If $A\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 4}\\1&1\\2&0\end{array}} \right]$ and $B = \left[ {\begin{array}{cccccccccccccccccccc}2&1&2\\1&2&4\end{array}} \right]$, then verify ${(BA)^2} \ne {B^2}{A^2}$.

We have, $A = {\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 4}\\1&1\\2&0\end{array}} \right]_{3 \times 2}}$and $B = {\left[ {\begin{array}{cccccccccccccccccccc}2&1&2\\1&2&4\end{array}} \right]_{2 \times 3}}$

$\therefore$ $BA = {\left[ {\begin{array}{llllllllllllllllllll}2&1&2\\1&2&4\end{array}} \right]_{2 \times 3}}{\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 4}\\1&1\\2&0\end{array}} \right]_{3 \times 2}}$

$= \left[ {\begin{array}{llllllllllllllllllll}{6 + 1 + 4}&{ - 8 + 1 + 0}\\{3 + 2 + 8}&{ - 4 + 2 + 0}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{11}&{ - 7}\\{13}&{ - 2}\end{array}} \right]$

and $(BA) \cdot (BA) = \left[ {\begin{array}{cccccccccccccccccccc}{11}&{ - 7}\\{13}&{ - 2}\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}{11}&{ - 7}\\{13}&{ - 2}\end{array}} \right]$

$\Rightarrow$ ${(BA)^2} = \left[ {\begin{array}{cccccccccccccccccccc}{121 - 91}&{ - 77 + 14}\\{143 - 26}&{ - 91 + 4}\end{array}} \right]$ $ = \left[ {\begin{array}{cccccccccccccccccccc}{30}&{ - 63}\\{117}&{ - 87}\end{array}} \right]$

…….(i)
Also ${B^2} = B \cdot B = {\left[ {\begin{array}{llllllllllllllllllll}2&1&2\\1&2&4\end{array}} \right]_{2 \times 3}}{\left[ {\begin{array}{llllllllllllllllllll}2&1&2\\1&2&4\end{array}} \right]_{2 \times 3}}$

Therefore, ${B^2}$ is not possible, since the B is not a square matrix.

Hence, ${(BA)^2} \ne {B^2}{A^2}$.