Answer
See answer below
Work Step by Step
We obtain $A$ as $A=\begin{bmatrix}
1 & 2 & 2 & 1\\
3 & 4 & 4 & 3\\
-2 & -1 & -1 & -2\\
-3 & 0 & 0 & 3 \\
2 & 0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
2 & 2 & 1 & 1\\
4 & 4 & 3 & 3\\
-1 & -1 & -2 & -2\\
-3 & 0 & 0 & 3 \\
2 & 0 & 0 & 0
\end{bmatrix} \approx \begin{bmatrix}
2 & 2 & 1 & 1\\
0 & 0 & -1 & -1\\
0 & 0 & -3 & -3\\
0 & 0 & 3 & -3 \\
0 & 0 & 0 & 2
\end{bmatrix}\approx \begin{bmatrix}
2 & 2 & 1 & 1\\
0 & 0 & -1 & -1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & -6 \\
0 & 0 & 0 & 2
\end{bmatrix} \approx \begin{bmatrix}
2 & 2 & 1 & 1\\
0 & 0 & -1 & -1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & -6 \\
0 & 0 & 0 & 0
\end{bmatrix}$
Since $rank (A)=3 \lt dim (M_{2}R)=4$, the vectors of $S$ do not span $M_2(R)$
