Answer
$ A(B+C)=\left[ \begin{array}{*{35}{l}}
-12 & -2 \\
-21 & -4 \\
3 & 1 \\
\end{array} \right]$
Work Step by Step
To solve the provided expression, follow the below methods:
Matrix addition: Consider two given matrices $ X,Y $
$ 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, $\begin{align}
& X+Y=\left[ \begin{matrix}
{{x}_{1}} & {{x}_{2}} \\
{{x}_{3}} & {{x}_{4}} \\
\end{matrix} \right]+\left[ \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
{{y}_{3}} & {{y}_{4}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
{{x}_{1}}+{{y}_{1}} & {{x}_{2}}+{{y}_{2}} \\
{{x}_{3}}+{{y}_{3}} & {{x}_{4}}+{{y}_{4}} \\
\end{matrix} \right]
\end{align}$
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, $\begin{align}
& X\times Y=\left[ \begin{matrix}
{{x}_{1}} & {{x}_{2}} \\
{{x}_{3}} & {{x}_{4}} \\
\end{matrix} \right]\times \left[ \begin{matrix}
{{y}_{1}} & {{y}_{2}} \\
{{y}_{3}} & {{y}_{4}} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
{{x}_{1}}{{y}_{1}}+{{x}_{2}}{{y}_{3}} & {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{4}} \\
{{x}_{3}}{{y}_{1}}+{{x}_{4}}{{y}_{3}} & {{x}_{3}}{{y}_{2}}+{{x}_{4}}{{y}_{4}} \\
\end{matrix} \right]
\end{align}$
Substitute the value of $ A=\left[ \begin{array}{*{35}{l}}
0 & 2 \\
-1 & 3 \\
1 & 0 \\
\end{array} \right]$, $ B=\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]$ and $ C=\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ in $ A(B+C)$
$\begin{align}
& A(B+C)=\left[ \begin{array}{*{35}{l}}
0 & 2 \\
-1 & 3 \\
1 & 0 \\
\end{array} \right]\left( \left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{array}{*{35}{l}}
0 & 2 \\
-1 & 3 \\
1 & 0 \\
\end{array} \right]\left( \left[ \begin{matrix}
4-1 & 1+0 \\
-6+0 & -2+1 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{array}{*{35}{l}}
0 & 2 \\
-1 & 3 \\
1 & 0 \\
\end{array} \right]\left( \left[ \begin{matrix}
3 & 1 \\
-6 & -1 \\
\end{matrix} \right] \right) \\
& =\left[ \begin{array}{*{35}{l}}
-12 & -2 \\
-21 & -4 \\
3 & 1 \\
\end{array} \right]
\end{align}$
Hence $ A(B+C)$ $=\left[ \begin{array}{*{35}{l}}
-12 & -2 \\
-21 & -4 \\
3 & 1 \\
\end{array} \right]$