Given, $A = \left[ {\begin{array}{llllllllllllllllllll}2&4&0\\3&9&6\end{array}} \right]$ and $B = \left[ {\begin{array}{llllllllllllllllllll}1&4\\2&8\\1&3\end{array}} \right]$. is ${(AB)^\prime } = {B^\prime }{A^\prime }$ ?

We have, $A = {\left[ {\begin{array}{llllllllllllllllllll}2&4&0\\3&9&6\end{array}} \right]_{2 \times 3}}$

and $B = {\left[ {\begin{array}{llllllllllllllllllll}1&4\\2&8\\1&3\end{array}} \right]_{3 \times 2}}$

$\therefore$ $AB = \left[ {\begin{array}{cccccccccccccccccccc}{2 + 8 + 0}&{8 + 32 + 0}\\{3 + 18 + 6}&{12 + 72 + 18}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{10}&{40}\\{27}&{102}\end{array}} \right]$

and ${(AB)^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}{10}&{27}\\{40}&{102}\end{array}} \right]$

…….(i)
Also ${B^\prime } = {\left[ {\begin{array}{llllllllllllllllllll}1&2&1\\4&8&3\end{array}} \right]_{2 \times 3}}$ and ${A^\prime } $ $= {\left[ {\begin{array}{llllllllllllllllllll}2&3\\4&9\\0&6\end{array}} \right]_{3 \times 2}}$

$\therefore$ ${B^\prime }{A^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}{2 + 8 + 0}&{3 + 18 + 6}\\{8 + 32 + 0}&{12 + 72 + 18}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{10}&{27}\\{40}&{102}\end{array}} \right]$

Thus, we see that, ${(AB)^\prime } = {B^\prime }{A^\prime }$

[using Eqs. (i) and (ii)]