Matrices notes and ncert solution class 12 cbse

5/27/2018 Matrices notes and ncert solution class 12 cbse

12 Math Ch-3 Matrices

GenextStudents 1

Chapter 3

Matrices

Exercise 3.1

Q 1. In a matrix

2 5 19 7

535 2 12

2

3 1 5 17

A

=

, write:

(i) The order of the matrix (ii) The number of elements,

(iii) Write the elements 13 21 33 24 23, , , ,a a a a a .

Ans.: (i) 3 4

(ii) 12

(iii) 13a = 19, 21a = 35, 33a = 5, 24a = 12, 23a =5

2

Q 2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13

elements?

Ans.: Possible order of matrix

1 24, 2 12, 3 8, 4 6, 6 4, 12 2, 24 1, 8 3

Possible order for 13 elements 1 13, 13 1

Q 3. If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5

elements?

Ans.: Possible order for 18 elements

1 8, 2 9, 3 6, 6 3, 9 2, 18 1

Possible order for 5 elements = 1 5, 5 1

Q 4. Construct a 2 2 matrix, A = [a ij], whose elements are given by:

(i)

2( )

2ij

i ja

+= (ii) ij

ia

j= (iii)

2( 2 )

2ij

i ja

+=

Ans.: A = [a ij]



GenextStudents 2

(i)

2( )

2ij

i ja

+=

A =

11 12

21 22

a a

a a

=

922

9 22

(ii) iji

aj

=

A =

11 12

21 22

a a

a a

=11

2

2 1

(iii)

2( 2 )

2ij

i ja

+=

A =11 12

21 22

a a

a a

=9 25

2 2

8 18

Q 5. Construct a 3 4 matrix, whose elements are given by:

(i)1

| 3 |2

ija i j= + (ii) 2ija i j=

Ans.: (i) 1 | 3 |2

ija i j= +

A =

11 12 13 14

21 22 23 24

31 32 33 34

a a a a

a a a a

a a a a



GenextStudents 3

=

1 11 0

2 2

5 32 1

2 2

7 54 3

2 2

(ii) 2ija i j=

A =

11 12 13 14

21 22 23 24

31 32 33 34

a a a a

a a a a

a a a a

=1 0 1 23 2 1 0

5 4 3 2

Q 6. Find the values ofx ,y and z from the following equations:

(i)4 3

5 1 5

y z

x

=

(ii)

2 6 2

5 5 8

x y

z xy

+ = +

(ii)

9

5

7

x y z

x z

y z

+ + + = +

Ans.: (i) 4 35 1 5

y zx

=

4y= , 3z= , 1x=

(ii)2 6 2

5 5 8

x y

z xy

+ = +

6x y+ = , 5 5z+ = 8xy=

86x x+ = 0z=

8

yx=

2 8 6x x+ =

2 6 8 0x x + =

2 2 4 8 0x x x + = ( 2) 4( 2) 0x x x =



GenextStudents 4

If 4x= , 2y=

2x= 4y=

(iii)95

7

x y zx z

y z

+ + + = +

5x y+ =

_ 7y z+ =

2x y =

From (1) & (2)

5 9y+ =

4y=

7y z+ =

3z=

5x yz+ =

3 5x + = 2x=

2x= , 4y= , 3z=

Q 7. Find the value ofa , b , c and d from the equation:

2 1 5

2 3 0 13

a b a c

a b c d

+ = +

Ans.:2

2 3

a b a c

a b c d

+ +

=1 5

0 13

1a b = (1)

2 5a c+ = ...(2)

2 0a b = (3)

3 13c d+ = .(4)



GenextStudents 5

Solve (1) & (3)

2 2 2a b =

_ 2 _ _ 0a b+ =

2b =

2b=

2 1a =

1 2 1a= + =

2 (1) + c = 5

C = 3

9 + d = 13

d = 4

Q 8. ij m nA a

= is a square matrix, if

(A) m n< (B) m n> (C) m n= (D) None of these

Ans.: ij m nA a

= , A is square matrix if m n=

Q 9. Which of the given values of x and y make the following pair of matrices equal

3 7 5

1 2 3

x

y x

+ +

(A)1

, 73

x y

= = (B) Not possible to find

(C)2

7,3

y x

= = (D)1 2

,3 3

x y

= =

Ans.: (B) Not possible to find.

Q 10. The number of all possible matrices of order 3 3 with each entry 0 or 1 is:

(A) 27 (B) 18 (C) 81 (D) 512

Ans.: No. of elements are 9 and each element can be filled by 0 or 1.



GenextStudents 6

Total no. of elements = 29= 512

D is correct



GenextStudents 7

Exercise 3.2

Q 1. In the matrix2 4

A3 2

=

,

1 3B

2 5

=

,2 5

C3 4

=

Find each of the following:

(i) A + B (ii) A B (iii) 3A C (iv) AB (v) BA

Ans.:2 4

A3 2

=

,

1 3B

2 5

=

,2 5

C3 4

=

(i) A + B =2 4 1 3

3 2 2 5

+

=2 1 4 3

3 2 2 5

+ + +

=3 7

1 7

(ii) A B =2 4 1 3

3 2 2 5

=2 1 4 3

3 ( 2) 2 5

=1 1

5 3

(iii)3A C = 3A =2 4

33 2

=3 2 4 3

3 3 3 2

=6 12

9 6

3A C =6 12 2 5

9 6 3 4

=6 2 12 5

9 3 6 4

+



GenextStudents 8

=8 7

6 2

(iv) AB =

2 4 1 3

3 2 2 5

=2(1) 4( 2) 2(3) 4(5)

3(1) 2( 2) 3(3) 2(5)

+ + + +

=2 8 6 20

3 4 9 10

+ +

=6 26

1 19

(v) BA =1 3 2 4

2 5 3 2

=1(2) 3(3) 1(4) 3(2)

2(2) 5(3) 2(4) 5(2)

+ + + +

=2 9 4 6

4 15 8 10

+ +

+ +

=11 10

11 2

Q 2. Compute the following:

(i)a b a b

b a b a

+

(ii)

2 2 2 2

2 2 2 2

2 2

2 2

ab bca b b c

ac aba c a b

+ + + + +

(iii)

1 4 6 12 7 6

8 5 16 8 0 5

2 8 5 3 2 4

+

(iv)

2 2 2 2

2 2 2 2

cos sin sin cos

sin cos cos sin

x x x x

x x x x

+

Ans.: (i)a b a b

b a b a

+



GenextStudents 9

=a a b b

b b a a

+ + + +

=

2 2

0 2

a b

a

(ii)

2 2 2 2

2 2 2 2

2 2

2 2

ab bca b b c

ac aba c a b

+ + + + +

=

2 2 2 2

2 2 2 2

( ) 2 ( ) 2

( ) 2 ( 2 )

a b ab b c bc

a b ac a b ab

+ + + +

+ +

=

2 2

2 2

( ) ( )

( ) ( )

a b b c

a c a b

+ +

(iii)

1 4 6 12 7 6

8 5 16 8 0 5

2 8 5 3 2 4

+

=

1 12 4 7 6 6

8 8 5 0 16 5

2 3 8 2 5 4

+ + + + + + + + +

=

11 11 0

16 5 21

5 10 9

(iv)

2 2 2 2

2 2 2 2

cos sin sin cos

sin cos cos sin

x x x x

x x x x

+

=

2 2 2 2

2 2 2 2

cos sin sin cos

sin cos cos sin

x x x x

x x x x

+ +

+ +

=1 1

1 1

Q 3. Compute the indicated products:



GenextStudents 10

(i)a b a b

b a b a

+

(ii) [ ]

1

2 2 3 4

3

(iii)1 2 1 2 3

2 3 2 3 1

(iv)

2 3 4 1 3 5

3 4 5 0 2 4

4 5 6 3 0 5

(v)

2 11 0 1

3 21 2 1

1 1

(vi)

2 33 1 3

1 01 0 2

3 1

Ans.: (i)a b a b

b a b a

+

=( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

a a b b a b b a

a b b a b b a a

+ + + +

=

2 2

2 2

a b ab ab

ab ab b a

+ +

+ + +

=

2 2

2 2

0

0

a b

a b

+

+

(ii) [ ]

