matrices and determinants

LEVEL - IModel Question :Problem based on operation on two matricesOne correct answer MCQ

1. If

0214222322360

2330164

7243

bcxy

czbbax

yzx

Then value of

a+b+c+x+y+ z is(A) -16 (B) - 11(C) 20 (D) - 52

2. If Y=

4123

and

2301

2 YX , then the sum of element of X is equal to

(A) - 5 (B) -6(B) - 10 (D) -12

3. Write the following as a single matrix : 752057420512

321

(A) 151221 (B) 101221 (C) 101521 (D) 151241

4. If

5207

YX and

3003

YX then the sum of the elements of the matrix 3X -4Y

is equal to (A) 14 (B) 16 (C) 12 (D) 25.5 If A and B are square matrices of the same order, then which of the following are not true ? (A) ''' BAAB (B) ''' ABAB (C) 0AB 0AB If 0A or 0B (D) AB=0 If A = I or B =I

6. If A =

0110

and B =

0

0i

i, then

(A) IA 2 (B) IA 2

(C) IB 2 (D) IB 2

7. If abccabbca

111

1 and 2

2

2

111

2

ccbbaa

, then which of the following is not true ?

(A) 21 (B) 21

(C) 21 2 (D) 12 28. For the equations 4955,232,132 zyxzyxzyx

(A) there is only one solution (B) there exist infinitely many solutions (C) there is no solution (D) the equations are consistent.

9. Statement 1 : The determinants

2

2

2

111

1

1

ccbbaa

andabccababca

are not identical.

because Statement 2 : The first two columns in both the determinants are identical and third columns are different. (A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.

10. Statement :1 If thenii

A

721212

det (A) is real .

Statement : 2 If A = ,2221

1211

aaaa

A ija being complex numbers then det (A) is always real.

(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.

Comprehension-1 321,.123012001

andUUifUA

are columns matrices satisfying

andAUAU

032

,001

21 UAU ,132

3

is 33 matrix whose columns are

321 ,, UUU then answer the following questions

11. The value of [U] is(A) 3 (B) -3(C) 3/2 (D) 2

12. The sum of the elements of 1U is(A) -1 (B) 0(C) 1 (D) 3 Comprehension

(A) 1 (B) 0(C) 2 (D)

13. The value of [3 2 0]

023

is

(A) [5] (B)

25

(C) [4) (D)

23

MatrixMatch TypeThis section contains 2questions. Each question contains statementsgiven in two column which have to be matched. Statements (A, B, C, D)in Column I have to be matched with statements (p, q, r, s) in Column II.The answers to these questions have to be appropriately bubblesas illustrated in the following example.If the correct matches are Ap, As, Bq, Br, Cp, Cq and Ds,then the correctly bubbled 4 4 matrix should be as follows :

p q r s

p q r s

p

p

p

q

q

q

r

r

r

s

s

s

A

B

C

D

14 Column-I Column-II

(A) Idempotent (p)

2 2 41 3 4

1 2 3

(B) Orthogonal (q)

4 1 43 0 43 1 3

(C) Unitary (r)

1 i i 12 2

1 i 1 i2 2

(D) Involutory (s)cos sinsin cos

PQ15 The matrix X is equal to such that 3A - 2B + X = 0 , where

2312

;3124

BA

(A)

31416

(B)

53416

(C)

5344

(D)

8344

16 Let and

02

tan2

tan0

I is the identity matrix of order 2. Then ( I -A).

cossinsincos

is equal to (A) 2I+2A (B) 2I -2B

(C) I - A (D) I +A

17. If

7101

A and

1001

I , then find k so that kIAA 82 .

(A) 7 (B) 8(C) 10 (D) none of these.

18 If A= '

2'

1 21&

21,

00

0AAQAAifQ

cbcaba

Then 21.QQ is equal to

(A) 3I (B) 3O(C) A (D) None of these.

19 If

221321121

,212101654

,131201321

CBA then A - 2B + 3C is equal to

(A)

311696261410

(B)

315696261410

(C)

315696261810

(D)

3156156261810

20. . The system of linear equations 023,02,0 yxzyxzyx (A) has no solution (B) has a unique solution (C) is consistent (D) an infinitely many solutions.

21 . The value of satisfying 0

11142782cos

323sin

is

(A) n (B) 6

n

(C) 62 n (D) 6

1 nn

22 Statement 1 : If A is a matrix of order ,22 then ladjA = A . because

Statement 2 : A = TA . A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.

23 Statement -1 : If areaaa n..............., 21 in G.P.. ( 0ia for all i ) 0

logloglogloglogloglogloglog

876

543

21

nnn

nnn

nnn

aaaaaaaaa

because Statement : The three elements in any row of the determinant are in A.P. (A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.

Comprehension - 2 is an imaginary cube root of unity..

24 is an imaginary cube root of unity . 1

11

45

43

53

(A) 3 (B) - 3 (C) 21 (D) 21

25 A root of the polynomial

xx

x

11

1

2

2

2

is

(A) 1 (B) 0 (C) 2 (D)

26 The value of Nnnn

nn

nn

?1

11

2

2

2

=

(A) (B) 2 (C) 1 (D) 0


p q r s

p q r s

p

p

p

q

q

q

r

r

r

s

s

s

A

B

C

D

27 If C & D are symmetric matrices of order 3 3Column-I Column-II(A) CD + DC (p) symmetric(B) C + D (q) Antisymmetric(C) CD DC(D) C D

Answers1.(A) 2.(A) 3.(C). 4(A). 5.(B)(C)(D) 6.(B) (C) 7.(B)(C)(D) 8(C) (D) 9.(D) 10.(D)11.(A) 12(B) 13(A) 14() 15(B) 16(D) 17(A) 18(B) 19.(A) 20(A) (D) 21.(A) (D) 22.(B)(A) (D)23. (A) 24.(AQ) 25.( B) 26(D) 27.(C)

Level - IM.Q Solution1. As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing

the corresponding elements we get)....(03 ix )....(64 iiz )....(2372 iiiyy

),.....(264 ivxx )....(31 va )....(220 vic )...(423 viibb ).....(213 viiib )...(02 ixcz

From (i) andyiiifromx 5)(,3 from (ii) z =2From (V) a = -2, from (viii) b = - 7 and from (vi) c- 1Thus a = - 2, b = -7, c = - 1, x = -3, y = - 5, z = 2

2.

