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Eighth Edition GATE Engineering Mathematics RK Kanodia Ashish Murolia NODIA & COMPANY
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Transcript of 24.Engineering maths by kanodia
Eighth Edition
GATE
Engineering Mathematics
RK Kanodia Ashish Murolia
NODIA & COMPANY
GATE Electronics & Communication Vol 2, 8eEngineering MathematicsRK 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.
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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 [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. KanodiaAshish 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
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
1.7.1 Elementary Transformations 4
1.7.2 Inverse of Matrix by Elementary Transformations 4
1.8 ECHELON FORM 5
1.9 NORMAL FORM 5
EXERCISE 6
SOLUTIONS 20
CHAPTER 2 SYSTEMS OF LINEAR EQUATIONS
2.1 INTRODUCTION 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 LINEAR EQUATIONS 40
2.4 SOLUTION OF A SYSTEM OF LINEAR EQUATIONS 40
EXERCISE 42
SOLUTIONS 51
CHAPTER 3 EIGENVALUES AND EIGENVECTORS
3.1 INTRODUCTION 65
3.2 EIGENVALUES AND EIGEN VECTOR 65
3.3 DETERMINATION OF EIGENVALUES AND EIGENVECTORS 66
3.4 CAYLEY-HAMILTON THEOREM 66
3.4.1 Computation of the Inverse Using Cayley-Hamilton Theorem 67
3.5 REDUCTION OF A MATRIX TO DIAGONAL FORM 67
3.6 SIMILARITY OF MATRICES 68
EXERCISE 69
SOLUTIONS 80
CHAPTER 4 LIMIT, CONTINUITY AND DIFFERENTIABILITY
4.1 INTRODUCTION 99
4.2 LIMIT OF A FUNCTION 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 CONTINUITY OF A FUNCTION 100
4.3.1 Continuity in an interval 100
4.4 DIFFERENTIABILITY 101
EXERCISE 102
SOLUTIONS 115
CHAPTER 5 MAXIMA AND MINIMA
5.1 INTRODUCTION 139
5.2 MONOTONOCITY 139
5.3 MAXIMA AND MINIMA 139
EXERCISE 140
SOLUTIONS 147
CHAPTER 6 MEAN VALUE THEOREM
6.1 INTRODUCTION 163
6.2 ROLLE’S THEOREM 163
6.3 LAGRANGE’S MEAN VALUE THEOREM 163
6.4 CAUCHY’S MEAN VALUES THEOREM 163
EXERCISE 164
SOLUTIONS 168
CHAPTER 7 PARTIAL DERIVATIVES
7.1 INTRODUCTION 175
7.2 PARTIAL DERIVATIVES 175
7.2.1 Partial Derivatives of Higher Orders 175
7.3 TOTAL DIFFERENTIATION 176
7.4 CHANGE OF VARIABLES 176
7.5 DIFFERENTIATION OF IMPLICIT FUNCTION 176
7.6 EULER’S THEOREM 176
EXERCISE 177
SOLUTIONS 182
CHAPTER 8 DEFINITE INTEGRAL
8.1 INTRODUCTION 191
8.2 DEFINITE INTEGRAL 191
8.3 IMPORTANT FORMULA FOR DEFINITE INTEGRAL 192
8.4 DOUBLE INTEGRAL 192
EXERCISE 193
SOLUTIONS 202
CHAPTER 9 DIRECTIONAL DERIVATIVES
9.1 INTRODUCTION 223
9.2 DIFFERENTIAL ELEMENTS IN COORDINATE SYSTEMS 223
9.3 DIFFERENTIAL CALCULUS 223
9.4 GRADIENT OF A SCALAR 224
9.5 DIVERGENCE OF A VECTOR 224
9.6 CURL OF A VECTOR 225
9.7 CHARACTERIZATION OF A VECTOR FIELD 225
9.8 LAPLACIAN OPERATOR 225
9.9 INTEGRAL 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
EXERCISE 227
SOLUTIONS 234
CHAPTER 10 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 249
10.5 VARIABLES SEPARABLE FORM 249
10.5.1 Equations Reducible to Variable Separable Form 250
10.6 HOMOGENEOUS EQUATIONS 250
10.6.1 Equations Reducible to Homogeneous Form 251
10.7 LINEAR DIFFERENTIAL EQUATION 252
10.7.1 Equations Reducible to Linear Form 253
10.8 BERNOULLI’S EQUATION 253
10.9 EXACT DIFFERENTIAL EQUATION 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
EXERCISE 257
SOLUTIONS 266
CHAPTER 11 HIGHER ORDER DIFFERENTIAL EQUATIONS
11.1 INTRODUCTION 283
11.2 LINEAR DIFFERENTIAL EQUATION 283
11.2.1 Operator 283
11.2.2 General Solution of Linear Differential Equation 283
11.3 DETERMINATION OF COMPLEMENTARY FUNCTION 284
11.4 PARTICULAR INTEGRAL 285
11.4.1 Determination of Particular Integral 285
11.5 HOMOGENEOUS LINEAR DIFFERENTIAL EQUATION 287
11.6 EULER EQUATION 288
EXERCISE 289
SOLUTIONS 300
CHAPTER 12 INITIAL AND BOUNDARY VALUE PROBLEMS
12.1 INTRODUCTION 317
12.2 INITIAL VALUE PROBLEMS 317
12.3 BOUNDARY-VALUE PROBLEM 317
EXERCISE 319
SOLUTIONS 325
CHAPTER 13 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
13.3 HOMOGENEOUS FUNCTIONS 339
13.4 EULER’S THEOREM 339
13.5 COMPOSITE FUNCTIONS 340
13.6 ERRORS AND APPROXIMATIONS 341
EXERCISE 342
SOLUTIONS 347
CHAPTER 14 ANALYTIC FUNCTIONS
14.1 INTRODUCTION 357
14.2 BASIC TERMINOLOGIES IN COMPLEX FUNCTION 357
14.3 FUNCTIONS OF COMPLEX VARIABLE 358
14.4 LIMIT OF A COMPLEX FUNCTION 358
14.5 CONTINUITY OF A COMPLEX FUNCTION 359
14.6 DIFFERENTIABILITY OF A COMPLEX FUNCTION 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 ANALYTIC FUNCTION 362
14.7.1 Required Condition for a Function to be Analytic 362
14.8 HARMONIC FUNCTION 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 SINGULAR POINTS 366
EXERCISE 367
SOLUTIONS 380
CHAPTER 15 CAUCHY’S INTEGRAL THEOREM
15.1 INTRODUCTION 405
15.2 LINE INTEGRAL OF A COMPLEX FUNCTION 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 INTEGRAL FORMULA 408
15.4.1 Cauchy’s Integral Formula for Derivatives 409
EXERCISE 410
SOLUTIONS 420
CHAPTER 16 TAYLOR’S AND LAURENT’ SERIES
16.1 INTRODUCTION 439
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 443
16.5.2 Evaluation of Definite Integral 443
EXERCISE 444
SOLUTIONS 453
CHAPTER 17 PROBABILITY
