Upload
psingh1766
View
53
Download
0
Embed Size (px)
DESCRIPTION
best notes and solution from CBSE
Citation preview
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]
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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 =
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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)
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
GenextStudents 6
Total no. of elements = 29= 512
D is correct
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
+
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
+
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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:
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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)
+ + + +
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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 =
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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+ =
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
+ + + + + + + + + + + +
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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:
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
+
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
= B = symmetric matrix
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
= B = symmetric matrix
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
Operating 2 2 13R R R 1 1 1 1
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
GenextStudents 52
Operating 1 1 2R R R 1 4 1 1
A1 2 0 1
=
Operating 2 2 1R R R 1 4 1 1
A0 2 1 2
=
Operating 2 21
2R R
1 11 4
A10 1 1
2
=
Operating 1 1 24R R R +
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
GenextStudents 53
Operating 1 1 2R R R + 1 1 1 1
A1 2 0 1
=
Operating 2 2 1R R R + 1 1 1 1
A0 1 1 2
=
Operating 1 1 2R R R + 1 0 2 3
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
=
Operating 2 2 12R R R 2 1 1 0
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
Operating 1 1 24R R R +
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
Operating 2 2 38R R R +
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 +
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
Operating 1 1 22R R R
1 1 2 2 1 0
2 0 1 1 0 0 A
0 1 3 0 0 1
=
Operating 2 2 12R R R
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
=
Operating 2 2 3R R R
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
=
Operating 2 2 32R R R
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
GenextStudents 58
Ans.: If B is the inverse of A if AB = BA = I
Part (D) is the correct answer.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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)
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
=
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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
5/27/2018 Matrices notes and ncert solution class 12 cbse
12 Math Ch-3 Matrices
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.