1

2 2 3 4

3

=

1(2) 1(3) 1(4)

2(2) 2(3) 2(4)

3(2) 3(3) 3(4)

=

2 3 4

4 6 8

6 9 12

(iii)1 2 1 2 3

2 3 2 3 1

=1(1) 2(2) 1(2) 2(3) 1(3) 2(1)

2(1) 3(2) 2(2) 3(3) 2(3) 3(1)

+ + + +



GenextStudents 11

=1 4 2 6 3 2

2 6 4 9 6 3

+ + +

=

3 4 1

8 13 9

(iv)

2 3 4 1 3 5

3 4 5 0 2 4

4 5 6 3 0 5

=

2(1) 3(0) 4(3) 2( 3) 3(2) 4(0) 2(5) 3(4) 4(5)

3(1) 4(0) 5(3) 3( 3) 4(2) 5(0) 3(5) 4(4) 5(5)

4(1) 5(0) 6(3) 4( 3) 5(2) 6(0) 4(5) 5(4) 6(5)

+ + + + + + + + + + + + + + + + + +

=

2 0 12 6 6 0 10 12 20

3 0 15 9 8 0 15 16 25

4 0 18 12 10 0 20 20 30

+ + + + + + + + + + + + + + + + + +

=

14 0 42

18 1 56

22 2 70

(v)

2 11 0 1

3 21 2 1

1 1

=

2(1) 1( 1) 2(0) 1(2) 2(1) 1(1)

3(1) 2( 1) 3(0) 2(2) 3(1) 2(1)

1(1) 1( 1) 1(0) 1(2) 1(1) 1(1)

+ + + + + + + + +

=

2 1 0 2 2 1

3 2 0 4 3 2

1 1 0 2 1 1

+ + + + + +

=

1 2 3

1 4 5

2 2 0



GenextStudents 12

(vi)

2 33 1 3

1 01 0 2

3 1

=3(2) ( 1)(1) 3(3) 3( 3) 1(0) 3(1)

1(2) 0(1) 2(3) 1( 3) 0(0) 2(1)

+ + + + + + +

=6 1 9 9 0 3

2 0 6 3 0 2

+ + + + + + +

=14 6

4 5

Q 4. If

1 2 3

5 0 2

1 1 1

A

=

,

3 1 2

4 2 5

2 0 3

B

=

and

4 1 2

0 3 2

1 2 3

C

=

, the compute (A + B) and (B

C). Also, verify that ( ) ( )A B C A B C+ = + .

Ans.: Given

1 2 3

5 0 2

1 1 1

A

=

,

3 1 2

4 2 5

2 0 3

B

=

,

4 1 2

0 3 2

1 2 3

C

=

(A + B) =

1 2 3

5 0 2

1 1 1

+

3 1 2

4 2 5

2 0 3

=

1 3 2 1 3 2

5 4 0 2 2 5

1 2 1 0 1 3

+ + + + + + + +

=

4 1 1

9 2 7

3 1 4

(B C) =

3 1 2

4 2 5

2 0 3

4 1 2

0 3 2

1 2 3



GenextStudents 13

=

3 4 1 1 2 2

4 0 2 3 5 2

2 1 0 2 3 3

+

=

1 2 0

4 1 3

1 2 0

A + (B C) =

1 2 3

5 0 2

1 1 1

+

1 2 0

4 1 3

1 2 0

=1 1 2 2 3 05 4 0 1 2 3

1 1 1 2 1 0

+ + + + + +

=0 0 39 1 5

2 1 1

(A + B) C =

4 1 1

9 2 7

3 1 4

4 1 2

0 3 2

1 2 3

=

4 1 2

0 3 21 2 3

=

0 0 3

9 1 52 1 1

Hence A + (B C) = (A + B) C Verfied.

Q 5. If

2 51

3 3

1 2 4

3 3 3

7 22

3 3

A

=

and B =

2 31

5 5

1 2 4

5 5 5

7 6 2

5 5 5

, then compute 3A 5B



GenextStudents 14

Ans.:

2 51

3 3

1 2 4

3 3 3

7 22

3 3

A

=

& B =

2 31

5 5

1 2 4

5 5 5

7 6 2

5 5 5

3A =

2 53 3 1 3

3 3

1 2 43 3 3

3 3 3

7 23 3 2 3

3 3

=

2 2 5

1 2 4

7 6 2

5B =

2 35 5 5 1

5 5

1 2 45 5 5

5 5 5

7 6 25 5 55 5 5

=

2 3 5

1 2 4

7 6 2

3A 5B =

0 0 0

0 0 0

0 0 0

Q 6. Simplifycos sin sin cos

cos sinsin cos cos sin

+

.

Ans.:cos sin

cossin cos

=

2

2

cos cos sin

cos sin cos



GenextStudents 15

sin cossin

cos sin

=

2

2

sin cos sin

sin cos sin

2

2cos cos sincos sin cos

+

2

2sin cos sincos sin sin

1 0

0 1

Q 7. Find X and Y, if

(i)7 0

2 5X Y

+ =

and

3 0X Y

0 3

=

(ii)2 3

2X 3Y4 0

+ =

and

2 23X 2Y

1 5

+ =

Ans.: (i)7 0

2 5X Y

+ =

&

3 0X Y

0 3

=

(X + Y) + (X Y) = 2X

=

7 3 0 0

2 0 5 3

+ +

+ + =

10 0

2 8

X =

10 0

2 2

2 8

2 2

=5 0

1 4

Let Y =1 2

3 4

y y

y y

=5 0

1 4

+1 2

3 4

y y

y y

=7 0

2 5

=7 0

2 5

5 0

1 4

=1 2

3 4

y y

y y



GenextStudents 16

=7 5 0 0

2 1 5 4

=1 2

3 4

y y

y y

=2 0

1 1

= 1 23 4

y y

y y

2 0

1 1Y

=

&

5 0

1 4X

=

(ii) 2X + 3Y =2 3

4 3

and 3 2X Y+ =2 2

1 5

2(2 3 )X Y+ =

4 6

8 0

3(3 2 )X Y+ =6 6

3 15

4X + 6Y =4 6

8 0

& 9X + 6Y =6 6

3 15

(9 6 ) (4 6 )X Y X Y+ + = 5X =6 6 4 6

3 15 8 0

5X =6 4 6 6

3 8 15 0

=2 12

11 15

X =

2 12

5 5

113

5

2X + 3Y =2 3

4 0



GenextStudents 17

3Y =

2 122 3 5 5

24 0 11

35

=

4 242 3 5 5

4 0 226

5

=

4 242 3

5 5

224 0 6

5

+

+

=

6 39

5 5

426

5

Y =

2 13

5 5

142

5

Q 8. Find X, if Y =3 2

1 4

and 2X + Y =1 0

3 2

Ans.: Y =3 2

1 4

, 2X + Y =1 0

3 2

, X = ?

2X =1 0

3 2

3 2

1 4

=1 3 0 2

3 1 2 4

=2 2

4 2



GenextStudents 18

X =1 1

2 1

Q 9. Ifx and y, if

1 3 0 5 6

2 0 1 2 1 8

y

x

+ = ?

Ans.:1 3 0

20 1 2

y

x

+

=

5 6

1 8

2 6 0 5 6

0 2 1 2 1 8

y

x

+ =

2 5y+ =

6 0 6+ =

0 1 1+ =

2 2 8x + =

2 5y+ =

3y=

2 6x=

3x=

3x y= =

Q 10. Solve the equation for , ,x y z and t, if1 1 3 5

2 3 30 2 4 6

x z

y t

+ =

Ans.:1 1

2 30 2

x z

y t

+

=3 5

34 6

2 2 3 3

2 2 0 6

x z

y t

+

=

9 15

12 18

2 3 9x + =

2 3 15x =



GenextStudents 19

2 0 12y + =

2 6 18t+ =

2 6x= 3x=

2 18z= 9z=

2 12y= 6y=

2 12t= 6t=

Q 11. If2 1 10

3 1 5x y

+ =

, find the value ofx andy .

Ans.:2 1 10

3 1 5x y + =

2 10

3 5

x y

x y

+ =

2 10x y =

_ 3 5x y+ =

5 15x=

3x=

6 10y =

6 10 y =

4y=

Q 12. Given 3x y

z w

=6 4

1 2 3

x x y

w z w

+ + +

, find the values of , ,x y z and w .

