# Fill in the blank

(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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