Set of vectors (elements - matrices to be precise - in your case) $\{A_1, A_2, \dots, A_n\}$ is said to be linearly independent iff
$$\sum_{i=1}^nc_iA_i = 0 \iff c_1, c_2, \dots, c_n = 0$$
$$
5\begin{bmatrix}1 & -3 \\ -4 & 0 \end{bmatrix}
+
3\begin{bmatrix} -3 & 7 \\ 6 & -2 \end{bmatrix}
-
2\begin{bmatrix} w & x \\ y & z \\ \end{bmatrix}
\neq 0$$
Multiplying each matrix by scalar we get
$$
\begin{bmatrix}5 & -15 \\ -20 & 0 \end{bmatrix}
+
\begin{bmatrix} -9 & 21 \\ 18 & -6 \end{bmatrix}
+
\begin{bmatrix} -2w & -2x \\ -2y & -2z \\ \end{bmatrix}
\neq 0$$
Performing matrix addition element-wise we obtain
$$
\begin{bmatrix}5 - 9 - 2w & -15 + 21 - 2x \\ -20 + 18 - 2y & -6 - 2z \end{bmatrix} = 0$$
This yields,
$5 - 9 - 2w \neq 0 \Longrightarrow w \neq -2$
$-15 + 21 - 2x \neq 0 \Longrightarrow x\neq 3$
$-20 + 18 - 2y \neq 0 \Longrightarrow y \neq -1$
$-6 - 2z \neq 0 \Longrightarrow z \neq -3$
Therefore if set
$$S = \left\{\begin{bmatrix}1 & -3 \\ -4 & 0 \end{bmatrix},
\begin{bmatrix} -3 & 7 \\ 6 & -2 \end{bmatrix},
\begin{bmatrix} w & x \\ y & z \\ \end{bmatrix}\right\}$$
is lineary independent then at leat one of following is satisfied
$$x\neq 3, \qquad y \neq -1, \qquad w \neq -2, \qquad z \neq -3$$
Note that it is necessary condition but not sufficient.
In short, lineary independence of set $S$ implies that
$$x\neq 3, \qquad y \neq -1, \qquad w \neq -2, \qquad z \neq -3$$
however, it does not hold the other way round, in general.
