Based on NCERT/CBSE, XI/XII class Syllabus

1. CHAPTER-03
Matrices
PUC-II MATHEMATICS
1 M 2M 3M 4M 5M 6M TOTAL
1 - 1 - 1 _ 9

2. One Mark questions
1. Define a scalar matrix. (K)
Scalar Matrix is a diagonal matrix whose principal diagonal
elements are same
4 0 0
0 4 0
0 0 4
 
 
 
  
Examples
5 0 0
0 5 0
0 0 5
 
  
  

3. One Mark questions
2. Define an Identity matrix. (K)
Identity Matrix is a scalar matrix whose principal diagonal
elements are equal to 1
 
   
  
3
1 0 0
0 1 0
0 0 1
I IExamples 1 0
I
0 1

4. One Mark questions
3. Define a diagonal matrix.
Diagonal matrix is a square matrix whose non- principal diagonal
elements are zero irrespective of principal diagonal elements.

5. One Mark questions
4. In the matrix 2 5 19 7
5
35 2 12
2
3 1 5 17
Write: (i) the order of the matrix
(ii) The number of elements
(iii) Write the elements , , , ,
R1
R2
R3
C1 C2 C3 C4
=12
a13 a21 a33 a42a23
a13
a21
a33a42
a23
a24
3 4
a24

6. One Mark questions
5. If a matrix has 24 elements, what are
the possible orders it can have? What,
if it has 13 elements?
Soln: 1 24 24 1
2 12 12 2
3 8 8 3
4 6 6 4

 

 

Soln: 1 13 13 1

7. One Mark questions
6. If a matrix has 18 elements, what are
the possible orders it can have? What,
if it has 5 elements?
Soln: 1 18 18 1
2 9 9 2
3 6 6 3

 

 
Soln: 1 5 5 1

8. One Mark questions
8. If a matrix has 8 elements, what are the
possible orders it can have?

9. One Mark questions
7. Find the number of all possible matrices
of order 3 ×3 with each entry 0 or 1 ?
Solution: since order is 3 ×3
∴ matrices having 9 elements (entries)
Given that: Each entry is either 0 or 1
i.e. TWO ways to be filled an entry .
∴According to Fundamental Principle of counting
9 entries can be filled in 2 × 2×2 ×…….. 9 times
∴ No. of possible matrices =
9
2

10. One Mark questions
9. Construct a 2 × 2 matrix,
11 12
21 22
a a
a a
 
 
 
A =
whose elements are given by ija i j 
Solution : Let 2 × 2 matrix
ija i j 
11 1 1 2a   
12 1 2 3a   
21 2 1 3a   
22 2 2 4a   
∴Required matrix is
2 3
3 4
 
 
 
A =
11For , put 1& 1a i j 
Given

11. One Mark questions
9. Construct a 2 × 2 matrix,
11 12
21 22
a a
a a
 
 
 
A =
whose elements are given by
Solution : Let 2 × 2 matrix
   11
1
3 1 1 2
2
a   
   12
1 5
3 1 2
2 2
a   
   21
1 7
3 2 1
2 2
a   
   22
1
3 2 2 4
2
a   
∴Required matrix is
5
2
2
7
4
2
 
 
 
 
  
A =
11For , put 1& 1a i j 
Given
1
3
2
ija i j 
1
3
2
ija i j 

12. One Mark questions
9. Construct a 2 × 2 matrix,   ijA = a
whose elements are given by;
) ij
i
iii a
j
 ) ijiv a i j 

13. One Mark questions
9. Construct a 2 × 2 matrix,   ijA = a
whose elements are given by;
) 2ijv a i j 
 
2
) ijvi a i j   
2
)
2
ij
i j
viii a


 
2
2
)
2
ij
i j
vii a



14. One Mark questions
9. Construct a 3 × 3 matrix,
11 12 13
21 22 23
31 32 33
A =
a a a
a a a
a a a
 
 
 
  
whose elements are given by
Solution : Let 2 × 2 matrix
   11
1
1 3 1 1
2
a       12
1 5
1 3 2
2 2
a   
   21
1 1
2 3 1
2 2
a       22
1
2 3 2 2
2
a   
∴Required matrix is
Given
1
3
2
 ija i j
5
1 4
2
1 7
2
2 2
3
0 3
2
 
 
 
 
 
 
 
  
A =
   13
1
1 3 3 4
2
a   
   23
1 7
2 3 3
2 2
a   
   31
1
3 3 1 0
2
a       32
1 3
3 3 2
2 2
a       32
1
3 3 3 3
2
a   
1
3
2
 ija i j

15. One Mark questions
11. Construct a 3 × 4 matrix, whose elements
are given by:
1
) 3
2
iji a i j   ) 2ijii a i j 

16. One Mark questions
12. Find the values of x, y and z if
4 3 y z
x 5 1 5
Solution: By definition of Equality of Two matrices, we have corresponding elements are equal
4 y 3 z 1x 
∴ 1x  4y  3z 

17. One Mark questions
12. Find the values of x, y and z if
Solution: By definition of Equality of Two matrices, we have corresponding elements are equal
9x y z   5x z  7y z 
7 9x 
4y 
3z 
x y z 9
x z 5
y z 7
2x 
2 5z 
3 7y  
∴ 2x  4y  3z 
7
y z


18. One Mark questions
13. Find x and y, if
1 3 y 0 5 6
2
0 x 1 2 1 8
Solution: By definition of Scalar Multiplication of a Matrix, Addition & Equality of Two matrices, we
have
2x  2 1
3x 
2 8 5y 
3y 