4123

2301

22 YYXX

2422

42132031

1211

2422

21X

3. 33

057420512

321

5.32)2()1(1)7(30.22.1[ 31]0.34)2(5.1

75231019752057420512

321

4.

3003

5207

2

2

YXYXX

XYXYX

82010

35020037

4105

82010

21X

Again

1102

2204

21

2204

35020037

3003

5207

)()(2

Y

YXYXY

5. )(),(),( dca

6

1001

0110

01102A

1001

00

00

00

2

22

ii

ii

ii

B )(),( cb

7. 111

1

111

2

2

2

2

2

2

ccbbaa

abcccabcbbabcaa

abcabccabbca

21

)(),(),( dcb

8. 0023112

111, Here

)(),( dc9. ANSWER = D

10. clearly 9721212

ii

A Which is real SI is not current

11.

zyx

U Given

001

123012001

001

1

zyx

AU

001

2302

1

zyxyx

x

02302

1

zyxyx

x

Solving we get 12

1

zyx

12

1

1U

41

2

032

22 UAU

and

31

2

132

33 UAU

Hence U=

341112

221U

12. Adj 3369357

021

UandU

369357

021

311

UadjUU

Sum of elements of 01 U

13.

023

341112

221023

023

023 U

= 583023

441

14. P,Q,R,S be the matrices corresponding to the options (P) (Q) , (R), (S) respectivelyP,Q,S, have real elements

TT

TT

TT

SS

QQ

PP

22

2

2

).........(..........PP

APPPP

PP

TT

).........(........... 3

2

2

DIQQQQ

QIQ

TT

)...(..........

.

2

22

22

2

CIRR

IRRIRRR

T

))((........... 22

22

2

CBISSSS

SSIS

T

15

2312

;3124

BA

Now 3A- 2B + X =0X = - 3A + 2B

2312

23124

3

= 53416

4624

93612

16.

222

2

2

2

12

2/tan12/tan2sin,

11

2/12/tan1cos

tt

tt

t

where t2tan

Also

11

00

1001

tt

tt

AI

And.

11

00

1001

tt

tt

AI

Now

cossinsincos

AI

2

2

2

22

2

11

12

12

11

11

tt

tt

tt

tt

tt

2

2

2

2

22

2

2

2

22

2

2

2

11

12

12

11

11

12

12

11

tt

tt

tt

ttt

ttt

tt

tt

tt

AItt

1

1

Hence,

cossinsincos

AIAI

P.Q solutions

17 Given , A =

71

01

49801

7101

71012 AAA

Again, 8A+ kI = 8

k

kk

kk

56808

00

56808

1001

7101

k

kkIAA

56808

49801

82

k 81 and 56 + k = 49 7 k

18

1121

21

00

0

AAAAA

cbcaba

A

000000000

000000000

21

00

0

00

0

21

1

cbcaba

cbcaba

Q

Similarly

00

0

21 1

1

cbcaba

AAQ 321 QQQ

19 A - 2B + 3C = 0

663963363

424202

12108

131201321

311696261410

20 Here ,

106

,21

321111

BA

If equations have no solution, then |A| = 0 and (adj A). 0BCalculating |A| = 303

Also, adj 01022010

..110220

110

BAadjA

da ,10

21 0162cos143142cos1123sin10 0202cos203sin10 022cos23sin 02sin42sin43sin 23

0sin0sin3sin4sin4 23 or

da

21sin

22 In fact adj A is got by keeping the diagonal elements as they are and changing the sign of theother two elements.

b a d

23.

rmarmarma

rmarmarmrmrmarma

log)7(loglog6loglog5loglog4loglog3loglog2log

log1loglog)(loglog1log

rrrrrr

rnarnarna

log3log3log3log3log3log3

log1loglogloglog1log Using 122( RRR and )233 RRR

= 0Hence = (A)

24. 3211454

443

5353

45

43

53

1111

1

11

1CCCC

.,,110

1212

43

22

etcas

= 12222

22

10012

CCR

= 32322 12 3122212 24222

25. 32112

22

22

2

2

2

1111

1

11

1CCCC

xxxx

x

x

x

0

011

1 222

xxxx

x

When x =0 0 is a root

26. nnnnnnn

nn

nn

nn

42223

2

2

2

11

11

110.111 3 nn = 0 answer D27.

28.

)(

05

05

CAnswer

yxx

yy

x

Level - II28 If A=

05y

xA and A = thenAT ,

(a) x=o. y = 5 (b) x + y = 5(c) x = y (d) none of these

29 If

126523

,652132

BA and A +B - C =0, then C =

(a)

778655

(b)

5344

11

(c)

776855

(d)

577855

30 If A and B are square matrices of order 3, then(A) adj (AB) = adjA + adj B (B) (A + B)1 = A1 + B1(C) AB = 0 |A| = 0 or |B| = 0 (d) AB = 0 |A| = 0 and |B| = 0

31. The inverse of the matrix iscb

aB

101001

(a) =

101001

cbaca (b) =

100001

cba

(c) =

101001

baca (d)

10010

1c

baca

32. The values of and for which the system of equations. zyxxyxzyx 2,1032,6 have no solution are

(A) 3 (B) 10 (C) 3 (D) 3

MQ solution

33. If rqp TTT ,, are the thth qp , and thr terms of an A.P ., then 11141 rqp

TTT rqp

cannot be equal to

(A) 1 (B) - 1 (C) 0 (D) p + q + r

34 Statement 1 : If A = 22

22

22

xcaxbcbxacaxbcxbxcabbxaccxabxa

and xabaxcbcx

B

, then .2BA


35 Let A be a 2 2 matrix with real entries. let I be the 2 2 identity matrix. Denote by tr(A), the sumof diagonal entries of A. Assume that A2 = I.Statement1: : If A I and A I, then det A = 1Statement2: : If A I and A I, then tr(A) 0.


