Suppose that there exist constants $a_1$, $a_2$, $a_3$, $a_4$, and $a_5$ so that
\begin{align*}
a_1 \begin{bmatrix} -2 & 3 & 4 \\ -1 & 3 & -2 \end{bmatrix} +
a_2 \begin{bmatrix} 4 & -2 & 2 \\ 0 & -1 & 1 \end{bmatrix} +
a_3 \begin{bmatrix} -1 & -2 & -2 \\ 2 & 2 & 2 \end{bmatrix} +
a_4 \begin{bmatrix} -1 & 1 & 0\\ -1 & 0 & -2 \end{bmatrix} +
a_5 \begin{bmatrix} -1 & 2 & -2\\ 0 & -1 & -2 \end{bmatrix}
&=
\begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0 \end{bmatrix}.
\end{align*}
Then, we have the matrix equality (Definition ME)
\begin{align*}
\begin{bmatrix}
-2a_1 + 4a_2 - a_3 - a_4 - a_5 & 3a_1 - 2a_2 - 2a_3 + a_4 + 2a_5 & 4a_1 + 2a_2 - 2a_3 - 2a_5\\
-a_1 + 2a_3 - a_4 & 3a_1 - a_2 + 2a_3 - a_5 & -2a_1 + a_2 + 2a_3 - 2a_4 - 2a_5
\end{bmatrix}
&=
\begin{bmatrix} 0 & 0 & 0\\0 & 0 & 0 \end{bmatrix},
\end{align*}
which yields the linear system of equations
\begin{align*}
-2a_1 + 4a_2 - a_3 - a_4 - a_5 &= 0\\
3a_1 - 2a_2 - 2a_3 + a_4 + 2a_5&= 0\\
4a_1 + 2a_2 - 2a_3 - 2a_5&=0\\
-a_1 + 2a_3 - a_4 &= 0\\
3a_1 - a_2 + 2a_3 - a_5 &= 0\\
-2a_1 + a_2 + 2a_3 + 2a_4 - 2a_5 &= 0.
\end{align*}
By row-reducing the associated $6\times 5$ homogeneous system, we see that the only solution is
$ a_1 = a_2 = a_3 = a_4 = a_5 = 0$, so these matrices are a linearly independent subset of $M_{23}$.