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GATE

ELECTRICAL ENGINEERINGVol 4 of 4


2/31

Second Edition

GATEELECTRICAL ENGINEERING

Vol 4 of 4

RK Kanodia

Ashish Murolia

NODIA & COMPANY


3/31

GATE Electrical Engineering Vol 4, 2eRK Kanodia & 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 ofuse of this information. This book is published with the understanding that NODIA & COMPANY and its author

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4/31

SYLLABUS

GENERAL ABILITY

Verbal Ability : English grammar, sentence completion, verbal analogies, word groups,instructions, critical reasoning and verbal deduction.

Numerical Ability :Numerical computation, numerical estimation, numerical reasoning anddata interpretation.

ENGINEERING MATHEMATICS

Linear Algebra:Matrix Algebra, Systems of linear equations, Eigen values and eigen vectors.

Calculus:Mean value theorems, Theorems of integral calculus, Evaluation of definite andimproper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series.Vector identities, Directional derivatives, Line, Surface and Volume integrals, Stokes, Gaussand Greens theorems.

Differential equations: First order equation (linear and nonlinear), Higher order lineardifferential equations with constant coefficients, Method of variation of parameters, Cauchysand Eulers equations, Initial and boundary value problems, Partial Differential Equations andvariable separable method.

Complex variables: Analytic functions, Cauchys integral theorem and integral formula,Taylors and Laurent series, Residue theorem, solution integrals.

Probability and Statistics:Sampling theorems, Conditional probability, Mean, median, mode andstandard deviation, Random variables, Discrete and continuous distributions, Poisson,Normaland Binomial distribution, Correlation and regression analysis.

Numerical Methods:Solutions of non-linear algebraic equations, single and multi-step methodsfor differential equations.

Transform Theory:Fourier transform,Laplace transform, Z-transform.

ELECTRICAL ENGINEERING

Electric Circuits and Fields:Network graph, KCL, KVL, node and mesh analysis, transientresponse of dc and ac networks; sinusoidal steady-state analysis, resonance, basic filter concepts;ideal current and voltage sources, Thevenins, Nortons and Superposition and MaximumPower Transfer theorems, two-port networks, three phase circuits; Gauss Theorem, electricfield and potential due to point, line, plane and spherical charge distributions; Amperes andBiot-Savarts laws; inductance; dielectrics; capacitance.


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Signals and Systems: Representation of continuous and discrete-time signals; shifting andscaling operations; linear, time-invariant and causal systems; Fourier series representation ofcontinuous periodic signals; sampling theorem; Fourier, Laplace and Z transforms.

Electrical Machines: Single phase transformer equivalent circuit, phasor diagram, tests,

regulation and efficiency; three phase transformers connections, parallel operation; auto-transformer; energy conversion principles; DC machines types, windings, generatorcharacteristics, armature reaction and commutation, starting and speed control of motors;three phase induction motors principles, types, performance characteristics, starting andspeed control; single phase induction motors; synchronous machines performance, regulationand parallel operation of generators, motor starting, characteristics and applications; servo andstepper motors.

Power Systems: Basic power generation concepts; transmission line models and performance;cable performance, insulation; corona and radio interference; distribution systems; per-unit

quantities; bus impedance and admittance matrices; load flow; voltage control; power factorcorrection; economic operation; symmetrical components; fault analysis; principles of over-current, differential and distance protection; solid state relays and digital protection; circuitbreakers; system stability concepts, swing curves and equal area criterion; HVDC transmissionand FACTS concepts.

Control Systems:Principles of feedback; transfer function; block diagrams; steady-state errors;Routh and Niquist techniques; Bode plots; root loci; lag, lead and lead-lag compensation; statespace model; state transition matrix, controllability and observability.

Electrical and Electronic Measurements:Bridges and potentiometers; PMMC, moving iron,dynamometer and induction type instruments; measurement of voltage, current, power, energyand power factor; instrument transformers; digital voltmeters and multimeters; phase, timeand frequency measurement; Q-meters; oscilloscopes; potentiometric recorders; error analysis.

Analog and Digital Electronics:Characteristics of diodes, BJT, FET; amplifiers biasing,equivalent circuit and frequency response; oscillators and feedback amplifiers; operationalamplifiers characteristics and applications; simple active filters; VCOs and timers;combinational and sequential logic circuits; multiplexer; Schmitt trigger; multi-vibrators;sample and hold circuits; A/D and D/A converters; 8-bit microprocessor basics, architecture,

programming and interfacing.

Power Electronics and Drives: Semiconductor power diodes, transistors, thyristors, triacs,GTOs, MOSFETs and IGBTs static characteristics and principles of operation; triggeringcircuits; phase control rectifiers; bridge converters fully controlled and half controlled;principles of choppers and inverters; basis concepts of adjustable speed dc and ac drives.

