Matrices Questions & Answers

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 400
040
004





Examples 500
050
005




 


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
II
Examples 10
I
01

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 25197
5
35212
2
31517





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 a42 a23
a13
a21
a33 a42
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
aa
aa



A=
whose elements are given by ij
a i j
Solution : Let 2 × 2 matrix ij
a i j 11
1 1 2a   12
1 2 3a   21
2 1 3a   22
2 2 4a  
∴Required matrix is 23
34



A= 11
For , put 1& 1a i j
Given

11. One Mark questions
9. Construct a 2 × 2 matrix, 11 12
21 22
aa
aa



A=
whose elements are given by
Solution : Let 2 × 2 matrix 
11
1
3 1 1 2
2
a   
12
15
3 1 2
22
a   
21
17
3 2 1
22
a   
22
1
3 2 2 4
2
a  
∴Required matrix is 5
2
2
7
4
2





A= 11
For , put 1& 1a i j
Given 1
3
2
ij
a i j 1
3
2
ij
a i j

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

13. One Mark questions
9. Construct a 2 × 2 matrix, 
ij
A = a
whose elements are given by; )2
ij
v a i j 
2
)
ij
vi a i j 
2
)
2
ij
ij
viii a

  
2
2
)
2
ij
ij
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
15
1 3 2
22
a   
21
11
2 3 1
22
a   
22
1
2 3 2 2
2
a  
∴Required matrix is
Given 1
3
2

ij
a i j 5
14
2
17
2
22
3
03
2








A= 
13
1
1 3 3 4
2
a   
23
17
2 3 3
22
a   
31
1
3 3 1 0
2
a   
32
13
3 3 2
22
a   
32
1
3 3 3 3
2
a   1
3
2

ij
a i j

15. One Mark questions
11. Construct a 3 × 4 matrix, whose elements
are given by: 1
)3
2
ij
i a i j   )2
ij
ii 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 4y 3z 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   5xz 7yz 79x 4y 3z x y z 9
x z 5
y z 7 2x 25z 37y
∴ 2x 4y 3z 7
yz


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 21 3x 2 8 5y 3y

19. One Mark questions
14. if 2 1 10
xy
3 1 5
Find the
values of x & y  2 1 10 ........ 1xy    3 1 5.......... 2xy 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 &
32
Y
14 10
2X Y
32 10
2X Y
32
Solution:
By properties of Matrix addition 10
2X Y
32
∴ 1 0 3 2
2X
3 2 1 4 22
2X
42
By Scalar Multiplication of a Matrix 11
X
21
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 16xy    
By definition of Scalar Multiplication of a Matrix, Addition & Equality of Two matrices,
we have 2 7 3 2 -6 216xy   
∴ 2 10xy

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........ 1ab   2 0 .... 2ab 2 5....... 3ac 3 13....... 4cd
From (2), 2 .... *ba
Substitute in (1),we get 21
1
aa
a
  

From (*) b = 2
From (3) c = 3
Substitute c=1 in (4) 4d 1, 2
3& 4
  

ab
cd

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 matrix A

25. One Mark questions
22. Given Find 1 2 3
A and
2 3 1 3 1 3
B
102 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 2 153
5 6 0

26. One Mark questions
23. Given 3 1 1
A and
230 2 5 1
B
1
23
2
Find AB

27. One Mark questions
25.Find AB 69
A
23 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 10
A
01
and 01
B
10 then prove that 01
i AB
10 01
ii BA
10

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
22
22      
     
cos sin cos sin sin cos
cos sin sin cos cos sin
22
22 10
01
I




Addition of Matrices
Scalar Multiplication of a Matrix 22
cos sin 1

30. One Mark questions
28. Find 1
P if it exists , given 10 2
P
51
Solution: 10 2
Since P 0
51 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 '
34
3 3 2
A A 3 2
4 2 0
20 '
21
2 1 2
B B 1 2
1 2 4
24 '
34
i A 3 2
20 ''
3 3 2
AA
4 2 0 Clearly '
55
5 3 1 4
A B A B 3 1 4
5 4 4
44 Clearly ' '
354 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 22
22
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 22
22
22
22
22
22
a b ab ba
ba ab b a
a b 0
0 a b
10
ab
01
a b I

35. 1
ii 2 2 3 4
3 32. Compute the indicated products:
Solution: 1
2 2 3 4
3 13 Order 3 1 33 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 21
1 0 1
v 3 2
1 2 1
11 23
3 1 3
vi 1 0
102
31