17.1 INTRODUCTION 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 DEFINITION OF PROBABILITY 471
17.4.1 Classical Definition 471
17.4.2 Statistical Definition 472
17.4.3 Axiomatic Definition 472
17.5 PROPERTIES OF PROBABILITY 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
EXERCISE 476
SOLUTIONS 491
CHAPTER 18 RANDOM VARIABLE
18.1 INTRODUCTION 515
18.2 RANDOM VARIABLE 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 VARIABLES AND VARIANCE 518
18.4.1 Moments about the Origin 518
18.4.2 Central Moments 518
18.4.3 Variance 518
18.5 BINOMIAL DISTRIBUTION 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 POISSON DISTRIBUTION 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 DISTRIBUTION 522
18.7.1 Mean and Variance of Normal Distribution 523
EXERCISE 526
SOLUTIONS 533
CHAPTER 19 STATISTICS
19.1 INTRODUCTION 543
19.2 MEAN 543
19.3 MEDIAN 544
19.4 MODE 545
19.5 MEAN DEVIATION 545
19.6 VARIANCE AND STANDARD DEVIATION 546
EXERCISE 547
SOLUTIONS 550
CHAPTER 20 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 555
20.3.2 Karl Pearson’s Coefficient of Correlation 556
20.3.3 Computation of Correlation Coefficient 557
20.4 RANK CORRELATION 558
20.5 REGRESSION 559
20.5.1 Lines of Regression 559
20.5.2 Angle between Two Lines of Regression 560
EXERCISE 562
SOLUTIONS 565
CHAPTER 21 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 599
22.4 NUMERICAL INTEGRATION 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
EXERCISE 604
SOLUTIONS 608
CHAPTER 23 SINGLE AND MULTI STEP METHODS FOR DIFFERENTIAL EQUATIONS
23.1 INTRODUCTION 617
23.2 PICARD’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 MILNE’S PREDICTOR AND CORRECTOR METHOD 620
23.6 TAYLOR’S SERIES METHOD 621
EXERCISE 623
SOLUTIONS 628
***********
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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 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 aij m n=
#6 @
and B bjk n p=
#6 @
then their product,
AB cC ik m p= =
#6 @
will be matrix of order m p# , where
cik a bijj
n
jk1
==/
PROPERTIES OF MATRIX MULTIPLICATION
If A, B and C are three matrices such that their product is defined, then1. 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 BAT T=
CHAPTER 1MATRIX ALGEBRA
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1.3 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# .
PROPERTIES OF TRANSPOSE OF MATRIX
Consider the two matrices A and B1. ( )A AT T =
2. ( )A B A BT T T! !=
3. ( )AB B AT T T=
4. ( ) ( )k kA AT T=5. ( ... ) ...A A A A A A A A A An n
TnT
nT T T T
1 2 3 1 1 3 2 1=- -
1.4 DETERMINANT OF A MATRIX
The determinant of square matrix A is defined as
A aaa
aaa
aaa
11
21
31
12
22
32
13
23
33
=
PROPERTIES OF DETERMINANT OF MATRIX
Consider the two matrices A and B .1. A exists A+ is a square matrix
2. AB A B=
3. A AT =
4. ,k kA An= 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 an1 2 then , ...a a aA n1 2= .
8. ,n NA An 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 MATRIX
The number, r with the following two properties is called the rank of the matrix1. 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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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 ( ) .Ar
Nullity of a Matrix
If A is a square matrix of a order ,n then ( )n Ar- 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 aA ij= 6 @ be a square matrix and Fij be the cofactor of aij in A , then adj FA ij
T+ 6 @ .
PROPERTIES OF ADJOINT MATRIX
If A, B are square matrices of order n ad In is corresponding unit matrix, then1. A (adj A) IA n= = (adj A)A2. A Aadj n 1= -
3. adj (adj A) A An 2= -
4. A Aadj(adj ) ( )n 1 2
= -
5. ) ( )A Aadj( adjT T=6. AB B Aadj( ) (adj )(adj )=
7. ) ( ) ,m NA Aadj( adjm m!=
8. ( ) ( ),k k k RA Aadj adjn 1 != -
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
A 1- AAadj=
Thus A 1- exists if A 0! and matrix A is called invertible.
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PROPERTIES OF INVERSE MATRIX
Let A and B are two invertible matrices of the same order, then1. ( ) ( )A AT T1 1=- -
2. AB B A( ) 1 1 1=- - -
3. ( ) ( ) ,k NA Ak k1 1 !=- -
4. A Aadj( ) (adj )1 1=- -
5. A A A11 1= =- -
6. If A = diag ( , ,..., )a a an1 2 , then A 1 =- diag ( , , ..., )a a an11
21 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 Ri j)
(2) The interchange of i th and j th columns is denoted by C Ci 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 kRi i"
(2) The multiplication of i th column by k is denoted by C kCi 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 kRi i j" + .
(2) The addition of k times the j th column to the i th column is denoted by C C kCi 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 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= . Post-multiplying by A 1- , we get
IA 1- BAA 1= -
or A 1- ( )B AA BI B1= = =-
or B A 1= -
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Sample Chapter of Engineering Mathematics1.8 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 000
200
610
120
3 4
=
#
R
T
SSSS
V
X
WWWW
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.