***********


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CONTENTS

EM ELECTRICAL MACHINES

EM 1 Transformer 3

EM 2 DC Generator 36

EM 3 DC Motor 57

EM 4 Synchronous Generator 87

EM 5 Synchronous Motor 119

EM 6 Induction Motor 139

EM 7 Single Phase Induction Motor & Special Purpose Machines 166

EM 8 Gate Solved Questions 181

PS POWER SYSTEM

PS 1 Fundamentals of Power System 3

PS 2 Transmission Lines 28

PS 3 Load Flow Studies 66

PS 4 Symmetrical Fault Analysis 82PS 5 Symmetrical Components and Unsymmetrical Fault Analysis 109

PS 6 Power System Stability and Protection 134

PS 7 Power System Control 162

PS 8 Gate Solved Questions 179

MA ENGINEERING MATHEMATICS

MA 1 Linear Algebra 3

MA 2 Differential Calculus 27

MA 3 Integral Calculus 51

MA 4 Directional Derivatives 73

MA 5 Differential Equation 85

MA 6 Complex Variable 110

MA 7 Probability & Statistics 132

MA 8 Numerical Methods 153

MA 9 Gate Solved Questions 171


7/31

VA VERBAL ABILITY

VA 1 Synonyms 3

VA 2 Antonyms 18

VA 3 Agreement 29

VA 4 Sentence Structure 42

VA 5 Spellings 65

VA 6 Sentence Completion 95

VA 7 Word Analogy 123

VA 8 Reading Comprehension 152

VA 9 Verbal Classification 168

VA 10 Critical Reasoning 174

VA 11 Verbal Deduction 190

QA QUANTITATIVE ABILITY

QA 1 Number System 3

QA 2 Surds, Indices and Logarithm 16

QA 3 Sequences and Series 30

QA 4 Averages, Mixture and Alligation 47

QA 5 Ratio, Proportion and Variation 61

QA 6 Percentage 78

QA 7 Interest 92

QA 8 Time, Speed & Distance 102

QA 9 Time, Work & Wages 116

QA 10 Data Interpretation 130

QA 11 Number Series 151

***********


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9/31

MA 10 Linear Algebra MA 1PE 1 Linear Algebra PE 10EF 1 Linear Algebra EF 10

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by R. K. Kanodia & Ashish Murolia

SampleChapterof

GATEElect

ricalEngine

ering,

Volum

e-4

MA 1.10 If Ais Skew-Hermitian, then iAis(A) Symmetric (B) Skew-symmetric

(C) Hermitian (D) Skew-Hermitian

MA 1.11 If A

1

2

2

2

1

2

2

2

1

=

- -

-

-

-

R

T

SSSS

V

X

WWWW, then adj. Ais equal to

(A) A (B) cT

(C) 3AT (D) 3A

MA 1.12 The inverse of the matrix1

3

2

5

-

-> His

(A)

5

3

2

1> H (B)5

2

3

1> H(C)

5

3

2

1

-

-

-

-> H (D) None of these

MA 1.13 Let A

1

5

3

0

2

1

0

0

2

=

R

T

SSSS

V

X

WWWW, then A 1- is equal to

(A) 4

14

101

0

21

0

02- -

R

T

S

SSS

V

X

W

WWW (B) 21

2

51

0

11

0

02-- -

R

T

S

SSS

V

X

W

WWW

(C)

1

10

1

0

2

1

0

0

2

-

- -

R

T

SSSS

V

X

WWWW (D) None of these

MA 1.14 If the rank of the matrix, A

2

4

1

1

7

4

3

5

l=

-R

T

SSSS

V

X

WWWWis 2, then the value of lis ____

MA 1.15 Let Aand Bbe non-singular square matrices of the same order. Consider the

following statements(I) ( )AB A BT T T=

(II) AB B A( ) 11 1=- - -

(III) adj AB A B( ) (adj. )(adj. )=

(IV) ( ) ( ) ( )AB A Br r r=

(V) AB A B.=

Which of the above statements are false ?(A) I, III & IV (B) IV & V

(C) I & II (D) All the above

MA 1.16 The rank of the matrix A

2

0

2

1

3

4

1

2

3

=

-

-

-

R

T

SSSS

V

X

WWWWis _____


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SampleChapte

rofGATEE

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gineering,V

olume-4

GATE EE vol-1

Electric circuit & Field, Electrical & electronic measurementGATE EE vol-2

Analog electronics, Digital electronics, Power electronics

GATE EE vol-3

Control systems, Signals & systems

GATE EE vol-4

Electrical machines, Power systemsEngineering mathematics, General Aptitude

MA 1.17 The system of equations 3 0x y z- + = , x y z15 6 5 0- + = , x y z2 2 0l - + = hasa non-zero solution, if lis ____

MA 1.18 The system of equation x y z2 0- + = , 2 3 0x y z- + = , 0x y zl + - = has the

trivial solution as the only solution, if lis

(A)54

!l - (B)34l=

(C) 2!l (D) None of these

MA 1.19 The system equations 6x y z+ + = , 2 3 10x y z+ + = , 2 12x y zl+ + = isinconsistent, if lis(A) 3 (B) 3-

(C) 0 (D) None of these

MA 1.20 The system of equations 5 3 7 4x y z+ + = , 3 26 2 9x y z+ + = , 7 2 10 5x y z+ + = has(A) a unique solution (B) no solution

(C) an infinite number of solutions (D) none of these

MA 1.21 If Ais an n-row square matrix of rank ( 1)n- , then(A) adj 0A = (B) adj 0A !