Ans.:3 3 6 4

3 3 1 2 3

x y x x y

z w w z w

+ = + +

4 3x x+ =



GenextStudents 20

6 3x y y+ + =

1 3w y z + + =

2 3 3w w+ =

2 4x= 2x=

6 2 3y y+ + =

8 2y= 4y=

w = 3

1 3 z + + = 3z

2 = 2z

3 =1

2x= , 4x= , 1z= & 3w=

Q 13. If

cos sin 0

( ) sin cos 0

0 0 1

x x

F x x x

=

, show that ( ) ( ) ( )F x F y F x y= + .

Ans.:

cos sin 0

( ) sin cos 0

0 0 1

x x

F x x x

=

, ( ) ( ) ( )F x F y F x y= +

cos sin 0

( ) sin cos 0

0 0 1

y y

F y y y

=

( ) ( )f x f y =

cos sin 0 cos sin 0

sin cos 0 sin cos 0

0 0 1 0 0 1

x x y y

x x y y

=

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

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

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

x y x y x y x y

x y x y x y x y

y y y y

+ + + + + + + + + + + +



GenextStudents 21

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

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

0(0) 0(0) 1(1)

x

x x

+ + +

+ +

=

cos cos sin sin cos sin sin cos 0

sin cos cos cos cos cos sin sin 0

0 0 1

x y x y x y x y

x y x y x y x y

+

=

cos( ) sin( ) 0

cos( ) cos( ) 0

0 0 1

x y x y

x y x y

+ + +

= ( )F x y+ = ( ) ( )f x f y+

=

cos( ) sin( ) 0

sin( ) cos( ) 0

0 0 1

x y x y

x y x y

+ + + +

( ) ( ) ( )f x y f x f y+ =

Q 14. Show that ;

(i)5 1 2 1

6 7 3 4

2 1 5 1

3 4 6 7

(ii)

1 2 3 1 1 0

0 1 0 0 1 1

1 1 0 2 3 4

1 1 0

0 1 1

2 3 4

1 2 3

0 1 0

1 1 0

Ans.: (i)5 1 2 1

6 7 3 4

2 1 5 1

3 4 6 7

5 1 2 1

6 7 3 4

=5(2) ( 1)(3) 5(1) 1(4)

6(2) 7(3) 6(1) 7(4)

+ + +

=10 3 5 4

12 21 6 28

+ +

=7 1

3 34



GenextStudents 22

2 1 5 1

3 4 6 7

=2(5) 1(6) 2( 1) 1(7)

3(5) 4(6) 3( 1) 4(7)

+ + + +

=

10 6 2 7

15 24 3 28

+ +

+ + =

16 5

39 25

Hence proved

(ii)

1 2 3 1 1 0

0 1 0 0 1 1

1 1 0 2 3 4

1 1 0

0 1 1

2 3 4

1 2 3

0 1 0

1 1 0

LHS

1( 1) 2(0) 3(2) 1(1) 2( 1) 3(3) 1(0) 2(1) 3(4)

0( 1) 1(0) 0(2) 0(1) 1( 1) 0(3) 0(0) 1(1) 0(4)

1( 1) 1(0) 0(2) 1(1) 1( 1) 0(3) 1(0) 1(1) 0(4)

+ + + + + + + + + + + + + + + + + +

=

1 0 6 1 2 9 0 2 12

0 0 0 0 1 0 0 1 0

1 0 0 1 1 0 0 1 0

+ + + + + + + + + + + + + + +

=

5 8 14

0 1 1

1 0 1

RHS

1(1) 1(0) 0(1) 1(2) 1(1) 0(1) 1(3) 1(0) 0(0)

0(1) ( 1)(0) 1(1) 0(2) ( 1)1 1(1) 0(3) ( 1)(0) 1(0)

2(1) 3(0) 4(1) 2(2) 3(1) 4(1) 2(3) 3(0) 4(0)

+ + + + + + + + + + + + + + + + + +

=

1 0 0 2 1 0 3 0 0

0 0 1 0 1 1 0 0 0

2 0 4 4 3 4 6 0 0

+ + + + + + + + + + + + + + + + + +

=

1 1 3

1 0 0

6 11 6

LHS RHS

Q 15. Find 2A 5A+6I , if

2 0 1

A = 2 1 3

1 1 0



GenextStudents 23

Ans.:2

A 5A+6I ,

2 0 1

A = 2 1 3

1 1 0

A2

= A A

=

2 0 1 2 0 1

2 1 3 2 1 3

1 1 0 1 1 0

=

2(2) 0(2) 1(1) 2(0) 0(1) 1( 1) 2(1) 0 0

2(2) 1(2) 3(1) 2(0) 1(1) 3( 1) 2(1) 1(3) 0

1(2) ( 1)(2) 0 0 ( 1)(1) 0 1(1) 1(3) 0

+ + + + + + + + + + + + + + + + +

=

4 0 1 0 0 1 2 0 0

4 2 3 0 1 3 2 3 0

2 2 0 0 1 0 1 3 0

+ + + + + + + + + + + + +

=

5 1 2

9 2 5

0 1 2

5A =

5(2) 5(0) 1(5)

5(2) 5(1) 5(3)

5(1) 5( 1) 5(0)

=

10 0 5

10 5 15

5 5 0

6I =

6(1) 6(0) 6(0)

6(0) 6(1) 6(0)

6(0) 6(0) 6(1)

=

6 0 0

0 6 0

0 0 6



GenextStudents 24

A2 5A =

5 1 2 10 0 5

9 2 5 10 5 15

0 1 2 5 5 0

=

4 10 1 0 2 5

9 10 2 5 5 15

0 5 1 5 2 0

+

=

5 1 3 6 0 0

1 7 10 0 6 0

5 4 2 0 0 6

+

=1 1 31 1 10

5 4 4

Q 16. If A =

1 0 2

0 2 1

2 0 3

, prove that A3 6A

2+ 7A + 2I = 0.

Ans.: A =

1 0 2

0 2 12 0 3

, A3

6A2

+ 7A + 2I = 0

A2=

1 0 2 1 0 2

0 2 1 0 2 1

2 0 3 2 0 3

=

1 0 4 0 0 0 2 0 6

0 0 2 0 4 0 0 2 3

2 0 6 0 0 0 4 0 9

+ + + + + + + + + + + + + + + + + +

=

5 0 8

2 4 5

8 0 13

6A2=

30 0 48

12 24 30

48 0 78



GenextStudents 25

A3 =

5 0 8 1 0 2

2 4 5 0 2 1

8 0 13 2 0 3

=

5 0 16 0 0 0 10 0 24

2 0 10 0 8 0 4 6 15

8 0 26 0 0 0 16 0 39

+ + + + + + + + + + + + + + + + + +

=

21 0 34

12 8 23

34 0 55

7A =1 0 2

7 0 2 1

2 0 3

=7 0 140 14 7

14 0 21

2I =

1 0 0

2 0 1 0

0 0 1

=

2 0 0

0 2 0

0 0 2

3 2

6A A =

21 0 34 30 0 48 7 0 14 2 0 0

12 8 23 12 24 30 0 14 7 0 2 034 0 55 48 0 78 14 0 21 0 0 2

+ +

=

21 30 7 2 0 0 0 0 34 48 14 0

12 12 0 0 8 24 14 2 23 30 7 0

34 48 14 0 0 0 0 0 55 78 21 2

+ + + + + + + + + + + + + + + + + +

=

0 0 0

0 0 0

0 0 0

Q 17. If A =3 2

4 2

and I =1 0

0 1

, find kso that 2 2A kA I=

Ans.: A =3 2

4 2

, I =1 0

0 1



GenextStudents 26

2 2A kA I=

A2 =

3 2 3 2

4 2 4 2

=9 8 6 4

12 8 6 4

+ +

=1 2

4 2

2kA I =3 2

4 2

k k

k k

2 0

0 2

=3 2 2

4 2 2

k k

k k

1 2

4 2

=3 2 2

4 2 2

k k

k k

3 2 1k = 3 3k= 1k=

Q 18. If A =

0 tan2

tan 02

and I is the identify matrix order 2, show that I + A = (I

A)cos sin

sin cos

Ans.: I + A =0 tan1 0 2

0 1 tan 02

+

=1 tan

2

tan 12

R.H.S. =cos sin

(I A)

sin cos

I A =1 tan

2

tan 12

=

sin21

cos2

sin2 1

cos2



GenextStudents 27

cos sin(I A)

sin cos

=

sin 21cos cos sin2

sin cossin2 1

cos2

=

sin sin sin2 2cos sin cos

cos cos2 2

sin sin2 2cos sin sin coscos cos

2 2

+ +

+ +

=

cos cos sin sin sin cos cos sin2 2 2 2

cos cos2 2

sin cos cos sin sin sin cos cos2 2 2 2

cos cos2 2

=

( ) ( )

( ) ( )

cos sin2 2

cos cos2 2

sin cos2 2

cos cos2 2

=

cos sin2 2

cos cos2 2

sin cos2 2

cos cos2 2

=

1 tan2

tan 12

Q 19. A trust fund has Rs 30,000 that must be invested in two different types of bonds. The

first bond pays 5% interest per year, and the second bond pays 7% interest per year.