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 6 From y x   
2
2
6 - 8
-6 8 0
4 2 =0
4 or 2
From (1) 2 or 4
xx
xx
xx
xx
yy



  
     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 9 x 2 3 15 z 2 0 12y 2 6 18t 3 x 9 z 6 y 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
xz
yt
   

   
   

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 4 y 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 1aa   From (1) we get 2b
Consider 3(1)+(2), we get

15c 3d 33
4c 3d 24
+ 19 57 3cc   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 11
23
Solution: Let given matrix be 11
A
23 Find , sing Elementary Row Operation(ERO )
1
AU Consider A IA Apply '
2 2 1
R R 2R 1 1 1 0
A
2 3 0 1 10
11
A21
01
55 1 1 1 0
A
0 5 2 1 Apply '
22
1
RR
5 Apply '
1 1 2
R R R 0
31
1 55
A
0 1 2 1
55 . . i e I BA
By definition of Invertible
Matrix, 1
BA

 1
31
55
21
55
A



 




44. Using elementary transformations, find the inverse of
Solution: Let given matrix be Find , sing Elementary Row Operation(ERO )
1
AU Consider A IA Apply
12
RR 2 6 1 0
A
1 2 0 1 01
12
A1
01 1
2 1 2 0 1
A
2 6 1 0 Apply '
22
1
RR
2 Apply '
1 1 2
R R 2R . . i e I BA
By definition of Invertible
Matrix, 1
BA

 26
12 26
A
12 Apply '
2 2 1
R R 2R 1 2 0 1
A
0 2 1 2 13
10
A1
01 1
2 1
13
A 1
1
2

45. Using elementary transformations, find the inverse of
Solution: Let given matrix be Find , sing Elementary Row Operation(ERO )
1
AU 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
BA

 Apply '
2 2 1
R R 3R 1
45
A
34 45
34 45
A
34 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
AU 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
BA

 Apply '
2 2 1
R R R 1
23
A
12 23
12 23
A
12 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
AU 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
BA

 Apply '
22
R 1 R 1
21
A
53 31
A
52 31
52 3 1 1 0
A
1 0 2 1 3 1 1 0
A
1 0 2 1 Apply
12
RR 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 70
XY
25 and 30
XY
03
Solution: Given .....
70
X Y 1
25 30
XY
03
Adding (1) & (2),We get 7 0 3 0
X Y X Y
2 5 0 3 10 0
2X
28 50
X
14
From (1) 70
YX
25 7 0 5 0
Y
2 5 1 4 20
Y
11

49. 9. Find X and Y, if 23
2X 3Y
40 22
3X 2Y
15
Solution: Given that ......
23
2X 3Y 1
40 ......
22
3X 2Y 2
15
Consider 3(1) – 2(2), We get 23
6X 9Y 3
40 22
6X 4Y 2
15
__
__
(--) (--) -
6 9 4 4
5Y
12 0 2 10 2 13
5Y
14 10 2 13
55
Y
14
2
5
Consider 2(1) – 3(2), We get 4 6 6 6
5X
8 0 3 15 2 12
2 12 55
5X X
11 15 11
3
5

50. 10. If 80
A 4 2
36 and 22
B 4 2
51
then find the matrix X, such that 2A 3X 5B
Solution : For Given 80
A 4 2
36 22
B 4 2
51
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
X4
3
31
7
3
3

51. Three Mark questions
11. Find X and Y, if 52
XY
09 and 36
XY
01
12. If 25
1
33
1 2 4
A
3 3 3
72
2
33 and 23
1
55
1 2 4
B
5 5 5
762
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
22
22
10
01
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
113
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 '
34
A 1 2
01
and 1 2 1
B
1 2 3
then verify that ' ' 'i A B A B ' ' 'ii A B A B

58. 19. if '
23
A
12
and 10
B
12 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
41
36  
43
2'
16
AB

  



59. 20. if 1 2 3
A
4 2 5

and 23
B 4 5
21 AB BA
then find AB, BA
Show that
Solution: Given that 1 2 3
A
4 2 5
=
23
1 2 3
AB 4 5
4 2 5
21
2 8 6 3 10 3
8 8 10 12 10 5
04
10 3 23
B 4 5
21
=
23
1 2 3
BA 4 5
4 2 5
21
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 11
By 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
BA
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
BA
A A B B
BA
AB AB