19. One Mark questions
14. if
2 1 10
x y
3 1 5
Find the
values of x & y
   2 1 10 ........ 1x y  
   3 1 5.......... 2x y 
5 15
3x
x 
 
Adding (1) and (2),we get
Substitute x=3 in eqn (2),we get
 3 3 5
4
y
y
 
  
Solution:

20. One Mark questions
15. Find X, if &
3 2
Y
1 4
1 0
2X Y
3 2
1 0
2X Y
3 2
Solution:
By properties of Matrix addition
1 0
2X Y
3 2
∴
1 0 3 2
2X
3 2 1 4
2 2
2X
4 2
By Scalar Multiplication of a Matrix
1 1
X
2 1
Given

21. One Mark questions
16. Find the values of x and y from the following equation
x 5 3 4 7 6
2
7 y 3 1 2 15 16
Solution :
 2 3 7 2 3 2 16x y    
By definition of Scalar Multiplication of a Matrix, Addition & Equality of Two matrices,
we have
2 7 3 2 -6 2 16x y   
∴ 2 10x y 

22. One Mark questions
17. Find the value of a, b, c and d from the equation:
a b 2a c 1 5
2a b 3c d 0 13
Solution : By definition of Equality of Two matrices, we have
 1........ 1a b  
 2 0 .... 2a b 
 2 5....... 3a c 
 3 13....... 4c d 
From (2),  2 .... *b a
Substitute in (1),we get
2 1
1
a a
a
  
 
From (*) b = 2
From (3) c = 3
Substitute c=1 in (4)
4d
1, 2
3& 4
  
 
a b
c d

23. One Mark questions
18. Show that the matrix
1 1 5
A 1 2 1
5 1 3
is a symmetric matrix
Solution: Since A'
1 1 5
1 2 1 A
5 1 3
A is a symmetric matrix
∴

24. One Mark questions
19. Show that the matrix
0 1 1
A 1 0 1
1 1 0
is a skew symmetric matrix
Solution:
∴
'
0 1 1
Since A 1 0 1
1 1 0
A is a skew symmetric matrixA

25. One Mark questions
22. Given Find
1 2 3
A and
2 3 1
3 1 3
B
1 0 2 2A B
Solution:
1 2 3 3 1 3
2A B 2
2 3 1 1 0 2
2 4 6 3 1 3
4 6 2 1 0 2
2 3 4 1 6 3
4 1 6 0 2 2
1 5 3
5 6 0

26. One Mark questions
23. Given
3 1 1
A and
2 3 0
2 5 1
B 1
2 3
2
Find A B

27. One Mark questions
25.Find AB
6 9
A
2 3
and
2 6 0
B
7 9 8
If
Solution :
6 9 2 6 0
AB
2 3 7 9 8
6 2 9 7 6 6 9 9 6 0 9 8
2 2 3 7 2 6 3 9 2 0 3 8
75 117 72
25 39 24

28. One Mark questions
26. if
1 0
A
0 1
and
0 1
B
1 0
then prove that
0 1
i AB
1 0
0 1
ii BA
1 0

29. One Mark questions
27. Simplify cos sin sin cos
cos sin
sin cos cos sin
   
 
   
Solution:
cos sin sin cos
cos sin
sin cos cos sin
   
 
   
     
     
cos cos sin sin sin cos
cos sin cos sin cos sin
2 2
2 2
     
     
cos sin cos sin sin cos
cos sin sin cos cos sin
2 2
2 2
1 0
0 1
I
 
  
 
Addition of Matrices
Scalar Multiplication of a Matrix
2 2
cos sin 1  

30. One Mark questions
28. Find
1
P if it exists , given 10 2
P
5 1
Solution: 10 2
Since P 0
5 1
does not exist
Hence P is a Singular Matrix
1
P

31. One Mark questions
Solution:
30.If
3 3 2
A
4 2 0
2 1 2
B
1 2 4
Verify that ' 'i A A ' ' 'ii A B A B
'
3 4
3 3 2
A A 3 2
4 2 0
2 0
'
2 1
2 1 2
B B 1 2
1 2 4
2 4
'
3 4
i A 3 2
2 0
' '
3 3 2
A A
4 2 0
Clearly '
5 5
5 3 1 4
A B A B 3 1 4
5 4 4
4 4
Clearly ' '
3 54 2 1 5
A B 3 2 1 2 3 1 4
2 0 2 4 4 4

32. Solution:
31. Compute the following
2 2 2 2
2 2 2 2
a b b c 2ab 2bc
i
2ac 2aba c a b
2 2 2 2
2 2 2 2
a b b c 2ab 2bc
2ac 2aba c a b
2 2 2 2
2 2 2 2
a b 2ab b c 2bc
a c 2ac a b 2ab
2 2
2 2
a b b c
a c a b

33. 31. Compute the following
a b a b
ii
b a b a
1 4 6 12 7 6
iii 8 5 16 8 0 5
2 8 5 3 2 4
cos sin sin cos
sin cos cos sin
2 2 2 2
2 2 2 2
x x x x
iv
x x x x

34. a b a b
i
b a b a
32. Compute the indicated products:
Solution:
a b a b
b a b a
2 2
2 2
2 2
2 2
2 2
2 2
a b ab ba
ba ab b a
a b 0
0 a b
1 0
a b
0 1
a b I