GenextStudents 28

Using matrix multiplication, determine how to divide Rs 30,000 among the two types of

bonds. If the trust fund must obtain an annual total interest of:

(a) Rs 1800 (b) Rs 2000

Ans.:I II 5%

(30, 000 ) 7%x x

=1800

2000

(i) 5% (30,000 )7%x x + = 1800

5 7(30,000 )

100 100x x + = 1800

5 (30, 000 )7x x+ = 1800 100

5 2,10,000 7x x+ = 1, 80, 000

2x = 1,80,000 2,10,000

= 30, 000

x = 15,000

(ii) 5% (30,000 )7%x x + = 2000

5 7(30,000 )100 100

x x + = 2000

5 2,10,000 7x x+ = 2000 100

2x = 2,00,000 2,10,000

2x = 10,000

x =10,000

2= 5000

(i) To get annual interest of 1800, he should invest Rs 15,000 in I type of bond and Rs

15,000 in II type of bond.

(ii) To get annual interest of Rs 2000 he should invest Rs 5000 & Rs 25000 respectively

in I and II type of bond.



GenextStudents 29

Q 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics

books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each

respectively. Find the total amount the bookshop will receive from selling all the books

using matrix algebra.

Ans.:

80che phy Eco

60120 96 120

40

Total SP = (120 80 + 90 60 + 120 40)

= (9600 + 5400 + 4800)

= (20160) = Rs 20, 160

Assume X, Y, Z, W and P are matrices of order 2 n , 3 k , 2 p , 3n and p k , respectively.

Choose the correct answer in Exercises 21 and 22.

Q 21. The restriction of n , kandpso that PY + WY will be defined are:

(A) k=3,p= n (B) kis arbitrary,p=2

(C)pis arbitrary, k= 3 (D) k=2,p=3

Ans.: Matrix Order

X 2 n

Y 3 k

Z 2 p

W 3n

(P K) (3 k) + (n 3) (3 k)

P K + n K

If K = 3 , P n=

Part (A) is correct.

Q 22. If n p= , then the order of the matrix 7X 5Z is:



GenextStudents 30

(A) 2p (B) 2 n (C) 3n (D) p n

Ans.: The order of matrix X is 2 n

The order of matrix Z is 2 p

7X 5Z is defined when X and Z an of the same order.

n p= (Given)

Thus the order of 7X 5Z is 2 n

Part (B) is the correct answer.



GenextStudents 31

Exercise 3.3

Q 1. Find the transpose of each of the following matrices:

(i)

51

2

1

(ii)1 1

2 3

(iii)

1 5 6

3 5 6

2 3 1

Ans.: (i) Transpose of

5

1

2

1

is1

5 12

(ii) Transpose of1 1

2 3

is1 2

1 3

(iii) Transpose of

1 5 6

3 5 6

2 3 1

is

1 3 2

5 5 5

6 6 1

Q 2. If

1 2 3

A 5 7 92 1 1

=

and B =

4 1 5

1 2 01 3 1

, then verify that

(i) ( )A B A B + = + , (ii) ( )A B A B =

Ans.: (i) A + B =

1 2 3

5 7 9

2 1 1

+

4 1 5

1 2 0

1 3 1

=

1 4 2 1 3 5

5 1 7 2 9 0

2 1 1 3 1 1

+

+ + + + + +

=

5 3 0

6 9 9

1 4 2

L.H.S. = (A B)+ =

5 3 2

6 9 9

1 4 2

=

5 6 1

3 9 4

2 9 2



GenextStudents 32

R.H.S. = A + B =

1 2 3 4 1 5

5 7 9 1 2 0

2 1 1 1 3 1

+

=

1 5 2 4 1 1

2 7 1 1 2 3

3 9 1 5 0 1

+

=

1 4 5 1 2 1

2 1 7 2 1 3

3 5 9 0 1 1

+ + + + + + +

=

5 6 1

3 9 4

2 9 2

= L.H.S.

Hence ( )A B A B + = +

(ii) A B =

1 2 3 4 1 5

5 7 9 1 2 0

2 1 1 1 3 1

=

1 ( 4) 2 1 3 ( 5)

5 1 7 2 9 0

2 1 1 3 1 1

=

3 1 8

4 5 9

3 2 0

L.H.S. =

3 1 8 3 4 3

(A B) 4 5 9 1 5 2

3 2 0 8 9 0

= =

R.H.S. =

1 2 3 4 1 5

A B 5 7 9 1 2 0

2 1 1 1 3 1

=

=

1 5 2 4 1 1

2 7 1 1 2 3

3 9 1 5 0 1



GenextStudents 33

=

1 ( 4) 5 1 2 1

2 1 7 2 1 3

3 ( 5) 9 0 1 1

=

3 4 2

1 5 2

8 9 0

= L.H.S.

Hence ( )A B A B =

Q 3. If

3 4

A = 1 2

0 1

and B =1 2 1

1 2 3

, then verify that

(i) ( )A B A B + = + , (ii) ( )A B A B =

Ans.: (i)

3 4

A = 1 20 1

A =3 1 0

4 2 1

A + B =3 1 0 1 2 1

4 2 1 1 2 3

+

=3 1 1 2 0 1

4 1 2 2 1 3

+ + + + +

=2 1 1

5 4 4

(A B)+ =2 52 1 11 4

5 4 41 4

=

R.H.S. = A B + =

3 41 2 1

1 21 2 3

0 1

+

=

3 4 1 1

1 2 2 20 1 1 3

+

=

3 1 4 1

1 2 2 20 1 1 3

+

+ + + +

=

2 5

1 4

1 4

= L.H.S.



GenextStudents 34

Hence ( )A B A B + = +

(ii)

3 4

A 1 2

0 1

=

3 1 0

A=

4 2 1

A B =3 1 0 1 2 1

4 2 1 1 2 3

=3 ( 1) 1 2 0 1 4 3 1

4 1 2 2 1 3 3 0 2

=

L.H.S. = ( )A B =

4 3

4 3 1 3 03 0 2

1 2

=

R.H.S. = A B =

3 41 2 1

1 21 2 3

0 1

=

3 4 1 1

1 2 2 2

0 1 1 3

=

3 ( 1) 4 1

1 2 2 2

0 1 1 3

=

4 3

3 0

1 2

= L.H.S.

Hence ( ) ' 'A B A B =

Q 4. If2 3

1 2

A

=

and B =1 0

1 2

, then find (A + 2B)

Ans.: A' =2 3

1 2

A =2 1

3 2

and B =1 0

1 2

A + 2B =2 1 1 0

23 2 1 2

+



GenextStudents 35

=2 1 2 0 2 2 1 0

3 2 2 4 3 2 2 4

+ + = + +

=

4 1

5 6

( 2 )A B + =4 1 4 5

5 6 1 6

=

Q 5. For the matrices A and B, verify that ( )AB B A = , where

(i) A =

1

4

3

, B = [ ]1 2 1 (ii) A =

0

1

2

, B = [ ]1 5 7

Ans.: (i) AB = [ ]

1

4 1 2 1

3

=

1 ( 1) 1 2 1 1

4 ( 1) 4 2 4 1

3 ( 1) 3 2 3 1

=

1 2 1

4 8 4

3 6 3

L.H.S. = (A B) =

1 4 3

2 8 6

1 4 3

R.H.S. = B A = [ ]

1

1 2 1 4

3

= [ ]

1

2 1 4 3

1

=

1 1 1 ( 4) 1 3

2 1 2 ( 4) 2 3

1 1 1 ( 4) 1 3



GenextStudents 36

=

1 4 3

2 8 6

1 4 3

L.H.S.

