Answer
a) $AB=\left[ \begin{matrix}
2 & -8 & 20 \\
8 & 3 & 5 \\
10 & 2 & 10 \\
\end{matrix} \right]$
b) $BA=\left[ \begin{matrix}
12 & 14 \\
9 & 3 \\
\end{matrix} \right]$
Work Step by Step
(a)
Consider,
$\begin{align}
& AB=\left[ \begin{array}{*{35}{l}}
2 & 4 \\
3 & 1 \\
4 & 2 \\
\end{array} \right]\left[ \begin{array}{*{35}{l}}
3 & 2 & 0 \\
-1 & -3 & 5 \\
\end{array} \right] \\
& =\left[ \begin{matrix}
2\left( 3 \right)+4\left( -1 \right) & 2\left( 2 \right)+4\left( -3 \right) & 2\left( 0 \right)+4\left( 5 \right) \\
3\left( 3 \right)+1\left( -1 \right) & 3\left( 2 \right)+1\left( -3 \right) & 3\left( 0 \right)+1\left( 5 \right) \\
4\left( 3 \right)+2\left( -1 \right) & 4\left( 2 \right)+2\left( -3 \right) & 4\left( 0 \right)+2\left( 5 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6-4 & 4-12 & 0+20 \\
9-1 & 6-3 & 0+5 \\
12-2 & 8-6 & 0+10 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
2 & -8 & 20 \\
8 & 3 & 5 \\
10 & 2 & 10 \\
\end{matrix} \right]
\end{align}$
(b)
$\begin{align}
& BA=\left[ \begin{array}{*{35}{l}}
3 & 2 & 0 \\
-1 & -3 & 5 \\
\end{array} \right]\left[ \begin{array}{*{35}{l}}
2 & 4 \\
3 & 1 \\
4 & 2 \\
\end{array} \right] \\
& =\left[ \begin{matrix}
3\left( 2 \right)+2\left( 3 \right)+0\left( 4 \right) & 3\left( 4 \right)+2\left( 1 \right)+0\left( 2 \right) \\
-1\left( 2 \right)-3\left( 3 \right)+5\left( 4 \right) & -1\left( 4 \right)-3\left( 1 \right)+5\left( 2 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6+6+0 & 12+2+0 \\
-2-9+20 & -4-3+10 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
12 & 14 \\
9 & 3 \\
\end{matrix} \right]
\end{align}$