III. For a given square matrix A, if there exists a matrix B such that AB = BA = I, then B is calledinverse of A. Every nonsingular square matrix passes inverse and it exists if |A| 0.

A1 = adj(A)det (A) adj A = |A| (A

1).

36 Let, a matrix A = 2 31 2

, then it will satisfy the equation

(A) A2 4A + I = 0 (B) A2 + 4A + I = 0(C) A2 4A 5I = 0 (D) A2 4A + 5I = 0

37 Let, a matrix A = 2 31 2

, then AA1 will be

(A) 2 31 2

(B) 3 2

2 1

(C) 1 2

2 3

(D) 2 3

1 2

38 Let matrix A = 3 21 1

satisfies the equation AA2 + aA + bI = 0, then the value of 4b

3

a

x . cosxdxequals

(A) a ba b

(B) a 2ba b

(C) a 4b4a b

(D) a 4b4a b


p q r s

p q r s

p

p

p

q

q

q

r

r

r

s

s

s

A

B

C

D

39. Let

1tan

tan1x

xA

COLUM I COLUM -II

(A) 1A (P)

1tan

tan1x

x

(B) 1AdjA (Q)

1tantan1

xx

(C) AdjAAdj (R)

xxxx2cos12sin

sin2cos121

(D) AAdj 2 (S)

xxxx2cos12sin

2sin2cos1

PQ solution

40 If A is

42476

268

is a singular matrix , then =

(A) 3 (B) 4(C) 2 (D) 5

41 If A = ,12100

,0201

B then

(A) AB = O, BA =O (B) OBAOAB ,(C) OBAOAB , (D) OBAOAB ,

42. If

5005

A and

dcba

B , then AB =

(A) B (B) 5B(C) 5B (D) none of these.

43 If A =

00

0

abacbc

A and

2

2

2

cbcacbcbabacaba

B , then AB =

(A) A (B) B(C) I (D) O

44. If F (X) =

yy

yyyGxx

xx

cos0sin010

sin0cos)(,

1000cossin0sincos

then [F(x). 1]yG is equal to

(A) F (X) G(-Y) (B) 11 yGxF(C) 11 xfyG (D) G (-y) F(-x)

45 IF

then,1sin1

sin1sin1sin1

lies in the interval

(A) [2,3] (B) [3,4](C) [2,4] (D) (2,4)

46 The determinant

cossincossincossin

2cossincos

is equal to

(A) (B) (C) and (D) Neither nor

47. If , CBA then the value of

0tan)cos(tan0sincossinsin

ABAAB

BCBA

is

(A) 0 (B) 1(C) 2 sinB tan A cosC (D) None of these.

48. Suppose x, y, z are positive and none of x,y, z is 1. If

thenzyxyxzxzy

yy

xx

,sincossinlog1logloglog1

2 is independent of

(A) x (B) y(C) y and z only (D) z

49. If

124

321612

33

22

nnznrrnnyr

nnxrSr , then the value

n

rrS

1 is independent of

(A) x (B) y(C) n (D) z

50 Statement 1 : If

012101210

, |A| = 0

BecauseStatement 2 : The value of the determinant of a skew symmetric matrix is zero


51. Statement 1 : The inverse of the matrix A=

113111

111

Does not exist

becauseStatement 2 : |A| =0


Paragraph Let

;..............;..............

22222121

11212111

bxaxaxabxaxaxa

nn

nn

nnnmnn bxaxaxa ....2211 be a system of n linear equations in n unknowns. Then this can bewritten in the matrix form as

AX = B Where A =

nn

ij

b

bbb

B

x

xxx

Xnma

.

.

.

.

.

.;][3

2

1

3

2

1

Then (I) If |A| 0, the system is consistent, and has a unique solution given by BAX 1 (II) If |A| =0 and (adj A) B =0, then the system is consistent and has infinitely many solutions. (III) If (A) = 0 and (adj A ) B 0, then the system is inconsistent.

52. The system of equations 1,52,132 zyxzyxzyx has(A) a unique solution (B) infinitely many solutions(C) no solutions (D) finite number of solutions.

53 Let 1023,1223,42 kzyxzyxzyx . The value of k in the above system ofequations so that system does not have a unique solution is(A) 2 (B) 3(C) -1 (D) -2

54. If yxzyxzyx 2,1032,6 , the values of and , for which the systemhas infinitely many solutions is(A) 9,3 (B) 10,3 (C) 10,2 (D) 3,10

ANSWERS

28(C) 29(B) 30 (C) 31(A) 32 (A)(B) 33 (A)(B) (D) 34 (A) 35 (C) 36 (A) 37 (A) 38(C)39 (A)-(R) ;(B)-(S); (C)-(P);(D)-(Q) 40 (A) 41(B)42 (B) 43 (D) 44(D) 45(C) 46 (C) 47 (A)48 (A)(B)(D) 49 (A) (B) (C) (D) 50 (C) 51 (A) 52 (A) 53 (B) 54(B) 55 (B) 56 (A) 57 (D)58 (C) 59(B) 60(A) (C) (D) 61 (C) 62(D) 63 (C) 64 (C) 65(A) 66 (A) (B)(C) (D)67(A) (B)(C) (D) 68 (B)69 (D) 70 (A)71(D)

ANSWERS & SOLUTIONS

LEVEL -II M.Q29. A+B - C = 0

)(0

50

5

BAnswer

yxx

yy

x

30. 00)( AABC or |B| =0

31. adj A =

101001

cbaca

|A| = 1 {Upper triangular matrix }

)(101001

1 AAnswercbac

aA

32. The required conditions are |A| = 0 and (adj A ) B=0

021

321111

and

000

106

110213

1262

i.e., 00362 and 0.6 - 10 + 0

10,3

33. ,1 dpaTp

Similarly, rqandTT

111

111rqp

dradqadpa

0111111111

rqpddd

rqprdqdpd

rqpaaa

)(),(),( DBAAnswer

34.