Hence (AB) B A =

(ii) A B = [ ]

0

1 1 5 7

2

=

0 1 0 5 0 7

1 1 1 5 1 7

2 1 2 5 2 7

=

0 0 0

1 5 7

2 10 14

L.H.S. = (AB) =

0 0 0 0 1 2

1 5 7 0 5 10

2 10 14 0 7 14

=

R.H.S. = B A = [ ]

0

1 5 7 1

2

= [ ]

1

5 0 1 2

7

=

1 0 1 1 1 2

5 0 5 1 5 2

7 0 7 1 7 2

=

0 1 2

0 5 10

0 7 14

= L.H.S.

Hence (A B) B A

=

Q 6. If (i) A =cos sin

sin cos

, then verify that A A I =

(ii) A =sin cos

cos sin

, then verify that A A = I



GenextStudents 37

Ans.: (i) L.H.S. = A A =cos sin

Asin cos

= cos sin cos sinsin cos sin cos

=

2 2

2 2

cos sin cos sin sin cos

sin cos cos sin sin cos

+

+

=1 0

I0 1

=

A A = I

(ii) A A =sin cos sin cos

cos sin cos sin

=sin cos sin cos

cos sin cos sin

=

2 2

2 2

sin cos sin cos cos sin

cos sin sin cos cos sin

+

+

=1 0

0 1

A A = I

Q 7. (i) Show that the matrix A =

1 1 5

1 2 1

5 1 3

is a symmetric matrix.

(ii) Show that the matrix A =

0 1 1

1 0 1

1 1 0

is a skew symmetric matrix.

Ans.: (i) For a symmetric matrix na a=



GenextStudents 38

Now A =

1 1 5

1 2 1

5 1 3

21 121a a= = , 31 135a a= =

32 231a a= = , 11 22 33, ,a a a are 1, 2, 3 respectively.

Henceij ija a= A is a symmetric matrix.

Alternatively: A =

1 1 5

1 2 1

5 1 3

=

1 1 5

1 2 1

5 1 3

= A

A = A A is a symmetric matrix.

(ii) For a skew symmetric matrix

ij ija a=

Now A =

0 1 1

1 0 1

1 1 0

21 1a = , 12 1a = 12 1a = or 21 12a a=

31 1a = , 13 1a = or 13 1a = 31 13a a=

32 1a = , 23 1a = or 23 1a = , 32 23a a=

11 0a = , 22 0a = , 33 0a =

Hence A skew-symmetric matrix

Alternatively: A =

0 1 1 0 1 1

1 0 1 1 0 1

1 1 0 1 1 0

=



GenextStudents 39

=

0 1 1

1 0 1

1 1 0

= A

A = A A is a skew symmetric matrix.

Q 8. For the matrix A =1 5

6 7

, verify that

(i) (A A )+ is a symmetric matrix

(ii) (A A ) is a skew symmetric matrix

Ans.: (i)

1 5

6 7A

=

1 6

A 5 7

=

1 1 5 6 2 11A+A

6 5 7 7 11 14

+ + = = + +

Here 21 1211a a= =

A + A is a symmetric matrix

(ii) A A =1 5 1 6

6 7 5 7

=

1 1 5 6

6 5 7 7

=0 1

1 0

Here 21 1a = , 12 1a = or 12 1a =

21 12a a= , 11 22 0a a= =

A A is skew-symmetric matric

Q 9. Find1

( )2

A A+ and1

( )2

A A , when A =

0

0

0

a b

a c

b c



GenextStudents 40

Ans.: A =

0

0

0

a b

a c

b c

A =

0

0

0

a b

a c

b c

A + A =

0 0

0 0

0 0

a b a b

a c a c

b c b c

+

=

0

0

0

a a b b

a a c c

b b c c

+ + +

=

0 0 0

0 0 0

0 0 0

1

(A + A )2

=0 0 0 0 0 010 0 0 0 0 0

20 0 0 0 0 0

=

A A =

0 0

0 0

0 0

a b a b

a c a c

b c b c

=

0 0 ( ) ( )

0 0 ( )0 0

a a b b

a a c cb b c c

=

0 2 2

2 0 22 2 0

a b

a cb c

1(A A )

2 =

0 2 21

2 0 22

2 2 0

a b

a c

b c

=

0

0

0

a b

a c

b c

Q 10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i) 3 51 1

(ii)

6 2 2

2 3 1

2 1 3

(iii)

3 3 1

2 2 1

4 5 2

(iv)1 5

1 2



GenextStudents 41

Ans.: (i) Let A =3 5

1 1

A =3 1

5 1

1 (A A )2 + =3 5 3 11

1 1 5 12

+

=3 3 5 1 6 61 1

1 5 1 1 6 22 2

+ + = +

=3 3

3 1

= B = symmetric matrix

1

(A A )2 =

3 5 3 11

1 1 5 12

=3 3 5 1 0 41 1

1 5 1 ( 1) 4 02 2

=

=0 2

2 0

= C= skew symmetric matrix

B + C =3 3 0 2 3 0 3 2

3 1 2 0 3 2 1 10

+ + + = +

=3 5

A1 1

=

(ii) A =

6 2 2

2 3 1

2 1 3

A =

6 2 2

2 3 1

2 1 3

A A+ =

6 6 2 2 2 2

2 2 3 3 1 1

2 2 1 1 3 3

+ + + + +

=

12 4 4

4 6 2

4 2 6



GenextStudents 42

6 2 21

(A A ) 2 3 12

2 1 3

+ =

= B = Symmetric matrix

A A =

6 2 2 6 2 2

2 3 1 2 3 1

2 1 3 2 1 3

A A =

6 6 2 ( 2) 2 2 0 0 0

2 ( 2) 3 3 1 ( 1) 0 0 0

2 2 1 ( 1) 3 3 0 0 0

=

1

(A A )2

=

0 0 0 0 0 01

0 0 0 0 0 02

0 0 0 0 0 0

=

= C = Skew symmetric matrix

B + C =

6 2 2 0 0 0

2 3 1 0 0 0

2 1 3 0 0 0

+

=

6 2 2

2 3 1

2 1 3

= A

(iii) A =

3 3 1

2 2 1

4 5 2

A =

3 2 4

3 2 5

1 1 2

A+A =

3 3 1 3 2 4

2 2 1 3 2 54 5 2 1 1 2

+

=

3 3 3 2 1 4

2 3 2 2 1 5

4 1 5 1 2 2

+ + + +

=

6 1 5

1 4 4

5 4 4



GenextStudents 43

1(A + A )

2 =

6 1 51

1 4 52

5 4 4

=

1 53

2 2

12 2

2

52 2

2


A A =

3 3 1 3 2 4

2 2 1 3 2 5

4 5 2 1 1 2

=3 3 3 2 1 42 3 2 2 1 5

4 1 5 1 2 2

+ + + +

=0 5 35 0 6

3 6 0

1(A A )

2 =

5 30

2 2

50 3

2

33 0

2

= C = Skew symmetrice matrix

B + C =

1 5 5 33 0

2 2 2 2

1 52 2 0 3

2 2

5 32 2 3 0

2 2

+ +

=

1 5 5 33 0

2 2 2 2

1 52 0 2 3

2 2

5 32 3 2 0

2 2

+ + +

+ + + +

=

3 3 1

2 2 1

4 5 2

= A



GenextStudents 44

(iv) A =1 5

1 2

A =1 1

5 2

A A+ =1 5 1 1

1 2 5 2

+ =

1 1 5 1

1 5 2 2

+

+ +

=2 4

4 4

1

(A A )2

+ =2 4 1 21

4 4 2 22

=


A A =

1 5 1 1

1 2 5 2

=

1 1 5 1

1 5 2 2

+

=

0 6

6 0

1

(A A )2

=0 6 0 31

6 0 3 02

=

= C (Skew symmetric matrix)

B + C =1 2 0 3 1 0 2 3 1 5

2 2 3 0 2 3 2 0 1 2

+ + + = = +

Choose the correct answer in the Exercises 11 and 12.