35. 1
ii 2 2 3 4
3
32. Compute the indicated products:
Solution: 1
2 2 3 4
3
1 3Order 3 1 3 3
2 3 4
4 6 8
6 9 12

36. 32. Compute the indicated products:
1 2 1 2 3
iii
2 3 2 3 1
2 3 4 1 3 5
iv 3 4 5 0 2 4
4 5 6 3 0 5
2 1
1 0 1
v 3 2
1 2 1
1 1
2 3
3 1 3
vi 1 0
1 0 2
3 1

37. Three Mark questions
1. Find the values of x, y and z from the following equations
x y 2 6 2
5 z xy 5 8
By definition of Equality of Two matrices, we have
   6..... 1 8..... 2x y xy  
Solution:

38. x y 2 6 2
5 z xy 5 8
Solution: Given
By definition of Equality of Two matrices, we have
   6..... 1 8..... 2x y xy  
 1 6From y x  
  
2
2
6 - 8
- 6 8 0
4 2 =0
4 or 2
From (1) 2 or 4
x x
x x
x x
x x
y y

 
 
  
   
   2 6 8becomes x x 

39. 2. Solve the equation for x, y, z and t, if
x z 1 1 3 5
2 3 3
y t 0 2 4 6
Solution: Given
x z 1 1 3 5
2 3 3
y t 0 2 4 6
By definition of Scalar Multiplication of a matrix & Addition ,we have
2 3 9x  2 3 15z   2 0 12y   2 6 18t  
3x  9z  6y  6t 
By definition of Equality of Two matrices, we have
2x 2z 3 3 9 15
2y 2t 0 6 12 18
2 3 2 3 9 15
2 0 2 6 12 18
x z
y t
    
       

40. 3. Given x y x 6 4 x y
3
z w 1 2w z w 3
find the values of x, y, z and w
By definition of Scalar Multiplication of a matrix, Addition & Equality of Two matrices,
we have
Solution: Given
x y x 6 4 x y
3
z w 1 2w z w 3
3 4 2x x x   
   3 6 using 2 we get 4y x y x y     
3 2 3 3w w w   
   3 1 Using w=3 we get 1z z w z     

41. 4. Find the values of a, b, c and d from the following equation
2a b a 2b 4 3
5c d 4c 3d 11 24
Solution: By definition of Equality of Two matrices, we have
......
....
2a b 4 1
a 2b 3 2
.....
.....
5c d 11 3
4c 3d 24 4
Consider 2(1)+(2), we get
4a 2b 8
a 2b 3
+
5 5 1a a  
From (1) we get 2b 
Consider 3(1)+(2), we get
15c 3d 33
4c 3d 24
+
19 57 3c c  
From (3) we get 4d 

42. 5. If
x 3 z 4 2y 7 0 6 3y 2
6 a 1 0 6 3 2c 2
b 3 21 0 2b 4 21 0
Find the values of a, b, c, x, y and z.
Solution: By definition of Equality of Two matrices, we have
3, 2, 5 2, 1, 7x z y a c b          

43. 6. Using elementary transformations, find the inverse of
1 1
2 3
Solution: Let given matrix be
1 1
A
2 3
Find , sing Elementary Row Operation(ERO)1
A U
Consider A IA
Apply '2 2 1
R R 2R
1 1 1 0
A
2 3 0 1
1 0
1 1
A2 1
0 1
5 5
1 1 1 0
A
0 5 2 1
Apply '2 2
1
R R
5
Apply '1 1 2
R R R
0
3 1
1 5 5
A
0 1 2 1
5 5
. .i e I BA
By definition of Invertible
Matrix, 1
B A

1
3 1
5 5
2 1
5 5
A
 
 
   
 
  

44. Using elementary transformations, find the inverse of
Solution: Let given matrix be
Find , sing Elementary Row Operation(ERO)1
A U
Consider A IA
Apply 1 2
R R
2 6 1 0
A
1 2 0 1
0 1
1 2
A1
0 1 1
2
1 2 0 1
A
2 6 1 0
Apply '2 2
1
R R
2
Apply '1 1 2
R R 2R
. .i e I BA
By definition of Invertible
Matrix, 1
B A

2 6
1 22 6
A
1 2
Apply '2 2 1
R R 2R
1 2 0 1
A
0 2 1 2
1 3
1 0
A1
0 1 1
2
1
1 3
A 1
1
2

45. Using elementary transformations, find the inverse of
Solution: Let given matrix be
Find , sing Elementary Row Operation(ERO)1
A U
Consider A IA
Apply '1 1 2
R R R
4 5 1 0
A
3 4 0 1
Apply '1 1 2
R R R
. .i e I BA
By definition of Invertible
Matrix, 1
B A

Apply '2 2 1
R R 3R
1
4 5
A
3 4
4 5
3 44 5
A
3 4
1 1 1 1
A
3 4 0 1
1 1 1 1
A
0 1 3 4
1 0 4 5
A
0 1 3 4

46. Using elementary transformations, find the inverse of
Solution: Let given matrix be
Find , sing Elementary Row Operation(ERO)1
A U
Consider A IA
Apply '1 1 2
R R R
2 3 1 0
A
1 2 0 1
Apply '1 1 2
R R R
. .i e I BA
By definition of Invertible
Matrix, 1
B A

Apply '2 2 1
R R R
1
2 3
A
1 2
2 3
1 22 3
A
1 2
1 1 1 1
A
1 2 0 1
1 1 1 1
A
0 1 1 2
1 0 2 3
A
0 1 1 2