, , ,IO
OO
IO I O I
r rr r> > 8 8H H B B
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 aA 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: 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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EXERCISE 1
QUE 1.1 Match the items in column I and II.
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
A32
20=
--> 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 B5= and A Ba =, then a is_______.
ARIHANT/286/26
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Sample Chapter of Engineering MathematicsQUE 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) AAT will be invertible
(D) A AT will be invertible
QUE 1.6 If An n# is a triangular matrix then det A is
(A) ( )a1 iii
n
1
-=% (B) aii
i
n
1=%
(C) ( )a1 iii
n
1
-=/ (D) aii
i
n
1=/
QUE 1.7 If ,RA 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) AT 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 513
325
2610
R
T
SSSS
V
X
WWWW
is_____
QUE 1.10 The value of the determinant
ahg
hbf
gfc
(A) abc fgh af bg ch2 2 2 2+ - - -
(B) ab a c d+ + +(C) abc ab bc cg+ - -
(D) a b c+ +
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QUE 1.11 The value of the determinant 673981
191324
211426
is______
QUE 1.12 If 102
357
268
26- = , then the determinant of the matrix 201
753
862
-
R
T
SSSS
V
X
WWWW
is____
QUE 1.13 The determinant of the matrix
0101
1102
0100
2311
-
-
R
T
SSSSS
V
X
WWWWW
is______
QUE 1.14 Let A be an m n# matrix and B an n m# matrix. It is given that determinant I ABm + =^ h determinant I BAn +^ h, where Ik is the k k# identity matrix. Using the above property, the determinant of the matrix given below is______2111
1211
1121
1112
R
T
SSSSSS
V
X
WWWWWW
QUE 1.15 Let ii
A3
1 21 2
2= --
> H, then
(1) ii
A3
1 21 2
2= +-
> H (2) * ii
A2
1 21 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?
8412
06
020
l R
T
SSSS
V
X
WWWW
QUE 1.17 If 012
102
23l
--
-R
T
SSSS
V
X
WWWW
is a singular, then l is______
ARIHANT/306/9
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Sample Chapter of Engineering MathematicsQUE 1.18 Multiplication of matrices E and F is G . matrices E and G are
cossin
sincosE
0 0
001
qq=
-R
T
SSSS
V
X
WWWW
and G100
010
001
=
R
T
SSSS
V
X
WWWW
What is the matrix F ?
(A) cossin
sincos
0 0
001
-R
T
SSSS
V
X
WWWW
(B) coscos
cossin
0 0
001
qq-
R
T
SSSS
V
X
WWWW
(C) cossin
sincos
0 0
001
qq-
R
T
SSSS
V
X
WWWW
(D) sincos
cossin
0 0
001
-R
T
SSSS
V
X
WWWW
QUE 1.19 Rank of matrix 12
20
35
47-= G is
QUE 1.20 The rank of the matrix 111
111
101
-
R
T
SSSS
V
X
WWWW
is______
QUE 1.21 Given,
A 112
246
325
=
R
T
SSSS
V
X
WWWW
(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, A202
134
123
=---
R
T
SSSS
V
X
WWWW 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 A462
231
140
371
=
R
T
SSSS
6
V
X
WWWW
@ , the rank of the matrix is_____
ARIHANT/306/8
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QUE 1.24 The rank of the matrix A241
174
3
5l =
-R
T
SSSS
V
X
WWWW
is 2. The value of l must be
QUE 1.25 The rank of matrix
1236
2428
3317
0235
R
T
SSSSS
V
X
WWWWW
is______
QUE 1.26 Given,
A
acac
bdbd
1010
0101
=
(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 pr
qs= > H;
p qpr qs
pr qsr s
B2 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) N2
QUE 1.28 If , ,x y z are in AP with common difference d and the rank of the matrix
k
xyz
456
56
R
T
SSSS
V
X
WWWW 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
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QUE 1.30 If xxx
yyy
A111
1
2
3
1
2
3
=
R
T
SSSS
V
X
WWWW
and the point ( , ),( , ),( , )x y y y x y1 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 nA 1ij # #= with n 3$ and .a i jij = 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 ( ) nPr = and ( ) pQr = , then rank (PQ)r is(A) n (B) p
(C) np (D) n p+
QUE 1.33 x x xx n1 2Tg= 8 B 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
QUE 1.34 If , ,x y z in A.P. with common difference d and the rank of the matrix
k
xyz
456
56
R
T
SSSS
V
X
WWWW
is 2, then the values of d and k are respectively
(A) x4
and 7 (B) 7, and x4
(C) x7
and 5 (D) 5, and x7
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) QQT will be invertible
