Answer
See answer below
Work Step by Step
We are given:
$A_1=\begin{bmatrix}
1 & 3\\
-1 & 2
\end{bmatrix}$
$A_2=\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix} $
$A_3=\begin{bmatrix}
-1 & 4\\
1 & 1
\end{bmatrix}$
$A_4=\begin{bmatrix}
5 & -6\\
-5 & 1
\end{bmatrix}$
From what we can see: $2A_1-3A_3=2\begin{bmatrix}
1 & 3\\
-1 & 2
\end{bmatrix}-3\begin{bmatrix}
-1 & 4\\
1 & 1
\end{bmatrix}=\begin{bmatrix}
2 & 6\\
-2 & 4
\end{bmatrix}+\begin{bmatrix}
3 & -12\\
-3 & -3
\end{bmatrix}=\begin{bmatrix}
5 & -6\\
-5 & 1
\end{bmatrix}=A_4$
Let $c_1,c_2,c_3$ and $c_4$ be scarlars, we obtain:
$c_1A_1+c_2A_2+c_3A_3+c_4A_4=c_1A_1+c_3A_3+c_4(2A_1-3A_3)=c_1A_1+c_3A_3+2c_4A_1-3c_4A_3=(c_1+2c_4)A_1+(c_3-3c_4)A_3$
Since any linear combination of $A_1, A_2, A_3, A_4$ can be written as a linear combination of $A_1$ and $A_3$, span $\{A_1,A_2,A_3,A_4\} \subset$ span $\{A_1,A_3\}$.
And $\{A_1,A_3\} \subset \{A_1,A_2,A_3,A_4\} \rightarrow $ span $\{A_1,A_3\} \subset$ span $\{A_1,A_2,A_3,A_4\} \rightarrow $ span $\{A_1,A_3\}=$ span $\{A_1,A_2,A_3,A_4\}$
We can see that $A_1$ and $A_3$ are linearly dependent in $M_2(R)$
Hence, $\{A_1,A_3\}$ is a linearly independent set for $S$ and $\{A_1,A_3\}$ is also a basis for $S$