47. Using elementary transformations, find the inverse of
Solution: Let given matrix be
Find , sing Elementary Row Operation(ERO)1
A U
Consider A IA
Apply '2 2 1
R R 2R
3 1 1 0
A
5 2 0 1
Apply '2 2 1
R R 3R
. .i e I BA
By definition of Invertible
Matrix, 1
B A

Apply '2 2
R 1 R
1
2 1
A
5 3
3 1
A
5 2
3 1
5 2
3 1 1 0
A
1 0 2 1
3 1 1 0
A
1 0 2 1
Apply 1 2
R R
1 0 2 1
A
3 1 1 0
1 0 2 1
A
0 1 5 3

48. 8. Find X and Y, if
7 0
X Y
2 5
and
3 0
X Y
0 3
Solution: Given
.....
7 0
X Y 1
2 5
3 0
X Y
0 3
Adding (1) & (2),We get
7 0 3 0
X Y X Y
2 5 0 3
10 0
2X
2 8
5 0
X
1 4
From (1)
7 0
Y X
2 5
7 0 5 0
Y
2 5 1 4
2 0
Y
1 1

49. 9. Find X and Y, if
2 3
2X 3Y
4 0
2 2
3X 2Y
1 5
Solution: Given that
......
2 3
2X 3Y 1
4 0
......
2 2
3X 2Y 2
1 5
Consider 3(1) – 2(2), We get
2 3
6X 9Y 3
4 0
2 2
6X 4Y 2
1 5
__
__
(--)(--)
-
6 9 4 4
5Y
12 0 2 10
2 13
5Y
14 10
2 13
5 5
Y
14
2
5
Consider 2(1) – 3(2), We get
4 6 6 6
5X
8 0 3 15
2 12
2 12 5 5
5X X
11 15 11
3
5

50. 10. If
8 0
A 4 2
3 6
and
2 2
B 4 2
5 1
then find the matrix X, such that 2A 3X 5B
Solution : For Given
8 0
A 4 2
3 6
2 2
B 4 2
5 1
By properties of Matrix Addition & Scalar Multification of a matrix
2A 3X 5B
3X 5B 2A
2 2 8 0
3X 5 4 2 2 4 2
5 1 3 6
10 10 16 0
3X 20 10 8 4
25 5 6 12
6 10
3X 12 14
31 7
10
6
3
3
14
X 4
3
31
7
3
3

51. Three Mark questions
11. Find X and Y, if 5 2
X Y
0 9
and
3 6
X Y
0 1
12. If
2 5
1
3 3
1 2 4
A
3 3 3
7 2
2
3 3
and
2 3
1
5 5
1 2 4
B
5 5 5
7 6 2
5 5 5
Then compute
3A 5B

52. 13. if
cos sin
( ) sin cos
x x 0
F x x x 0
0 0 1
Show that
F x F y F x y
Solution:
LHS= F x F y
cos sin cos sin
sin cos sin cos
x x 0 y y 0
x x 0 y y 0
0 0 1 0 0 1
cos cos sin sin cos siny sinxcosy
sin cos cos sin sin sin cos cos
cos sin
sin cos
x y x y x 0
x y x y x y x y 0
0 0 1
x y x y 0
x y x y 0
0 0 1
F x y RHS

53. 15.(i) if
cos sin
sin cos
A
 
 
then verify that
'A A I
Solution: Given that
 
 
cos sin
sin cos
A
 
 
cos sin
'
sin cos
A
   
   
cos sin cos sin
'
sin cos sin cos
LHS A A
    
    
cos sin sin coscos sin
sin cossin cos cos sin
2 2
2 2
1 0
0 1
I RHS

54. 14. Show that
5 1 2 1 2 1 5 1
6 7 3 4 3 4 6 7
(ii) if
sin cos
cos sin
A
 
 
then verify that
'A A I

55. 16.Show that
1 2 3 1 1 0 1 1 0 1 2 3
0 1 0 0 1 1 0 1 1 0 1 0
1 1 0 2 3 4 2 3 4 1 1 0
Solution:
1 2 3 1 1 0
LHS 0 1 0 0 1 1
1 1 0 2 3 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 6 14
0 1 1
1 0 1
1 1 0 1 2 3
RHS 0 1 1 0 1 0
2 3 4 1 1 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
By Definition of equality of Two Matrices LHS RHS

56. 17. if
1 2 3
A 5 7 9
2 1 1
and
4 1 5
B 1 2 0
1 3 1
then verify that
' ' 'i A B A B
' ' 'ii A B A B

57. 18. if '
3 4
A 1 2
0 1
and
1 2 1
B
1 2 3
then verify that
' ' 'i A B A B
' ' 'ii A B A B

58. 19. if '
2 3
A
1 2
and
1 0
B
1 2
then find 'A 2B
Solution: Given that
Consider
2 1 1 0
A 2B 2
1 2 1 2
2 1 2 0
1 2 2 4
2 2 1 0
1 2 2 4
A+2B
4 1
3 6
 
4 3
2 '
1 6
A B
 
    
 

59. 20. if
1 2 3
A
4 2 5
and
2 3
B 4 5
2 1 AB BA
then find AB, BA
Show that
Solution: Given that
1 2 3
A
4 2 5
=
2 3
1 2 3
AB 4 5
4 2 5
2 1
2 8 6 3 10 3
8 8 10 12 10 5
0 4
10 3
2 3
B 4 5
2 1
=
2 3
1 2 3
BA 4 5
4 2 5
2 1
2 12 4 6 6 15
4 20 8 10 12 25
2 4 4 2 6 5
10 2 21
16 2 37
2 2 11By definition of Equality of Two matrices
AB ≠ BA

