - 2 =

1 >1 7 H> 5 - 1H - 13 1 20 - 3 - 2

=

1 >26 - 13H - 13 65 - 39

2 1 == G 5 3

SOL 1.64

Correct answer is 0.35. We have Now

1 2 - 0.1 2 a -1 and = A A == G > 0 3 0 bH

AA-1 = I


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Mathematics

or

2 - 0.1 12 a 1 0 >0 3 H> H = >0 1H 0 b


or

1 2a - 0.1b 1 0 => H >0 H 3b 0 1

2a - 0.1 = 0 and 3b = 1 Thus solving above we have b = 1 and a = 1 3 60 Therefore a+b = 1+ 1 = 7 3 60 20 or

SOL 1.65

Correct option is (A). We know that AA-1 = I 2 - 0.1 12 a 1 2a - 0.1b 1 0 or => => H >0 H H > H 3 0 b 0 3b 0 1 We get 3b = 1 or b = 1 3 and


Thus

SOL 1.66

2a - 0.1b = 0 or a = b 20

a+b = 1+1 1 = 7 3 3 20 20

Correct option is (A). We have We get or

We get and

2 A == 3 0 2 6 3 x =3 9G=y 2 G = =0 6 + 6y 2x + 12 0 =9 + 9y 3x + 18G = =0

3 x 6 and B = > H, AB = 0 G y 2 9 0 0G 0 0G

6 + 6y = 0 or y =- 1 2x + 12 = 0 or x =- 6

SOL 1.67


SOL 1.68

Correct option is (A).

R V 1 1 1 0S W S0 W AB = = 1 0 1GS W S1 W T X(1)( 0) + (0)( 1) (1)( 1) + 1 == G = = (1)( 1) + (0)( 0) + (1)( 1) 2G

We have

V R 1 2W 1 1 2S 1 4 S 2 0W = = AB = = G 4 4G 2 1 0S W S- 1 1 W X 1 4 T T (AB) = = 4 4G


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Mathematics

SOL 1.69

Correct option is (C). V R S 2 - 1W 1 - 2 - 5 AB = S 1 0 W= G SS- 3 4 WW 3 4 0 V X TR S(2)( 1) + (- 1)( 3) (2)( - 2) + (- 1)( 4) (2)( - 5) + (- 1)( 0)W = S (1)( 1) + (0)( 3) (1)( - 2) + (0)( 4) (1)( - 5) + (0)( 0) W SS(- 3)( 1) + (4)( 3) (- 3)( - 2) + (4)( 4) (- 3)( - 5) + (4)( 0)WW X V RT S- 1 - 8 - 10 W = S 1 -2 -5 W SS 9 22 15 WW X T

SOL 1.70

Correct answer is 3.

Matrix Algebra

R V S0 1 0 0W S0 0 1 0W W We have X =S S0 0 0 1W S0 0 0 0W TR XV 0 0 0 0 S W S1 0 0 0W W and transpose of X , XT = S S0 1 0 0W S0 0 1 0W Here, we can see that rank ofT matrix Xx = 3 , hence, we can determine the rank of XT X . Let Y = X $ XT , the rank of Y # rank of X . Also, X-1 Y = XT and so we have Rank X = Rank XT # Rank of Y Hence, from Eqs. (1) and (2), we get Rank of X = Rank of Y Hence, rank of X $ XT is 3

i. n

o .c

a i d

o n

. w w

w SOL 1.71

Correct option is (A). X 4 # 3 " X cT# 4 X T3 # 4 Y4 # 3 " (XT Y) 3 # 3 (XY Y) 3 # 3 " (XT Y) -13 # 3 P2 # 3 " P 3T# 2 1 T -1 T (XT Y) -3 # 3 P 3 # 2 " "(X Y) PT ,3 2 # T -1 T P2 # 3 {(XT ) Y-1 PT } 3 # 2 " 6P (X Y) P @2 # 2 [P (XT Y) -1 PT ]T2 # 2 " [P (XT Y) -1 PT ] 2 # 2