(D) Q QT will be invertible
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QUE 1.36 The adjoint matrix of 10
42
-= G is
(A) 40
21= G (B)
14
02= G
(C) 21
40= G (D)
20
41= G
QUE 1.37 If xz
ybA = > H, then adj(adj A) is equal to
(A) by
zx-
-= G (B)
by
zx= G
(C) xb yz
by
yx
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) 400
040
004
R
T
SSSS
V
X
WWWW
(B) 200
020
002
R
T
SSSS
V
X
WWWW
(C) 100
010
001
R
T
SSSS
V
X
WWWW
(D) 00
0
0
00
21
21
21
R
T
SSSS
V
X
WWWW
QUE 1.39 If A is a 2 2# non-singular square matrix, then adj(adj A) is(A) A2 (B) A(C) A 1- (D) None of the above
Common Data For Q. 40 to 42If 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 Aadj(ajd ) is equal to(A) 3 (B) 9
(C) 27 (D) 81
QUE 1.42 The value of Aadj(adj )2 is equal to(A) 34 (B) 38
(C) 316 (D) 332
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Sample Chapter of Engineering MathematicsQUE 1.43 The rank of an n row square matrix A is ( )n 1- , then
(A) adjA 0! (B) adjA 0=(C) adj IA n= (D) adj IA n 1= -
QUE 1.44 The adjoint of matrix A122
212
221
=- -
-
--
R
T
SSSS
V
X
WWWW
is equal to
(A) A (B) 3A(C) 3AT (D) At
QUE 1.45 The matrix, that has an inverse is
(A) 36
12= G (B)
52
21= G
(C) 69
23= G (D)
84
21= G
QUE 1.46 The inverse of the matrix A13
25=
--> H is
(A) 53
21= G (B)
53
31
-= G
(C) 53
21
--
--= G (D)
52
31= G
QUE 1.47 The inverse of the 2 2# matrix 15
27= G is
(A) 31 7
521
--= G (B)
31 7
521= G
(C) 31 7
521-
-= G (D)
31 7
521
--
--= G
QUE 1.48 If B is an invertible matrix whose inverse in the matrix 35
46= G, then B is
(A) 65
46-
-= G (B) 5
431
61> H
(C) 3 2
25
23
--= G (D) 3
1
51
41
61> H
QUE 1.49 Matrix AC
BM 0= > H is an orthogonal matrix. The value of B is
(A) 21 (B)
21
(C) 1 (D) 0
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QUE 1.50 If A121
211
=
R
T
SSSS
V
X
WWWW
then A 1- is
(A) 232
317
R
T
SSSS
V
X
WWWW
(B) 121
212
--R
T
SSSS
V
X
WWWW
(C) 132
425
R
T
SSSS
V
X
WWWW
(D) Undefined
QUE 1.51 The inverse of matrix A153
021
002
=
R
T
SSSS
V
X
WWWW
is equal to
(A) 41
251
021
002
-- -
R
T
SSSS
V
X
WWWW
(B) 21
251
011
002
-- -
R
T
SSSS
V
X
WWWW
(C) 1101
021
002
-- -
R
T
SSSS
V
X
WWWW
(D) 41
4101
021
002
-- -
R
T
SSSS
V
X
WWWW
QUE 1.52 If det A = 7, where adg
beb
cfc
A =
R
T
SSSS
V
X
WWWW
then det( )A2 1- is_____
QUE 1.53 If R122
013
112
=--
R
T
SSSS
V
X
WWWW
, 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 14
27
--= G, the matrix B
is
(A) 1
74
72
71> H (B)
74
21= G
(C) 1
72
74
71> H (D)
72
41= G
QUE 1.55 If A
0001
1234
0021
0100
=
R
T
SSSSSS
V
X
WWWWWW
, then det A( )1- is equal to_____
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QUE 1.56 Given an orthogonal matrix A
1110
1110
1101
1101
= -- -
-
R
T
SSSSSS
V
X
WWWWWW
, AAT 1-6 @ is
(A) 000
0
00
00
0
000
41
41
21
21
R
T
SSSSSS
V
X
WWWWWW
(B) 000
0
00
00
0
000
21
21
21
21
R
T
SSSSSS
V
X
WWWWWW
(C)
1000
0100
0010
0001
R
T
SSSSS
V
X
WWWWW
(D) 000
0
00
00
0
000
41
41
41
41
R
T
SSSSSS
V
X
WWWWWW
QUE 1.57 Given an orthogonal matrix
A
1110
1110
1101
1101
= -- -
R
T
SSSSS
V
X
WWWWW
AAT 1-6 @ is
(A) 000
0
00
00
0
000
41
41
21
21
R
T
SSSSSS
V
X
WWWWWW
(B) 000
0
00
00
0
000
21
21
21
21
R
T
SSSSSS
V
X
WWWWWW
(C)
1000
0100
0010
0001
R
T
SSSSS
V
X
WWWWW
(D) 000
0
00
00
0
000
41
41
41
41
R
T
SSSSSS
V
X
WWWWWW
QUE 1.58 A is m n# full rank matrix with m n> and I is identity matrix. Let matrix ( )A A A AT T1= -l , Then, which one of the following statement is FALSE ?
(A) AA A A=l (B) ( )AA 2l
(C) IA A =l (D) AA A A=l l
QUE 1.59 For a matrix xM 53
54
53=6 >@ H, the transpose of the matrix is equal to the
inverse of the matrix, M MT 1= -6 6@ @ . The value of x is given by
(A) 54- (B) 5
3-
(C) 53 (D) 5
4
QUE 1.60 If xx xA2 0
= > H and A11
02
1
--> H then the value of x is_____
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QUE 1.61 The value of 23
12
21
-= =G G is
(A) 83= G (B)
38= G
(C) ,3 8- -8 B (D) [ , ]3 8
QUE 1.62 If A13
21
04= -> H, then AAT is
(A) 11
34-= G (B)
11
02
13-= G
(C) 51
126= G (D) Undefined
QUE 1.63 If A23
15
11
720
-=
--> >H H, then the matrix A is equal to
(A) 13
25= G (B)
25
13= G
(C) 52
31= G (D)
52
31
-= G
QUE 1.64 Let, .
A20
0 13=
-> H and
abA 0
1 21
=-> H. Then ( )a b+ =_____
QUE 1.65 Let .
