Engineering Mathematics Kanodia

Eighth Edition
GATE
Engineering Mathematics
RK Kanodia
Ashish Murolia
NODI A & COMPANY
GATE Electronics & Communication Vol 2, 8e
Engineering Mathematics
RK Kanodia and Ashish Murolia
Copyright © By NODIA & COMPANY
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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 MATRI X ALGEBRA
1.1 I NTRODUCTI ON 1
1.2 MULTI PLI CATI ON OF MATRI CES 1
1.3 TRANSPOSE OF A MATRI X 2
1.4 DETERMI NANT OF A MATRI X 2
1.5 RANK OF MATRI X 2
1.6 ADJ OI NT OF A MATRI X 3
1.7 I NVERSE OF A MATRI X 3
1.7.1 Elementary Transformations 4
1.7.2 Inverse of Matrix by Elementary Transformations 4
1.8 ECHELON FORM 5
1.9 NORMAL FORM 5
EXERCI SE 6
SOLUTI ONS 20
CHAPTER 2 SYSTEMS OF LI NEAR EQUATI ONS
2.1 I NTRODUCTI ON 39
2.2 VECTOR 39
2.2.1 Equality of Vectors 39
2.2.2 Null Vector or Zero Vector 39
2.2.3 A Vector as a Linear Combination of a Set of Vectors 40
2.2.4 Linear Dependence and Independence of Vectors 40
2.3 SYSTEM OF LI NEAR EQUATI ONS 40
2.4 SOLUTI ON OF A SYSTEM OF LI NEAR EQUATI ONS 40
EXERCI SE 42
SOLUTI ONS 51
CHAPTER 3 EI GENVALUES AND EI GENVECTORS
3.1 I NTRODUCTI ON 65
3.2 EI GENVALUES AND EI GEN VECTOR 65
3.3 DETERMI NATI ON OF EI GENVALUES AND EI GENVECTORS 66
3.4 CAYLEY-HAMI LTON THEOREM 66
3.4.1 Computation of the Inverse Using Cayley-Hamilton Theorem 67
3.5 REDUCTI ON OF A MATRI X TO DI AGONAL FORM 67
3.6 SI MI LARI TY OF MATRI CES 68
EXERCI SE 69
SOLUTI ONS 80
CHAPTER 4 LI MI T, CONTI NUI TY AND DI FFERENTI ABI LI TY
4.1 I NTRODUCTI ON 99
4.2 LI MI T OF A FUNCTI ON 99
4.2.1 Left Handed Limit 99
4.2.2 Right Handed Limit 99
4.2.3 Existence of Limit at Point 100
4.2.4 L’ hospital’s Rule 100
4.3 CONTI NUI TY OF A FUNCTI ON 100
4.3.1 Continuity in an interval 100
4.4 DI FFERENTI ABI LI TY 101
EXERCI SE 102
SOLUTI ONS 115
CHAPTER 5 MAXI MA AND MI NI MA
5.1 I NTRODUCTI ON 139
5.2 MONOTONOCI TY 139
5.3 MAXI MA AND MI NI MA 139
EXERCI SE 140
SOLUTI ONS 147
CHAPTER 6 MEAN VALUE THEOREM
6.1 I NTRODUCTI ON 163
6.2 ROLLE’S THEOREM 163
6.3 LAGRANGE’S MEAN VALUE THEOREM 163
6.4 CAUCHY’S MEAN VALUES THEOREM 163
EXERCI SE 164
SOLUTI ONS 168
CHAPTER 7 PARTI AL DERI VATI VES
7.1 I NTRODUCTI ON 175
7.2 PARTI AL DERI VATI VES 175
7.2.1 Partial Derivatives of Higher Orders 175
7.3 TOTAL DI FFERENTI ATI ON 176
7.4 CHANGE OF VARI ABLES 176
7.5 DI FFERENTI ATI ON OF I MPLI CI T FUNCTI ON 176
7.6 EULER’S THEOREM 176
EXERCI SE 177
SOLUTI ONS 182
CHAPTER 8 DEFI NI TE I NTEGRAL
8.1 I NTRODUCTI ON 191
8.2 DEFI NI TE I NTEGRAL 191
8.3 I MPORTANT FORMULA FOR DEFI NI TE I NTEGRAL 192
8.4 DOUBLE I NTEGRAL 192
EXERCI SE 193
SOLUTI ONS 202
CHAPTER 9 DI RECTI ONAL DERI VATI VES
9.1 I NTRODUCTI ON 223
9.2 DI FFERENTI AL ELEMENTS I N COORDI NATE SYSTEMS 223
9.3 DI FFERENTI AL CALCULUS 223
9.4 GRADI ENT OF A SCALAR 224
9.5 DI VERGENCE OF A VECTOR 224
9.6 CURL OF A VECTOR 225
9.7 CHARACTERI ZATI ON OF A VECTOR FI ELD 225
9.8 LAPLACI AN OPERATOR 225
9.9 I NTEGRAL THEOREMS 226
9.9.1 Divergence theorem 226
9.9.2 Stoke’s Theorem 226
9.9.3 Green’s Theorem 226
9.9.4 Helmholtz’s Theorem 226
EXERCI SE 227
SOLUTI ONS 234
CHAPTER 10 FI RST ORDER DI FFERENTI AL EQUATI ONS
10.1 I NTRODUCTI ON 247
10.2 DI FFERENTI AL EQUATI ON 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 DI FFERENTI AL EQUATI ON OF FI RST ORDER AND FI RST DEGREE 248
10.4 SOLUTI ON OF A DI FFERENTI AL EQUATI ON 249
10.5 VARI ABLES SEPARABLE FORM 249
10.5.1 Equations Reducible to Variable Separable Form 250
10.6 HOMOGENEOUS EQUATI ONS 250
10.6.1 Equations Reducible to Homogeneous Form 251
10.7 LI NEAR DI FFERENTI AL EQUATI ON 252
10.7.1 Equations Reducible to Linear Form 253
10.8 BERNOULLI ’S EQUATI ON 253
10.9 EXACT DI FFERENTI AL EQUATI ON 254
10.9.1 Necessary and Sufficient Condition for Exactness 254
10.9.2 Solution of an Exact Differential Equation 254
10.9.3 Equations Reducible to Exact Form: Integrating Factors 255
10.9.4 Integrating Factors Obtained by Inspection 255
EXERCI SE 257
SOLUTI ONS 266
CHAPTER 11 HI GHER ORDER DI FFERENTI AL EQUATI ONS
11.1 I NTRODUCTI ON 283
11.2 LI NEAR DI FFERENTI AL EQUATI ON 283
11.2.1 Operator 283
11.2.2 General Solution of Linear Differential Equation 283
11.3 DETERMI NATI ON OF COMPLEMENTARY FUNCTI ON 284
11.4 PARTI CULAR I NTEGRAL 285
11.4.1 Determination of Particular Integral 285
11.5 HOMOGENEOUS LI NEAR DI FFERENTI AL EQUATI ON 287
11.6 EULER EQUATI ON 288
EXERCI SE 289
SOLUTI ONS 300
CHAPTER 12 I NI TI AL AND BOUNDARY VALUE PROBLEMS
12.1 I NTRODUCTI ON 317
12.2 I NI TI AL VALUE PROBLEMS 317
12.3 BOUNDARY-VALUE PROBLEM 317
EXERCI SE 319
SOLUTI ONS 325
CHAPTER 13 PARTI AL DI FFERENTI AL EQUATI ON
13.1 I NTRODUCTI ON 337
13.2 PARTI AL DI FFERENTI AL EQUATI ON 337
13.2.1 Partial Derivatives of First Order 337
13.2.2 Partial Derivatives of Higher Order 338
13.3 HOMOGENEOUS FUNCTI ONS 339
13.4 EULER’S THEOREM 339
13.5 COMPOSI TE FUNCTI ONS 340
13.6 ERRORS AND APPROXI MATI ONS 341
EXERCI SE 342
SOLUTI ONS 347
CHAPTER 14 ANALYTI C FUNCTI ONS
14.1 I NTRODUCTI ON 357
14.2 BASI C TERMI NOLOGI ES I N COMPLEX FUNCTI ON 357
14.3 FUNCTI ONS OF COMPLEX VARI ABLE 358
14.4 LI MI T OF A COMPLEX FUNCTI ON 358
14.5 CONTI NUI TY OF A COMPLEX FUNCTI ON 359
14.6 DI FFERENTI ABI LI TY OF A COMPLEX FUNCTI ON 359
14.6.1 Cauchy-Riemann Equation: Necessary Condition for Differentiability of a Complex
Function 360
14.6.2 Sufficient Condition for Differentiability of a Complex Function 361
14.7 ANALYTI C FUNCTI ON 362
14.7.1 Required Condition for a Function to be Analytic 362
14.8 HARMONI C FUNCTI ON 363
14.8.1 Methods for Determining Harmonic Conjugate 363
14.8.2 Milne-Thomson Method 364
14.8.3 Exact Differential Method 366
14.9 SI NGULAR POI NTS 366
EXERCI SE 367
SOLUTI ONS 380
CHAPTER 15 CAUCHY’S I NTEGRAL THEOREM
15.1 I NTRODUCTI ON 405
15.2 LI NE I NTEGRAL OF A COMPLEX FUNCTI ON 405
15.2.1 Evaluation of the Line Integrals 405
15.3 CAUCHY’S THEOREM 406
15.3.1 Cauchy’s Theorem for Multiple Connected Region 407
15.4 CAUCHY’S I NTEGRAL FORMULA 408
15.4.1 Cauchy’s Integral Formula for Derivatives 409
EXERCI SE 410
SOLUTI ONS 420
CHAPTER 16 TAYLOR’S AND LAURENT’ SERI ES
16.1 I NTRODUCTI ON 439
16.2 TAYLOR’S SERI ES 439
16.3 MACLAURI N’S SERI ES 440
16.4 LAURENT’S SERI ES 441
16.5 RESI DUES 443
16.5.1 The Residue Theorem 443
16.5.2 Evaluation of Definite Integral 443
EXERCI SE 444
SOLUTI ONS 453
CHAPTER 17 PROBABI LI TY
17.1 I NTRODUCTI ON 469
17.2 SAMPLE SPACE 469
17.3 EVENT 469
17.3.1 Algebra of Events 470
17.3.2 Types of Events 470
17.4 DEFI NI TI ON OF PROBABI LI TY 471
17.4.1 Classical Definition 471
17.4.2 Statistical Definition 472
17.4.3 Axiomatic Definition 472
17.5 PROPERTI ES OF PROBABI LI TY 472
17.5.1 Addition Theorem for Probability 472
17.5.2 Conditional Probability 473
17.5.3 Multiplication Theorem for Probability 473
17.5.4 Odds for an Event 473
17.6 BAYE’S THEOREM 474
EXERCI SE 476
SOLUTI ONS 491
CHAPTER 18 RANDOM VARI ABLE
18.1 I NTRODUCTI ON 515
18.2 RANDOM VARI ABLE 515
18.2.1 Discrete Random Variable 516
18.2.2 Continuous Random Variable 516
18.3 EXPECTED VALUE 517
18.3.1 Expectation Theorems 517
18.4 MOMENTS OF RANDOM VARI ABLES AND VARI ANCE 518
18.4.1 Moments about the Origin 518
18.4.2 Central Moments 518
18.4.3 Variance 518
18.5 BI NOMI AL DI STRI BUTI ON 519
18.5.1 Mean of the Binomial distribution 519
18.5.2 Variance of the Binomial distribution 519
18.5.3 Fitting of Binomial Distribution 520
18.6 POI SSON DI STRI BUTI ON 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
18.7 NORMAL DI STRI BUTI ON 522
18.7.1 Mean and Variance of Normal Distribution 523
EXERCI SE 526
SOLUTI ONS 533
CHAPTER 19 STATI STI CS
19.1 I NTRODUCTI ON 543
19.2 MEAN 543
19.3 MEDI AN 544
19.4 MODE 545
19.5 MEAN DEVI ATI ON 545
19.6 VARI ANCE AND STANDARD DEVI ATI ON 546
EXERCI SE 547
SOLUTI ONS 550
CHAPTER 20 CORRELATI ON AND REGRESSI ON ANALYSI S
20.1 I NTRODUCTI ON 555
20.2 CORRELATI ON 555
20.3 MEASURE OF CORRELATI ON 555
20.3.1 Scatter or Dot Diagrams 555
20.3.2 Karl Pearson’s Coefficient of Correlation 556
20.3.3 Computation of Correlation Coefficient 557
20.4 RANK CORRELATI ON 558
20.5 REGRESSI ON 559
20.5.1 Lines of Regression 559
20.5.2 Angle between Two Lines of Regression 560
EXERCI SE 562
SOLUTI ONS 565
CHAPTER 21 SOLUTI ONS OF NON-LI NEAR ALGEBRAI C EQUATI ONS
21.1 I NTRODUCTI ON 569
21.2 SUCCESSI VE BI SECTI ON METHOD 569
21.3 FALSE POSI TI ON METHOD (REGULA-FALSI METHOD) 569
21.4 NEWTON - RAPHSON METHOD (TANGENT METHOD) 570
EXERCI SE 21 571
SOLUTI ONS 21 579
CHAPTER 22 I NTEGRATI ON BY TRAPEZOI DAL AND SI MPSON’S RULE
22.1 I NTRODUCTI ON 597
22.2 NUMERI CAL DI FFERENTI ATI ON 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 MAXI MA AND MI NI MA OF A TABULATED FUNCTI ON 599
22.4 NUMERI CAL I NTEGRATI ON 600
22.4.1 Newton-Cote’s Quadrature Formula 600
22.4.2 Trapezoidal Rule 601
22.4.3 Simpson’s One-third Rule 601
22.4.4 Simpson’s Three-Eighth Rule 602
EXERCI SE 604
SOLUTI ONS 608
CHAPTER 23 SI NGLE AND MULTI STEP METHODS FOR DI FFERENTI AL EQUATI ONS
23.1 I NTRODUCTI ON 617
23.2 PI CARD’S METHOD 617
23.3 EULER’S METHOD 618
23.3.1 Modified Euler’s Method 618
23.4 RUNGE-KUTTA METHODS 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 MI LNE’S PREDI CTOR AND CORRECTOR METHOD 620
23.6 TAYLOR’S SERI ES METHOD 621
EXERCI SE 623
SOLUTI ONS 628
***********
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Chap 1
Matrix Algebra
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Chap 1
Matrix Algebra
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1.1 I NTRODUCTI ON
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 MULTI PLI CATI ON OF MATRI CES
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 a
ij
m n
=
#
6 @
and B b
jk
n p
=
#
6 @
then their product,
AB c C
ik
m p
= =
#
6 @
will be matrix of order m p # , where
c
ik
a b
ij
j
n
jk
1
=
=
/
PROPERTI ES OF MATRI X MULTI PLI CATI ON
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 BA
T T
=
CHAPTER 1
MATRI X ALGEBRA
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Chap 1
Matrix Algebra
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Chap 1
Matrix Algebra
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1.3 TRANSPOSE OF A MATRI X
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 A
T
. From the definition it is obvious that if order of A is m n # , then
order of A
T
is n m # .
PROPERTI ES OF TRANSPOSE OF MATRI X
Consider the two matrices A and B
1. ( ) A A
T T
=
2. ( ) A B A B
T T T
! ! =
3. ( ) AB B A
T T T
=
4. ( ) ( ) k k A A
T T
=
5. ( ... ) ... A A A A A A A A A A
n n
T
n
T
n
T T T T
1 2 3 1 1 3 2 1
=
- -
1.4 DETERMI NANT OF A MATRI X
The determinant of square matrix A is defined as