Q 11. If A, B are symmetric matrices of same order, then AB BA is a

(A) Skew symmetric matrix (B) Symmetric matrix

(C) Zero matrix (D) Identify matrix

Ans.: A and B are symmetric matrices

A A, B B= =

Now (AB BA) ( ) ( )AB BA = = B A A B

= BA AB ,B B A A = =

= (AB BA)

AB BA is a skew symmetric matrix.



GenextStudents 45

Part (A) is the correct answer.

Q 12. If A =cos sin

sin cos

, then A + A = I, if the value of is

(A)6

(B)

3

(C) (D)

3

2

Ans.: A =cos sin

sin cos

, A =cos sin

sin cos

A A+ =cos cos sin sin

sin sin cos cos

+ + +

=2cos 0

0 2cos

=1 0

0 1

1

2cos 1,cos2

= =

3

=



GenextStudents 46

Exercise 3.4

Using elementary transformations, find the inverse of each of the matrices, if it exists in

Exercise 1 to 17

Q 1.1 1

2 3

Ans.: A = I2A

or1 1

2 3

=1 0

0 1

A, where A =1 1

2 3

Applying 2 2 12R R R

1 1

0 5

=1 0

2 1

A

Applying 2 21

5R R

1 1

0 1

=

1 0

2 1

5 5

A

Applying 1 1 2R R R +

1 0

0 1

=

3 1

5 5

2 1

5 5

A

A1

=

3 1

5 5

2 15 5

Q 2.2 1

1 1

Ans.: A = I2A



GenextStudents 47

Or1 1 1 0

2 1 0 1

=

A

Apply 1 2R R 1 1 0 1

A2 1 1 0

=

Applying 1 2 12R R R

1 1 0 1A

0 1 1 2

=

Applying 2 2( 1)R R

1 1 0 1A0 1 1 2

=

Applying 1 1 2R R R

1 0 1 1A

0 1 1 2

=

Hence A1

=1 1

1 2

Q 3.1 3

2 7

Ans.: A =I2A 1 3 1 0

A2 7 0 1

=

Apply 1 2 12R R R 1 3 1 0

A0 1 2 1

=

Apply 1 1 23R R R 1 0 7 3

A0 1 2 1

=

A1 =7 3

2 1



GenextStudents 48

Q 4.2 3

5 7

Ans.: A = I2A

2 3 1 0

A5 7 0 1

=

Apply 1 2R R ,5 7 0 1

A2 3 1 0

=

Apply 1 1 22R R R ,1 1 2 1

A2 3 1 0

=

Apply 2 2 12R R R ,1 1 2 1

A0 1 5 2

=

Apply 1 1 2R R R ,1 0 7 3

A0 1 5 2

=

A1 =

7 3

5 2

Q 5. 2 17 4

Ans.: A = I2A

2 1 1 0

A7 4 0 1

=

Apply 1 2R R ,7 4 0 1

A2 1 1 0

=

Apply 1 1 23R R R ,1 1 3 1

A2 1 1 0

=

Apply 2 2 12R R R ,1 1 3 1

A0 1 7 2

=



GenextStudents 49

Apply 2 2R R ,1 1 3 1

A0 1 7 2

=

Apply 1 1 2R R R ,1 0 4 1

A0 1 7 2

=

A 1 =4 1

7 2

Q 6.2 5

1 3

Ans.: A = I2A 2 5 1 0

A

1 3 0 1

=

Apply 1 1 2R R R 1 2 1 1

A1 3 0 1

=

Apply 2 2 1R R R 1 2 1 1

A0 1 1 2

=

Apply 1 1 22R R R 1 0 3 5

A0 1 1 2

=

A1=3 5

1 2

Q 7.3 1

5 2

Ans.: A = I2A 3 1 1 0

A5 2 0 1

=

Apply 1 1 22R R R 1 0 2 1

A5 2 0 1

=

Apply 2 2 15R R R 1 0 2 1

A0 2 10 6

=



GenextStudents 50

Apply 2 21

2R R

1 0 2 1A

0 1 5 3

=

A1

=

2 1

5 3

Q 8.4 5

3 4

Ans.: A = I2A 4 5 1 0

A3 4 0 1

=

Apply 1 1 2R R R 1 1 1 1

A

3 4 0 1

=

Apply 2 2 13R R R 1 1 1 1

A0 1 3 4

=

Apply 1 1 2R R R 1 0 4 5

A0 1 3 4

=

A1 =4 5

3 4

Q 9.3 10

2 7

Ans.: A = I2A 3 10 1 0

A2 7 0 1

=

Apply 1 1 2R R R 1 3 1 1

A2 7 0 1

=

Apply 2 2 12R R R 1 3 1 1

A0 1 2 3

=

Apply 1 1 23R R R 1 0 7 10

A0 1 2 3

=



GenextStudents 51

A1 =7 10

2 3

Q 10.

3 1

4 2

Ans.: A = I2A 3 1

4 2

=1 0

A0 1

Operating 1 2R R 4 2 0 1

A3 1 1 0

=

Operating 1 1( 1)R R 4 2 0 1

A

3 1 1 0

=

Operating 1 1 2R R R 1 1 1 1

A3 1 1 0

=


A0 2 4 3

=

Operating 2 21

2

R R

1 11 1

A3

0 1 2 2

=

Operating 1 1 2R R R +

11

1 0 2 A0 1 3

22

=

A1 =

11

23

2 2

Q 11.2 6

1 2

Ans.: A = I2A 2 6 1 0

A1 2 0 1

=



GenextStudents 52


A1 2 0 1

=


A0 2 1 2

=

Operating 2 21

2R R

1 11 4

A10 1 1

2

=


1 31 0

A10 1 1

2

=

A1 =

1 3

11

2

Q 12.6 3

2 1

Ans.: A = I2A 6 3 1 0

A2 1 0 1

=

Operating 1 2R R 2 1 0 1

A6 3 1 0

=

2 2 13R R R + 2 1 0 1

A0 0 1 3

=

In second row of L.H.S., elements are zero.

A1does not exist

Q 13.2 3

1 2

Ans.: A = I2A 2 3 1 0

A1 2 0 1

=



GenextStudents 53

Operating 1 1 2R R R + 1 1 1 1

A1 2 0 1

=


A0 1 1 2

=


A0 1 1 2

=

A1=2 3

1 2

Q 14.2 1

4 2

Ans.: A = I2A 2 1 1 0

A4 2 0 1

=


A0 0 2 1

=

In second row of L.H.S. each element is zero.

A1does not exist.

Q 15.

2 0 1

5 1 0

0 1 3

Ans.: A = I3A

2 3 3 1 0 0

2 2 3 0 1 0 A

3 2 2 0 0 1

=

Operating 1 3R R

3 2 2 0 0 1

2 2 3 0 1 0 A

2 3 3 1 0 0

=



GenextStudents 54

Operating 1 1 2R R R

1 4 1 0 1 1

2 2 3 0 1 0 A

2 3 3 1 0 0

=

Operating 2 2 12R R R and 3 12R R

1 4 1 0 1 1

0 10 5 0 3 2 A

0 5 5 1 2 2

=

Operating 2 3R R

1 4 1 0 1 1

0 5 0 1 1 0 A

0 5 5 1 2 2

=

Operating 2 21

5R R=

0 1 11 4 1

1 10 1 0 0 A

5 51 5 5

1 2 2

=

Operating3 3

1

5R R

0 1 11 4 1

1 1

0 1 0 0 A5 51 2 1

1 2 2

5 5 5

=


4 11

5 51 0 11 1

0 1 0 05 5

0 1 12 2 2

5 5 3

=

=

4 1 51

1 1 0 A5

1 2 2



GenextStudents 55

Operating 3 3 2R R R =

1 0 1 4 1 51

0 1 0 1 1 0 A5

0 0 1 2 1 2

=

1 1 3R R R + =

1 0 0 2 0 31

0 1 0 1 1 0 A5

0 0 1 2 1 2

=

Hence-1

2 30

5 5

1 1A 0

5 5

2 1 2

5 5 5

=

Q 16.

