If possible, find the value of BA and AB, where
$A = \left[ {\begin{array}{llllllllllllllllllll}2&1&2\\1&2&4\end{array}} \right]$ and $B = \left[ {\begin{array}{llllllllllllllllllll}4&1\\2&3\\1&2\end{array}} \right]$.

Therefore, AB and BA both are possible.

[since, in both $A \cdot B$ and $B \cdot A$,

the number of columns of first is equal to the number of rows of second.]

$\therefore$ $AB = {\left[ {\begin{array}{llllllllllllllllllll}2&1&2\\1&2&4\end{array}} \right]_{2 \times 3}} \cdot {\left[ {\begin{array}{llllllllllllllllllll}4&1\\2&3\\1&2\end{array}} \right]_{3 \times 2}}$

$= \left[ {\begin{array}{llllllllllllllllllll}{8 + 2 + 2}&{2 + 3 + 4}\\{4 + 4 + 4}&{1 + 6 + 8}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{12}&9\\{12}&{15}\end{array}} \right]$

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

$= \left[ {\begin{array}{cccccccccccccccccccc}{4 \times 2 + 1}&{4 + 2}&{8 + 4}\\{4 + 3}&{2 + 6}&{4 + 12}\\{2 + 2}&{1 + 4}&{2 + 8}\end{array}} \right] = \left[ {\begin{array}{llllllllllllllllllll}9&6&{12}\\7&8&{16}\\4&5&{10}\end{array}} \right]$