A

a
a
a
a
a
a
a
a
a
11
21
31
12
22
32
13
23
33
=
PROPERTI ES OF DETERMI NANT OF MATRI X
Consider the two matrices A and B .
1. A exists A + is a square matrix
2. AB A B =
3. A A
T
=
4. , k k A A
n
= 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 ( , ,...., ) a a a
n 1 2
then , ... a a a A
n 1 2
= .
8. ,n N A A
n n
! = .
9. If A 0 = , then matrix is called singular.
Singular Matrix
A square matrix A is said to be singular if A 0 = and non-singular if
. A 0 !
1.5 RANK OF MATRI X
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
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Matrix Algebra
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Chap 1
Matrix Algebra
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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 ( ). A r
Nullity of a Matrix
If A is a square matrix of a order , n then ( ) n A r - 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 ADJ OI NT OF A MATRI X
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 A
ij
=
6 @
be a square matrix and F
ij

be the cofactor of a
ij
in A , then adj
F
A ij
T
+6 @ .
PROPERTI ES OF ADJ OI NT MATRI X
If A , B are square matrices of order n ad I
n
is corresponding unit matrix,
then
1. A (adj A ) I A
n
= = (adj A )A
2. A A adj
n 1
=
-
3. adj (adj A ) A A
n 2
=
-
4. A A adj (adj )
( ) n 1
2
=
-
5. ) ( ) A A adj ( adj
T T
=
6. AB B A adj ( ) (adj )(adj ) =
7. ) ( ) ,m N A A adj ( adj
m m
! =
8. ( ) ( ), k k k R A A adj adj
n 1
! =
-
1.7 I NVERSE OF A MATRI X
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
A
1 -