60. 21. If A and B are symmetric matrices of the same order, then show that
AB is symmetric if and only if A and B are commutative, that is AB = BA.
Solution: A and B are symmetric matrices of the same order
First, We consider AB is symmetric
To Prove : AB = BA
 'AB AB
Now Consider
' and 'A A B  B
 '
' '
But ' & '
=
Hence proved
LHS AB AB
B A
A A B B
BA
 

 

Secondly,We consider AB BA
To Prove : AB is symmetric
Now Consider  
 
'
' '
But ' & '
=
'=
Hence AB is symmetric matrix
LHS AB
B A
A A B B
BA
AB AB


 



61. 22. If
3 2
A
4 2
and 1 0
I
0 1
Find k so that
2
A kA 2I
Solution: Given that
3 2
A
4 2
1 0
I
0 1
&
We have
2
=
A =
2
A AA
3 2 3 2
4 2 4 2
9 8 6 4
12 8 8 4
1 2
4 4
Given that 2
A kA 2I
1 2 3 2 1 0
k 2
4 4 4 2 0 1
1 2 3k 2 2k 0
4 4 4k 0 2k 2
3k 2 1
k 1

62. 23. For the matrix
1 5
A
6 7
verify that
'i A A is a symmetric matrix
'ii A A is a skew symmetric matrix
Solution:Given that '
1 5 1 6
A A
6 7 5 7
'
'
1 5 1 6
i A A
6 7 5 7
2 11
A A X say
11 14
2 11
Since '
11 14
' is symmetric
X X
X A A
 
  
 
  
'
'
1 5 1 6
ii A A
6 7 5 7
0 1
A A X say
1 0
0 1
' -
-1 0
' is skew-symmetric
Since X X
X A A
 
  
 
  

63. Solution:Given that
24. Find '
1
A A
2
'
1
and A A
2
When
0 a b
A a 0 c
b c 0

64. Solution: Let A be the given matrix
25. Express the matrix as the sum of a symmetric and
skew symmetric matrix:
6 2 2
2 3 1
2 1 3
'
6 2 2 6 2 2
A 2 3 1 A 2 3 1
2 1 3 2 1 3
We have    
 
 
 
1 1
' '
2 2
= *
6 2 2
1
' = 2 3 1
2
2 1 3
0 0 0
1
& ' 0 0 0
2
0 0 0
A A A A A
P Q say
where P A A
Q A A
   

 
     
  
 
     
  
6 2 2
Clearly ' 2 3 1 =P
2 1 3
P is a symmetric matrix
P
 
    
  

0 0 0
' 0 0 0
0 0 0
is a skew-symmetric matrix
Clearly Q Q
Q
 
   
  

From * , matrix A sum of symmetric &
skew-symmetric matrices
is

65. Solution: Let A be the given matrix
25. Express the matrix as the sum of a symmetric and
skew symmetric matrix:
'
2 2 2 2 1 1
B 1 3 4 B 2 3 2
1 2 3 2 4 3
We have    
 
 
 
 
      
      
      
          
  
 
 
 
  
1 1
B = B + B' + B - B'
2 2
= P + Q say *
-3 32
2 22 -2 2 2 -1 1
1 1 -3where P = B + B' = -1 3 4 + -2 3 - 2 P = 3 1
22 2
1 -2 -3 2 4 - 3 3 1 - 3
2
2 -2 2 2 -1
1 1
& Q = B - B' = -1 3 4 -
2 2
1 -2 -3
1
1
 
    
    
    
      
 
10
2 21
-2 3 - 2 Q = 0 3
2
2 4 - 3 -1 -3 0
2
2 2 2
B 1 3 4
1 2 3
Clearly ' and ' is symmetric and is skew-symmetricP P Q Q P Q   
Therefore from * B= symmetric matrix +skew-symmetric matrix

66. 25. Express the following matrix as the sum of a
symmetric and skew symmetric matrix: 3 5
1 1Solution:

67. 30.If A and B are invertible matrices of the same order,
then prove that
1 1 1
AB B A
Proof: Given that A and B are invertible matrices
1 1
AA A A I 
  1 1
BB B B I 
 And
    
 
1 1 1 1
1 1
1
1
Consider
=
= is invertible.
=
=
AB B A A B B A
A BB A
A IA B
A A
I
   
 


   
 
 
  
  
    
 
1 1 1 1
1 1
1
1
Consider
matrix multiplication is associative
=
= B is invertible.
=
B A AB B A A B
B A A B
B I A
B
   
 


   
 
 
  

=
B
I

Therefore AB is invertible &
1 1 1
AB B A

68. 31.Prove that for any square matrix A with real number entries,
A+ A’ is a symmetric matrix and A- A’ is a skew symmetric matrix.
Solution : Let A be a square matrix
To show : A+A’ is a symmetric matrix
Let 'X A A 
 
 
  
Consider ' ' '
= ' ' '
= ' ' '
= '
'=
' is symmetric matrix
X A A
A A
A A A A
A A
X X
X A A
 

 

  
'Y A A 
 
 
  
 
Consider ' ' '
= ' ' '
= ' ' '
= - '
'=
' is skew-symmetric matrix
Y A A
A A
A A A A
A A
Y Y
Y A A
 

 


