Given Is (AB)’ = B’ A’ ?

Given, $A=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right]_{2 \times 3}$ and $B=\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]_{3 \times 2}$

So, their product is

$A B=\left[\begin{array}{lll}2 & 4 & 0 \\ 3 & 9 & 6\end{array}\right]\left[\begin{array}{ll}1 & 4 \\ 2 & 8 \\ 1 & 3\end{array}\right]=\left[\begin{array}{cc}2+8+0 & 8+32+0 \\ 3+18+6 & 12+72+18\end{array}\right]=\left[\begin{array}{cc}10 & 40 \\ 27 & 102\end{array}\right]$

And, $(A B)^{\prime}=\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right]$ $\ldots$ (i)

Also,

$B^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right]_{2 \times 3}$ and $A^{\prime}=\left[\begin{array}{ll}2 & 3 \\ 4 & 9 \\ 0 & 6\end{array}\right]_{3 \times 2}$

Therefore, $B^{\prime} A^{\prime}=\left[\begin{array}{lll}1 & 2 & 1 \\ 4 & 8 & 3\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 4 & 9 \\ 0 & 6\end{array}\right]$

$=\left[\begin{array}{cc}2+8+0 & 3+18+6 \\ 8+32+0 & 12+72+18\end{array}\right]$

$=\left[\begin{array}{cc}10 & 27 \\ 40 & 102\end{array}\right]$ $\ldots$ (ii)

From (i) and (ii), we have

(AB)’ = B’ A’