35 Answer (C)

36

2132

21322A

=

74127

43226634

20000

1001

1001

O

37.

AIAAIA

IAAAAAIAA

IAA

444

404

11

1121

2

2

2132

4004

2132

1001

41A

=

2132

38. Clearly a = -4, b = 1

4

4

3 0cos xdxx

Also 014444

4

ba

aba Answer (C)

39

1tantan1

)(1tan

tan1)()();();()();()(

xx

Aadjx

xA

QDCSBRA

xxxxxx

xx

xAAadjA 2

2

21

coscossincossincos

1tantan1

tan11

xxxx

2cos12sin2sin2cos1

21

RA )(

Ax

xAdjAAdj

1tantan1

xxxxA

AdjAAdjAAdjAdjA

2cos12sin2sincos1

21

sec21

PCandSB ;

QDxx

1tantan1

2

40 0A ( A is singular)

042

476268

A

)(3

6020020483612856

AAnswer

41.

002500

00000

BA

AB

Answer (B)

42.

dcba

AB5050

0505 Answer (B)

43. 0000000000

33

AB

Answer - (D)

44. ifxFyGyGxF 111 .)().({ 0.0 xGxf

Here , 1.1 yGxf

Also , xFxxxx

xF

1000cossin0sincos

1

and

yGyy

yxyG

cos0sin010

sin0cos1

xFyGygxf 111. xFyG 1

45.

)(:

4,2sin22

1sin1sinsinsinsin112

22

CAnswer

46

cossincossincossinsinsincossinsin

cossincossincossin

coscossincoscos 22

21

Apply 321 cos RRR 22111 sin inRRRand

)(000CAnswer

47.

0)tan(sincos)tan(cossin0tancos

tan0sincossin0

ABCACBAC

ABCB

Answer = (A)

48.

0

sin)cos(sinloglogloglogloglog

loglog1

2

zyxyxzyxzyx

yx Using

abba log

loglog

Answer (A) ,(B) ,(d)

49.

0113232

11

124

3216

12

33

22

3

1

3

2

1

2

1

1

nnznnnnynn

nnxnn

nnznrr

nnyr

nnxr

Sn

r

n

r

n

rn

rr

( as 1C and 3C are same)Answer (A) ,(B) (C) (D)

50. Because the value of the determinants of skew - symmetric matrix of odd orderis zeroAnswe (C)

51 Answer (A)

52

0113211212111211

312

AA The solution is unique

Answer (A)53. If the system does not have a unique solution the value of the determinant of coefficients =0

023

231112

k

54. The required conditions are 0A and (Adj A) B = 0

,021

321111

and

000

106

110213

1262

i.e., 00362 and 0.6 - 10 + 010,3 Answer - (B)

Level -IIIMQ55 If A is the diagonal matrix diag ( ndddd ..............., 321 ) then ,, NnAn is

(A) diag nndndndnd ........,, 321 (B) diag nnnnn dddd ....,.........,, 321(C) diag 1131211 ....,.........,, nnnnn dddd (D) none of these.

56 If

abba

1tantan1

1tantan1

, then

(A) a = 2sin,2cos ba (B) a =1, b= 1(C) 2cos,2sin ba (D) none of these.

57 If A and B are any aa matrices , then det (A+B) = 0 implies(A) det A + det B = 0 (B) det A = 0 or det B = 0(C) det A =0 and B = 0 (D) None of these.

58 If

23213

xxx

A is a symmetric matrix, then x =

(A) 4 (B) 3(C) -4 (D) -3

59 If

13321210

aA and 1A

2/12/32/534

2/12/12/1c then

(A) 2/1,2 ca (B) 1,1 ca(C) 1,1 ca (D) 2/1,2/1 ca

60. If f(x) is a polynomial satisfying

xf

xffxfxf

11

21

21

and f (2) = 17 , then f (5)

(A) 126 (B) 626(C) 124 (B) 624

61. The number of distinct real roots of 0sincoscoscossincoscoscossin

xxxxxxxxx

in the interval isx 440

(A) 0 (B) 2(C) 1 (D) 3

LEVE L IIIP.Q62 For what value of x, the matrix,

isx

xx

142142223

singular

(A) x =1,2 (B) x =0,2(C) x= 0,1 (D) x =0,3

63 If the trace of the matrix : A =

6402132114235201

2

2

xx

xx

is o then x is equal to

(A) (-2, 3) (B) (2,-3)(C) (-3, 2) (D) (3,-2)

64 If A and B are square matrices of order 3, then

(A) adj (AB) = adjA+adjB (B) 111 BABA(C) 0|| AOAB or |B| (D) OAOAB or B =O

65 If andA

496121132

thenB ,103122131

(A) AB = BA (B) BAAB

(C) BAAB 21

(D) none of these.

66 If A=

1000sincos0cossin

then which of the following are true ?

(A) 1TA (B) 11 A

(C) adjAA 1 (D) 1TAA

67 If matrix ,

bacacbcba

A where a.b. c are real positive number , abc = 1 and ,1AAT

then which of the following is/are true.(A) a + b+ c =1 (B) 1222 cba(C) ab + bc+ ca =0 (D) 4333 cba

68 Statement - 1 : If a, b, c are real positive numbers with abc = 1 and ,1TAA where A =

bacacbcba

, then 4333 cba because

Statement - 2 : 0111

cba(A) Statement -1 is true, statement=2 is true ; statement -2 is a correct explanation for statement -1 (B) Statement -1 is true, statement-2 is true; statement-2 is not a correct explanation for statement-1 (C) Statement -1 is true, statement- is false. (D) Statement-1 is false, statement -2 is true.

r us called the rank of the matrix. A if there exists at least one nonzero minor or order r and everyminor of order r +1 of the matrix equals zero. Again if A is a square matrix and I is the unit matrixof the same order then the equation |a-xI| = 0 is called the characteristic equation and the rootsof the characteristic equation are called eigenvalues of the matrix.A.