  
To show : A-A’ is skew symmetric matrix

69. 32. Prove that any square matrix can be expressed as
the sum of symmetric and skew symmetric matrix
Solution:
1
2
2
1
2
1
' '
2
1 1
' '
2 2
=
A A
A A
A A A
A A A A A
A A A A A
P Q
Let A be the square matrix
   
   
   
   
 
1 1
' and '
2 2
1 1
' ' ' ' ' '
2 2
1 1
= ' ' ' ' ' ' '
2 2
1 1
= ' ' '
2 2
1
' = ' '
2
where P A A Q A A
P A A Q A A
A A Q A A
A A Q A A
P P Q A A
   
   
        
  
  
'Q Q
Any Square matrix can be expressed as the sum of symmetric and skew -symmetric matrix

70. 33.Prove inverse of a square matrix, if it exist, is unique
Solution: Let A be a square matrix
To prove : Uniqueness of the Inverse of a Matrix
This can be proved by Contradiction Method
Let us suppose that If possible matrix A has two inverses B & C
Since B is a inverse of A ..... *AB BA I
Since C is a inverse of A .... * *AC CA I
matrix multiplication is associative
Identity law
B BI
B AC AC I
IC BA I
C
BA C
Now , We have
This is the contrdiction to our suppose therefore Matrix A has only one inverse

71. Five Marks questions

72. 1. If  
2
4 1 3 6
5
A and B
 
    
  
Verify that ( )AB B A  
Solution: Consider
 
2 2 6 12
4 1 3 6 4 12 24
5 5 15 30
AB
     
         
      
 
2 4 5
' 6 12 15
12 24 30
LHS AB
 
    
   
   
2 1
4 ' 2 4 5 and 1 3 6 ' 3
5 6
A A B B
   
             
      
 
1
' ' 3 2 4 5
6
2 4 5
= 6 12 15
12 24 30
RHS B A
 
    
  
 
  
   
LHS = RHS
( )AB B A  

73. 2. If
1 2 3 3 1 2 4 1 2
5 0 2 , 4 2 5 0 3 2
1 1 1 2 0 3 1 2 3
A B and C
      
            
           
then compute that (A+B) & (B-C)Also verify that A+(B-C)=(A+B)-C
Solution:
1 2 3 3 1 2 4 1 1
5 0 2 4 2 5 9 2 7
1 1 1 2 0 3 3 1 4
A B
       
             
           
3 1 2 4 1 2 1 2 0
, 4 2 5 0 3 2 4 1 3
2 0 3 1 2 3 1 2 0
B C
       
              
          
 
1 2 3 1 2 0 0 0 3
5 0 2 4 1 3 9 1 5
1 1 1 1 2 0 2 1 1
LHS A B C
        
                 
          
 
4 1 1 4 1 2 0 0 3
9 2 7 0 3 2 9 1 5
3 1 4 1 2 3 2 1 1
A BRHS C
      
             
         
 


 
LHS RHS 

74. 3. Find
2
A 5A 6I 2 0 1
if A 2 1 3
1 1 0
Solution:
2 0 1
A 2 1 3
1 1 0
2
3 3 2 2 0 1 1 0 0
A 5A 6I 9 2 5 5 2 1 3 6 0 1 0
0 1 2 1 1 0 0 0 1
=2
2 0 1 2 0 1 2 0 1 4 0 1 2 0 0 3 3 2
A AA 2 1 3 2 1 3 4 2 3 0 1 3 2 3 0 9 2 5
1 1 0 1 1 0 2 2 0 0 1 0 1 3 0 0 1 2
3 3 2 10 0 5 6 0 0 3 10 6 3 0 0 2 5 0
9 2 5 10 5 15 0 6 0 9 10 0 2 5 6 5 15 0
0 1 2 5 5 0 0 0 6 0 5 0 1 5 0 2 0 6
1 3 3
1 1 10
5 4 4

75. Five Mark questions
5.If
0 6 7
A 6 0 8
7 8 0
0 1 1
B 1 0 2
1 2 0
2
C 2
3
Calculate AC, BC and (A + B) C. Also,
verify that (A + B)C = AC + BC.

76. Five Mark questions
6. if
1 1 1
A 1 2 3
2 1 3
then show that
3
A 23A 40I O

77. Five Mark questions
7. If
1 0 2
A 0 2 1
2 0 3
then show that
3 2
A 6A 7A 2I O

78. 8. Let
2 1
A
3 4
5 2
B
7 4
2 5
C
3 8
Find a matrix D such that CD AB 0
Solution: Since matrices A & B are of order 2×2
therefore AB is also a matrix of order 2×2
Since C is the matrix of order 2×2 therefore D must be the matrix of 2×2. Let
a b
D
c d
 
  
 
0CD AB 
2 5 2 1 5 2
0
3 8 3 4 7 4
a b
c d
       
        
       
2 5 2 5 10 7 4 4
0
3 8 3 8 15 28 6 16
a c b d
a c b d
      
          
2 5 3 2 5 0 0 0
3 8 43 3 8 22 0 0
a c b d
a c b d
      
         
2 5 3 0
2 5 0 0
3 8 43 0
3 8 22 0
a c
b d
a c
b d
  
  
  
  
Solving these
equations by
Cross
Multiplication
Method

79. 9. If tan
tan
0
2A
02


and I is identity matrix of order 2, Show that
cos sin
sin cos
tan cos sin
tan sin cos
tan cos sin
sin costan
cos tan sin sin tan cos
tan cos
RHS I A
01 0
2
0 1 02
1
2
1
2
2 2
2
 
