(i)For any matrices A and B of suitable orders, we have
(A′)′ = A,
(ii)
(kA)′ = kA′ (where k is any constant)
(iii)
(A + B)′ = A′ + B′
(iv)
(A B)′ = B′ A′
Example 20 If 
2
1
2
3
3
2
A
and B
1
2
4
4
2
0


−


=
=








, verify that
(i)
(A′)′ = A,
(ii)
(A + B)′ = A′ + B′,
(iii)
(kB)′ = kB′, where k is any constant Rationalised 2023-24
 62
MATHEMATICS
Solution
(i)
We have
A =
(
)
3
4
3
3
2
3
3
2
A
3 2
A
A
4
2
0
4
2
0
2
0








′
′
′
⇒
=
⇒
=
=














Thus
(A′)′ = A
(ii)
We have
A = 3
3
2 ,
4
2
0






 B =
2
1
2
5
3
1
4
A
B
1
2 4
5
4
4


−


−
⇒
+
= 







Therefore
(A + B)′ =
5
5
3
1
4
4
4




−






Now
A′ =
3
4
2 1
3 2 , B
1 2 ,
2
0
2 4








′ = −












So
A′ + B′ =
5
5
43 1 4
4




−






Thus
(A + B)′ = A′ + B′
(iii)
We have
kB = k 2
1
2
2
2
1
2
4
2
4
k
k
k
k
k
k
−
−




=








Then
(kB)′ =
2
2 1
2
1 2
B
2
4
2 4
k
k
k
k
k
k
k
k








′
−
=
−
=












Thus
(kB)′ = kB′
Rationalised 2023-24
MATRICES         63
Example 21 If 
[
]
2
A
4 , B
1
3
6
5
−



=
=
−






, verify that (AB)′ = B′A′ Solution We have
A =
[
]
2
4 , B
1
3
6
5
−



=
−






then
AB =
[
]
2
4
1
3
6
5
−



−






 = 
2
6
12
4
12
24
5
15
30
−
−




−




−


Now
A′ = [–2 4 5] , 
1
B
3
6




′ = 


−

B′A′ =
[
]
1
2
4
5
3
2
4
5
6
12
15
(AB)
6
12
24
30
−








′
−
= −
=








−
−
−




Clearly
(AB)′ = B′A′
3 6  Symmetric and Skew Symmetric Matrices
Definition 4 A square matrix A = [aij] is said to be symmetric if A′ = A, that is,
[aij] = [aji] for all possible values of i and j