Let

442331

311A

69 The characteristic equation of A is(A) 045183 xxx (B) 0496 23 xxx(C) 04518 23 xxx (D) 08203 xx

70 The inverse of matrix A is

(A) 281

25 AI (B) AI 8

125

(C) 261

25 AI (D) None of these.

71 The sum of elements of the matrix A is(A) 5/3 (B) 5/4(C) 5/2 (D) None of these.

ANSWERS

55 (B)56 (A) 57 (D)58 (C)59 (B) 60(A) (C) (D) 61 (C) 62(D) 63 (C) 64 (C) 65(A) 66 (A) (B)(C) (D)67(A) (B)(C) (D) 68 (B)69 (D) 70 (A)71(D)

Level - IIIM.Q Solutions55. Given A = diag ndddd .......,, 321 Now AAA 2

nn d

dd

d

d

dd

d

.......000...............................................................0...........000.........000........00

.......000...............................................................0...........000.........000........00

3

2

1

3

2

1

56.

adjAA

A 11tan

tan1tan11

1tantan1 1

2

=

1tantan1

tan11

2

abba

1tantan1

1tantan1

abba

1tantan1

tan11

1tantan1

2

abba

2

2

2 tan1tan2tan2tan1

tan11

cos

tan1tan1

2

2

a and sin

tan1tan2

2 b

57. A is symmetric AA 1

23213

421323

xxx

xxxx

4321 xxxHence (c) is the correct answer

58. Det (A+B) cannot be expressed in terms of det A and det B . Hence the given equation givesno inference.Hence (d) is the correct answer

59. If we must have IAA 1

(3.1)th entry of 01 AA = (3,1)th entry of 1AA

02514

213 a

2121

210 c

044 a and c+1 =0 a =1 and c = - 1.Hence (b) is the correct answer

60.

x

fxfx

fxfxfx

fx

fxfxf 11112

nxxf 1Now, as f(2) = 17

62651 4 fxxfAnswer (A),(C) (D)

61

3211sinsincoscoscossincos

cos2sincos2sincos2sinRRRRgU

xxxxxx

xxxxxx

0cos2sincossin0cos

0cossincos001

xxxxx

xxx

322sin CCCgU and 133 CCC

0cos2sincossin 2 xxxx xx cossin 0cos2sin xx

only one solution.Answer ( C)

62. Since , the given matrix is singular

0142142223

xx

x

0142

110223

0142

2223

32

x

x

xxx

xRR

3,0033

0}2220413{

xxxxxxx

63 Trace of matrix is defined as 012222

1

xxan

iii

2,3 xHence (C) is the correct answer.

64. If AB=O then either of A and B are necessarily singular.Hence (C) is the correct answer.

65.

103122131

496121132

AB

.100010001

49612113201818043066

121863413621

AB

Hence (A) is the correct answer

66 TA transpose of A=

1000sincos0cossin

1sincoscossin

1000sincos0cossin

TA

Also, 1.A ASoAT

adjAAadjAA 1

1000sincos0cossin

1

adjAA .

11000sincos0cossin

1

A

67. Here ,

bacacbcba

A So,

bacacbcba

AT , interchanging rows and columns.

2

2

Abacacbcba

AAT

;2AAAT but IAAT (given)

Now, |I| = 22 1 AA

Now, bacacbcba

bacacbcba

A111

3211 RRRR

cbcacbabcbcba

bacacbcba

A

001

133

122

CCCCCC

}{ cabacbbccba bcabacabccbcba 222 2 abcabccbacba 222 abccba 3333

13333 abccba )1........(131 3332 cbaA

As a,b, c are positive, 133

333333

abccbacba

.3333 cba 13)1( 333 cba

4333 cba

71. Answer = (A)

73.The characteristic equation is 0442

331311

i.e., 08203

74. By Cayley- Hamilton theorem, 08203 IAA where 21

81

25 AIA

75. 21 81

25 AIA

4/14/14/14/34/14/5

2/313

2222622101284

81

100010001

25

ADDITIONAL QUESTIONS

72. If

cfgfbhgha

Then which of the following is are true ? where A,B,C,F,G,H are the

co factor & of a,b,c,f,g,h(A) aFBC 2 (B) bGCA 2

(C) cHAB 2 (D) None of these.

73 The value of

CABABCBCA

2sinsinsinsin2sinsinsinsin2sin

is

(A) 1 (B) 4 sin A. SinB . Sin C(C) 0 (D) None of these.

74. If

cfgfbhgha

& If A, B, C, F, G, H, are the co factors of a,b,c,f,g,h then which of the

following is (are) true ?(A) FAFGH (B) GBGHF(C) HCHFG (D) None of these

75. Let hgxfxexdxcxbxax 234567

thanxxxx

xxxxxxxxx

1221

21

22

22

22

(a) g=3 and h = -5 (b) g= -3 and h = -5 (c) g = -3 and h = -9 (d) none of these.

76.If ,111))()((

32

32

32

zyxxzxyyx

zzzyyyxxx

nnn

nnn

nnn

then n equals

(a) 1 (b) 2 (c) 3 (d) none o these

77 Let{ k .....,3,2,1 }be the set of third-order determinants that can be made with the distinct nonzero

real numbers 9,3,2,1 .....aaaa .Than

(a) k = 9! (b)

k

ii

10 (c) at least one 0i (d) none of these

78 Let ,21

)(

22

11

22

11

nn

nn

nn

nn

nn

nn

CCCPPP

nnnnf where the symbols have their usual meanings.

the )(nf is divisible by(a) 12 nn (b) (n + 1)! (c) n! (d) none of these.