1 3 2

3 0 5

2 5 0

Ans.: A = I3A

1 3 2 1 0 0

3 0 5 0 1 0 A

2 5 0 0 0 1

=

Operating 2 2 13R R R + , 3 3 12R R R

1 3 2 1 0 0

0 9 11 3 1 0 A

0 1 4 2 0 1

=


1 3 2 1 0 0

0 1 21 13 1 8 A

0 1 4 2 0 1

=

Operating 1 1 23R R R , 3 3 2R R R +



GenextStudents 56

1 0 65 40 3 24

0 1 21 13 1 8 A

0 0 25 15 1 9

=

Operating 3 31

25R R

1 0 65 40 3 24

0 1 21 13 1 8 A

0 0 1 15 1 9

25 25 25

=

=

1000 75 6001

325 25 200 A25

15 1 9

Operating 1 1 365R R R + , 2 321R R

1 0 0

0 1 0

0 0 1

=

25 10 151

10 4 11 A25

15 1 9

1

2 31

3 5

2 4 11A5 25 25

3 1 9

5 25 25

=

Q 17.

2 0 1

5 1 0

0 1 3

Ans.: A = I3A 2 0 1 1 0 05 1 0 0 1 0 A

0 1 3 0 0 1

=



GenextStudents 57

Operating 1 2R R

5 1 0 0 1 0

2 0 1 1 0 0 A

0 1 3 0 0 1

=


1 1 2 2 1 0

2 0 1 1 0 0 A

0 1 3 0 0 1

=


1 1 2

0 2 5

0 1 3

=

2 1 0

5 2 0 A

0 0 1

Operating 2 2( 1)R R 1 1 2 2 1 00 2 5 5 2 1 A

0 1 3 0 0 1

=


1 1 2 2 1 0

0 1 2 5 2 1 A

0 1 3 0 0 1

=

Operating 1 1 2R R R , 3 2 2R R R

1 0 0 3 1 1

0 1 2 5 2 1 A

0 0 1 5 2 2

=


1 0 0 3 1 1

0 1 2 5 2 1 A

0 0 1 5 2 2

=

A1 =3 1 115 6 5

5 2 2

Q 18. Matrices A and B will be inverse of each other only if

(A) AB = BA (B) AB = BA = 0 (C) AB = 0, BA = I (D) AB = BA = I



GenextStudents 58

Ans.: If B is the inverse of A if AB = BA = I

Part (D) is the correct answer.



GenextStudents 59

Miscellaneous Exercise

Q 1. Let A =0 1

0 0

, show that1( )n n naI bA a I na bA

+ = + , where I is the identity matrix of

order 2 and N.n

Ans.: L.H.S. =1 0 0 1

[ ]0 1 0 0

n

naI bA a b

+ = +

=0 0

0 0 0

n

a b

a

= +

=

0

na b

a

R.H.S. =1n n

a I na bA+

=1

1 0 0 1

0 1 0 0

n na na b

+

=

10 0

0 0 0

n n

n

a na b

a

+

=

1

0

n n

n

a na b

a

We have to prove that

1

0 0

n n n

n

a b a na b

a a

=

Apply principle of Mathematical Induction

Put P(n):

1

0 0

n n n

n

a b a na b

a a

=

For 1n= , P (1) :0 0

a b a b

a a

=

P (n) is true for 1n=

Let P (n) be true for n k= .

P( k) :1

0 0

k k k

k

a b a ka b

a a

=



GenextStudents 60

Multiple both sides by0

a b

a

L.H.S. =

1

0 0 0

k na b a b a b

a a a

+

+ =

R.H.S. =

1

00

k n

k

a ba ka b

aa

=

1

10

k k k

k

a ba ka b

a

+

+

+

=

1

1

( 1)

0

k k

k

a k a b

a

+

+

+

This shows P(n) is true for n= k+ 1 then by principle of mathematical induction, P(n)

is true for all positive integral values of n.

Q 2. If A =

1 1 1

1 1 1

1 1 1

, prove that

1 1 1

1 1 1

1 1 1

3 3 3

3 3 3

3 3 3

n n n

n n n n

n n n

A

=

, Nn .

Ans.: Let P( )n : An=

1 1 1

1 1 1

1 1 1

3 3 3

3 3 3

3 3 3

n n n

n n n

n n n

where A =

1 1 1

1 1 1

1 1 1

For n= 1

L.H.S. = An= A=

1 1 1

1 1 1

1 1 1

R.H.S. = An=

0 0 0

0 0 0

0 0 0

3 3 3

3 3 3

3 3 3

=

1 1 1

1 1 1

1 1 1

P (n) is true for n= 1.

Let it be true on n= k



GenextStudents 61

Ak=

1 1 1

1 1 1

1 1 1

3 3 3

3 3 3

3 3 3

k k k

k k k

k k k

Multiplying both sides by A

L.H.S. = AkA = A

k+1

R.H.S. =

1 1 1

1 1 1

1 1 1

3 3 3

3 3 3 A

3 3 3

k k k

k k k

k k k

1kA + =

1 1 1

1 1 1

1 1 1

3 3 3 1 1 1

3 3 3 1 1 1

3 3 3 1 1 1

k k k

k k k

k k k

=

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3

k k k k k k k k k

k k k k k k k k k

k k k k k k k k k

+ + +

+ + +

+ + +

+ + + + + +

+ + + + + + + + + + + +

=

1 1 1

1 1 1

1 1 1

3.3 3.3 3.3

3.3 3.3 3.3

3.3 3.3 3.3

k k k

k k k

k k k

=

3 3 3

3 3 3

3 3 3

k k k

k k k

k k k

=

( 1) 1 ( 1) 1 ( 1) 1

( 1) 1 ( 1) 1 ( 1) 1

( 1) 1 ( 1) 1 ( 1) 1

3 3 3

3 3 3

3 3 3

k k k

k k k

k k k

+ + +

+ + +

+ + +

P (n) is true for n= k+ 1

By principle of mathematical induction that ( )P n is true for all n, Nn .

Q 3. If A =3 4

1 1

, then prove that An =1 2 4

1 2

n n

n n

+

, where nis any positive integer.

Ans.: Let P(n) : An=1 2 4

1 2

n n

n n

+

, where A =3 4

1 1



GenextStudents 62

Put n = 1,

A =1 2 4 3 4

1 1 2 1 1

+ =

P(n) is true for 1n=

Let P(n) be true for n = k

Ak =1 2 4

1 2

k k

k k

+

Multiplying both sides by A

1A A = Ak k+ = 1 2 4 A1 2k k

k k

+

=1 2 4 3 4

1 2 1 1

k k

k k

+

=3(1 2 ) 4 4 (1 2 ) 4

3 (1 2 ) 4 (1 2 )

k k k k k

k k k k

+ + + +

=

3 2 4 4

1 1 2

k k

k k

+

+ =

1 2( 1) 4( 1)

1 1 2( 1)

k k

k k

+ + +

+ +

This proves that P(n) is true for 1n k= +

Hence P(n) is true for all Nn

Q 4. If A and B are symmetric matrices, prove that AB BA is a skew symmetric matrix.

Ans.: A and B are symmetric matrix is A = A and B = B Also (AB) = BA.

Now ( )AB BA = ( ) ( )AB BA

= B A A B = BA AB

= (AB BA)

Hence AB BA is skew symmetric matrix.



GenextStudents 63

Q 5. Show that the matrix B AB is symmetric or skew symmetric according as A is symmetric

or skew symmetric.

Ans.: (i) Let A be symmetric matrix

A A=

(B'AB) = (B (AB)) = (AB) (B) = (B A )B

= B AB A A=

B AB is a symmetric matrix

(ii) Let A be skew-symmetric matrix

A A=

Now (B (AB)) = (AB) (B ) = (B A )B

= B (A)B = B AB A A=

BAB is skew symmetric matrix.

Q 6. Find the value of , ,x y z if the matrix A =

0 2y z

x y z

x y z

satisfy the equation A A = I .

