Answer
a) $AB=\left[ \begin{matrix}
6 & 8 & 16 \\
11 & 16 & 24 \\
1 & -1 & 12 \\
\end{matrix} \right]$
b) $BA=\left[ \begin{matrix}
38 & 27 \\
-16 & -4 \\
\end{matrix} \right]$
Work Step by Step
(a)
Consider,
$\begin{align}
& AB=\left[ \begin{array}{*{35}{l}}
4 & 2 \\
6 & 1 \\
3 & 5 \\
\end{array} \right]\left[ \begin{array}{*{35}{l}}
2 & 3 & 4 \\
-1 & -2 & 0 \\
\end{array} \right] \\
& =\left[ \begin{matrix}
4\left( 2 \right)+2\left( -4 \right) & 4\left( 3 \right)+2\left( -2 \right) & 4\left( 4 \right)+2\left( 0 \right) \\
6\left( 2 \right)+1\left( -1 \right) & 6\left( 3 \right)+1\left( -2 \right) & 6\left( 4 \right)+1\left( 0 \right) \\
3\left( 2 \right)+5\left( -1 \right) & 3\left( 3 \right)+5\left( -2 \right) & 3\left( 4 \right)+5\left( 0 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
8-2 & 12-4 & 16+0 \\
12-1 & 18-2 & 24+0 \\
6-5 & 9-10 & 12+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
6 & 8 & 16 \\
11 & 16 & 24 \\
1 & -1 & 12 \\
\end{matrix} \right]
\end{align}$
(b)
Consider,
$\begin{align}
& BA=\left[ \begin{array}{*{35}{l}}
2 & 3 & 4 \\
-1 & -2 & 0 \\
\end{array} \right]\left[ \begin{array}{*{35}{l}}
4 & 2 \\
6 & 1 \\
3 & 5 \\
\end{array} \right] \\
& =\left[ \begin{matrix}
2\left( 4 \right)+3\left( 6 \right)+4\left( 3 \right) & 2\left( 2 \right)+3\left( 1 \right)+4\left( 5 \right) \\
-1\left( 4 \right)-2\left( 6 \right)+0\left( 3 \right) & -1\left( 2 \right)-2\left( 1 \right)+0\left( 5 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
8+18+12 & 4+3+20 \\
-4-12+0 & -2-2+0 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
38 & 27 \\
-16 & -4 \\
\end{matrix} \right]
\end{align}$