79 if A+B+CiciAiB

iAiBiC

iBiCiA

i

eeeeeeeee

andzie2

2

2

sincos,

than

(a) Re(z) = 4 (b) Im(z) = 0 (c) Re(z) = - 4 (d) Im(z) = -1

COMPRIHENSIONLet A be a square matrix of order 2 ot 3 and I be the identity matrix of the same order then the

matrix A I is called charactristic matrix of the matrix A where is same complex no. The determi-nant of the characteristic matrix is called characteristic determinant of the matrix A which willof coursebe a polynomial of degree 3 in . The equation Det(A I ) = 0 is called characheristic equation ofthe matrix A and its roots(the values of ) are called characteristic roots or eigen values. It is alsoknown that every square matrix has its characteristic equation.

80. The eigen values of the matrix 211

432112

A are

(a) 2,1,1 (b) 2,3,-2 (c) -1,1,3 (d) none of these

81. which of the following matrices do not have eigen values as 1 and -1?

(a)

0110

(b)

00i

i (c)

1001

(d)

1001

82.If one of the eigen values of a square matrix A order 3 * 3 is zero then,(a) det A must be non -zero (b) det A must be zero (c) adj A must be a zore matrix (d) none of these

ANSWERS

72(A) (B)(C) 73 (C)74 (A)(B) (C) 75 (D) 76 (D) 77 (A) (B) 78(A) (C) 79 (B)(C)80 (C) 81(D) 82 (B)

72.

cfgfbhgha

if A,B,C,D,F,G,H are the cofactors of a,b.c,d,e,f,g,h then(i) ;,, 222 cHABbGCAaFBC(ii) hCHFGgBGHFfAFGH ,, .it will be sufficient to prove one of each set. Consider first

bGCA 2 .

Now 2GCA COGFIHGaA

CGGA

COGFIHGOA

cfgfbhgha

GCA )( 2

cCfFgGfcGfHgAfCbFhGbfGbHhAgChFaGhgGhHaA

2

bfo

obooh

Hence bGCA 2 .

73.

Consider the expression:

CxyBzxAyzCzByAxE sin2sin2sin22sin2sin2sin 222 We prove the result by showing that E can be expressed as the product of two factors linear in x,y,zNow in any triangle ABC

kc

Cb

Ba

A

sinsinsin

using this property and the identity cos.sin22sin we obtain

cxybzxayzCcxBbyAaxk

E coscoscos

2222

Again

BaAbcAcCabBcCba

coscoscoscoscoscos

zxAcCayzBcCbCcxBbyAaxk

E coscoscoscoscoscoscos2

222

xyBaAb )coscos(

)coscoscos)(( CzByAxczbyax Hece E can be written as the product of two factors linear in x,y,z. Thuis

02sinsinsin

sin2sinsinsinsin2sin

CABABCBCA

74Next consider . gBGFH

Now CFFBGH

FBGH

BGHF100

Hence CFFBGH

cfgfbhgha

BGHF100

)(

cCfFgGcFfBgHcfCbFhGfFbBhHfgChFaGgFHbaHg

2

0000

gcfg

Hence gBGHF

Similarly, fAFGH and hCHFG

75 (d) : By putting x = 0 an both sides of the equation we have

g = 9102210021 differentiaing both sides and then putting x =0, we get f= -5.

76.(d): Degree of L.H.S. = n+n +2+n+3 and that of R.H.S.= 2 1 n

77. The number of third-order determinant = the number of arrangements of nine different numbers in nineplaces =9!. Corresponding to each determinant made, there is a determinant obtained by interchangingtwo consecutive rows (or columns). so, the sum of this pair will be 0.

the sum of all the determinants 02!9.....000 timesto

78. 00.1

!1.1!.!11

111!2!1!

21)(

nnnnnn

nnnnnn

nf

(using 122233 , CCCCCC )

}.1{!!!11 2 nnnnnnn

79.

iCCAiCBi

BAiiBBCi

ABiACiiA

iCiBiA

eeeeeeeee

eeez

because

,1sincos)( iee iCBAi etc.

iCiBiA

iA

iB

eeee

e

002020

(using 322311 , RRRRRR )

.444}2{2 )()( iCBAiCAiiB eeee

80.

211432112

A

000000

I

211432112

IA

)3)(1)(1()det( IA

this the characteristic roots are -1,1&3.

81. (a) is not correct since its characteristic determinant

11

characteristic equation is 012 1,1 eigen values are 1 & -1we similarly note that matrices given in choice (b) and (b) and (c) have eigen values 1 and -1.

they are not correct. (d) has characteristic equation 2)1( = 0 eigen values are not 1 & -1 choice (d) is correct

82. Let A=

333

222

111

cbacbacba

A than

333

222

111

cbacbacba

IA

]))()[(()det( 23321 cbcbaIA )].([])[( 233213231 babaccccbnow one of the eigen values is zero, one root of should be zero constant term in the above polynomial is zero

0231321321321321321 bacaaaacbcabcbacba (collecting constant terms).But this value is value of determinant of A. det A = 0A should be non-singular..

SUBJECTIVE QUESTIONS 83. If A is non-singular matrix, then show that adj (adj A) = AA n 1

84. If A is a square matrix of order n, prone that 21)( nAadjAadj

85. If A and B are two square matrixs. such that BAAB 1 , then prove that 222)( BABA

86. Find the value of )( 1p interms of p where p is non-singular matrix and hence show that

PAQBPQadj )( 11 given taht 1& QPAB

87. If hgfcd

aB

cbb

aA 0

11,

11

01

zyx

Xa

Vhgf

U ,00,

2

and AX = U has infinitely many solutions, prove that if 0agf , then BX = V has no solution.