A20
0 13=
-> H and
abA 0
1 21
=-> H, Then ( )a b+ is
(A) 207 (B)
203
(C) 2019 (D)
2011
QUE 1.66 If A23
69= > H and y
xB
32= > 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 A11
10
01= > H and B
101
=
R
T
SSSS
V
X
WWWW
, the product of A and B is
(A) 10= G (B)
10
01= G
(C) 12= G (D)
10
02= G
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QUE 1.68 If A12
11
20= > H and B
121
201
=-
R
T
SSSS
V
X
WWWW
, then ( )AB T is
(A) 14
44= G (B)
11
44= G
(C) 14
41= G (D)
14
14= G
QUE 1.69 If A213
104
=-
-R
T
SSSS
V
X
WWWW
and B13
24
50=
- -> H then AB is
(A) 119
8222
10515
--
--
-R
T
SSSS
V
X
WWWW
(B) 010
0221
10515
- ----
R
T
SSSS
V
X
WWWW
(C) 119
8222
10515
- --
--
R
T
SSSS
V
X
WWWW
(D) 019
8221
10515
--
--
R
T
SSSS
V
X
WWWW
QUE 1.70 If X
0000
1000
0100
0010
=
R
T
SSSSSS
V
X
WWWWWW
, then the rank of X XT , where XT 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 PT T T T will be(A) ( )2 2# (B) ( )3 3#
(C) ( )4 3# (D) ( )3 4#
QUE 1.72 If cos
sinsincosA
a aa= -a > H, then consider the following statements :
1. .A A A=a b ab 2. .A A A( )=a b a b+
3. ( )cossin
sincosA n
n
n
n
n
aa
aa= -a > H 4. ( )
cossin
sincos
nn
nnA n
aa
aa= -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 tantan
A0
02
2=-
a
a
> H then ( )cossin
sincosI A 2a
a a-- a
> H is equal to
(A) I A2- (B) I A-(C) I A2+ (D) I A+
ARIHANT/287/32
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QUE 1.74 Let cossin
sincosA
qq= -> H
(1) AA10
01
T = > H (2) AA 1T =
(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 sincos sin
sin
2
2q
q qq q
q= = G and
B cos
cos sincos sin
sin
2
2f
f ff 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 A11
11- =- -> >H H and
A12 2
12- =- -> >H H. The matrix A is
(A) A21
11
10
02
11
12= - -
-- - -> > >H H H
(B) A11
12
10
02
21
11= - - - -> > >H H H
(C) A11
12
10
02
21
11= - -
-- - -> > >H H H
(D) 01
23
--= G
QUE 1.77 If A31
41=
--> H, then for every positive integer ,n An is equal to
(A) n
nn
n1 2 4
1 2+
+= G (B) n
nnn
1 2 41 2
- -+= G
(C) n
nn
n1 2 4
1 2-
+= G (D) n
nnn
1 2 41 2
+ --= G
ARIHANT/306/13
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QUE 1.78 For which values of the constants b and c is the vector bc
3R
T
SSSS
V
X
WWWW
a linear
Combination of ,132
264
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
and 132
---
R
T
SSSS
V
X
WWWW
(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 adf
bkg
ceh
R
T
SSSS
V
X
WWWW
is invertible for all real numbers k .
(A) finite soln (B) infinite soln
(C) 0 (D) none
***********
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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 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 IT =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 AT=- and B BT=- ...(1)
Also, given that AB BA= B AT T= - -^ ^h h [from Eq. (1)]
B AT 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 kA An= .
A B5= & A B B B5 5 6254= = =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 An n# triangular matrix det A is equal to the product of the diagonal entries of A.
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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- .
513
325
2610
5(20 30) 3(10 18) 2(5 6)= +- - - -
50 24 2 28=- + - =-
SOL 1.10 Correct option is (A).
ahg
hbf
gfc
aaf
fc h
hg
fc g
hg
bf= - +
a bc f h hc fg g hf gb2= - - - + -^ ^ ^h h h
abc af h c hfg ghf g b2 2 2= = - + + - abc fgh af bg ch2 2 2 2= + - - -
SOL 1.11 Correct answer is 43- .
Determinant 671324
1426 19
3981
1426 21
3981
1324= - +
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
0101
1102
0100
2311
=-
-
R
T
SSSSS
V
X
WWWWW
Expanding cofactor of a34
A 1011
112
010
=- --
[ ( ) ]0 1 0 1 0=- - - + 1=-
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SOL 1.14 Correct answer is 5.Consider the given matrix be
I ABm +
2111
1211
1121
1112
=
R
T
SSSSSS
V
X
WWWWWW
where m 4= so, we obtain
AB
2111
1211
1121
1112
1000
0100
0010
0001
= -
R
T
SSSSSS
R
T
SSSSSS
V
X
WWWWWW
V
X
WWWWWW
1111
1111
1111
1111
=
R
T
SSSSSS
V
X
WWWWWW
1111
=
R
T
SSSSSS
V
X
WWWWWW
1 1 1 16 @
Hence, we get A
1111
=
R
T
SSSSSS
V
X
WWWWWW
, B 1 1 1 1= 6 @
Therefore, BA 1 1 1 1
1111
=
R
T
SSSSSS
8
V
X
WWWWWW
B 4= 6 @
From the given property,
Det I ABm +^ h Det I BAn= +^ h
Det
2111
1211
1121
1112
R
T
SSSSSS
V
X
WWWWWW
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
ii3
1 21 2
2= ++
> H
and A A* T= =^ h transpose of A
ii3
1 21 2
2= -+
> H
Since, A* A=Hence, A is hermitian matrix.
SOL 1.16 Correct answer is 4.For singularity of matrix,
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8412
06
020
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=
012
102
23l
--
-R
T
SSSS
V
X
WWWW
0=
or ( )112
22
10
23 0
02
3l l- - -
-+
-+ - 0=
or ( ) ( )4 2 3l - + 0=or l 2=-
SOL 1.18 Correct option is (C).
Given EF G= where G I= = Identity matrix
cossin
sincos F
0 0
001#
-R
T
SSSS
V
X
WWWW
0 01
00
10
01
=
R
T
SSSS
V
X
WWWW
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 @
adjE ( )cos
sinsincos
0 0
001
Tqq
qq=
-R
T
SSSS
V
X
WWWW
cossin
sincos
0 0
001
qq= -
R
T
SSSS
V
X
WWWW
E ( )cos cos sin sin0 0 0# #q q q q= - - - - +^ ^h h6 8@ B
cos sin 12 2q q= + =
Hence, F .adj
EE
= 6 @
cossin
sincos
0 0
001
qq= -
R
T
SSSS
V
X
WWWW
SOL 1.19 Correct answer is 2.
We have A 12
20
35
47= -= G
It is a 2 4# matrix, thus )A(r 2#
The second order minor
12
20- 4 0!=
Hence, ( )Ar 2=
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SOL 1.20 Correct answer is 2.We have
A 111
111
101
110
110
100
+= - -
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
R R3 1-
Since one full row is zero, ( )A 3<r
Now 11
11- 2 0!=- , thus ( )A 2r =
SOL 1.21 Correct option is (B).