A
A adj
=
Thus A
1 -
exists if A 0 ! and matrix A is called invertible.
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Chap 1
Matrix Algebra
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Chap 1
Matrix Algebra
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PROPERTI ES OF I NVERSE MATRI X
Let A and B are two invertible matrices of the same order, then
1. ( ) ( ) A A
T T 1 1
=
- -
2. AB B A ( )
1 1 1
=
- - -
3. ( ) ( ) ,k N A A
k k 1 1
! =
- -
4. A A adj ( ) (adj )
1 1
=
- -
5. A
A
A
1 1 1
= =
- -
6. If A = diag ( , ,..., ) a a a
n 1 2
, then A
1
=
-
diag ( , ,..., ) a a a
n 1
1
2
1 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
(1) The interchange of i
th
and j
th
rows is denoted by R R
i j
)
(2) The interchange of i
th
and j
th
columns is denoted by C C
i j
) .
2. Multiplication of (each element ) a row or column by a . k
(1) The multiplication of i
th
row by k is denoted by R kR
i i
"
(2) The multiplication of i
th
column by k is denoted by C kC
i i
"
3. Addition of k times the elements of a row (or column) to the corresponding
elements of another row (or column), k 0 !
(1) The addition of k times the j
th
row to the i
th
row is denoted by
R R kR
i i j
" + .
(2) The addition of k times the j
th
column to the i
th
column is denoted by
C C kC
i i 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 + .
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 I A =
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 = . Post-
multiplying by A
1 -
, we get
I A
1 -
BAA
1
=
-

or A
1 -
( ) B AA BI B
1
= = =
-
or B A
1
=
-
Page 5
Chap 1
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Page 5
Chap 1
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1.8 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,
the rank of the matrix A
0
0
0
2
0
0
6
1
0
1
2
0
3 4
=
#
R
T
S
S
S
S
V
X
W
W
W
W
is 2
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.
, , ,
I
O
O
O
I
O
I O I
r r
r r > >
8 8
H H
B B
I
r
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 A
ij
=
6 @
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: REDUCTI ON 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.
***********
Page 6
Chap 1
Matrix Algebra
Page 6
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EXERCI SE 1
QUE 1.1 Match the items in column I and I I.
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
(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
QUE 1.2 Cayley-Hamilton Theorem states that a square matrix satisfies its own
characteristic equation. Consider a matrix
A
3
2
2
0
=
-
-
> H
If A be a non-zero square matrix of orders n, then
(A) the matrix A A + l is anti-symmetric, but the matrix A A - l is
symmetric
(B) the matrix A A + l is symmetric, but the matrix A A - l is anti-
symmetric
(C) Both A A + l and A A - l are symmetric
(D) Both A A + l and A A - l are anti-symmetric
QUE 1.3 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 B 5 = and A B a =
, then a is_ _ _ _ _ _ _ .
ARIHANT/286/26
ARIHANT/285/3
Page 7
Chap 1
Matrix Algebra
Page 7
Chap 1
Matrix Algebra
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QUE 1.5 If the rank of a 5 6
# ^ h matrix A is 4, then which one of the following
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) AA
T
will be invertible
(D) A A
T
will be invertible
QUE 1.6 If A
n n #
is a triangular matrix then det A is
(A) ( )a 1
ii
i
n
1
-
=
%
(B) a
ii
i
n
1 =
%
(C) ( )a 1
ii
i
n
1
-
=
/ (D) a
ii
i
n
1 =
/
QUE 1.7 If , R A
n 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
QUE 1.8 Square matrix A of order n over R has rank n. Which one of the following
statement is not correct?
(A) A
T
has rank n
(B) A has n linearly independent columns
(C) A is non-singular
(D) A is singular
QUE 1.9 Determinant of the matrix
5
1
3
3
2
5
2
6
10
R
T
S
S
S
S
V
X
W
W
W
W
is_ _ _ _ _
QUE 1.10 The value of the determinant

a
h
g
h
b
f
g
f
c
(A) abc fgh af bg ch 2
2 2 2
+ - - -
(B) ab a c d + + +
(C) abc ab bc cg + - -
(D) a b c + +
ARIHANT/292/117
ARIHANT/286/28
ARIHANT/305/7
Page 8
Chap 1
Matrix Algebra
Page 8
Chap 1
Matrix Algebra
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QUE 1.11 The value of the determinant
67
39
81
19
13
24
21
14
26
is_ _ _ _ _ _
QUE 1.12 If
1
0
2
3
5
7
2
6
8
26 - = , then the determinant of the matrix
2
0
1
7
5
3
8
6
2
-
R
T
S
S
S
S
V
X
W
W
W
W
is_ _ _ _
QUE 1.13 The determinant of the matrix
0
1
0
1
1
1
0
2
0
1
0
0
2
3
1
1
-
-
R
T
S
S
S
S
S
V
X
W
W
W
W
W
is_ _ _ _ _ _
QUE 1.14 Let A be an m n
#
matrix and B an n m
#
matrix. It is given that
determinant I AB
m
+ = ^ h determinant I BA
n
+ ^ h, where I
k
is the k k
#

identity matrix. Using the above property, the determinant of the matrix
given below is_ _ _ _ _ _
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
QUE 1.15 Let
i
i
A
3
1 2
1 2
2
=
-
-
> H, then
(1)
i
i
A
3
1 2
1 2
2
=
+
-
> H (2) *
i
i
A
2
1 2
1 2
2
=
-
+
> H
(3) * A A = (4) A is hermitian matrix
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 1.16 For which value of l will the matrix given below become singular?

8
4
12
0
6
0
2
0
l
R
T
S
S
S
S
V
X
W
W
W
W
QUE 1.17 If
0
1
2
1
0
2
2
3
l
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
is a singular, then l is_ _ _ _ _ _
ARIHANT/306/9
ARIHANT/306/14
ARIHANT/292/110
Page 9
Chap 1
Matrix Algebra
Page 9
Chap 1
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QUE 1.18 Multiplication of matrices E and F is G . matrices E and G are

