Fill in the blanks in each of the
If A and B are square matrices of the same order, then
(i) (AB)’ = ________.
(ii) (kA)’ = ________. (k is any scalar)
(iii) [k (A – B)]’ = ________.
(i) (AB)’ = ________.
(AB)’ = B’A’
Let A be matrix of order m× n and B be of n× p.
A’ is of order n× m and B’ is of order p× n.
Hence B’ A’ is of order p× m.
So, AB is of order m× p.
And (AB)’ is of order p× m.
We can see (AB)’ and B’ A’ are of same order p× m.
Hence (AB)’ = B’ A’
Hence proved.
(ii) (kA)’ = ________. (k is any scalar)
If a scalar “k” is multiplied to any matrix the new matrix becomes
K times of the old matrix.
Eg: A = ![]()
2A = ![]()
=![]()
(2A)’ = ![]()
A’ = ![]()
Now 2A’ = ![]()
= ![]()
Hence (2A)’ =2A’
Hence (kA)’ = k(A)’
(iii) [k (A – B)]’ = ________.
A = ![]()
A’ = ![]()
2A’ = 2![]()
= ![]()
B= ![]()
B’ = ![]()
2B’ = ![]()
= ![]()
A-B = ![]()
Now Let k =2
2(A-B) = ![]()
= ![]()
[2(A-B)]’ = ![]()
2A’ – 2B’ = ![]()
=![]()
A’ – B’ = ![]()
=![]()
2(A’ – B’) = 2 ![]()
=![]()
Hence we can see [k (A – B)]’= k(A)’- k(B)’= k(A’-B’)
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