(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’)