Here, A 112
246
325
=
R
T
SSSS
V
X
WWWW
Performing operation R 131 -^ h, we get A111
244
322
+
R
T
SSSS
V
X
WWWW
By operation R 132 -^ h, we get A110
240
320
+
R
T
SSSS
V
X
WWWW
A 0=
and 11
24 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 20
13 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
462
231
140
0= , 231
140
371
0= , 462
140
371
0= and 462
231
371
0=
Since, all 3 3# minors are zero. Now, we consider 2 2# minors
46
23 0= ,
23
14 8 3 5 0= - = =Y
Hence, rank 2=
SOL 1.24 Correct answer is 13.
Since ( )Ar 2 <= order of matrix
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Thus A 241
174
3
50l=
-=
or ( ) ( ) ( )235 4 1 20 3 16 7l l- + - + - 0=or 70 8 20 27l 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
1236
2428
3317
0235
=
1230
2420
3310
0230
= applying ( )R R R R R4 1 2 3 4"- + +
1000
2040
3380
0230
= --- applying
R R RR R R
23
2 1 2
3 1 3
"
"
--
The only fourth order minor is zero.
Since the third order minor ( )( )( )100
240
383
1 4 3 12 0!- --
= - - =
Therefore its rank is ( )A 3r =
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 10
01 0=Y
Hence, Rank A^ h 2=
SOL 1.27 Correct option is (C).
Given the two matrices,
A pr
qs= > H
and B p qpr qs
pr qsr s
2 2
2 2=++
++> H
To determine the rank of matrix A, we obtain its equivalent matrix using
the operation, a i2 a aa ai i2
11
211! - as
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A p q
s pr q0= -> H
If s pr 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 qpr qs
pr qsr s
2 2
2 1=++
++> H
To determine the rank of matrix B , we obtain its equivalent matrix using
the operation, a i2 a aa ai i2
11
211! - as
B p q pr qs
r sp qpr qs
0
2 2
2 22 2
2=+ +
+ -++
^^
hh
R
T
SSSS
V
X
WWWW
If r sp qpr qs2 2
2 2
2
+ -++
^^
hh
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 d2= +
Let A k
xyz
456
56=
k
xx dx d
456
56
2= +
+
k
xdd
411
51
6=
-Applying R R R2 1 2- = and R R R3 2 3- =
k
xd
410
51
7 0=
- R R R3 3 2= -
A 0= k d x7 4 0& - - =^ ^h h
d x4= , k 7= .
SOL 1.29 Correct option is (B).
Let A abc
1
1
1
=
R
T
SSSS
V
X
WWWW
, a b cB 2 2 2= 6 @
Let C AB=
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abc
a b ca ab ac a
a bb bc b
a cb cc 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
SSSS
R
T
SSSS
8
V
X
WWWW
V
X
WWWW
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.
Cr 6 @ 1=
SOL 1.30 Correct option is (A).Since all point are collinear,
Thus xxx
yyy
111
1
2
3
1
2
3
0=
Therefore A( )r 3<
SOL 1.31 Correct option is (B).
Let n 3=
Then A 123
246
369
=
R
T
SSSS
V
X
WWWW
and A 123
246
369
= 100
200
300
= applying R R RR R R
32
3 1 3
2 1 2
"
"
--
Thus rank if n 3= then ( )A 1r = so possible answer is (B).
SOL 1.32 Correct option is (B).If P is a matrix of order m n# and ( ) nPr = then n m#
In the normal form of P only n rows are non-zeroNow Q is a matrix of order n p# and ( ) pPQr = 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.
(PQ)r p=
SOL 1.33 Correct option is (D).
x x x xn1 2Tg= 8 B
V xxT=
xx
x
xx
xn n
1
2
1
2
h h=
R
T
SSSSSS
R
T
SSSSSS
V
X
WWWWWW
V
X
WWWWWW
So rank of V is n .
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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 d2= +
Now A k
xdd k
xx dx d k
xdd
411
51
6
456
56
2
411
51
6=
-= +
+=
-
Apply R R R2 1 2- = and R R R3 2 3- =
k
xd
410
51
7 0=
- applying R R R3 2 3"-
Thus A ( )( )k d x7 4= - -Since ( )A 2 <r = order of matrix, Thus A 0= or we get
( )( )k d x7 4- - 0=
or d ,x k4
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 4r =So Q will have 4 linearly independent rows and flour independent columns.
SOL 1.36 Correct option is (D).
We have A 10
42=
-= G
C11 , , ( )C C2 0 4 412 21= = =- - = , and C 122 =
C 24
01= = G
adj A CT=
20
41= = G
SOL 1.37 Correct option is (D).
A xz
yb=
adj A bz
yx= -
-= G
adj(adj A) xz
yb= = G
SOL 1.38 Correct option is (B).
Since, A(adjA) A In=
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We have A(adjA) 2100
010
001
200
020
002
= =
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
SOL 1.39 Correct option is (B).
We know that
adj(adj(A) A An 2= -
Here n 2= so we get
adj(adjA) A A2 2= -
A A IA A0= = =
SOL 1.40 Correct option is (A).
We know that
adj(adj A) A An 2= -
Putting n 3= and A 3= . so we get
adj(adj A) A A3 2= -
A A A3= =
SOL 1.41 Correct option is (D).
We have Aadj(adj ) A ( )n 1 2
= -
Putting n 3= and A 3= we get
Aadj(adj ) A 3 814 4= = =
SOL 1.42 Correct option is (B).Let B Aadj 2= then B is also a 3 3# matrix.
Aadj(adj )2 B B Badj 3 1 2= = =-
Aadj 2 2=
A A A 32 (3 1) 2 2 4 8 8= = = =-8 B
SOL 1.43 Correct option is (A).Since ( ) nA 1r = - , 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 [ ]aA ij n n= # then det [ ]CA ij n n
T= # where Cij is the cofactor of aij
Also ( ) ,C M1iji j
ij= - + 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.
