Answer
$ A(BC)=\left[ \begin{matrix}
12 & -4 \\
22 & -7 \\
-4 & 1 \\
\end{matrix} \right]$
Work Step by Step
Matrix multiplication: Consider two matrices ${{X}_{2\times 2}},{{Y}_{2\times 2}}$.
The multiplication of two matrices is possible only if the number of rows of the first matrix equals the number of columns of the second matrix.
$ X=\left[ \begin{matrix}
{{x}_{1}} & {{x}_{2}} \\
{{x}_{3}} & {{x}_{4}} \\
\end{matrix} \right]$ and $ Y=\left[ \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
{{y}_{3}} & {{y}_{4}} \\
\end{matrix} \right]$
Now, $\begin{align}
& X\times Y=\left[ \begin{matrix}
{{x}_{1}} & {{x}_{2}} \\
{{x}_{3}} & {{x}_{4}} \\
\end{matrix} \right]\times \left[ \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
{{y}_{3}} & {{y}_{4}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
{{x}_{1}}{{y}_{1}}+{{x}_{2}}{{y}_{3}} & {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{4}} \\
{{x}_{3}}{{y}_{1}}+{{x}_{4}}{{y}_{3}} & {{x}_{3}}{{y}_{2}}+{{x}_{4}}{{y}_{4}} \\
\end{matrix} \right]
\end{align}$
Use the above method to solve the provided expression as below:
$\begin{align}
& (BC)=\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
4\cdot -1+1\cdot 0 & 4\cdot 0+1\cdot 1 \\
-6\cdot -1+-2\cdot 0 & -6\cdot 0+-2\cdot -2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-4 & 1 \\
6 & -2 \\
\end{matrix} \right]
\end{align}$
Now,
$\begin{align}
& A(BC)=\left[ \begin{matrix}
0 & 2 \\
-1 & 3 \\
1 & 0 \\
\end{matrix} \right]\left[ \begin{matrix}
-4 & 1 \\
6 & -2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
12 & -4 \\
22 & -7 \\
-4 & 1 \\
\end{matrix} \right]
\end{align}$
Hence, $ A(BC)=\left[ \begin{matrix}
12 & -4 \\
22 & -7 \\
-4 & 1 \\
\end{matrix} \right]$
