Answer
$ X=\left[ \begin{matrix}
0.5 & 0.5 \\
-3 & 0.5 \\
\end{matrix} \right]$
Work Step by Step
Given $ C=\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right]$ and $ B=\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]$.
Since both matrices are $2\times 2$, the matrix operations are valid.
Now
$\begin{align}
& 2X-3C=B \\
& 2X=B+3C \\
& X=\frac{B+3C}{2}
\end{align}$
Substitute the corresponding values to get:
$\begin{align}
& X=\frac{\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-1 & 0 \\
0 & 1 \\
\end{matrix} \right]}{2}=\frac{\left[ \begin{matrix}
4 & 1 \\
-6 & -2 \\
\end{matrix} \right]+\left[ \begin{matrix}
-3 & 0 \\
0 & 3 \\
\end{matrix} \right]}{2} \\
& =\frac{\left[ \begin{matrix}
4-3 & 1+0 \\
-6+0 & -2+3 \\
\end{matrix} \right]}{2}=\frac{\left[ \begin{matrix}
1 & 1 \\
-6 & 1 \\
\end{matrix} \right]}{2} \\
& =\frac{1}{2}\left[ \begin{matrix}
1 & 1 \\
-6 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{2} & \frac{1}{2} \\
\frac{-6}{2} & \frac{1}{2} \\
\end{matrix} \right]
\end{align}$
Therefore, $ X=\left[ \begin{matrix}
0.5 & 0.5 \\
-3 & 0.5 \\
\end{matrix} \right]$