Now, M11 = minor of a11 i.e. ( )112
21- -
- 3=-
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Similarly
M12 22
21 6=
-= ; M
22
12 613 = - =-
M21 22
21 6=
--
-=- ; M
12
21 322 = =
- - ;
M23 12
22 6=
- -- = ; M
21
22 631 =
- -- = ;
M32 ;12
22 6=
- -- = M
12
21 333 =
- -=
C11 ( ) ;M1 31 111= - =-+ ( ) ;C M1 612
1 212= - =-+
C13 ( ) ;M1 61 313= - =-+ ( ) ;C M1 621
2 121= - =+
C22 ( ) ;M1 32 222= - =+ ( ) ;C M1 623
2 323= - =-+
C31 ( ) ;M1 63 131= - =+ ( ) ;C M1 632
3 232= - =-+
C33 ( ) M1 33 333= - =+
Thus adj A 366
636
663
T
=- -
-
--
R
T
SSSS
V
X
WWWW
A3122
212
221
3
T
T=- -
-
-- =
R
T
SSSS
V
X
WWWW
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 ( )( ) ( )( )52
21 5 1 2 2 1= = - =
SOL 1.46 Correct option is (A).
We know A 1- A A1 adj=
Here A 13
25 1=
-- =-= G
Also, Aadj 53
21=
--
--= G
A 1- 11 5
321
53
21=
---
-- == =G G
SOL 1.47 Correct option is (A).
We know A 1- A A1 adj=
Here A 15
27 3= =-
Also, Aadj 75
21= -
-= G
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A 1- 31 7
521 3
1 75
21=
- --
=-
-= =G G
SOL 1.48 Correct option is (C).
Let B 1- 35
46= = G A= and B A 1= -
We know A 1- A A1 adj=
Here A 35
46 2= =-
Also, adj A 65
43= -
-
A 1- 21 6
543=
- --
= G
3 2
25
23=
--= G
SOL 1.49 Correct option is (C).For orthogonal matrix det M 1= and M MT1 =- ,
MT AB
C0= = G
BC CB
AM 1 01= = - -
--> H
This implies B BCC=
--
or B B1= 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 A1 adj=
A 153
021
002
4 0!= = ,
adj A 400
1020
101
2
410
1
021
002
T
=-- =
- -
R
T
SSSS
V
X
WWWW
A 1- 41
4101
021
002
=- -
R
T
SSSS
V
X
WWWW
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SOL 1.52 Correct answer is 0.01786.
( )det A2 1- A A21
21
n= =
Since A is a 3 3# matrix, therefore n 3= and we have A 7= .
( )det A2 1- A21
8 71
561
3 #= = =
SOL 1.53 Correct option is (B).
C11 ( )2 3 5= - - = C21 [ ( )]0 3 3=- - - =- C31 [ ( )]1 1= - - = R ( )C C C1 2 211 21 31= + + 5 6 2= - + 1=
SOL 1.54 Correct option is (A).
Let B(7 ) 1- A14
27= =
--> H and B A7 1= -
We know A 1- A A1 adj=
Here A 14
27 1=
-- =-
Also, adj A 74
21=
--
--= G
A 1- 11 7
421=
---
--= G
or B7 A74
21
1= =-> H
or B 1
74
72
71= > H
SOL 1.55 Correct answer is 0.5.
We have A
0001
1234
0021
0100
=
R
T
SSSSSS
V
X
WWWWWW
A ( )001
021
100
1 0 2 2=- =- - =
Now Adet( )1- detA1
21= =
SOL 1.56 Correct option is (C).For orthogonal matrix we know that
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AAT I=and [ ]AAT 1- I I1= =-
SOL 1.57 Correct option is (C).From orthogonal matrix
[ ]AAT I=Since the inverse of I is I , thus
[ ]AAT 1- I I1= =-
SOL 1.58 Correct option is (D).
Al ( )A A AT T1= -
( )A A AT T1 1= - - A I1= -
Put A A I1= -l in all option.
option (A) AA Al A= AA A1- A= A A= (true)
option (B) ( )AA 2l I= ( )AA I1 2- I= I 2
^ h I= (true)
option (C) A Al I= A IA1- I= I I= (true)
option (D) AA Al A= l
AA IA1- A A= = lY (false)
SOL 1.59 Correct option is (A).
Given : M x53
54
53= > H
And [ ]M T [ ]M 1= -
We know that when A A 1=T -6 6@ @ then it is called orthogonal matrix.
M T6 @ M
I=6 @
M MT6 6@ @ I=
Substitute the values of M and MT , we get
x
x53
54
53
53
54
53.> >H H 1
10
0= > H
x
x
x53
53
54
53
53
53
54
53
54
54
53
53
2#
#
#
# #
+
+
+
+
b
b
b
b b
l
l
l
l l
R
T
SSSS
V
X
WWWW
110
0= > H
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xx
x1
259 2
2512
53
2512
53+
++
> H 110
0= > H
Comparing both sides a12 element,
x2512
53+ 0= " x 25
1235
54
#=- =-
SOL 1.60 Correct answer is 0.5.
xx x2 0 1
102-> >H H
10
01= = G
or x
x20
02= G
10
01= = G
So, x x2 121
&= =
SOL 1.61 Correct option is (B).
23
12
21
-> >H H
( )( ) ( )( )( )( ) ( )( )2 2 1 13 2 2 1=
+ -+> H
38= = G
SOL 1.62 Correct option is (C).
AAT 13
21
04
120
314
= - -
R
T
SSSS
>
V
X
WWWW
H
( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )
( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )
1 1 2 2 0 03 1 1 2 4 0
1 3 2 1 0 43 3 1 1 4 4=
+ ++ - +
+ - ++ - - +> H
51
126= > H
SOL 1.63 Correct option is (B).
A 11
720
23
15
1
=--
- -
> >H H
11
720 13
1 53
12=
-- - -
--b l> >H H
131 1
1720
53
12= - -
--> >H H
131 26
651339= -
--> H
25
13= = G
SOL 1.64 Correct answer is 0.35.
We have A .2
00 13=
-= G and
abA 0
1 21
=-> H
Now AA 1- I=
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or . a
b20
0 13 0
21-
> >H H 10
01= > H
or .a b
b10
2 0 13-
> H 10
01= > H
or .a2 0 1- 0= and b3 1=
Thus solving above we have b31= and a
601=
Therefore a b+ 31
601
207= + =
SOL 1.65 Correct option is (A).