cos
sin
sin
cos E
0 0
0
0
1
q
q
q
q =
-
R
T
S
S
S
S
V
X
W
W
W
W
and G
1
0
0
0
1
0
0
0
1
=
R
T
S
S
S
S
V
X
W
W
W
W
What is the matrix F ?
(A)
cos
sin
sin
cos
0 0
0
0
1
q
q
q
q
-
R
T
S
S
S
S
V
X
W
W
W
W
(B)
cos
cos
cos
sin
0 0
0
0
1
q
q
q
q -
R
T
S
S
S
S
V
X
W
W
W
W
(C)
cos
sin
sin
cos
0 0
0
0
1
q
q
q
q -
R
T
S
S
S
S
V
X
W
W
W
W
(D)
sin
cos
cos
sin
0 0
0
0
1
q
q
q
q
-
R
T
S
S
S
S
V
X
W
W
W
W
QUE 1.19 Rank of matrix
1
2
2
0
3
5
4
7 -
= G is
QUE 1.20 The rank of the matrix
1
1
1
1
1
1
1
0
1
-
R
T
S
S
S
S
V
X
W
W
W
W
is_ _ _ _ _ _
QUE 1.21 Given,
A
1
1
2
2
4
6
3
2
5
=
R
T
S
S
S
S
V
X
W
W
W
W
(1)
A
0 = (2) A 0 =Y
(3) rank A 2 = ^ h (4) rank A 3 = ^ h
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
QUE 1.22 Given, A
2
0
2
1
3
4
1
2
3
=
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
is
(1) A 0 = (3) A 0 =Y
(3) rank A 2 = ^ h (4) rank A 5 = ^ h
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
QUE 1.23 Given matrix A
4
6
2
2
3
1
1
4
0
3
7
1
=
R
T
S
S
S
S
6
V
X
W
W
W
W
@ , the rank of the matrix is_ _ _ _ _
ARIHANT/306/8
ARIHANT/306/11
ARIHANT/292/111
Page 10
Chap 1
Matrix Algebra
Page 10
Chap 1
Matrix Algebra
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QUE 1.24 The rank of the matrix A
2
4
1
1
7
4
3
5
l =
-
R
T
S
S
S
S
V
X
W
W
W
W
is 2. The value of l must be
QUE 1.25 The rank of matrix
1
2
3
6
2
4
2
8
3
3
1
7
0
2
3
5
R
T
S
S
S
S
S
V
X
W
W
W
W
W
is_ _ _ _ _ _
QUE 1.26 Given,
A
a
c
a
c
b
d
b
d
1
0
1
0
0
1
0
1
=
(1)
A
0 = (2) Two rows are identical
(3) rank A 2 = ^ h (4) rank A 3 = ^ h
Which of above statement is/ are correct ?
(A) 1, 3 and 4 (B) 1 and 3
(C) 1, 2 and 3 (D) 2 and 4
QUE 1.27 Two matrices A and B are given below :
A
p
r
q
s
=> H;
p q
pr qs
pr qs
r s
B
2 2
2 2
=
+
+
+
+
> H
If the rank of matrix A is N, then the rank of matrix B is
(A) / N 2 (B) N 1 -
(C) N (D) N 2
QUE 1.28 If , , x y z are in AP with common difference d and the rank of the matrix
k
x
y
z
4
5
6
5
6
R
T
S
S
S
S
V
X
W
W
W
W
is 2, then the value of d and k are
(A) / d x 2 = ; k is an arbitrary number
(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
ARIHANT/306/10
ARIHANT/286/27
Page 11
Chap 1
Matrix Algebra
Page 11
Chap 1
Matrix Algebra
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QUE 1.30 If
x
x
x
y
y
y
A
1
1
1
1
2
3
1
2
3
=
R
T
S
S
S
S
V
X
W
W
W
W
and the point ( , ),( , ),( , ) x y y y x y
1 1 2 2 3 2
are collinear, then the
rank of matrix A is
(A) less than 3 (B) 3
(C) 1 (D) 0
QUE 1.31 Let [ ], , a i j n A 1
ij
# # = with n 3 $ and . a i j
ij
= Then the rank of A is
(A) 0 (B) 1
(C) n 1 - (D) n
QUE 1.32 Let P be a matrix of order m n # , and Q be a matrix of order , n p n p # ! .
If ( ) n P r = and ( ) p Q r = , then rank (P Q) r is
(A) n (B) p
(C) np (D) n p +
QUE 1.33 x x x x
n 1 2
T
g =
8 B
is an n-tuple nonzero vector. The n n # matrix
V xx
T
=
(A) has rank zero (B) has rank 1
(C) is orthogonal (D) has rank n
QUE 1.34 If , , x y z in A.P. with common difference d and the rank of the matrix
k
x
y
z
4
5
6
5
6
R
T
S
S
S
S
V
X
W
W
W
W
is 2, then the values of d and k are respectively
(A)
x
4
and 7 (B) 7, and
x
4
(C)
x
7
and 5 (D) 5, and
x
7
QUE 1.35 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
(C)
QQ
T
will be invertible
(D) Q Q
T
will be invertible
Page 12
Chap 1
Matrix Algebra
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QUE 1.36 The adjoint matrix of
1
0
4
2
-
= G is
(A)
4
0
2
1
= G (B)
1
4
0
2
= G
(C)
2
1
4
0
= G (D)
2
0
4
1
= G
QUE 1.37 If
x
z
y
b
A => H, then adj(adj A ) is equal to
(A)
b
y
z
x -
-
= G (B)
b
y
z
x
= G
(C)
xb yz
b
y
y
x
1
- -
= G (D) None of these
QUE 1.38 If A is a 3 3 # matrix and A 2 = then A (adj A ) is equal to
(A)
4
0
0
0
4
0
0
0
4
R
T
S
S
S
S
V
X
W
W
W
W
(B)
2
0
0
0
2
0
0
0
2
R
T
S
S
S
S
V
X
W
W
W
W
(C)
1
0
0
0
1
0
0
0
1
R
T
S
S
S
S
V
X
W
W
W
W
(D) 0
0
0
0
0
0
2
1
2
1
2
1
R
T
S
S
S
S
V
X
W
W
W
W
QUE 1.39 If A is a 2 2 # non-singular square matrix, then adj(adj A ) is
(A) A
2
(B) A
(C) A
1 -
(D) None of the above
Common Data For Q. 40 to 42
If A is a 3 - rowed square matrix such that A 3 = .
QUE 1.40 The adj(adj A ) is equal to
(A) 3A (B) 9A
(C) 27A (D) 81A
QUE 1.41 The value of A adj (ajd ) is equal to
(A) 3 (B) 9
(C) 27 (D) 81
QUE 1.42 The value of A adj (adj )
2
is equal to
(A) 3
4
(B) 3
8
(C) 3
16
(D) 3
32
Page 13
Chap 1
Matrix Algebra
Page 13
Chap 1
Matrix Algebra
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QUE 1.43 The rank of an n row square matrix A is ( ) n 1 - , then
(A) adjA 0 ! (B) adjA 0 =
(C) adj I A
n
= (D) adj I A
n 1
=
-
QUE 1.44 The adjoint of matrix A
1
2
2
2
1
2
2
2
1
=
- -
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
is equal to
(A) A (B) 3A
(C) 3A
T
(D) A
t
QUE 1.45 The matrix, that has an inverse is
(A)
3
6
1
2
= G (B)
5
2
2
1
= G
(C)
6
9
2
3
= G (D)
8
4
2
1
= G
QUE 1.46 The inverse of the matrix A
1
3
2
5
=
-
-
> H is
(A)
5
3
2
1
= G (B)
5
3
3
1
-
= G
(C)
5
3
2
1
-
-
-
-
= G (D)
5
2
3
1
= G
QUE 1.47 The inverse of the 2 2 # matrix
1
5
2
7
= G is
(A)
3
1
7
5
2
1
-
-
= G (B)
3
1
7
5
2
1
= G
(C)
3
1
7
5
2
1 -
-
= G (D)
3
1
7
5
2
1
-
-
-
-
= G
QUE 1.48 If B is an invertible matrix whose inverse in the matrix
3
5
4
6
= G, then B is
(A)
6
5
4
6 -
-
= G (B)
5
4
3
1
6
1 > H
(C)
3 2
2
5
2
3
-
-
= G (D)
3
1
5
1
4
1
6
1 > H
QUE 1.49 Matrix
A
C
B
M
0
=> H is an orthogonal matrix. The value of B is
(A)
2
1
(B)
2
1
(C) 1 (D) 0
Page 14
Chap 1
Matrix Algebra
Page 14
Chap 1
Matrix Algebra
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QUE 1.50 If A
1
2
1
2
1
1
=
R
T
S
S
S
S
V
X
W
W
W
W
then A
1 -
is
(A)
2
3
2
3
1
7
R
T
S
S
S
S
V
X
W
W
W
W
(B)
1
2
1
2
1
2
-
-
R
T
S
S
S
S
V
X
W
W
W
W
(C)
1
3
2
4
2
5
R
T
S
S
S
S
V
X
W
W
W
W
(D) Undefined
QUE 1.51 The inverse of matrix A
1
5
3
0
2
1
0
0
2
=
R
T
S
S
S
S
V
X
W
W
W
W
is equal to
(A)
4
1
2
5
1
0
2
1
0
0
2
-
- -
R
T
S
S
S
S
V
X
W
W
W
W
(B)
2
1
2
5
1
0
1
1
0
0
2
-
- -
R
T
S
S
S
S
V
X
W
W
W
W
(C)
1
10
1
0
2
1
0
0
2
-
- -
R
T
S
S
S
S
V
X
W
W
W
W
(D)
4
1
4
10
1
0
2
1
0
0
2
-
- -
R
T
S
S
S
S
V
X
W
W
W
W
QUE 1.52 If det A = 7, where
a
d
g
b
e
b
c
f
c
A =
R
T
S
S
S
S
V
X
W
W
W
W
then det( ) A 2
1 -
is_ _ _ _ _
QUE 1.53 If R
1
2
2
0
1
3
1
1
2
=
-
-
R
T
S
S
S
S
V
X
W
W
W
W
, the top of R
1 -
is
(A) [ , , ] 5 6 4 (B) [ , , ] 5 3 1 -
(C) [ , , ] 2 0 1 - (D) [ , , ] 2 1
2
1
-
QUE 1.54 Let B be an invertible matrix and inverse of 7B is
1
4
2
7
-
-
= G, the matrix B
is
(A)
1
7
4
7
2
7
1 > H (B)
7
4
2
1
= G
(C)
1
7
2
7
4
7
1 > H (D)
7
2
4
1
= G
QUE 1.55 If A
0
0
0
1
1
2
3
4
0
0
2
1
0
1
0
0
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
, then det A ( )
1 -
is equal to_ _ _ _ _
Page 15
Chap 1
Matrix Algebra
Page 15
Chap 1
Matrix Algebra
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QUE 1.56 Given an orthogonal matrix A
1
1
1
0
1
1
1
0
1
1
0
1
1
1
0
1
=
-
- -
-
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
, AA
T 1 -
6 @ is
(A)
0
0
0
0
0
0
0
0
0
0
0
0
4
1
4
1
2
1
2
1
R
T
S
S
S
S
S
S
V
X
W
W
W
W
W
W
(B)
0
0
0
0
0
0
0
0
0
0
0
0
2
1
2
1
2
1
2
1
R
T
S
S
S
S
S
S
V
X
W
W
W
W
W
W
(C)
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
R
T
S
S
S
S
S
V
X
W
W
W
W
W
(D)
0
0
0
0
0
0
0
0
0
0
0
0
4
1
4
1
4
1
4
1
R
T
S
S
S
S
S
S
V
X
W
W
W
W
W
W
QUE 1.57 Given an orthogonal matrix
A
1
1
1
0
1
1
1
0
1
1
0
1
1
1
0
1
=
-
- -
R
T
S
S
S
S
S
V
X
W
W
W
W
W