61. 22. If 32
A
42
and 10
I
01 Find k so that 2
A kA 2I
Solution: Given that 32
A
42 10
I
01 &
We have 2
=
A =
2
A AA
3 2 3 2
4 2 4 2
9 8 6 4
12 8 8 4
12
44
Given that 2
A kA 2I
1 2 3 2 1 0
k2
4 4 4 2 0 1
1 2 3k 2 2k 0
4 4 4k 0 2k 2
3k 2 1
k1

62. 23. For the matrix 15
A
67 verify that 'i A A
is a symmetric matrix 'ii A A
is a skew symmetric matrix
Solution:Given that '
1 5 1 6
AA
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
XX
X A A




   '
'
1 5 1 6
ii A A
6 7 5 7
01
A A X say
10 01
'-
-1 0
' is skew-symmetric
Since X X
X A A




  

63. Solution:Given that
24. Find '
1
AA
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   
 
 
 
11
''
22
= *
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   
 
 
 

    
    
    
       






11
B = B + B' + B -B'
22
= P +Q say *
-3 3
2
222 -2 2 2 -1 1
11
-3
where P = B + B' = -1 3 4 + -2 3 - 2 P = 3 1
2
22
1 -2 -3 2 4 - 3
3
1 - 3
2
2 -2 2 2 -1
11
& Q = B -B' = -1 3 4 -
22
1 -2 -3
1
1

  
  
  
  

1
0
22 1
-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-sy mmetricP 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: 35
11
Solution:

67. 30.If A and B are invertible matrices of the same order,
then prove that 1
11
AB B A
Proof: Given that A and B are invertible matrices 11
AA A A I

 11
BB B B I


And    

1 1 1 1
11
1
1
Consider
=
= is invertible.
=
=
AB B A A B B A
A BB A
A IA B
AA
I
   










   

1 1 1 1
11
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
11
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
AA
A A A A
AA
XX
X A A




   'Y A A  

 
 
Consider ' ' '
= ' ' '
= ' ' '
= - '
'=
' is skew-symmetric matrix
Y A A
AA
A A A A
AA
YY
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
11
''
22
=
AA
AA
A A A
A A A A A
A A A A A
PQ
Let A be the square matrix    
   
 
   
 
11
' and '
22
11
' ' ' ' ' '
22
11
= ' ' ' ' ' ' '
22
11
= ' ' '
22
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
   
   
     
   
  
  
'QQ
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


  


    
21
4 ' 2 4 5 and 1 3 6 ' 3
56
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
AB
       
     
   
     
     
      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
BC
       
     
    
     
     
       
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
ABRHS 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
C2
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 32
A 6A 7A 2I O

78. 8. Let 21
A
34 52
B
74 25
C
38
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 ab
D
cd



 0CD AB 2 5 2 1 5 2
0
3 8 3 4 7 4
ab
cd
       

       
        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
ac
bd
ac
bd
  
  
  
  
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 cos
tan
cos tan sin sin tan cos
tan cos
RHS I A
0
10
2
01
02
1
2
1
2
22
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
22
c
2
22
2 2 1
2 2 2
22
22
2 1 2
2 2 2
22
1 2 2
2

  

  



  



sin
sin sin cos
cos
sin
sin cos sin sin sin
cos
tan
tan
22
2
2
2 2 2
2
2
2 2 1 2
2 2 2 2
2
1
2
12





   




Solution: 0
tan10
2
tan01
02
1 tan
2
=
tan 1
2
LHS I A




 
   



 




80. 

cos sin
sin cos
RHS I A 22
Useful sults
22
2 2 1 1 2



  
  
Re
sin
tan
cos
sin sin cos
cos cos sin 



tan cos sin
tan sin cos
0
10
2
01
02 

  
tan
cos sin
sin cos
tan
1
2
1
2 
   

   
cos tan sin sin tan cos
tan cos sin tan sin os
22
c
22 
  



  


sin sin
cos sin cos sin cos
cos cos
sin sin
cos sin sin cos cos
cos cos
2
2
22
2 2 1
2 2 2
22
22
2 1 2
2 2 2
22 
   



   


sin
sin sin sin sin cos
cos
sin
sin cos sin sin sin
cos
22
22
2
1 2 2 2
2 2 2 2
2
2
2 2 1 2
2 2 2 2
2 tan
tan
1
2LHS
12