88. find the invrse of the matrix

011101110

S and show that 1SAS is a diagonal matrix

where

bacacbbaacbcabaccb

A21

89 If A & B are different matrices satisfying ABBABA 2233 & Find 22 BA 90 Prove that If A is nn matrix with real entries then get 02 nFA91 Show that if A & B are nn matrices with realentries and AB = nO then get

0222 BAI Pn for any positive integer P & Q92 Let A,B, C be nn real matrices that are pairwise communthetics nOCBA ,, Prove that

Get getCBA 333 (A + B+ C) 093 Let A,B be two & square matrices such that A+B = AB . Prove that AB =BA

94. Show that

2

sin2

sin2

sin2

sin2

sin2

sinsinsinsin23sin2sinsin3sin2sinsin3sin2sinsin

6

95. Find all the values of t for which the system of equations.02)13()1( tzytxt 0)3()24()1( ztytxt 0)1(3)13(2 ztytx

has a non-trivial solution.

96. Prove that

3

222

222

222

)(2 cbaabcbacc

bacbaacb

97 Show that

)(2sin

2sin0sinsincossin

cossincos

2

2

yxyxyxyxyxayxyxyx

98. Evaluate the determinant1111

23

23

23

23

daacaabaaaaa

99. solve

0333

333

222222

cxbxaxcxbxaxcxbxax

100. show that -(a+b+c) is one root of the equation

0

bxacacxbcbax

and solve the equation completely..

101. solve the equation ccacbbabxxax

233

233

233

= 0 where b, c are unequal.

102. Solve the equation 0

4321432143214321

xx

xx

ANSWERS

86

Solution:83. Let nIBadjBB )( Replacing B by adj AA

(adj A)(adj (adj A)) = nIadjA or nn IAadjAadjadjA 1))()(( or n

n IAAadjAadjadjAA 1))()((

or nn AIAadjAadjA 1))(( or AAadjAadjA n 1))(( or AAadjAadj n 2)(

84. adjAadjAadj )( 1)( adjA = 1111111 AAAAAAAA n }{ AkkAas n AA n 2

AAAas 111

85. BAAB 1 BABAAAAB 1 0 BAAB and

22222 ))(()( BABBAABABABABA Hence Proved.

86. 1 XXadjX 1111 PPPadj }1{ 11P

PPas 1111 adjPPP

Also 1111111 )( BPQBPQBPQadj 11111111 QBPPPBQ

= QPBPBQ 111 }1{ PQas PAQQadjBPQBBP )(1 {as adj B = A}A}

87. Consider[A:U]=

hcbgdbfa

11

01

applying 233 RRR

ghdcbgdbfa

11

01

AX =U has infinitely many solutions c= d and h = g ...........(i)

it is obvious that if (i) holds then, B = 0 BX = V has no unique solution.

Now consider [B:V] =

000

11 2

hgfcd

aa

apply 133 RafRR

=

afafh

afgf

cdaa00

11 2

apply 2331 R

afg

dRR

=

afafg

dc

afh

cdaa

00

0011 2

faafd ,0 and d are nonzero af is nonzero

Also,as (i) holds 0

afg

dc

afh BX = V has no solution if 0afd .

88. 2s and adj

111111111

21

111111111

1SS

also,

022202220

21

011101110

21

ccbbaa

bacacbbaacbcabaccb

SA

and

cb

a

cb

a

ccbbaa

SAS00

0000

400040004

41

111111111

022202220

411

which is a diagonal matrix.

89. We have 322322 ))(( BABBAABABA . since BA , this shows that 22 BA has a zero divisor. Hence it is not invertible, so its determinant is 0.

90. Write nnn iIAiIAIA 2 , where i is the imaginary unit. If n ......., 21 are the eigenvlues ofA, then the eigen values of niIA are iii n ......., 21 ,Hence

d e t )).......((),( 212 iiiiIA nn . S i m i l a r l y

det )).......((),( 212 iiiiIA nn

. Subce A has reak entries, its complex eigen values come in pairs of conjugate

numbers. By using the formulas aibaibiaibia 2)1())(( 22 and

aibaibiaibia 2)1())(( 22

, we see that det niIA and det niIA can be written as products of terms that are complex conjugate of each other. Hence the determinats themselves are complex conjugates of

each other, which implies that their product is nonnegative real number.

91. By the previous problem, we have det 02 pn AI and det 02 qn BI . from nAB 0 We ob

tain nqp BA 022 .thus )det()det( 222222 qpqpn

qpn BABAIBAI

)))(det(( 22 qnp

n BIAI

)det()det( 22 qnp

n BIAI

92. Let 1 be a third root of unit. Since A,B,C commute and nABC 0 , we can write, ABCCBACBA 3333333

)( 222 CABCABCBACBA ))(( 22 CBACBACBA ))(( 22 CBACBACBA

Hence CBACBA det)det( 333

)det()det()det( 222 CBACBACBA

)det()det()(det 222 CBACBACBA .0)det()(det 222 CBACBA

93. Since ,0nBAAB by adding nI to both sides and factoring we obtain nnn IBIAI . follows that AIn is invertible, and its inverse is BIn . Hence nnn IBIAI , which implies ,0nBAAB . Consequently, ABBABA .

94. )1cos4(sincossin2sin)1cos4(sincossin2sin)1cos4(sincossin2sin

2

2

2

3sin4sin3,3sin

)]cos1(43[sin 2 sin)1cos4( 2

Take sin , sin , sin common from 321, andRRR .

sin sin sin1cos4cos211cos4cos211cos4cos21

2

2

2

=8 sin sin sin

2

2

2

coscos1coscos1coscos1

95. For non-triival solution 0)1(3132

32412131

ttttt

ttt

303

3302131

tttt

ttt

)(

133

122

RRRRRR

0

3011102131

)3( 2

t

tttt

0)]1(2)1)(13()1)(1[()3( 2 tttt 0]2131[)3( 3 tttt 0 t or 3t96. Subtracting the second column from the third and the first from the second

222

22222

22

)(0)()(

0)(

cbacacbbacb

acb

))((0))(())((

0))(()(

2

2

2

cbacbaccbacbacbabacb

cbacbacb

cbaccbacbab

cbacbcba

0

0)()(

2

2

2

2

subtract from the first row the sum of the second and third rows. then

cbaccbacbabcbcbc

cba

0

2222)(

2

22

cbaccbacbab

cbcbccba

0)(2

2

22

Now add the second column on to the third and we have

cbac

cbabbcbc

cba

00)(2

2

22

maltiply the third column by c and add to the first.

cbabaccbab

bccba

0)(0

0)(2 22

Finding along the frist column.