Ans.: A =

0 2y z

x y z

x y z

A =

0

2

x x

y y y

z z z

A A =

0 0 2

2

x x y z

y y y x y z

z z z x y z



GenextStudents 64

=

2 2

2 2 2

2 2 2

0 0

0 4 2

0 2

x x xy xy xz xz

yx yx y y y yz yz yz

zx zx yz zy zy z z z

+ + + +

+ + + + + +

=

2

2

2

2 0 0

0 6 0

0 0 3

x

y

z

=

1 0 0

0 1 0

0 0 1

3A A I =

Equating corresponding elements

2

2 1x =

,

1

2x=

26 1y = ,1

6y=

2 1z = ,1

3z=

Q 7. For what values of [ ]

1 2 0 0

: 1 2 1 2 0 1 2 O

1 0 2

x

x

=

?

Ans.: L.H.S. =

1 2 0 0

[1 2 1] 2 0 1 2

1 0 2 x

=

0 4 0

[1 2 1] 0 0

0 0 2

x

x

+ + + + + +

=

4

[1 2 1]

2

x

x

= [4 2 2 ] [4 4]x x x+ + = + = [0]

4 4x + 1x=

Q 8. If A =3 1

1 2

, show that A2 5A + 7I = 0.



GenextStudents 65

Ans.: A2 =

3 1 3 1 9 1 3 2

1 2 1 2 3 2 1 4

+ = +

=

8 5

5 3

L.H.S. =2A 5A + 7I

=8 5 3 1 1 0

5 75 3 1 2 0 1

+

=8 5 15 5 7 0

5 3 5 10 0 7

+

=8 5 7 5 5 0

5 5 0 3 10 7 + +

+ + + =

0 0

0 0

R.H.S.

Hence2A 5A + 7I = O

Multiple both the sides by A ...(1)

3 2A 5A + 7IA = O

Or3 2A 5A + 7A = O

3A = 25A 7A =8 5 3 1

5 75 3 1 2

=40 25 21 7

25 15 7 14

+

=40 21 25 7

25 7 15 14

+

=19 18

18 1

A3 =19 18

18 1

Again multiple (1) by A

4 3 2A 5A + 7A = O



GenextStudents 66

A4 = 5A37A2=19 18 8 5

5 718 1 5 3

=

95 90 56 35

90 5 35 21

+

=95 56 90 35

90 35 5 21

+

=39 55

55 16

Hence A4 =

39 55

55 16

Q 9. Findx, if [ ]

1 0 2

5 1 0 2 1 4 O

2 0 3 1

x

x

=

Ans.: L.H.S. = [ ]

1 0 2

5 1 0 2 1 4

2 0 3 1

x

x

= [ ]

0 2

5 1 0 8 1

2 0 3

x

x

x

+ + + + + +

= [ ]

2

5 1 9

2 3

x

x

x

+ +

= [ ]( 2) 45 (2 3)x x x+ +

= 2 2 45 2 3x x x +

= 2 48x

R.H.S. = [0]

2 48 0x = , 4 3x=



GenextStudents 67

Q 10. A manufacturer produces three product x, y, zwhich he sells in two markets. Annual

sales are indicated below:

Market Products

I 10,000 2,000 18,000

II 6,000 20,000 8,000

(a) If unit sales prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find

the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are Rs 2.00, Rs.1.00 and 50 paise

respectively. Find the gross profit.

Ans.: (a) Matrix for the products x, y, z is

Matrix corresponding to sale price of each product

2.50

1.50

1.00

x

y

z

The revenue collected by market is given by

2.5010,000 2,000 18,0001.50

6,000 20,000 8,0001.00

=25,000 3,000 18,000

15,000 30,000 8,000

+ + + +

=46,000

53,000

Revenue in each market Rs 46,000 and 53,000

Total revenue = Rs (46,000 + 53,000) Rs 99,000

(b) The cost price of commodities x, y and z are respectively Rs 2.00, Rs 1.00 and Rs 0.50

cost price of each market is given below

2.0010,000 2000 18000

1.006000 20000 8000

0.50



GenextStudents 68

=10000 2 2000 1 18000 0.50

6000 2 20000 1 8000 0.50

+ + + +

=

20000 2000 9000

12000 20000 4000

+ +

+ + =

31000

36000

Total cost prices of the commodities each market I and II are Rs 31,000, Rs 36,000

Total cost price = Rs (31000 + 36000) = Rs 67000

Gross profit = Revenue Cost price

= (99000 67000) = Rs 32000

Q 11. Find the matrix X so that 1 2 3 7 8 9X4 5 6 2 4 6

=

Ans.: Let X be of order m n and1 2 3

4 5 6

is of the order 2 3 is n= 2.

7 8 9

2 4 6

is of the order 2 3 is m= 2

X is of the order 2 2

Let X =a b

c d

1 2 3 7 8 9X

4 5 6 2 4 6

=

1 2 3

4 5 6

a b

c d

=4 2 5 3 6

4 2 5 3 6

a b a b a b

c d c d c d

+ + + + + +

=7 8 9

2 4 6

Equating the corresponding elements

4 7a b+ = .(1)



GenextStudents 69

2 5 8a b+ = .(2)

3 6 9a b+ = ..(3)

Multiple (1) by 2

2 8 14a b+ = .(4)

And 2 5 8a b+ =

Subtracting (2) from (4)

3 6b= 2b=

Putting value of b in (1)

8 7a = 8 7 1a= =

1, 2a b= = Satisfy eq. (3) also

Equating the elements of second row

4 2c d+ = (5)

2 5 4c d+ = ..(6)

3 6 6c d+ = ..(7)

Multiple eq. (5) by 2

2 8 4c d+ = (8)

Subtracting (6) from (8)

3 4 4 0d= = 0d=

From 4 2c d+ = 2c=

2c= , d = 0 satisfy eq (7) also

Thus 1, 2, 2, 0a b c d = = =

Hence1 2

X2 0

=



GenextStudents 70

Q 12. If A and B are square matrices of the same that AB = BA, then prove by induction that

AB B An n= . Further, prove that (AB) A Bn n n= for all Nn .

Ans.: Let ( ) : n nP n AB B A=

But 1n= , AB = BA (given)

( )P n is true for n= 1 Let ( )P n be true for n= k.

k kAB B A=

Multiplying both side by B

L.H.S. =1( )k k kAB B A B B AB + = = Associateive property

R.H.S. = ( ) ( )k kB A B B AB=

= ( )kB BA AB BA=

= ( )k

B B A

=1kB A+ is P(n) is true for 1n k= +

By principle of mathematical induction

( )P n is true for all Nn

(ii) ( ) : ( )n n n

P n AB A B=

For 1n= L.H.S. = AB and R.H.S. = AB

( )P n be true for n = 1

Let ( )P n be true for n k=

( )k k k

AB A B=

Multiply both sides by AB

L.H.S. =1( ) ( ) ( )k kAB AB AB +=

R.H.S. =k k

A B (AB) ( )k kA B BA= AB = BA



GenextStudents 71

=1( ) ( )k k k k A B B A A B A

+ = n nAB B A =

=1 1 1

( )k k k k A AB A B+ + +=

( )P n is true for 1n k= +

By principle of mathematical induction

( )P n is true for all n N .

Choose the correct answer in the following questions:

Q 13. If A =

is such that2

A I= , then

(A)21 0 + + = (B) 21 0 + =

(C)21 0 = (D) 21 0 + =

Ans.: A2=

=

2

2

+

+ =

1 0

0 1

2A 1=

2 1 + = or 21 0 =

Part (C) is the required answer.

Q 14. If the matrix A is both symmetric and skew symmetric, then

(A) A is a diagonal matrix (B) A is a zero matrix

(C) A is a square matrix (D) none of these

Ans.: A is symmetric matrix if ij jia a=

A is a skew symmetric matrix if ij jia a=

Ifij ij jia a a= = 0ija =

A is zero matrix



GenextStudents 72

Part (B) is the correct answer.

Q 15. If A is square matrix such that A2= A, then (I + A)

3 7 A is equal to

(A) A (B) I A (C) I (D) 3A

Ans.:3 2A = A .A but 2A =A

=2A.A = A = A

3(1 + A) 7A = 2 3(1 3 3 ) 7A A A A+ + +

= (1 3 3 ) 7A A A A+ + +

= (1 7 ) 7 1A A+ =

Hence, Part (C) is the correct answer.

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Matrices notes and ncert solution class 12 cbse

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