AA
T 1 -
6 @ is
(A)
0
0
0
0
0
0
0
0
0
0
0
0
4
1
4
1
2
1
2
1
R
T
S
S
S
S
S
S
V
X
W
W
W
W
W
W
(B)
0
0
0
0
0
0
0
0
0
0
0
0
2
1
2
1
2
1
2
1
R
T
S
S
S
S
S
S
V
X
W
W
W
W
W
W
(C)
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
R
T
S
S
S
S
S
V
X
W
W
W
W
W
(D)
0
0
0
0
0
0
0
0
0
0
0
0
4
1
4
1
4
1
4
1
R
T
S
S
S
S
S
S
V
X
W
W
W
W
W
W
QUE 1.58 A is m n # full rank matrix with m n > and I is identity matrix. Let matrix
( ) A A A A
T T 1
=
-
l , Then, which one of the following statement is FALSE ?
(A) AA A A = l (B) ( ) AA
2
l
(C) I A A = l (D) AA A A = l l
QUE 1.59 For a matrix
x
M
5
3
5
4
5
3
= 6 > @ H, the transpose of the matrix is equal to the
inverse of the matrix,
M M
T 1
=
-
6 6 @ @ . The value of x is given by
(A)
5
4
- (B)
5
3
-
(C)
5
3
(D)
5
4
QUE 1.60 If
x
x x
A
2 0
=> H and A
1
1
0
2
1
-
-
> H then the value of x is_ _ _ _ _
Page 16
Chap 1
Matrix Algebra
Page 16
Chap 1
Matrix Algebra
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QUE 1.61 The value of
2
3
1
2
2
1
-
= = G G is
(A)
8
3
= G (B)
3
8
= G
(C) , 3 8 - -
8 B
(D) [ , ] 3 8
QUE 1.62 If A
1
3
2
1
0
4
=
-
> H, then AA
T
is
(A)
1
1
3
4 -
= G (B)
1
1
0
2
1
3 -
= G
(C)
5
1
1
26
= G (D) Undefined
QUE 1.63 If A
2
3
1
5
1
1
7
20
-
=
-
-
> > H H, then the matrix A is equal to
(A)
1
3
2
5
= G (B)
2
5
1
3
= G
(C)
5
2
3
1
= G (D)
5
2
3
1
-
= G
QUE 1.64 Let,
.
A
2
0
01
3
=
-
> H and
a
b
A
0
1 2
1
=
-
> H. Then ( ) a b + =_ _ _ _ _
QUE 1.65 Let
.
A
2
0
01
3
=
-
> H and
a
b
A
0
1 2
1
=
-
> H, Then ( ) a b + is
(A)
20
7
(B)
20
3
(C)
20
19
(D)
20
11
QUE 1.66 If A
2
3
6
9
=> H and
y
x
B
3
2
=> H, then in order that AB 0 = , the values of x and
y will be respectively
(A) 6 - and 1 - (B) 6 and 1
(C) 6 and 3 - (D) 5 and 14
QUE 1.67 If A
1
1
1
0
0
1
=> H and B
1
0
1
=
R
T
S
S
S
S
V
X
W
W
W
W
, the product of A and B is
(A)
1
0
= G (B)
1
0
0
1
= G
(C)
1
2
= G (D)
1
0
0
2
= G
Page 17
Chap 1
Matrix Algebra
Page 17
Chap 1
Matrix Algebra
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QUE 1.68 If A
1
2
1
1
2
0
=> H and B
1
2
1
2
0
1
=
-
R
T
S
S
S
S
V
X
W
W
W
W
, then ( ) AB
T
is
(A)
1
4
4
4
= G (B)
1
1
4
4
= G
(C)
1
4
4
1
= G (D)
1
4
1
4
= G
QUE 1.69 If A
2
1
3
1
0
4
=
-
-
R
T
S
S
S
S
V
X
W
W
W
W
and B
1
3
2
4
5
0
=
- -
> H then AB is
(A)
1
1
9
8
2
22
10
5
15
-
-
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
(B)
0
1
0
0
2
21
10
5
15
- -
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
(C)
1
1
9
8
2
22
10
5
15
- -
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
(D)
0
1
9
8
2
21
10
5
15
-
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
QUE 1.70 If X
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
, then the rank of X X
T
, where X
T
denotes the transpose
of X , is_ _ _ _ _ _
QUE 1.71 Consider the matrices , X Y
( ) ( ) 4 3 4 3 # #
and P
( ) 2 3 #
. The order of [ ( ) ] P X Y P
T T T T

will be
(A) ( ) 2 2 # (B) ( ) 3 3 #
(C) ( ) 4 3 # (D) ( ) 3 4 #
QUE 1.72 If
cos
sin
sin
cos
A
a a
a
=
-
a > H, then consider the following statements :
1. . A A A =
a b ab
2. . A A A
( )
=
a b a b +
3. ( )
cos
sin
sin
cos
A
n
n
n
n
n
a
a
a
a
=
-
a > H 4. ( )
cos
sin
sin
cos
n
n
n
n
A
n
a
a
a
a
=
-
a > H
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 If
tan
tan
A
0
0
2
2
=
-
a
a
> H then ( )
cos
sin
sin
cos
I A
2
a
a a
-
-
a
> H is equal to
(A) I A 2 - (B) I A -
(C) I A 2 + (D) I A +
ARIHANT/287/32
Page 18
Chap 1
Matrix Algebra
Page 18
Chap 1
Matrix Algebra
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QUE 1.74 Let
cos
sin
sin
cos
A
q
q
q
q
=
-
> H
(1) AA
1
0
0
1
T
=> H (2) AA 1
T
=
(3) A is orthogonal matrix (4) A is not a orthogonal matrix
Which of above statement is/ are correct ?
(A) 1, 3 and 4
(B) 2 and 3
(C) 1, 2 and 3
(D) 2 and 4
QUE 1.75 If the product of matrices
A
cos
cos sin
cos sin
sin
2
2
q
q q
q q
q
== G and
B
cos
cos sin
cos sin
sin
2
2
f
f f
f f
f
== G
is a null matrix, then q and f differ by
(A) an even multiple
2
p

(B) an even multiple p
(C) an odd multiple of
2
p

(D) an odd multiple of p
QUE 1.76 For a given 2 2 # matrix A , it is observed that A
1
1
1
1 -
=-
-
> > H H and
A
1
2
2
1
2 -
=-
-
> > H H. The matrix A is
(A) A
2
1
1
1
1
0
0
2
1
1
1
2
=
- -
-
- - -
> > > H H H
(B) A
1
1
1
2
1
0
0
2
2
1
1
1
=
- - - -
> > > H H H
(C) A
1
1
1
2
1
0
0
2
2
1
1
1
=
- -
-
- - -
> > > H H H
(D)
0
1
2
3
-
-
= G
QUE 1.77 If A
3
1
4
1
=
-
-
> H, then for every positive integer , n A
n
is equal to
(A)
n
n
n
n
1 2 4
1 2
+
+
= G (B)
n
n
n
n
1 2 4
1 2
- -
+
= G
(C)
n
n
n
n
1 2 4
1 2
-
+
= G (D)
n
n
n
n
1 2 4
1 2
+ -
-
= G
ARIHANT/306/13
Page 19
Chap 1
Matrix Algebra
Page 19
Chap 1
Matrix Algebra
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QUE 1.78 For which values of the constants b and c is the vector b
c
3
R
T
S
S
S
S
V
X
W
W
W
W
a linear
Combination of ,
1
3
2
2
6
4
R
T
S
S
S
S
R
T
S
S
S
S
V
X
W
W
W
W
V
X
W
W
W
W
and
1
3
2
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
(A) 9, 6 (B) 6, 9
(C) 6, 6 (D) 9, 9
QUE 1.79 The values of non zero numbers , , , , , , , a b c d e f g h such that the matrix
a
d
f
b
k
g
c
e
h
R
T
S
S
S
S
V
X
W
W
W
W

is invertible for all real numbers k.
(A) finite sol
n
(B) infinite sol
n
(C) 0 (D) none
***********
Page 20
Chap 1
Matrix Algebra
Page 20
Chap 1
Matrix Algebra
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SOLUTI ONS 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 non-
square 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
AA I
T
=
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 A + l
is symmetric, but A A - l will be anti-symmetric.
SOL 1.3 Correct option is (C).
If A and B are both order skew-symmetric matrices, then
A A
T
=- and B B
T
=- ...(1)
Also, given that AB BA =
B A
T T
= - - ^ ^ h h [from Eq. (1)]
B A
T T
= AB
T
=^ h
AB AB
T
=^ h 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 k k A A
n
= .
A B 5 = & A B B B 5 5 625
4
= = =
or a 625 =
SOL 1.5 Correct option is (A).
If rank of 5 6
# ^ h 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 A
n n #
triangular matrix det A is equal to the product
of the diagonal entries of A .
Page 21
Chap 1
Matrix Algebra
Page 21
Chap 1
Matrix Algebra
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SOL 1.7 Correct option is (B).
If det , A 0 ! then A
n n #
is non-singular, but if A
n 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
1
3
3
2
5
2
6
10
5(20 30) 3(10 18) 2(5 6) = + - - - -
50 24 2 28 =- + - =-
SOL 1.10 Correct option is (A).

a
h
g
h
b
f
g
f
c
a
a
f
f
c
h
h
g
f
c
g
h
g
b
f
= - +
a bc f h hc fg g hf gb
2
= - - - + - ^ ^ ^ h h h
abc af h c hfg ghf g b
2 2 2
= = - + + -
abc fgh af bg ch 2
2 2 2
= + - - -
SOL 1.11 Correct answer is 43 - .
Determinant 67
13
24
14
26
19
39
81
14
26
21
39
81
13
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 - .
We have A
0
1
0
1
1
1
0
2
0
1
0
0
2
3
1
1
=
-
-
R
T
S
S
S
S
S
V
X
W
W
W
W
W
Expanding cofactor of a
34

A 1
0
1
1
1
1
2
0
1
0
=- -
-
[ ( ) ] 0 1 0 1 0 =- - - + 1 =-
Page 22
Chap 1
Matrix Algebra
Page 22
Chap 1
Matrix Algebra
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SOL 1.14 Correct answer is 5.
Consider the given matrix be
I AB
m
+
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
where m 4 = so, we obtain
AB
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
= -
R
T
S
S
S
S
SS
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
V
X
W
W
W
W
WW

1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW

1
1
1
1
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW

1 1 1 1 6 @
Hence, we get A
1
1
1
1
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
, B 1 1 1 1 =6 @
Therefore, BA 1 1 1 1
1
1
1
1
=
R
T
S
S
S
S
SS
8
V
X
W
W
W
W
WW
B
4 =6 @
From the given property,
Det I AB
m
+ ^ h Det I BA
n
= + ^ h
Det
2
1
1
1
1
2
1
1
1
1
2
1
1
1
1
2
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
Det 1 4 = + 6 6 @ @ " ,
Det 5 = 6 @ 5 =
NOTE :
Determinant of identity matrix is always 1.
SOL 1.15 Correct option is (D).
A = conjugate of A

i
i 3
1 2
1 2
2
=
+
+
> H
and A A
* T
= = ^ h transpose of A

i
i 3
1 2
1 2
2
=
-
+
> H
Since, A
*
A =
Hence, A is hermitian matrix.
SOL 1.16 Correct answer is 4.
For singularity of matrix,
Page 23
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Chap 1
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8
4
12
0
6
0
2
0
l
0 =
8 0 12 0 2 12 0
#
l - - - = ^ ^ h h 4 &l =
SOL 1.17 Correct answer is 2 - .
Matrix A is singular if A 0 =