(C) adj IA n= (D) None of these

MA 1.22 The system of equations 4 7 14x y z- + = , 3 8 2 13x y z+ - = , 7 8 26 5x y z- + = has(A) a unique solution

(B) no solution

(C) an infinite number of solution

(D) none of these

MA 1.23 The eigen values of A3

9

4

5

=

-> Hare

(A) 1! (B) 1, 1

(C) 1, 1- - (D) None of these

MA 1.24 The eigen values of A

8

6

2

6

7

4

2

4

3

= -

-

-

-

R

T

SSSS

V

X

WWWWare

(A) 0,3, 15- (B) 0, 3, 15- -

(C) 0,3,15 (D) 0, 3,15-

MA 1.25 If the eigen values of a square matrix be 1, 2- and 3, then the eigen values ofthe matrix 2Aare(A) , 1,2

123- (B) 2, 4,6-

(C) 1, 2,3- (D) None of these


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MA 1.26 If Ais a non-singular matrix and the eigen values of Aare , ,2 3 3- then theeigen values of A 1- are(A) 2,3, 3- (B) , ,2

131

31-

(C) 2 ,3 , 3A A A- (D) None of these

MA 1.27 If , ,1 2 3- are the eigen values of a square matrix Athen the eigen values ofA2are(A) 1,2,3- (B) 1,4, 9

(C) 1,2,3 (D) None of these

MA 1.28 If ,2 4- are the eigen values of a non-singular matrix Aand 4A = , then theeigen values of adj Aare

(A) , 1

2

1- (B) 2, 1-

(C) 2, 4- (D) 8, 16-

MA 1.29 If 2 and 4 are the eigen values of Athen the eigen values of AT are(A) ,2

141 (B) 2, 4

(C) 4, 16 (D) None of these

MA 1.30 If 1 and 3 are the eigen values of a square matrix Athen A3is equal to(A) 13( )A I 2- (B) A I13 12 2-

(C) A I12( )2- (D) None of these

MA 1.31 If Ais a square matrix of order 3 and 2A = then A(adjA) is equal to

(A)

2

0

0

0

2

0

0

0

2

R

T

SSSS

V

X

WWWW (B)

21

0

0

0

21

0

0

0

21

R

T

SSSSSS

V

X

WWWWWW

(C)

1

0

0

0

1

0

0

0

1

R

T

SSSS

V

X

WWWW (D) None of these

MA 1.32 The sum of the eigenvalues of A

8

4

2

2

5

0

3

9

5

=R

T

SSSS

V

X

WWWWis equal to ____

MA 1.33 If 1, 2 and 5 are the eigen values of the matrix Athen A is equal to ____

MA 1.34 If the product of matrices

Acos

cos sin

cos sin

sin

2

2

q

q q

q q

q= > Hand B cos

cos sin

cos sin

sin

2

2

f

f f

f f

f= > H

is a null matrix, then qand fdiffer by(A) an odd multiple of p (B) an even multiple of p

(C) an odd multiple of /2p

(D) an even multiple /2p


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GATE EE vol-3


GATE EE vol-4


MA 1.35 If Aand Bare two matrices such that A B+ and ABare both defined, then Aand Bare(A) both null matrices

(B) both identity matrices

(C) both square matrices of the same order

(D) None of these

MA 1.36 If A3

1

4

1=

-

-> H, then for every positive integer ,n Anis equal to

(A)n

n

n

n

1 2 4

1 2

+

+> H (B) n

n

n

n

1 2 4

1 2

+ -

-> H

(C)n

n

n

n

1 2 4

1 2

-

+> H (D) None of these

MA 1.37 Ifcos

sin

sin

cosA

a

a

a

a=

-a > H, then consider the following statements :

I. A A A: =a b ba II. A A A( ): =a b a b+

III. ( )cos

sin

sin

cosA n

n

n

n

n

a

a

a

a=

-a > H IV. ( ) cossin sincosnn nnA n aa aa= -a > H

Which of the above statements are true ?(A) I and II (B) I and IV

(C) II and III (D) II and IV

MA 1.38 If Ais a 3-rowed square matrix such that 3A = , then adj(adjA) is equal to :(A) 3A (B) 9A

(C) 27A (D) none of these

MA 1.39 If Ais a 3-rowed square matrix, then adj(adj A) is equal to(A) A 6 (B) A 3

(C) A 4 (D) A 2

MA 1.40 If Ais a 3-rowed square matrix such that 2A = , then Aadj(adj )2

is equal to(A) 24 (B) 28

(C) 216 (D) None of these

MA 1.41 If A

1

2

1

2

1

1

=

R

T

SSSS

V

X

WWWWthen A 1- is

(A)

1

3

2

4

2

5

R

T

SS

SS

V

X

WW

WW

(B)

1

2

1

2

1

2

-

-R

T

SS

SS

V

X

WW

WW

(C)

2

3

2

3

1

7

R

T

SSSS

V

X

WWWW (D) Undefined


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SampleChapterof

GATEElect

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MA 1.42 If Ax

x x

2 0= > Hand A 11 021 = -- > H, then the value of xis ____

MA 1.43 If A11

9

82

22

105

15

=--

--

-R

T

SSSS

V

X

WWWWand B

13

24

50

=- -> Hthen ABis

(A)