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

87. One Mark questions
12. Find the values of x, y and z from the
following equations: 43yz
i
x515 xyz9
iixz5
yz7

88. One Mark questions
13. Find x and y, if 13y056
2
0x1218
14. if 2110
xy
315 Find the
values of x & y

89. One Mark questions
15. Find X, if &
32
Y
14 10
2XY
32
16. Find the values of x and y from the following equation x53476
2
7y3121516

90. One Mark questions
17. Find the value of a, b, c and d from the equation: ab2ac15
2ab3cd013
18. Show that the matrix 115
A121
513 is a symmetric matrix

91. One Mark questions
19. Show that the matrix 011
A101
110 is a skew symmetric matrix
20. let ,,
241325
A B C
322534
Find each of the following

92. One Mark questions
22. Given 311
A and
230 251
B
1
23
2
Find AB
23. Given
Find 123
A and
231 313
B
102 2AB

93. One Mark questions
25.Find AB If 69
A
23 and 260
B
798
26. if 10
A
01
and 01
B
10 then prove that 01
iAB
10 01
iiBA
10

94. One Mark questions
27. Simplify cossin sincos
cos sin
sincos cossin
 

 
28. Find 1
P if it exists , given 102
P
51

95. One Mark questions
29. Find the transpose of each of the following matrices 332
A
420
and 212
B
124
30.If 332
A
420 212
B
124
Verify that ''iAA '''iiABAB

96. One Mark questions
31. Compute the following abab
i
baba 2 2 2 2
2 2 2 2
abbc2ab2bc
ii
2ac2abacab 1461276
iii8516805
285324 cossinsincos
sincoscossin
2 2 2 2
2 2 2 2
xxxx
iv
xxxx

97. One Mark questions
32. Compute the indicated products: abab
i
baba 1
ii2234
3 12123
iii
23231 234135
iv345024
456305

98. One Mark questions 21
101
v32
121
11 23
313
vi 10
102
31
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 11
i
23 21
ii
11 13
iii
27 23
iv
57 21
v
74 25
vi
13

105. Three Mark questions 31
vii
52 45
viii
34 310
ix
27 26
x
12 63
xi
21 23
xii
12
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 12
A
21
8. Find X and Y, if 70
XY
25 and 30
XY
03

107. Three Mark questions
9. Find X and Y, if 23
2X3Y
40 and 22
3X2Y
15
10. If 80
A42
36 and 22
B42
51
then find the matrix X, such that 2A3X5B

108. Three Mark questions
11. Find X and Y, if 52
XY
09 and 36
XY
01
12. If 25
1
33
124
A
333
72
2
33 and 23
1
55
124
B
555
762
555 Then compute 3A5B

109. Three Mark questions
13. if cossin
()sincos
xx0
Fxxx0
001
Show that FxFyFxy
14. Show that 51212151
67343467

111. Three Mark questions
16.Show that 123110110123
010011011010
110234234110
17. if 123
A579
211
and 415
B120
131
then verify that '''iABAB '''iiABAB

112. Three Mark questions
18. if '
34
A12
01
and 121
B
123
then verify that '''iABAB '''iiABAB

113. Three Mark questions
19. if '
23
A
12 and 10
B
12 then find 'A2B
20. if 123
A
425
and 23
B45
21 ABBA
then 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 32
A
42 and 10
I
01
Find k so that 2
AkA2I

116. Three Mark questions
23. For the matrix 15
A
67 verify that 'iAA
is a symmetric matrix 'iiAA
is a skew symmetric matrix

117. Three Mark questions
24. Find '
1
AA
2 '
1
andAA
2
When 0ab
Aa0c
bc0

118. Three Mark questions
25. Express the following matrices as the sum of a
symmetric and skew symmetric matrix: 35
11 622
231
213

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
11
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
4 136
5
AandB


 


 Verify that ()ABBA

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
A5A6I 201
ifA213
110

126. Five Mark questions
5.If 067
A608
780 011
B102
120 2
C2
3
Calculate AC, BC and (A + B) C. Also,
verify that (A + B)C = AC + BC.

127. Five Mark questions
6. if 111
A123
213
then show that 3
A23A40IO

128. Five Mark questions
7. If 102
A021
203 then show that 32
A6A7A2IO

129. Five Mark questions
8. Let 21
A
34 52
B
74 25
C
38
Find a matrix D such that CDAB0

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

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: [email protected]
Contact : 9591319967