 
  
  

 
  
    
  sin tan sin os
sin sin
cos sin cos sin cos
cos cos
sin sin
cos sin sin cos cos
cos cos
sin sin
2
2
2 2
c
2
2 22 2 1
2 2 2
2 2
2 22 1 2
2 2 2
2 2
1 2 2
2
  
 
   
 
 
   
 
 
sin
sin sin cos
cos
sin
sin cos sin sin sin
cos
tan
tan
2 2
22
2 2 2
2
22 2 1 2
2 2 2 2
2
1
2
12

 


   



Solution:
0 tan1 0
2
tan0 1 02
1 tan
2
=
tan 1
2
LHS I A




  
     
    
 
 
 
 

80.  
 
cos sin
sin cos
RHS I A
2 2
Useful sults
2 2
2 2 1 1 2



  
  
Re
sin
tan
cos
sin sin cos
cos cos sin
  
  
tan cos sin
tan sin cos
01 0
2
0 1 02

 
  
tan cos sin
sin costan
1
2
1
2
    
    
cos tan sin sin tan cos
tan cos sin tan sin os
2 2
c
2 2
 
   
 
 
   
 
sin sin
cos sin cos sin cos
cos cos
sin sin
cos sin sin cos cos
cos cos
2
2
2 22 2 1
2 2 2
2 2
2 22 1 2
2 2 2
2 2

   


   

sin
sin sin sin sin cos
cos
sin
sin cos sin sin sin
cos
2 2
2 2
21 2 2 2
2 2 2 2
2
22 2 1 2
2 2 2 2
tan
tan
1
2 LHS
12



86. One Mark questions
10. Construct a 3 × 3 matrix whose
elements are given by
1
3
2
 ija i j
11. Construct a 3 × 4 matrix, whose elements
are given by:
1
) 3
2
iji a i j   ) 2ijii a i j 

87. One Mark questions
12. Find the values of x, y and z from the
following equations:
4 3 y z
i
x 5 1 5
x y z 9
ii x z 5
y z 7

88. One Mark questions
13. Find x and y, if
1 3 y 0 5 6
2
0 x 1 2 1 8
14. if
2 1 10
x y
3 1 5
Find the
values of x & y

89. One Mark questions
15. Find X, if &
3 2
Y
1 4
1 0
2X Y
3 2
16. Find the values of x and y from the following equation
x 5 3 4 7 6
2
7 y 3 1 2 15 16

90. One Mark questions
17. Find the value of a, b, c and d from the equation:
a b 2a c 1 5
2a b 3c d 0 13
18. Show that the matrix
1 1 5
A 1 2 1
5 1 3
is a symmetric matrix

91. One Mark questions
19. Show that the matrix
0 1 1
A 1 0 1
1 1 0
is a skew symmetric matrix
20. let , ,
2 4 1 3 2 5
A B C
3 2 2 5 3 4
Find each of the following

92. One Mark questions
22. Given
3 1 1
A and
2 3 0
2 5 1
B 1
2 3
2
Find A B
23. Given
Find
1 2 3
A and
2 3 1
3 1 3
B
1 0 2
2A B

93. One Mark questions
25.Find AB If
6 9
A
2 3
and
2 6 0
B
7 9 8
26. if
1 0
A
0 1
and
0 1
B
1 0
then prove that
0 1
i AB
1 0
0 1
ii BA
1 0

94. One Mark questions
27. Simplify cos sin sin cos
cos sin
sin cos cos sin
   
 
   
28. Find
1
P if it exists , given
10 2
P
5 1

95. One Mark questions
29. Find the transpose of each of the following matrices
3 3 2
A
4 2 0
and
2 1 2
B
1 2 4
30.If
3 3 2
A
4 2 0
2 1 2
B
1 2 4
Verify that ' 'i A A ' ' 'ii A B A B

96. One Mark questions
31. Compute the following
a b a b
i
b a b a
2 2 2 2
2 2 2 2
a b b c 2ab 2bc
ii
2ac 2aba c a b
1 4 6 12 7 6
iii 8 5 16 8 0 5
2 8 5 3 2 4
cos sin sin cos
sin cos cos sin
2 2 2 2
2 2 2 2
x x x x
iv
x x x x

97. One Mark questions
32. Compute the indicated products:
a b a b
i
b a b a
1
ii 2 2 3 4
3
1 2 1 2 3
iii
2 3 2 3 1
2 3 4 1 3 5
iv 3 4 5 0 2 4
4 5 6 3 0 5

98. One Mark questions
2 1
1 0 1
v 3 2
1 2 1
1 1
2 3
3 1 3
vi 1 0
1 0 2
3 1
32. Compute the indicated products:

99. Three Mark questions
1. Find the values of x, y and z from the following equations
x y 2 6 2
5 z xy 5 8

100. Three Mark questions
2. Solve the equation for x, y, z and t, if
x z 1 1 3 5
2 3 3
y t 0 2 4 6

101. Three Mark questions
3. Given
x y x 6 4 x y
3
z w 1 2w z w 3
find the values of x, y, z and w

102. Three Mark questions
4. Find the values of a, b, c and d from the following equation
2a b a 2b 4 3
5c d 4c 3d 11 24

103. Three Mark questions
5. If
x 3 z 4 2y 7 0 6 3y 2
6 a 1 0 6 3 2c 2
b 3 21 0 2b 4 21 0
Find the values of a, b, c, x, y and z.