0
1
2
1
0
2
2
3
l
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
0 =
or ( ) 1
1
2
2
2
1
0
2
3
0
0
2
3
l l
- -
-
-
+
-
+
-
0 =
or ( ) ( ) 4 2 3 l - + 0 =
or l 2 =-
SOL 1.18 Correct option is (C).
Given EF G = where G I = = Identity matrix

cos
sin
sin
cos F
0 0
0
0
1
#
q
q
q
q
-
R
T
S
S
S
S
V
X
W
W
W
W

0 0 1
0
0
1
0
0
1
=
R
T
S
S
S
S
V
X
W
W
W
W
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
F E
1
=
-

. adj E
E
=
6 @
adj E
( ) cos
sin
sin
cos
0 0
0
0
1
T
q
q
q
q =
-
R
T
S
S
S
S
V
X
W
W
W
W

cos
sin
sin
cos
0 0
0
0
1
q
q
q
q = -
R
T
S
S
S
S
V
X
W
W
W
W
E ( ) cos cos sin sin 0 0 0
# #
q q q q = - - - - +
^ ^ h h 6 8 @ B
cos sin 1
2 2
q q = + =
Hence, F
. adj
E
E
=
6 @

cos
sin
sin
cos
0 0
0
0
1
q
q
q
q = -
R
T
S
S
S
S
V
X
W
W
W
W
SOL 1.19 Correct answer is 2.
We have A
1
2
2
0
3
5
4
7
=
-
= G
It is a 2 4 # matrix, thus ) A ( r 2 #
The second order minor

1
2
2
0 -
4 0 ! =
Hence, ( ) A r 2 =
Page 24
Chap 1
Matrix Algebra
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SOL 1.20 Correct answer is 2.
We have
A
1
1
1
1
1
1
1
0
1
1
1
0
1
1
0
1
0
0
+ = - -
R
T
S
S
S
S
R
T
S
S
S
S
V
X
W
W
W
W
V
X
W
W
W
W
R R
3 1
-
Since one full row is zero, ( ) A 3 < r
Now
1
1
1
1 -
2 0 ! =- , thus ( ) A 2 r =
SOL 1.21 Correct option is (B).
Here, A
1
1
2
2
4
6
3
2
5
=
R
T
S
S
S
S
V
X
W
W
W
W
Performing operation R 1
31
- ^ h, we get A
1
1
1
2
4
4
3
2
2
+
R
T
S
S
S
S
V
X
W
W
W
W
By operation R 1
32
- ^ h, we get A
1
1
0
2
4
0
3
2
0
+
R
T
S
S
S
S
V
X
W
W
W
W
A 0 =
and
1
1
2
4
0 =Y
Rank
A
^ h 2 =
SOL 1.22 Correct option is (B).
Here, A 2 9 8 2 2 3 = - + + - + ^ ^ h h 0 =
But
2
0
1
3
0 =Y
Hence, Rank A ^ h 2 =
SOL 1.23 Correct answer is 2.
Consider 3 3
#
minors, maximum possible rank is 3.
Now we can obtain

4
6
2
2
3
1
1
4
0
0 = ,
2
3
1
1
4
0
3
7
1
0 = ,
4
6
2
1
4
0
3
7
1
0 = and
4
6
2
2
3
1
3
7
1
0 =
Since, all 3 3
#
minors are zero. Now, we consider 2 2
#
minors

4
6
2
3
0 = ,
2
3
1
4
8 3 5 0 = - = =Y
Hence, rank 2 =
SOL 1.24 Correct answer is 13.
Since ( ) A r 2< = order of matrix
Page 25
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Thus A
2
4
1
1
7
4
3
5
0 l =
-
=
or ( ) ( ) ( ) 235 4 1 20 3 16 7 l l - + - + - 0 =
or 70 8 20 27 l l - + - + 0 0 = =
or, l 13 =
SOL 1.25 Correct answer is 3.
It is 4 4 # matrix, So its rank ( ) A 4 # r
We have A
1
2
3
6
2
4
2
8
3
3
1
7
0
2
3
5
=

1
2
3
0
2
4
2
0
3
3
1
0
0
2
3
0
= applying ( ) R R R R R
4 1 2 3 4
" - + +

1
0
0
0
2
0
4
0
3
3
8
0
0
2
3
0
=
-
-
-
applying
R R R
R R R
2
3
2 1 2
3 1 3
"
"
-
-
The only fourth order minor is zero.
Since the third order minor ( )( )( )
1
0
0
2
4
0
3
8
3
1 4 3 12 0 ! - -
-
= - - =
Therefore its rank is ( ) A 3 r =
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
1
0
0
1
0 =Y
Hence, Rank A ^ h 2 =
SOL 1.27 Correct option is (C).
Given the two matrices,
A
p
r
q
s
=> H
and B
p q
pr qs
pr qs
r s
2 2
2 2
=
+
+
+
+
> H
To determine the rank of matrix A , we obtain its equivalent matrix using
the operation, a
i 2
a
a
a
a
i i 2
11
21
1
! - as
Page 26
Chap 1
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A
p q
s
p
r
q 0
=
-
> H
If s
p
r
q - 0 =
or ps rq - 0 =
then rank of matrix A is 1, otherwise the rank is 2.
Now, we have the matrix
B
p q
pr qs
pr qs
r s
2 2
2 1
=
+
+
+
+
> H
To determine the rank of matrix B , we obtain its equivalent matrix using
the operation, a
i 2
a
a
a
a
i i 2
11
21
1
! - as
B
p q pr qs
r s
p q
pr qs
0
2 2
2 2
2 2
2
=
+ +
+ -
+
+
^
^
h
h
R
T
S
S
S
S
V
X
W
W
W
W
If r s
p q
pr qs
2 2
2 2
2
+ -
+
+
^
^
h
h
ps rq
2
= - ^ h 0 =
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.
SOL 1.28 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 d 2 = +
Let A
k
x
y
z
4
5
6
5
6
=
k
x
x d
x d
4
5
6
5
6
2
=
+
+

k
x
d
d
4
1
1
5
1
6
=
-
Applying R R R
2 1 2
- = and R R R
3 2 3
- =

k
x
d
4
1
0
5
1
7 0
=
-
R R R
3 3 2
= -
A 0 = k d x 7 4 0 & - - = ^ ^ h h
d
x
4
= , k 7 = .
SOL 1.29 Correct option is (B).
Let A
a
b
c
1
1
1
=
R
T
S
S
S
S
V
X
W
W
W
W
, a b c B
2 2 2
=6 @
Let C AB =
Page 27
Chap 1
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a
b
c
a b c
a a
ba
c a
a b
bb
c b
a c
bc
c c
1
1
1
2 2 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
1 2
#
= =
R
T
S
S
S
S
R
T
S
S
S
S
8
V
X
W
W
W
W
V
X
W
W
W
W
B
The 3 3
#
minor of this matrix is zero and all the 2 2
#
minors are also
zero. So the rank of this matrix is 1, i.e.
C r 6 @ 1 =
SOL 1.30 Correct option is (A).
Since all point are collinear,
Thus
x
x
x
y
y
y
1
1
1
1
2
3
1
2
3
0 =
Therefore A ( ) r 3 <
SOL 1.31 Correct option is (B).
Let n 3 =
Then A
1
2
3
2
4
6
3
6
9
=
R
T
S
S
S
S
V
X
W
W
W
W
and
A

1
2
3
2
4
6
3
6
9
=
1
0
0
2
0
0
3
0
0
= applying
R R R
R R R
3
2
3 1 3
2 1 2
"
"
-
-
Thus rank if n 3 = then ( ) A 1 r = so possible answer is (B).
SOL 1.32 Correct option is (B).
If P is a matrix of order m n # and ( ) n P r = 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 ( ) p P Q r = 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 P Q only p rows are non zero.
(P Q) r p =
SOL 1.33 Correct option is (D).
x x x x
n 1 2
T
g =
8 B
V xx
T
=

x
x
x
x
x
x
n n
1
2
1
2
h h
=
R
T
S
S
S
S
SS
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
V
X
W
W
W
W
WW
So rank of V is n.
Page 28
Chap 1
Matrix Algebra
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SOL 1.34 Correct option is (A).
Given that , , x y z are in A.P. with common differences d.
Thus y x d = + ,
z x d 2 = +
Now A
k
x
d
d k
x
x d
x d k
x
d
d
4
1
1
5
1
6
4
5
6
5
6
2
4
1
1
5
1
6
=
-
= +
+
=
-
Apply R R R
2 1 2
- = and R R R
3 2 3
- =

k
x
d
4
1
0
5
1
7 0
=
-
applying R R R
3 2 3
" -
Thus A ( )( ) k d x 7 4 = - -
Since ( ) A 2< r = order of matrix, Thus A 0 = or we get
( )( ) k d x 7 4 - - 0 =
or d ,
x
k
4
7 = =
SOL 1.35 Correct option is (A).
Rank of a matrix is no. of linearly independent rows and columns of the
matrix.
Here Rank ( ) Q 4 r =
So Q will have 4 linearly independent rows and flour independent columns.
SOL 1.36 Correct option is (D).
We have A
1
0
4
2
=
-
= G
C
11
, , ( ) C C 2 0 4 4
12 21
= = =- - = , and C 1
22
=
C
2
4
0
1
== G
adj A C
T
=

