gate ee vol-4.pdf
TRANSCRIPT
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GATE
ELECTRICAL ENGINEERINGVol 4 of 4
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Second Edition
GATEELECTRICAL ENGINEERING
Vol 4 of 4
RK Kanodia
Ashish Murolia
NODIA & COMPANY
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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
are supplying information but are not attempting to render engineering or other professional services.
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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
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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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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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lectricalEn
gineering,V
olume-4
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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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SampleChapterof
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ricalEngine
ering,
Volum
e-4
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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lectricalEn
gineering,V
olume-4
GATE EE vol-1
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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 1.35 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
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ricalEngine
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Volum
e-4
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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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 1.49 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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ricalEngine
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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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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.
MA 1.4 Correct option is (B).
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.
MA 1.5 Correct option is (B).
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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SampleChapte
rofGATEE
lectricalEn
gineering,V
olume-4
GATE EE vol-1
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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.
MA 1.8 Correct option is (C).
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.
MA 1.11 Correct option is (C).
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= - =-
+ ;
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ricalEngine
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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
MA 1.12 Correct option is (A).
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
MA 1.13 Correct option is (A).
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
MA 1.14 Correct answer is 13.
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=
MA 1.15 Correct option is (A).
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.
MA 1.16 Correct answer is 2.
Since
A 2( 9 8) 2( 2 3) 2 2 0= - + + - + =- + = & ( ) 3A
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SampleChapterof
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ricalEngine
ering,
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e-4
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=
MA 1.20 Correct option is (B).
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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gineering,V
olume-4
GATE EE vol-1
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MA 1.23 Correct option is (C).
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- -
MA 1.24 Correct option is (C).
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=
MA 1.25 Correct option is (B).
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
MA 1.26 Correct option is (B).
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- .
MA 1.27 Correct option is (B).
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.
MA 1.28 Correct option is (B).
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- .
MA 1.29 Correct option is (B).
Since, the eigen values of Aand AT are square so the eigen values of AT are 2
and 4.
MA 1.30 Correct option is (B).
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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ricalEngine
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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= -
MA 1.31 Correct option is (A).
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= + + =
MA 1.33 Correct answer is 10.
Since the product of the eigen values is equal to the determinant of the matrix
so 1 2 5 10A # #= =
MA 1.34 Correct option is (C).
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 .
MA 1.35 Correct option is (C).
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.
MA 1.36 Correct option is (B).
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= .
MA 1.37 Correct option is (D).
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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SampleChapte
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lectricalEn
gineering,V
olume-4
GATE EE vol-1
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( )A na cos
sin
sin
cos
n
n
n
n
a
a
a
a=
-> H
MA 1.38 Correct option is (A).
We know that adj(adj A) .A An 2 := -
Here 3n= , and A 3=
So, ( )Aadj adj 3 3A A( )3 2 := =- .
MA 1.39 Correct option is (C).
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&= = .
MA 1.43 Correct option is (D).
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
MA 1.44 Correct option is (C).
AAT 1
3
2
1
0
4
1
2
0
3
1
4
=-
-
R
T
SSSS
>V
X
WWWW
H
-
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ricalEngine
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e-4 ( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
( )( ) ( )( ) ( )( )
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= = - =
MA 1.46 Correct option is (A).
A skew symmetric matrix An n# is a matrix with A AT =- . The matrix of (A)satisfy this condition.
MA 1.47 Correct option is (C).
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
MA 1.48 Correct option is (C).
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 =
MA 1.49 Correct option is (B).
From linear algebra for An n# triangular matrix det ,aA iii
n
1
==
% . The product ofthe diagonal entries of A
MA 1.50 Correct option is (C).
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
MA 1.51 Correct option is (A).
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 =
MA 1.52 Correct option is (B).
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
MA 1.53 Correct option is (C).
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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SampleChapterof
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ricalEngine
ering,
Volum
e-4 A
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.
MA 1.55 Correct option is (C).
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.
MA 1.56 Correct option is (A).
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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SampleChapte
rofGATEE
lectricalEn
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
K
K
2-> Hor 21-> HSince option A
2
1-> His in the same ratio of x1and x2. Therefore option (A) is aneigen vector.
MA 1.57 Correct option is (A).
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
#=- =-
MA 1.58 Correct option is (C).
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= -
MA 1.59 Correct option is (B).
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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SampleChapterof
GATEElect
ricalEngine
ering,
Volum
e-4 :A B6 @
:
:
: 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=
MA 1.60 Correct option is (B).
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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SampleChapte
rofGATEE
lectricalEn
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
x l- y!=
l x y!=
So, eigen values are real if matrix is real and symmetric.
MA 1.62 Correct option is (B).
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.
MA 1.63 Correct option is (B).
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=
MA 1.64 Correct option is (C).
Eigen value of a Skew-symmetric matrix are either zero or pure imaginary in
conjugate pairs.
MA 1.65 Correct option is (D).
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=- +
--
-+
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GATE Electrical Engineering-2015in 4 Volumes
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ampleChapterof
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ricalEngine
ering,
Volum
e-4
MA 1.66 Correct option is (D).
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.
MA 1.67 Correct option is (C).
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=
MA 1.68 Correct option is (B).
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