1

1

9

8

2

22

10

5

15

-

-

-

-

-R

T

SSSS

V

X

WWWW (B)

0

1

0

0

2

21

10

5

15

- -

-

-

-

R

T

SSSS

V

X

WWWW

(C)

1

1

9

8

2

22

10

5

15

- -

-

-

-

R

T

SSSS

V

X

WWWW (D)

0

1

9

8

2

21

10

5

15

-

-

-

-

R

T

SSSS

V

X

WWWW

MA 1.44 If A1

3

2

1

0

4=

-> H, then AAT is

(A)1

1

3

4-> H (B)1

1

0

2

1

3-> H

(C)5

1

1

26> H (D) UndefinedMA 1.45 The matrix, that has an inverse is

(A) 36 12> H (B) 52 21> H(C)

6

9

2

3> H (D) 84 21> HMA 1.46 The skew symmetric matrix is

(A)

0

2

5

2

0

6

5

6

0-

-

-

R

T

SSSS

V

X

WWWW (B)

1

6

2

5

3

4

2

1

0

R

T

SSSS

V

X

WWWW

(C)

0

1

3

1

0

5

3

5

0

R

T

SSSS

V

X

WWWW (D)

0

2

1

3

0

1

3

2

0

R

T

SSSS

V

X

WWWW

MA 1.47 If1

1

1

0

0

1A = > Hand

1

0

1

B =

R

T

SSSS

V

X

WWWW, the product of Aand Bis

(A)1

0> H (B) 10 01> H(C)

1

2= G (D)1

0

0

2= G

MA 1.48 Matrix Dis an orthogonal matrixA

C

B

0D = > H. The value of B is

(A)21 (B)

21

(C) 1 (D) 0


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GATE EE vol-3


GATE EE vol-4


MA 1.49 If An n# is a triangular matrix then det Ais

(A) ( )a1 iii

n

1

-=

% (B) ai ii

n

1=

%

(C) ( )a1 iii

n

1

-=/

(D) ai ii

n

1=/MA 1.50 If

cos

sin

t

e

t

tA t

2

= > H, thendt

dA will be

(A)sin

sin

t

e

t

tt

2> H (B) cossinte tt2t> H(C)

sin

cos

t

e

t

t

2t

-> H (D) UndefinedMA 1.51 If , 0detA R An n !! # , then

(A) Ais non singular and the rows and columns of Aare linearly independent.

(B) Ais non singular and the rows Aare linearly dependent.

(C) Ais non singular and the Ahas one zero rows.

(D) Ais singular.

MA 1.52 For the matrix3

A5

1

3= > H, ONE of the normalized eigen vectors given as

(A)21

23> H (B) 21

2

1-> H(C) 10

3

10

1-> H (D) 515

2> H

MA 1.53 The system of algebraic equations x y z2+ + 4= , x y z2 2+ + 5= andx y z- + 1= has(A) a unique solution of 1, 1 1andx y z= = = .

(B) only the two solutions of ( 1, 1, 1) ( 2, 1, 0)andx y z x y z = = = = = =

(C) infinite number of solutions

(D) no feasible solution

MA 1.54 Eigen values of a real symmetric matrix are always(A) positive (B) negative

(C) real (D) complex

MA 1.55 Consider the following system of equations

x x x2 1 2 3+ + 0=

x x2 3- 0=

x x1 2+ 0=This system has(A) a unique solution (B) no solution

(C) infinite number of solutions (D) five solutions


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SampleChapterof

GATEElect

ricalEngine

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Volum

e-4 MA 1.56 One of the eigen vectors of the matrix A

2

1

2

3= > His

(A)2

1-> H (B)

2

1> H(C)

4

1> H (D) 11-> H

MA 1.57 For a matrix Mx53

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 xis given by

(A)54- (B)

53-

(C)

5

3 (D)

5

4

MA 1.58 The matrix

4

p

1

3

1

2

0

1

6

R

T

SSSS

V

X

WWWWhas one eigen value equal to 3. The sum of the

other two eigen value is(A) p (B) 1p-

(C) 2p- (D) 3p-

MA 1.59 For what value of a, if any will the following system of equation in , andx y zhave

a solution ? 2 3 4x y+ =

4x y z+ + =

3 2x y z a + - =(A) Any real number (B) 0

(C) 1 (D) There is no such value

MA 1.60 The eigen vector of the matrix2

1

0

2> Hare written in the form anda b

1 1> >H H. Whatis a b+ ?

(A) 0 (B)21

(C) 1 (D) 2

MA 1.61 If a square matrix A is real and symmetric, then the eigen values(A) are always real

(B) are always real and positive

(C) are always real and nonnegative

(D) occur in complex conjugate pairs

MA 1.62 The number of linearly independent eigen vectors of2

0

1

2> His

(A) 0 (B) 1

(C) 2 (D) infinite


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SampleChapterof

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e-4 SOLUTIONS

MA 1.1 Correct answer is 2- .