)})(()({)(2 22 cbabaccbacbcba

)})(()({)(2 2 cbabacbabcbabc 3)(2 cbaabc

97. Subtracting the third column from the first,

yyxyxyxyxyxyx

2sin02sin2sin)sin()cos(0)cos()sin()cos(2

yyxyxyxyxyxyxyx

2sin0)sin()cos(2)sin()cos(0)cos()sin()cos(2

yyxyxyxyxyx

yx2sin0)sin(

)sin()cos(0)cos()sin(1

)cos(2

multiply the first row by sin(x-y) and subtract from the third, then

)sin()sin(2sin)sin()sin(0

)sin()cos(0)cos()sin(1

)cos(2yxyxyyxyx

yxyxyxyx

yx

)cos()sin()sin()sin(0)sin()cos(0

)cos()sin(1)cos(2

yxyxyxyxyxyx

yxyxyx

)cos()sin()sin()cos(

)sin()cos(2yxyxyxyx

yxyx

).(2sin)}(sin)(){cos(2sin 22 yxyxyxyx

98. Subtract the first row from the second, third and fourth rows and them expand along the fourth column. thus

adadabacacacababab

adadabacacacababab

aaa

2233

2233

2233

2233

2233

2233

23

0001

111

))()((22

22

22

dadadacacacabababa

adacab

Now subtract the first row from the second and third fourth rows and then expand along the fourth . Thus

001

))()((22

22

22

bdadadbdcaabacbcbababa

adacab

bddbabdbccbabc

adacab

)()()()(

))()((

= IdbaIcba

bdbcadacab

))()()()((

).)()()()()(( dcdbcbdacaba 99. For x where a,b,c, are unequal. adding the second row to the third.

)3(2)3(2)3(2

)()()(222222

333

222222

cxxbxxaxxcxbxaxcxbxax

222222

333

222222

333)()()(2cxbxax

cxbxaxcxbxax

x

adding the first row to the third,

222222

333

222222

)()()(4cxbxax

cxbxaxcxbxax

x

222222

2222223

222222

)333)(()333)(()(4accbax

aaccbxaxxbaaabbbxaxxbaaxcabaax

x

)()(

333333)())((422

2222223

22

acbaaxaabcbxaxxaabbbxaxxax

cabaaxcabax

00333333)())((4

22

2222223

22

axaabcbxaxxaabbbxaxxax

cabaaxcabax

2222223

333333))((8

aabcbxaxxaabbbxaxxacba

cabax

acbabccxbxaabbbxaxxbcba

cabax

22222

3

33333))((8

cbaxaabbbxaxxba

bccabax

33331

))()((8 2223

).3)()()((8 23 cabcabxbccabax

100. Adding the second and third rows to the first,

bxac

acxbcbaxcbaxcbax

bxac

acxbcbax

111

Hence x = - (a+b+c) is one solution of the equation. Now subtract the first column from the second and fromthe third. we have

cbxcacbabcxbcbax

001

))(( 222 cabcabbaxcbax

101. Subtract the first row from the second and third. then

xcxcaxxbxbax

xxax

2222

2233

233

11))((

22

22

233

xcxcxcxbxbxb

xxaxxcxb

subtract the second row from the third.

01))((

22

22

233

xcbxcxbcxbxbxb

xxaxxcxb

011))()(( 22

233

xcbxbxbxb

xxaxbcxcxb

Expanding along the third row,

)}())(){()()(( 32233 xbxxbaxbxxcbbcxcxb

= ).)()()(( 3 bcxabcxcxb Hence the solutions of 0 are cbx , or ./3 bca

102. Subtract the second row from the first and from the thirds and fourth. Then

xx

xxx

xx

0000432100

1010011043210011

3

x

x

Now subtract the first rowfromthe second. then

1010011043300011

3

x

x 101011433

3

xx

And the first column to the third and expand along the third row. thus

101111

7333

xxx = )10()73( 33 xxxx

Hence x= 0,0,0 or -10.

94. Statement 1 : The order of the matrix A is 35 and that of the matrix B is 54 , then the product BA is not possible

becauseStatement 2 : Number of columns of A number of rows of B.


ComprehensionLet A be a square matrix of order 2 or 3 and I be the identity matrix of the same order, then the matrix IA is called characteristic matrix of the matrix A, where is some complex number. The determinant of the characteristic matrix is called characteristic determinant of the matrix A which will of course be a polynomial of degree 3 in The equation det 0)( IA is called characteristic equation of the matrix A and its roots arecalled characteristic roots or eigen values. It is known that every square matrix satisfied itscharacteristic equation.

95 If one of the eigen values of a square matrix A or order 33 is zero , then(A) det A must be non -zero (B) det A must be zero(C) adj A must be a zero matrix (D) none of these.

96 If 'A denotes transpose of matrix A, A A = 1 and det A = 1, Then det (A-I) must be equalto(A) 0 (B) -1(C) 1 (D) none of these.

97 If A=

100010001

420110001

I and dIcAAA 21 61

, then the value of c and d are

(A) - 6, - 11 (B) 6, 11(C) -6, 11 (D) 6, -11

98 .If

0333

333

222222

cxbxaxcxbxaxcxbxax

Then value of x are where a,b, c, are unequal

(A) 0 (B) 2

(C) 3 (D) cabcab

31

1 EXTRA QUESTION

cossinsincos

cossinsincos

cossinsincos

2A

A

2cos2sinsin2cos

cossincos.sinsin.coscos.sinsin.cossincos

2

2222

2

cossinsincos

2cos2sin2sin2cos23

AAA

coscossin.sinsin.coscos.sincos.sinsin.cossin.sincos.2cos

2222

22

coscossin.sinsin.coscos.sincos.sinsin.cossin.sincos.2cos

2222

22

matrices and determinants

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