SOL 1.72

Correct option is (C). cos a sin a cos b sin b A a .A b = > - sin a cos aH>- sin b cos bH cos (a + b) sin (a + b) => = Aa + b - sin (a + b) cos (a + b)H


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Mathematics

Also, it is easy to prove by induction that cos na sin na (A a) n = = - sin na cos na G

SOL 1.73


Correct option is (D). Let tan a = t 2 Then,

cos a =

2 1 - tan2 a2 = 1 - t2 2 a 1 + tan 2 t+t

and

sin a =

2 tan a2 = 2t 2 2 a 1 + tan 2 1+t

cos a - sin a cos a - sin a 1 tan a2 (I - A) > => H # = H a sin a cos a G sin a cos a 1 - tan 2 R V 2 - 2t W S 1 - t2 1 t S 1 + t (1 + t2) W == - t 1G # S 2t 1 - t2 W S W 2 S(1 + t ) 1 + t2 W T X 1 -t 1 - tan a2 => = = (I + A) t 1 H >tan a2 1 H

SOL 1.74



cos q sin q cos q - sin q H> H AAT = > - sin q cos q sin q cos q

cos2 q + sin2 q - cos q sin q + sin q cos q H => - sin q cos q + cos q sin q sin2 q + cos2 q 1 0 => H 0 1

= 1, Hence A is orthogonal matrix.

SOL 1.75

Correct option is (C). cos q cos f cos (q - f) cos a sin f cos (q - f) AB = = cos f sin a cos (a - f) sin q sin f cos (q - f) G Is a null matrix when cos (q - f) = 0 , this happens when (q - f) is an odd multiple of p . 2

SOL 1.76

Correct option is (C). a b Let matrix A be = c dG 1 1 a b 1 a-b 1 From A > H =-> H we get = == =-= G G G G = -1 -1 c d -1 c-d -1 We have and

a - b =- 1 c-d = 1

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


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Mathematics

1 1 From A > H =- 2 > H we get -2 -2


a b 1 a - 2b 1 =c d G=- 2G = =c - 2d G =- 2 =- 2G a - 2b c - 2d Solving equation (1) Solving equation (2)

=- 2 =4 and (3) a = 0 and b = 1 and (4) c =- 2 and d =- 3 0 1 A == - 2 - 3G

we have

Thus

...(2) ...(4)

If we check all option then result of option C after multiplication gives result.

SOL 1.77

Correct option is (D). 3 -4 3 -4 5 -8 A2 = = = 1 - 1G=1 - 1G =2 - 3G

i. n

If we put n = 2 in option, then only D satisfy.

SOL 1.78

Correct option is (A).

o .c

a i d

R V R V R V R V S- 1W S2 W S3 W S1 W Sb W = k1 S3 W + k2 S6 W + k3 S- 3 W SS- 2 WW SS4 WW SSc WW SS2 WW R V R V T X T X TR VX R TV X S1 W S3 W S1 W S1 W k1 S3 W + 2k2 S3 W - k3 S3 W = Sb W SS2 WW SSc WW SS2 WW SS2 WW RT VX RT VX T X T X S1 W S3 W (k1 + 2k2 - k3) S3 W = Sb W SS2 WW SSc WW k1 + 2k2 -Tk3X = T3 X 3k1 + 6k2 - 3k3 = b 2k1 + 6k2 - 2k3 = c b = 9,

o n

. w w

w

&

c =6

SOL 1.79

Correct option is (B). det (A) = (ah - cf) k + bef + cdg - aeg - bdh Thus matrix A is invertible for all k if (and only if) the coefficient (ah - cf) of k is 0, while the sum bef + cdg - aeg - bdh is non zero. & Thus infinitely many other sol n ***********


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Engineering Mathematics Kanodia

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