Ais singular if 0A =

&

0

1

2

1

0

2

2

3

l

-

-

-R

T

SSSS

V

X

WWWW 0=

& ( 1) 2 01

2

2 1

0

2

3

0

2

3

l l- -

-

-+

-+

- 0=

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

& 4 6l - + 0= 2& l= -

MA 1.2 Correct answer is 625.

If kis a constant and Ais a square matrix of order n n# then k kA An- .

A B A B B B5 5 5 6254&= = = =

& a 625=

MA 1.3 Correct option is (B).

Ais singular, if A 0=

Ais Idempotent, if A A2 =

Ais Involutory, if IA2 =

Now, A2 AA AB A A BA AB A( ) ( )= = = = =

and B2 BB BA B AB BA B( ) ( )= = = = =

& A2 A= and B B2 = ,

Thus A B& both are Idempotent.


Since, A2

5

3

1

8

5

2

0

0

1

5

3

1

8

5

2

0

0

1

=

- -

-

- -

-

R

T

SS

SS

R

T

SS

SS

V

X

WW

WW

V

X

WW

WW

1

0

0

0

1

0

0

0

1

I= =

R

T

SSSS

V

X

WWWW

, IA A2 &= is involutory.


Let aA ij= be a skew-symmetric matrix, then

AT A=- , a ai j i j & =- ,

if i j= then 2 0 0a a a a i i i i i i i i & &=- = =

Thus diagonal elements are zero.

MA 1.6 Correct option is (C).

Ais orthogonal if AA IT =

Ais unitary if AA IQ = , where AQis the conjugate transpose of Ai.e., ( )A AQ T= .


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GATE EE vol-3


GATE EE vol-4


Here,

AAQ i

i

i

i

21

2

2

2

12

1

2

2

2

11

0

0

1I2=

- - - -= =

R

T

SSSSS

R

T

SSSSS

>V

X

WWWWW

V

X

WWWWW

HThus Ais unitary.

MA 1.7 Correct option is (A).

A square matrix A is said to Hermitian if A AQ = . So a ai j ji= . If i j= then

a aii ii= i.e. conjugate of an element is the element itself and aii is purely real.


A square matrix A is said to be Skew-Hermitian if A AQ =- . If A is Skew-

Hermitian then A AQ =-

& aj i aij=- ,

If i j= then 0a a a a i i i i i i i i&=- + = it is only possible when aii is purely

imaginary.

MA 1.9 Correct option is (D).

Ais Hermitian then A AQ =

Now, (iA)Q ( )i i i i A A A, A A(i )Q Q Q&= =- =- =-

Thus iAis Skew-Hermitian.

MA 1.10 Correct option is (C).Ais Skew-Hermitian then A AQ =-

Now, ( ) ( )i i iA A A AQ Q= =- - = then iAis Hermitian.


If [ ]aA i j n n = # then det [ ]cA i j n n T= # where cij is the cofactor of aij

Also ( 1)c Mi ji j

ij= - + , where Mij is the minor of aij , obtained by leaving the

row and the column corresponding to ai jand then take the determinant of the

remaining matrix.

Now, M11=minor of a11i.e. 1 312 21- =

-- =-

Similarly

M12 62

2

2

1=

-= ; 6M

2

2

1

213=

- =-

M21 62

2

2

1=

-

-

-=- ; 3M

1

2

2

122=

- -= ;

M23 61

2

2

2=

- -

- = ; 6M

2

1

2

231=

- -

- = ;

M32 61

2

2

2=- -

- = ; 3M1

2

2

133 =- -

=

C11 ( 1) 3;M1 1

11= - =-+ ( 1) 6;C M12

1 212= - =-

+

C13 ( 1) 6;M1 3

13= - =-+ ( 1) 6;C M21

2 121= - =

+

C22 ( 1) 6;M3 1

31= - =+ ( 1) 6C M23

2 323= - =-

+ ;


19/31


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C31 ( 1) 6;M3 1

31= - =+ ( 1) 6;C M32

3 232= - =-

+

C33 ( 1) 3M3 3

33= - =+

detA

C

C

C

C

C

C

C

C

C

T11

21

31

12

22

32

13

23

33

=

R

T

SSSS

V

X

WWWW

3 3A

3

6

6

6

3

6

6

6

3

1

2

2

2

1

2

2

2

1

T=

- -

-

-

- =

- -

-

-

- =

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW


Since A 1- A

A1 adj=

Now, Here A 11

3

2

5=

-

- =-

Also, adj A A5

2

3

1

5

3

2

1adj

T

&=-

-

-

- =

-

-

-

-

> >H H A 1- 11 5

3

2

1

5

3

2

1=

-

-

-

-

- => >H H


Since, A 1- A

A1 adj=

A 4 0,

1

5

3

0

2

1

0

0

2

!= =

adj A

4

0

0

10

2

0

1

1

2

4

10

1

0

2

1

0

0

2

T

=

-

- =

- -

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

A 1- 41

4

10

1

0

2

1

0

0

2

=

- -

R

T

SSSS

V

X

WWWW


A matrix A( )m n# is said to be of rank r if

(i) it has at least one non-zero minor of order r , and

(ii) all other minors of order greater than r ,

if any; are zero. The rank of Ais denoted by ( )Ar . Now, given that ( ) 2A "r =

minor of order greater than 2 i.e., 3 is zero.

