Answer
$4B-3C=\left[ \begin{matrix}
17 & 7 \\
-5 & -11 \\
\end{matrix} \right]$
Work Step by Step
Consider,
$\begin{align}
& 4B-3C=4\left[ \begin{matrix}
5 & 1 \\
-2 & -2 \\
\end{matrix} \right]-3\left[ \begin{matrix}
1 & -1 \\
-1 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4\times 5 & 4\times 1 \\
4\times \left( -2 \right) & 4\times \left( -2 \right) \\
\end{matrix} \right]-\left[ \begin{matrix}
3\times 1 & 3\times \left( -1 \right) \\
3\times \left( -1 \right) & 3\times 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
20 & 4 \\
-8 & -8 \\
\end{matrix} \right]-\left[ \begin{matrix}
3 & -3 \\
-3 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
20-3 & 4+3 \\
-8+3 & -8-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
17 & 7 \\
-5 & -11 \\
\end{matrix} \right]
\end{align}$