104. Three Mark questions
6. Using elementary transformations, find the inverse of each of the matrices
1 1
i
2 3
2 1
ii
1 1
1 3
iii
2 7
2 3
iv
5 7
2 1
v
7 4
2 5
vi
1 3

105. Three Mark questions
3 1
vii
5 2
4 5
viii
3 4
3 10
ix
2 7
2 6
x
1 2
6 3
xi
2 1
2 3
xii
1 2
Using elementary transformations, find the inverse of each of the matrices

106. Three Mark questions
7. By using elementary operations,
find the inverse of the matrix
1 2
A
2 1
8. Find X and Y, if
7 0
X Y
2 5
and
3 0
X Y
0 3

107. Three Mark questions
9. Find X and Y, if
2 3
2X 3Y
4 0
and
2 2
3X 2Y
1 5
10. If
8 0
A 4 2
3 6
and
2 2
B 4 2
5 1
then find the matrix X, such that 2A 3X 5B

108. Three Mark questions
11. Find X and Y, if 5 2
X Y
0 9
and
3 6
X Y
0 1
12. If
2 5
1
3 3
1 2 4
A
3 3 3
7 2
2
3 3
and
2 3
1
5 5
1 2 4
B
5 5 5
7 6 2
5 5 5
Then compute
3A 5B

109. Three Mark questions
13. if
cos sin
( ) sin cos
x x 0
F x x x 0
0 0 1
Show that
F x F y F x y
14. Show that
5 1 2 1 2 1 5 1
6 7 3 4 3 4 6 7

111. Three Mark questions
16.Show that
1 2 3 1 1 0 1 1 0 1 2 3
0 1 0 0 1 1 0 1 1 0 1 0
1 1 0 2 3 4 2 3 4 1 1 0
17. if
1 2 3
A 5 7 9
2 1 1
and
4 1 5
B 1 2 0
1 3 1
then verify that
' ' 'i A B A B
' ' 'ii A B A B

112. Three Mark questions
18. if '
3 4
A 1 2
0 1
and
1 2 1
B
1 2 3
then verify that
' ' 'i A B A B
' ' 'ii A B A B

113. Three Mark questions
19. if '
2 3
A
1 2
and
1 0
B
1 2
then find
'A 2B
20. if
1 2 3
A
4 2 5
and
2 3
B 4 5
2 1
AB BAthen find AB, BA Show that

114. Three Mark questions
21. If A and B are symmetric matrices
of the same order, then show that
AB is symmetric if and only if A and
B are commutative,
that is AB = BA.

115. Three Mark questions
22. If
3 2
A
4 2
and
1 0
I
0 1
Find k so that
2
A kA 2I

116. Three Mark questions
23. For the matrix
1 5
A
6 7
verify that
'i A A is a symmetric matrix
'ii A A is a skew symmetric matrix

117. Three Mark questions
24. Find '
1
A A
2
'
1
and A A
2
When
0 a b
A a 0 c
b c 0

118. Three Mark questions
25. Express the following matrices as the sum of a
symmetric and skew symmetric matrix:
3 5
1 1
6 2 2
2 3 1
2 1 3

119. Three Mark questions
26. Express the matrix
2 2 2
B 1 3 4
1 2 3
as the sum of a symmetric and a
skew symmetric matrix.

120. Three Mark questions
27. If A and B are symmetric matrices of
the same order, then show that AB is
symmetric if and only if AB = BA.
28.If A and B are invertible matrices of the
same order, then prove that
1 1 1
AB B A

121. Three Mark questions
30.Prove that for any square matrix A
with real number entries, A+ A’ is
a symmetric matrix and A- A’ is a
skew symmetric matrix.

122. Three Mark questions
32. Prove that any square matrix can be
expressed as the sum of symmetric
and skew symmetric matrix
33.Prove inverse of a square matrix, if
it exist, is unique

123. Five Marks questions
1. If  
2
4 1 3 6
5
A and B
 
    
  
Verify that
( )AB B A  

124. Five Marks questions
2. If
1 2 3 3 1 2 4 1 2
5 0 2 , 4 2 5 0 3 2
1 1 1 2 0 3 1 2 3
A B and C
      
            
           
then compute that (A+B) & (B-C)
Also verify that A+(B-C)=(A+B)-C

125. Five Mark questions
3. Find 2
A 5A 6I
2 0 1
if A 2 1 3
1 1 0

126. Five Mark questions
5.If
0 6 7
A 6 0 8
7 8 0
0 1 1
B 1 0 2
1 2 0
2
C 2
3
Calculate AC, BC and (A + B) C. Also,
verify that (A + B)C = AC + BC.

127. Five Mark questions
6. if
1 1 1
A 1 2 3
2 1 3
then show that
3
A 23A 40I O

128. Five Mark questions
7. If
1 0 2
A 0 2 1
2 0 3
then show that
3 2
A 6A 7A 2I O

129. Five Mark questions
8. Let
2 1
A
3 4
5 2
B
7 4
2 5
C
3 8
Find a matrix D such that CD AB 0

130. Five Mark questions
9. If
tan
tan
0
2A
02


and I is identity matrix of order 2, Show that
cos sin
sin cos
I A I A
 
 

131. Five Marks questions

133. CET questions
(A) (B)
(C) 3 (D) 2

134. Mr. Santosh T. M.Sc
BasavaJyoti Science & Commerce P.U. College, Jamkhandi
E-mail: santa1.6180@gmail.com
Contact : 9591319967