2
0
4
1
== G
SOL 1.37 Correct option is (D).
A
x
z
y
b
=
adj A
b
z
y
x
=
-
-
= G
adj(adj A )
x
z
y
b
== G
SOL 1.38 Correct option is (B).
Since, A (adjA ) A I
n
=
Page 29
Chap 1
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We have A (adjA ) 2
1
0
0
0
1
0
0
0
1
2
0
0
0
2
0
0
0
2
= =
R
T
S
S
S
S
R
T
S
S
S
S
V
X
W
W
W
W
V
X
W
W
W
W
SOL 1.39 Correct option is (B).
We know that
adj(adj(A ) A A
n 2
=
-
Here n 2 = so we get
adj(adjA ) A A
2 2
=
-
A A I A A
0
= = =
SOL 1.40 Correct option is (A).
We know that
adj(adj A ) A A
n 2
=
-
Putting n 3 = and A 3 = . so we get
adj(adj A ) A A
3 2
=
-
A A A 3 = =
SOL 1.41 Correct option is (D).
We have A adj (adj ) A
( ) n 1
2
=
-
Putting n 3 = and A 3 = we get
A adj (adj ) A 3 81
4 4
= = =
SOL 1.42 Correct option is (B).
Let B A adj
2
= then B is also a 3 3 # matrix.
A adj (adj )
2
B B B adj
3 1 2
= = =
-
A adj
2 2
=
A A A 3
2 (3 1)
2
2 4 8 8
= = = =
-
8 B
SOL 1.43 Correct option is (A).
Since ( ) n A 1 r = - , 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 A
ij n n
=
#
then det [ ] C A
ij n n
T
=
#
where C
ij
is the cofactor of a
ij
Also ( ) , C M 1
ij
i j
ij
= -
+
where M
ij
is the minor of a
ij
, obtained by leaving the
row and the column corresponding to a
ij
and then taken the determinant of
the remaining matrix.
Now, M
11
= minor of a
11
i.e. ( ) 1
1
2
2
1
-
-
-
3 =-
Page 30
Chap 1
Matrix Algebra
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Similarly
M
12

2
2
2
1
6 =
-
= ; M
2
2
1
2
6
13
=
-
=-
M
21

2
2
2
1
6 =
-
-
-
=- ; M
1
2
2
1
3
22 = =
- -
;
M
23

1
2
2
2
6 =
- -
-
= ; M
2
1
2
2
6
31
=
- -
-
= ;
M
32
;
1
2
2
2
6 =
- -
-
= M
1
2
2
1
3
33
=
- -
=
C
11
( ) ; M 1 3
1 1
11
= - =-
+
( ) ; C M 1 6
12
1 2
12
= - =-
+
C
13
( ) ; M 1 6
1 3
13
= - =-
+
( ) ; C M 1 6
21
2 1
21
= - =
+
C
22
( ) ; M 1 3
2 2
22
= - =
+
( ) ; C M 1 6
23
2 3
23
= - =-
+
C
31
( ) ; M 1 6
3 1
31
= - =
+
( ) ; C M 1 6
32
3 2
32
= - =-
+
C
33
( ) M 1 3
3 3
33
= - =
+
Thus adj A
3
6
6
6
3
6
6
6
3
T
=
- -
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
A 3
1
2
2
2
1
2
2
2
1
3
T
T
=
- -
-
-
- =
R
T
S
S
S
S
V
X
W
W
W
W
SOL 1.45 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.
A ( )( ) ( )( )
5
2
2
1
5 1 2 2 1 = = - =
SOL 1.46 Correct option is (A).
We know A
1 -

A
A
1
adj =
Here A
1
3
2
5
1 =
-
-
=- = G
Also, A adj
5
3
2
1
=
-
-
-
-
= G
A
1 -

1
1
5
3
2
1
5
3
2
1
=
-
-
-
-
-
= = = G G
SOL 1.47 Correct option is (A).
We know A
1 -

A
A
1
adj =
Here A
1
5
2
7
3 = =-
Also, A adj
7
5
2
1
=
-
-
= G
Page 31
Chap 1
Matrix Algebra
Page 31
Chap 1
Matrix Algebra
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Sample Chapter of Engineering Mathematics
A
1 -

3
1
7
5
2
1 3
1
7
5
2
1
=
- -
-
=
-
-
= = G G
SOL 1.48 Correct option is (C).
Let B
1 -

3
5
4
6
== G A = and B A
1
=
-
We know A
1 -

A
A
1
adj =
Here A
3
5
4
6
2 = =-
Also, adj A
6
5
4
3
=
-
-
A
1 -

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

3 2
2
5
2
3
=
-
-
= G
SOL 1.49 Correct option is (C).
For orthogonal matrix det M 1 = and M M
T 1
=
-
,
M
T

A
B
C
0
== G

BC C
B
A
M
1
0
1
= =
- -
-
-
> H
This implies B
BC
C
=
-
-
or B
B
1
= or B 1 ! =
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 -

A
A
1
adj =
A
1
5
3
0
2
1
0
0
2
4 0 ! = = ,
adj A
4
0
0
10
2
0
10
1
2
4
10
1
0
2
1
0
0
2
T
=
-
- =
- -
R
T
S
S
S
S
V
X
W
W
W
W
A
1 -

4
1
4
10
1
0
2
1
0
0
2
=
- -
R
T
S
S
S
S
V
X
W
W
W
W
Page 32
Chap 1
Matrix Algebra
Page 32
Chap 1
Matrix Algebra
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SOL 1.52 Correct answer is 0.01786.
( ) det A 2
1 -

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

A 2
1
8 7
1
56
1
3
#
= = =
SOL 1.53 Correct option is (B).
C
11
( ) 2 3 5 = - - =
C
21
[ ( )] 0 3 3 =- - - =-
C
31
[ ( )] 1 1 = - - =
R ( )C C C 1 2 2
11 21 31
= + + 5 6 2 = - + 1 =
SOL 1.54 Correct option is (A).
Let B (7 )
1 -
A
1
4
2
7
= =
-
-
> H and B A 7
1
=
-
We know A
1 -

A
A
1
adj =
Here A
1
4
2
7
1 =
-
-
=-
Also, adj A
7
4
2
1
=
-
-
-
-
= G
A
1 -

1
1
7
4
2
1
=
-
-
-
-
-
= G
or B 7 A
7
4
2
1
1
= =
-
> H
or B
1
7
4
7
2
7
1
= > H
SOL 1.55 Correct answer is 0.5.
We have A
0
0
0
1
1
2
3
4
0
0
2
1
0
1
0
0
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
A ( )
0
0
1
0
2
1
1
0
0
1 0 2 2 =- =- - =
Now A det( )
1 -

det A
1
2
1
= =
SOL 1.56 Correct option is (C).
For orthogonal matrix we know that
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Chap 1
Matrix Algebra
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Chap 1
Matrix Algebra
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Sample Chapter of Engineering Mathematics
AA
T
I =
and [ ] AA
T 1 -
I I
1
= =
-
SOL 1.57 Correct option is (C).
From orthogonal matrix
[ ] AA
T
I =
Since the inverse of I is I , thus
[ ] AA
T 1 -
I I
1
= =
-
SOL 1.58 Correct option is (D).
Al ( ) A A A
T T 1
=
-

( ) A A A
T T 1 1
=
- -
A I
1
=
-
Put A A I
1
=
-
l in all option.
option (A) AA A l A =
AA A
1 -
A =
A A = (true)
option (B) ( ) AA
2
l I =
( ) AA I
1 2 -
I =
I
2
^ h I = (true)
option (C) A A l I =
A I A
1 -
I =
I I = (true)
option (D) AA A l A = l
AA I A
1 -
A A = = l Y (false)
SOL 1.59 Correct option is (A).
Given : M
x
5
3
5
4
5
3
=> H
And [ ] M
T
[ ] M
1
=
-
We know that when
A A
1
=
T -
6 6 @ @ then it is called orthogonal matrix.

M
T
6 @
M
I
=
6 @
M M
T
6 6 @ @ I =
Substitute the values of M and M
T
, we get

x
x
5
3
5
4
5
3
5
3
5
4
5
3
. > > H H
1
1
0
0
=> H

x
x
x
5
3
5
3
5
4
5
3
5
3
5
3
5
4
5
3
5
4
5
4
5
3
5
3
2
#
#
#
# #
+
+
+
+
b
b
b
b b
l
l
l
l l
R
T
S
S
S
S
V
X
W
W
W
W

1
1
0
0
=> H
Page 34
Chap 1
Matrix Algebra
Page 34
Chap 1
Matrix Algebra
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x
x
x
1
25
9 2
25
12
5
3
25
12
5
3
+
+
+
> H
1
1
0
0
=> H
Comparing both sides a
12
element,
x
25
12
5
3
+ 0 = " x
25
12
3
5
5
4
#
=- =-
SOL 1.60 Correct answer is 0.5.

x
x x
2 0 1
1
0
2 -
> > H H
1
0
0
1
== G
or
x
x
2
0
0
2
= G
1
0
0
1
== G
So, x x 2 1
2
1
& = =
SOL 1.61 Correct option is (B).

