Answer
$X=\left[ \begin{array}{*{35}{l}}
\frac{29}{2} & \frac{25}{2} \\
-3 & \frac{27}{2} \\
-\frac{27}{2} & 8 \\
\end{array} \right]$
Work Step by Step
Consider the matrix equation, $4B+3A=-2X$.
This implies that, $X=-\frac{1}{2}\left[ 4B+3A \right]$
Now we will consider,
$\begin{align}
& X=-\frac{1}{2}\left[ 4B+3A \right] \\
& =-\frac{1}{2}\left( 4\left[ \begin{matrix}
-5 & -1 \\
0 & 0 \\
3 & -4 \\
\end{matrix} \right]+3\left[ \begin{matrix}
-3 & -7 \\
2 & -9 \\
5 & 0 \\
\end{matrix} \right] \right) \\
& =-\frac{1}{2}\left( \left[ \begin{matrix}
-20 & -4 \\
0 & 0 \\
12 & -16 \\
\end{matrix} \right]+\left[ \begin{matrix}
-9 & -21 \\
6 & -27 \\
15 & 0 \\
\end{matrix} \right] \right) \\
& =-\frac{1}{2}\left( \left[ \begin{array}{*{35}{l}}
-29 & -25 \\
6 & -27 \\
27 & -16 \\
\end{array} \right] \right)
\end{align}$
That is,
$\begin{align}
& X=-\frac{1}{2}\left( \left[ \begin{array}{*{35}{l}}
-29 & -25 \\
6 & -27 \\
27 & -16 \\
\end{array} \right] \right) \\
& =\left[ \begin{array}{*{35}{l}}
\frac{29}{2} & \frac{25}{2} \\
-3 & \frac{27}{2} \\
-\frac{27}{2} & 8 \\
\end{array} \right]
\end{align}$
