Answer
$$x=3, \quad y=2, \quad z=1.$$
Work Step by Step
We have
\begin{align*}
&4\left[\begin{array}{rrr}{x} & {y} \\ {z} & {-1} \end{array}\right]=2\left[\begin{array}{rrr}{y} & {z} \\ {-x} & {1} \end{array}\right]+2\left[\begin{array}{rrr}{4} & {x} \\ {5} & {-x} \end{array}\right]\\
\Longrightarrow &
\left[\begin{array}{rrr}{4x} & {4y} \\ {4z} & {-4} \end{array}\right]=\left[\begin{array}{rrr}{2y} & {2z} \\ {-2x} & {2} \end{array}\right]+\left[\begin{array}{rrr}{8} & {2x} \\ {10} & {-2x} \end{array}\right]\\
\Longrightarrow &
\left[\begin{array}{rrr}{4x} & {4y} \\ {4z} & {-4} \end{array}\right]=\left[\begin{array}{rrr}{2y+8} & {2z+2x} \\ {-2x+10} & {2-2x} \end{array}\right].
\end{align*}
By comparing the corresponding entries of the matrices in both sides of the above equation, we get the following system of equations
\begin{align*}
4x&=2y+8\\
4y&=2z+2x\\
4z&=-2x+10\\
-4&=2-2x.
\end{align*}
From which, we have the solution
$$x=3, \quad y=2, \quad z=1.$$