2
3
1
2
2
1
-
> > H H
( )( ) ( )( )
( )( ) ( )( )
2 2 1 1
3 2 2 1
=
+ -
+
> H
3
8
== G
SOL 1.62 Correct option is (C).
AA
T

1
3
2
1
0
4
1
2
0
3
1
4
=
-
-
R
T
S
S
S
S
>
V
X
W
W
W
W
H

( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
1 1 2 2 0 0
3 1 1 2 4 0
1 3 2 1 0 4
3 3 1 1 4 4
=
+ +
+ - +
+ - +
+ - - +
> H

5
1
1
26
=> H
SOL 1.63 Correct option is (B).
A
1
1
7
20
2
3
1
5
1
=
-
-
-
-
> > H H

1
1
7
20 13
1
5
3
1
2
=
-
- - -
-
-
b l
> > H H

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

13
1
26
65
13
39
=
-
-
-
> H

2
5
1
3
== G
SOL 1.64 Correct answer is 0.35.
We have A
. 2
0
01
3
=
-
= G and
a
b
A
0
1 2
1
=
-
> H
Now AA
1 -
I =
Page 35
Chap 1
Matrix Algebra
Page 35
Chap 1
Matrix Algebra
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Sample Chapter of Engineering Mathematics
or
. a
b
2
0
01
3 0
2
1
-
> > H H
1
0
0
1
=> H
or
. a b
b
1
0
2 01
3
-
> H
1
0
0
1
=> H
or . a 2 01 - 0 = and b 3 1 =
Thus solving above we have b
3
1
= and a
60
1
=
Therefore a b +
3
1
60
1
20
7
= + =
SOL 1.65 Correct option is (A).
We know that AA
1 -
I =
or
. a
b
2
0
01
3 0
2
1
-
> > H H
. a b
b
1
0
2 01
3
1
0
0
1
=
-
= > > H H
We get b 3 1 = or b
3
1
=
and . a b 2 01 - 0 = or a
b
20
=
Thus a b +
3
1
3
1
20
1
20
7
= + =
SOL 1.66 Correct option is (A).
We have A
2
3
6
9
== G and ,
y
x
B AB
3
2
0 = = > H
We get
y
x 2
3
6
9
3
2
= = G G
0
0
0
0
== G
or
y
y
x
x
6 6
9 9
2 12
3 18
+
+
+
+
= G
0
0
0
0
== G
We get y 6 6 + 0 = or y 1 =-
and x 2 12 + 0 = or x 6 =-
SOL 1.67 Correct option is (C).
AB
1
1
1
0
0
1
1
0
1
=
R
T
S
S
S
S
=
V
X
W
W
W
W
G

( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
1 1 1 0 0 1
1 1 0 0 1 1
1
2
+ +
+ +
= == = G G
SOL 1.68 Correct option is (A).
We have AB
1
2
1
1
2
0
1
2
1
2
0
1
=
-
R
T
S
S
S
S
=
V
X
W
W
W
W
G
1
4
4
4
== G
( ) AB
T

1
4
4
4
== G
Page 36
Chap 1
Matrix Algebra
Page 36
Chap 1
Matrix Algebra
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SOL 1.69 Correct option is (C).
AB
2
1
3
1
0
4
1
3
2
4
5
0
=
-
-
- -
R
T
S
S
S
S
=
V
X
W
W
W
W
G

( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )
2 1 1 3
1 1 0 3
3 1 4 3
2 2 1 4
1 2 0 4
3 2 4 4
2 5 1 0
1 5 0 0
3 5 4 0
=
+ -
+
- +
- + -
- +
- - +
- + -
- +
- - +
R
T
S
S
S
S
V
X
W
W
W
W

1
1
9
8
2
22
10
5
15
=
- -
-
-
-
R
T
S
S
S
S
V
X
W
W
W
W
SOL 1.70 Correct answer is 3.
We have X
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
and transpose of X , X
T

0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
=
R
T
S
S
S
S
SS
V
X
W
W
W
W
WW
Here, we can see that rank of matrix x 3 = , hence, we can determine the
rank of X X
T
.
Let Y X X
T
$ = , the rank of Y # rank of X . Also, X Y X
T 1
=
-
and so we
have
Rank X = Rank X
T
# Rank of Y
Hence, from Eqs. (1) and (2), we get
Rank of X = Rank of Y
Hence, rank of X X
T
$ is 3
SOL 1.71 Correct option is (A).
X
4 3 #
X
c
T
4
"
#
X Y
T
3 4 4 3 # #
( ) X Y
T
3 3
"
#
( ) X Y
Y
3 3 #
( ) X Y
T 1
3 3
"
#
-
P
2 3 #
P
T
3 2
"
#
( ) X Y P
T T
3 3
1
3 2 # #
-
( ) X Y
T
PT
1
3 2
"
#
-
" ,
{( ) } P X Y P
T T
2 3
1
3 2 # #
-
( ) P X Y P
T T 1
2 2
"
#
-
6 @
[ ( ) ] P X Y P
T T T 1
2 2 #
-
[ ( ) ] P X Y P
T T 1
2 2
"
#
-
SOL 1.72 Correct option is (C).
. A A
a b

cos
sin
sin
cos
cos
sin
sin
cos
a
a
a
a
b
b
b
b
=
- -
> > H H

( )
( )
( )
( )
cos
sin
sin
cos
A
a b
a b
a b
a b
=
+
- +
+
+
=
a b + > H
Page 37
Chap 1
Matrix Algebra
Page 37
Chap 1
Matrix Algebra
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Also, it is easy to prove by induction that
( ) A
n
a

cos
sin
sin
cos
n
n
n
n
a
a
a
a
=
-
= G
SOL 1.73 Correct option is (D).
Let tan t
2
a
=
Then, cosa
tan
tan
t t
t
1
1
1
2
2
2
2
2
2
=
+
-
=
+
-
a
a
and sina
tan
tan
t
t
1
2
1
2
2
2
2
2
=
+
=
+
a
a
( )
cos
sin
sin
cos
I A
a
a
a
a
-
-
> H
tan
tan cos
sin
sin
cos
1
1
2
2
#
a
a
a
a
=
-
-
a
a
> = H G

( )
( )
t
t
t
t
t
t
t
t
t
t
1
1
1
1
1
2
1
2
1
1
2
2
2
2
2
2
# =
-
+
-
+
+
-
+
-
R
T
S
S
S
S
S
=
V
X
W
W
W
W
W
G
(
tan
tan
t
t
I A
1
1
1
1
)
2
2
=
-
=
-
= +
a
a
> > H H
SOL 1.74 Correct option is (C).
AA
T

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

cos sin
sin cos cos sin
cos sin sin cos
sin cos
2 2
2 2
q q
q q q q
q q q q
q q
=
+
- +
- +
+
> H

1
0
0
1
=> H
1 = , Hence A is orthogonal matrix.
SOL 1.75 Correct option is (C).
AB
( )
( )
( )
( )
cos cos cos
cos sin cos
cos sin cos
sin sin cos
q f q f
f a a f
a f q f
q f q f
=
-
-
-
-
= G
Is a null matrix when ( ) cos 0 q f - = , this happens when ( ) q f - is an odd
multiple of
2
p
.
SOL 1.76 Correct option is (C).
Let matrix A be
a
c
b
d
= G
From A
1
1
1
1 -
=-
-
> > H H we get
a
c
b
d
a b
c d
1
1
1
1 -
=
-
-
=-
-
= = = = G G G G
We have a b - 1 =- ...(1)
and c d - 1 = ...(2)
Page 38
Chap 1
Matrix Algebra
Page 38
Chap 1
Matrix Algebra
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From A
1
2
2
1
2 -
=-
-
> > H H we get

a
c
b
d
= G
1
2 -
= G
a b
c d
2
2
2
1
2
=
-
-
=-
-
= = G G
we have a b 2 - 2 =- ...(2)
c d 2 - 4 = ...(4)
Solving equation (1) and (3) a 0 = and b 1 =
Solving equation (2) and (4) c 2 =- and d 3 =-
Thus A
0
2
1
3
=
- -
= G
If we check all option then result of option C after multiplication gives result.
SOL 1.77 Correct option is (D).
A
2

3
1
4
1
3
1
4
1
5
2
8
3
=
-
-
-
-
=
-
-
= = = G G G
If we put n 2 = in option, then only D satisfy.
SOL 1.78 Correct option is (A).
b
c
3
R
T
S
S
S
S
V
X
W
W
W
W
k k k
1
3
2
2
6
4
1
3
2
1 2 3
= + +
-
-
-
R
T
S
S
S
S
R
T
S
S
S
S
R
T
S
S
S
S
V
X
W
W
W
W
V
X
W
W
W
W
V
X
W
W
W
W
k k k
1
3
2
2
1
3
2
1
3
2
1 2 3
+ -
R
T
S
S
S
S
R
T
S
S
S
S
R
T
S
S
S
S
V
X
W
W
W
W
V
X
W
W
W
W
V
X
W
W
W
W
b
c
3
=
R
T
S
S
S
S
V
X
W
W
W
W
( ) k k k 2
1
3
2
1 2 3
+ -
R
T
S
S
S
S
V
X
W
W
W
W
b
c
3
=
R
T
S
S
S
S
V
X
W
W
W
W
k k k 2
1 2 3
+ - 3 =
k k k 3 6 3
1 2 3
+ - b =
k k k 2 6 2
1 2 3
+ - c =
& b 9 = ,
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
***********