Thus A 0

2

4

1

1

7

4

3

5

l=

-

=

R

T

SSSS

V

X

WWWW

& 2(35 4 ) 1(20 ) 3(16 7)l l- + - + - 0=

& 70 8 20 27l l- + - + 0= ,

& 9 117 &l l= 13=


The correct statements are

( )AB T B AT T= , ( ) ,AB B A1 1 1=- - -


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adj ( )AB ( ) ( )B Aadj adj=

( )ABr ( ) ( ),A B AB A . B!r r =

Thus statements I, III, and IV are wrong.


Since

A 2( 9 8) 2( 2 3) 2 2 0= - + + - + =- + = & ( ) 3A


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1

0

0

1

1

0

1

2

3

6

4

2l

=

-

R

T

SSSS

V

X

WWWW ( )R R R3 2 3&-

& ( : )A Br 3=

As one of the minor 0

1

0

0

1

1

0

6

4

2

!

Now, system is inconsistent if

( )Ar ( : )A B!r i.e. ( ) 3A !r It is possible only when 3 0l - = i.e. 3l=


The system xA B= is consistent (has solution) if ( ) ( : )A A Br t= Also if

( ) ( )A ABr r= .no= of unknowns, then system has a unique solution and if

( ) ( : ) .noA A B


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The characteristic equation of a matrix Ais given as 0A Il- = .

The roots of the characteristic equation are called

Now, here 0A Il- =

&3

4

5

5

l

l

-

- - - 0=

& (3 )( 5 ) 16l l- - - + 0 15 2 16 02& l l= - + + + =

& 2 12l l+ + 0 ( 1) 0 1, 12& &l l= + = =- -

Thus eigen values are 1, 1- -


Characteristic equation is 0A Il- =

&

8

6

2

6

7

4

2

4

3

l

l

l

-

-

-

-

-

-

-

0=

& 18 452 2l l l- + 0=

& ( 3)( 15)l l l- - 0= , ,0 3 15& l=


If eigen values of Aare , ,1 2 3l l l then the eigen values of kAare , ,k k k1 2 3l l l . So

the eigen values of 2A are 2, 4- and 6


If , , ...., n1 2l l l are the eigen values of a non-singular matrix A, then A1- has the

eigen values , , ....,1 1 1n1 2l l l. Thus eigen values of A 1- are , ,

21

31

31- .


If , , ..., n1 2l l l are the eigen values of a matrix A, then A2has the eigen values

, , ..., n12

22 2l l l . So, eigen values of A2are 1, 4, 9.


If , , ..., n1 2l l l are the eigen values of Athen the eigen values adj Aeigen values

adj Aare , , ..., ; 0A A A

An1 2

!l l l

. Thus eigen values of

adj Aare ,24

44- i.e. 2 and 1- .


Since, the eigen values of Aand AT are square so the eigen values of AT are 2

and 4.


Since 1 and 3 are the eigen values of Aso the characteristic equation of Ais

( 1)( 3)l l- - 0= 4 3 02& l l- + =

Also, by Cayley-Hamilton theorem, every square matrix satisfies its own


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characteristic equation so

4 3A A I2 2- + 0=

& A2 4 3A I2= -

&

A

3

4 3 4(4 3 ) 3A A A I A

2

= - = - -& A3 13 12A I 2= -


Since A(adj A) A I 3=

& A(adj A) 2

1

0

0

0

1

0

0

0

1

2

0

0

0

2

0

0

0

2

= =

R

T

SSSS

R

T

SSSS

V

X

WWWW

V

X

WWWW

MA 1.32 Correct answer is 18.Since the sum of the eigen values of an n-square matrix is equal to the trace of

the matrix (i.e. sum of the diagonal elements)

So, required sum 8 5 5 18= + + =


Since the product of the eigen values is equal to the determinant of the matrix

so 1 2 5 10A # #= =


AB( )

( )

( )

( )

cos cos cos

cos sin cos

cos sin cos

sin sin cos

q f q f

f q q f

q f q f

q f q f=

-

-

-

-> H

Null matrix when ( ) 0cos q f- =

This happens when ( )q f- is an odd multiple of /2p .


Since A B+ is defined, Aand Bare matrices of the same type, say m n# . Also,

ABis defined. So, the number of columns in Amust be equal to the number of

rows in B, i.e. n m= . Hence, Aand Bare square matrices of the same order.


A23

1

4

1

3

1

4

1

5

2

8

3=

-

-

-

- =

-

-> > >H H H

,n

n

n

n

1 2 4

1 2=

+ -

-> H where 2n= .


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

n

n

n

n

a

a

a

a=

-> H A= a b+

Also, it is easy to prove by induction that


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( )A na cos

sin

sin

cos

n

n

n

n

a

a

a

a=

-> H


We know that adj(adj A) .A An 2 := -

Here 3n= , and A 3=

So, ( )Aadj adj 3 3A A( )3 2 := =- .


We have ( )adj adjA A ( )n 12

= -

Putting 3n= , we get ( )adj adjA A 4=

MA 1.40 Correct option is (C).Let B ( )Aadj adj 2= .

Then, Bis also a 3 3# matrix.