We know that AA 1- I=
or . a
b20
0 13 0
21-
> >H H .a b
b10
2 0 13
10
01=
-=> >H H
We get b3 1= or b31=
and .a b2 0 1- 0= or a b20
=
Thus a b+ 31
31
201
207= + =
SOL 1.66 Correct option is (A).
We have A 23
69= = G and ,y
xB AB
32 0= => H
We get yx2
369
32= =G G
00
00= = G
or yy
xx
6 69 9
2 123 18
++
++= G
00
00= = G
We get y6 6+ 0= or y 1=-and x2 12+ 0= or x 6=-
SOL 1.67 Correct option is (C).
AB 11
10
01
101
=
R
T
SSSS
=
V
X
WWWW
G
( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )1 1 1 0 0 11 1 0 0 1 1
12
+ +
+ +== = =G G
SOL 1.68 Correct option is (A).
We have AB 12
11
20
121
201
=-
R
T
SSSS
=
V
X
WWWW
G 14
44= = G
( )AB T 14
44= = G
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SOL 1.69 Correct option is (C).
AB 213
104
13
24
50=
-
- - -R
T
SSSS
=
V
X
WWWW
G
( )( ) ( )( )( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )( )( ) ( )( )
( )( ) ( )( )
( )( ) ( )( )( )( ) ( )( )
( )( ) ( )( )
2 1 1 31 1 0 33 1 4 3
2 2 1 41 2 0 43 2 4 4
2 5 1 01 5 0 03 5 4 0
=+ -+
- +
- + -- +
- - +
- + -- +
- - +
R
T
SSSS
V
X
WWWW
119
8222
10515
=- -
---
R
T
SSSS
V
X
WWWW
SOL 1.70 Correct answer is 3.
We have X
0000
1000
0100
0010
=
R
T
SSSSSS
V
X
WWWWWW
and transpose of X , XT
0100
0010
0001
0000
=
R
T
SSSSSS
V
X
WWWWWW
Here, we can see that rank of matrix x 3= , hence, we can determine the rank of X XT .Let Y X XT$= , the rank of Y # rank of X . Also, X Y XT1 =- and so we have
Rank X = Rank XT # Rank of YHence, from Eqs. (1) and (2), we get
Rank of X = Rank of YHence, rank of X XT$ is 3
SOL 1.71 Correct option is (A).
X4 3# XcT
4" #
X YT3 4 4 3# # ( )X YT
3 3" #
( )X YY3 3# ( )X YT 1
3 3" #-
P2 3# PT3 2" #
( )X Y PT T3 3
13 2# #
- ( )X YTPT
13 2
"#
-" ,
{( ) }P X Y PT T2 3
13 2# #
- ( )P X Y PT T12 2" #
-6 @
[ ( ) ]P X Y PT T T12 2#
- [ ( ) ]P X Y PT T12 2" #
-
SOL 1.72 Correct option is (C).
.A Aa b cossin
sincos
cossin
sincos
aa
aa
bb
bb= - -> >H H
( )( )
( )( )
cossin
sincos A
a ba b
a ba b=
+- +
++ = a b+> H
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Also, it is easy to prove by induction that
( )A na
cossin
sincos
nn
nn
aa
aa= -= G
SOL 1.73 Correct option is (D).
Let tan t2a =
Then, cosa tantan
t tt
11 1
22
22
2
2=
+-
=+-
a
a
and sina tantan
tt
12
12
22
22=
+=
+a
a
( )cossin
sincosI A
aa
aa-
-> H tan
tan cossin
sincos
112
2#
aa
aa= -
-a
a
> =H G
( )
( )t
t tt
tt
tt
tt
11
11
12
12
11
2
2
2
2
2
2#= -+-
+
+-
+-
R
T
SSSSS
=
V
X
WWWWW
G
(tantan
tt
I A1
11
1 )2
2=-
=-
= +a
a
> >H H
SOL 1.74 Correct option is (C).
AAT cossin
sincos
cossin
sincos
qq= -
-> >H H
cos sin
sin cos cos sincos sin sin cos
sin cos
2 2
2 2
q qq q q q
q q q qq q=
+- +
- ++
> H
10
01= > H
1= , Hence A is orthogonal matrix.
SOL 1.75 Correct option is (C).
AB ( )( )
( )( )
cos cos coscos sin cos
cos sin cossin sin cos
q f q ff a a f
a f q fq f q f=
--
--= G
Is a null matrix when ( )cos 0q f- = , this happens when ( )q f- is an odd
multiple of 2p .
SOL 1.76 Correct option is (C).
Let matrix A be ac
bd= G
From A11
11- =- -> >H H we get
ac
bd
a bc d
11
11- =
-- =- -= = = =G G G G
We have a b- 1=- ...(1)
and c d- 1= ...(2)
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From A12 2
12- =- -> >H H we get
ac
bd= G
12-= G
a bc d
22 2
12=
-- =- -= =G G
we have a b2- 2=- ...(2)
c d2- 4= ...(4)
Solving equation (1) and (3) a 0= and b 1=Solving equation (2) and (4) c 2=- and d 3=-
Thus A 02
13= - -= G
If we check all option then result of option C after multiplication gives result.
SOL 1.77 Correct option is (D).
A2 31
41
31
41
52
83=
--
-- =
--= = =G G G
If we put n 2= in option, then only D satisfy.
SOL 1.78 Correct option is (A).
bc
3R
T
SSSS
V
X
WWWW
k k k132
264
132
1 2 3= + +---
R
T
SSSS
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
V
X
WWWW
k k k132
2132
132
1 2 3+ -
R
T
SSSS
R
T
SSSS
R
T
SSSS
V
X
WWWW
V
X
WWWW
V
X
WWWW
bc
3=
R
T
SSSS
V
X
WWWW
( )k k k2132
1 2 3+ -
R
T
SSSS
V
X
WWWW
bc
3=
R
T
SSSS
V
X
WWWW
k k k21 2 3+ - 3= k k k3 6 31 2 3+ - b= k k k2 6 21 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 soln
***********
