If possible,

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

So. $A B$ and $B A$ both are defined

Now,

$A B=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]\left[\begin{array}{ll}4 & 1 \\ 2 & 3 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}8+2+2 & 2+3+4 \\ 4+4+4 & 1+6+8\end{array}\right]=\left[\begin{array}{cc}12 & 9 \\ 12 & 15\end{array}\right]$

And, 

$B A=\left[\begin{array}{ll}4 & 1 \\ 2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]$

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