Answer
See answer below
Work Step by Step
Since $S$ has 4 elements and $dim M_{2 \times3}(R)=6 \rightarrow S$ do not span $ M_{2 \times3}(R) $
We obtain $A$ as $A=\begin{bmatrix}
-1 & 0 & 0 & 0 & 1 & 1\\
3 & 2 & 1 & 1 & 2 & 3\\
-1 & 2 & -3 & 3 & 2 & 1\\
-11 & -6 & -5 & 1 & -2 & -5
\end{bmatrix} \approx\begin{bmatrix}
-1 & 0 & 0 & 0 & 1 & 1\\
1 & 2 & 1 & 3 & 2 & 3\\
3 & 2 & -3 & -1 & 2 & 1\\
1 & -6 & -5 & -11 & -2 & -5
\end{bmatrix} \approx \begin{bmatrix}
0& 0 & 0 & -1 & 1 & 1\\
0 & -8 & -6 & -14 & -4 & -8\\
0 & -20 & -16 & -36 & -10 & -16\\
1 & -6 & -5 & -11 & -2 & -5
\end{bmatrix} \approx \begin{bmatrix}
0& 0 & 0 & -1 & 1 & 1\\
0 & -4& -3 & -7 & -2 & -4\\
0 & 10 & 8 & 16 & 5 & 8\\
1 & -6 & -5 & -11 & -2 & -5
\end{bmatrix} \approx \begin{bmatrix}
0& 0 & 0 & -1 & 1 & 1\\
0 & 0& 1 & -3 & 0 & -4\\
0 & 10 & 8 & 16 & 5 & 8\\
1 & -6 & -5 & -11 & -2 & -5
\end{bmatrix}$
Since $rank (A)=4$, the vectors of $S$ are linearly independent.
