We have, $A=\left[\begin{array}{lll}2 & 1 & 2 \\ 1 & 2 & 4\end{array}\right]_{2 \times 3}$ and
$$
\mathrm{B}=\left[\begin{array}{ll}
4 & 1 \\
2 & 3 \\
1 & 2
\end{array}\right]_{3 \times 2}
$$
In both $\mathrm{AB}$ and $\mathrm{BA}$ the number of columns of first is equal to the number of rows of second.
So, $\mathrm{AB}$ and $\mathrm{BA}$ both are possible.
$$
\begin{aligned}
&\therefore \mathrm{AB}=\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]_{2 \times 3} \cdot\left[\begin{array}{ll}
4 & 1 \\
2 & 3 \\
1 & 2
\end{array}\right]_{3 \times 2} \\
&=\left[\begin{array}{ll}
8+2+2 & 2+3+4 \\
4+4+4 & 1+6+8
\end{array}\right]=\left[\begin{array}{cc}
12 & 9 \\
12 & 15
\end{array}\right]
\end{aligned}
$$
$$
\begin{aligned}
&\text { and } B A=\left[\begin{array}{ll}
4 & 1 \\
2 & 3 \\
1 & 2
\end{array}\right]_{3 \times 2}\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]_{2 \times 3} \\
&=\left[\begin{array}{ccc}
4 \times 2+1 & 4+2 & 8+4 \\
4+3 & 2+6 & 4+12 \\
2+2 & 1+4 & 2+8
\end{array}\right]=\left[\begin{array}{lll}
9 & 6 & 12 \\
7 & 8 & 16 \\
4 & 5 & 10
\end{array}\right]
\end{aligned}
$$