( )}Aadj{adj adj 2 adj B BB 3 3 1 2= = =-

( )adj adjA 2 2= 2A A( )2 3 12

16 162

= = =-9 C

... A28 A 2= B

MA 1.41 Correct option is (D).Inverse matrix defined for square matrix only.

MA 1.42 Correct answer is 0.5.

x

x x

2 0 1

1

0

2-> >H H1

0

0

1= > H

&x

x

2

0

0

2> H ,10 01= > H So, 2 1x x 21&= = .


AB

2

1

3

1

0

4

1

3

2

4

5

0=

-

-- -

R

T

SSSS

>V

X

WWWW

H

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )( )

( )( ) ( )

( ) ( )( )

( )( ) ( )( )

( )( ) ( )( )

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

SSSS

V

X

WWWW

1

1

9

8

2

22

10

5

15

=

- -

-

-

-

R

T

SS

SS

V

X

WW

WW


AAT 1

3

2

1

0

4

1

2

0

3

1

4

=-

-

R

T

SSSS

>V

X

WWWW

H


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( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

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= > HMA 1.45 Correct option is (B).

If A is zero, A 1- does not exist and the matrix Ais said to be singular. Only

(B) satisfy the condition.

A (5)(1) (2)(2) 15

2

2

1= = - =


A skew symmetric matrix An n# is a matrix with A AT =- . The matrix of (A)satisfy this condition.


AB( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

1

1

1

0

0

1

1

0

1

1 1 1 0 0 1

1 1 0 0 1 1

1

2= =

+ +

+ + =

R

T

SSSS

> > >V

X

WWWW

H H H


For orthogonal matrix det 1M

= andM MT1

=

-

, therefore HenceD DT1

=

-

DT A

B

C

B C C

B

AD

01 01= = =

- -

--> >H H

This implies 1BB C

CB

B B

1& & !=

-- = =

Hence 1B =


From linear algebra for An n# triangular matrix det ,aA iii

n

1

==

% . The product ofthe diagonal entries of A


dtdA

( )

( )

( )

( )

cos

sinsin

cosdt

d t

dt

d edt

d t

dt

d t

t

e

t

t

2t t

2

= =-

R

T

SSSSS

>V

X

WWWWW

H


If det 0A ! , 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 0A =


Given A 3

5

1

3= > H


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For finding eigen values, we write the characteristic equation as

A Il- 0=

5

1

3

3

l

l

-

- 0=

& ( )( )5 3 3l l- - - 0=

8 122l l- + 0= & l ,2 6=

Now from characteristic equation for eigen vector.

xA Il-6 @" , 0= 6 @For 2l=

X

X

5 2

1

3

3 21

2

-

-> >H H 00= > H

&1

X

X

3

1

3 1

2> >H H 00= > H

X X1 2+ 0= X X1 2& =-

So eigen vector1

1=

-* 4

Magnitude of eigen vector ( ) ( )1 1 22 2= + =

Normalized eigen vector2

1

21

=-

R

T

SSSSS

V

X

WWWWW


For given equation matrix form is as follows

A

1

2

1

2

1

1

1

2

1

=

-

R

T

SSSS

V

X

WWWW, B

4

5

1

=

R

T

SSSS

V

X

WWWW

The augmented matrix is

:A B8 B:

:

:

1

2

1

2

1

1

1

2

1

4

5

1

=

-

R

T

SSSS

V

X

WWWW ,R R R22 2 1" - R R R3 3 1" -

:

:

:

1

0

0

2

3

3

1

0

0

4

3

3

+ -

-

-

-

R

T

SSSS

V

X

WWWW

R R R3 3 2" -

:

:

:

1

0

0

2

3

0

1

0

0

4

3

0

+ - -

R

T

SSSS

V

X

WWWW / 3R R2 2" -

:

:

:

1

0

0

2

1

0

1

0

0

4

1

0

+

R

T

SSSS

V

X

WWWW

This gives rank of A , ( )Ar 2= and Rank of : : 2A B A B r= =8 8B BWhich is less than the number of unknowns (3)

Ar

6 @ :A B 2 3


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x

y

y

x= > H

We know that the characteristic equation for the eigen values is given by

A Il- 0=

x

y

y

x

l

l

-

- 0=

( )x y2 2l- - 0=

( )x 2l- y2=

x l- y!= &l x y!=

So, eigen values are real if matrix is real and symmetric.


Given system of equations are, x x x2 1 2 3+ + 0= ...(i)

x x2 3- 0= ...(ii)

x x1 2+ 0= ...(iii)

Adding the equation (i) and (ii) we have

x x2 21 2+ 0=

x x1 2+ 0= ...(iv)

We see that the equation (iii) and (iv) is same and they will meet at infinite

points. Hence this system of equations have infinite number of solutions.


Let, A 3

2

1

2= > H

And 1l and 2l are the eigen values of the matrix A .

The characteristic equation is written as

A Il- 0=

2

1

2

3

1

0

0

1l-> >H H 0=

21

23

ll

--

0= ...(i)

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

5 42l l- + 0= &l &1 4=

Putting 1l= in equation (i),

x

x

2 1

1

2

3 11

2

-

-> >H H 00= > H where xx12> His eigen vector

x

x

1

1

2

2

1

2> >H H

0

0

=

> H x x21 2+ 0= or x x21 2+ 0=Let x2 K=

Then x K21 + 0= &x1 K2=-

So, the eigen vector is


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K

K

2-> Hor 21-> HSince option A

2

1-> His in the same ratio of x1and x2. Therefore option (A) is aneigen vector.


Given : M x53

54

53= > H

And [ ]M T [ ]M 1= -

We know that when A A1

=T -6 6@ @ then it is called orthogonal matrix.

M T

6 @ MI

= 6 @ M MT6 6@ @ I=Substitute the values of M and M T , we get

x

x53

54

53

53

54

53.> >H H 110 0= > H

x

x

x53

53

5

4

5

3

5

353

54

53

5

4

5

4

5

3

5

3

2#

#

#

# #

+

+

+

+

b

b

b

b b

l

l

l

l l

R

T

SSSS

V

X

WWWW

1

1

0

0= > H

x

x

x

1259 2

2512

53

2512

53+

+

+> H 110 0= > HComparing both sides a12element,

x2512

53

+ 0= "x2512

35

54

#=- =-


Let, A

2 4

p

1

31 01 6=

R

T

S

SSS

V

X

W

WWWLet the eigen values of this matrix are , &1 2 3l l l

Here one values is given so let 31l =

We know that

Sum of eigen values of matrix= Sum of the diagonal element of matrix A

1 2 3l l l+ + p1 0= + +

2 3l l+ p1 1l= + - p1 3= + - p 2= -


Given : x y2 3+ 4=

x y z+ + 4=

x y z2+ - a=

It is a set of non-homogenous equation, so the augmented matrix of this system

is


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:

:

: a

2

1

1

3

1

2

0

1

1

4

4=-

R

T

SSSS

V

X

WWWW

:

:

: a

2

0

2

3

1

3

0

2

0

4

4

4+ -

+

R

T

SSSS

V

X

WWWW R3 R R3 2" + , R R R2 22 1" -

:

:

: a

2

0

0

3

1

0

0

2

0

4

4+ -

R

T

SSSS

V

X

WWWW R3 R R3 1" -

So, for a unique solution of the system of equations, it must have the condition

[ : ]A Br [ ]Ar=

So, when putting a 0=

We get [ : ]A Br [ ]Ar=


Let A 2

2

1

0= > H 1l and 2l is the eigen values of the matrix.

For eigen values characteristic matrix is,

A Il- 0=

2

2

0

1

1

0

1

0l-> >H H 0=

( )

( )

1

0

2

2

l

l

-

- 0= ...(i)

( )( )1 2l l- - 0= & l &1 2=

So, Eigen vector corresponding to the 1l= is,

1 a

0

0

2 1> >H H 0= a a2 + 0= 0a& =

Again for 2l=

2

0 b

1

0

1-> >H H 0=

b

1 2- + 0= b

2

1

=

Then sum of &a b a b 021

21

& + = + =

MA 1.61 Option (A) is correct

Let square matrix

A x

y

y

x= > H

The characteristic equation for the eigen values is given by

A Il- 0=

x

y

y

x

l

l

-

- 0=

( )x y2 2l- - 0=

( )x 2l- y2=


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x l- y!=

l x y!=

So, eigen values are real if matrix is real and symmetric.


Let, A 2

0

1

2= > H

Let lis the eigen value of the given matrix then characteristic matrix is

A Il- 0= Here I1

0

0

1= > H= Identity matrix

2

0

1

2

l

l

-

- 0=

( )2 2l- 0=

l 2= , 2So, only one eigen vector.


Writing :A Bwe have

:

:

:

1

1

1

1

4

4

1

6

6

20

l m

R

T

SSSS

V

X

WWWW

Apply R R R3 3 2" -

:

:

: 20

1

1

0

1

4

0

1

6

6

6

20

l m- -

R

T

SSSS

V

X

WWWW

For equation to have solution, rank of A and :A Bmust be same. Thus for no

solution; 6, 20!ml=


Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in

conjugate pairs.


We have ( )f x sinx

xp

=-

Substituting x p- y= ,we get

( )f y p+ ( )sin sin

y

y

y

yp=

+=- ( )sin

y y1=-

! !

...y

y y y1

3 5=- - + -c m

or ( )f y p+ ! !

...y y13 52 4

=- + - +

Substituting x yp- = we get

( )f x !

( )!

( )...

x x1

3 5

2 4p p=- +

--

-+


31/31


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ampleChapterof

GATEElect

ricalEngine

ering,

Volum

e-4


Sum of the principal diagonal element of matrix is equal to the sum of Eigen

values. Sum of the diagonal element is 1 1 3 1- - + = .In only option (D), the

sum of Eigen values is 1.


The product of Eigen value is equal to the determinant of the matrix. Since one

of the Eigen value is zero, the product of Eigen value is zero, thus determinant

of the matrix is zero.

Thusp p p p 11 22 12 21- 0=


The given system is

xy

42

21= =G G 76= = G

We have A 4

2

2

1= = G

and A 4

2

2

10= = Rank of matrix ( )A 2

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