Answer
See below
Work Step by Step
Given: $V=M_2(R)$
and $v_1=\begin{bmatrix}
1 & 2 \\ 3 & 4
\end{bmatrix}\\
v_2=\begin{bmatrix}
-1 & 2 \\ 5 & 7
\end{bmatrix}\\
v_3=\begin{bmatrix}
3 & 2\\ 1 & 1
\end{bmatrix}$
Obtain the system
$av_1+bv_2+cv_3=0\\
\begin{bmatrix}
1 & -1 & 3 \\ 2 & 2 & 2\\ 3 & 5 & 1\\4 & 7 & 1
\end{bmatrix}\begin{bmatrix}
a \\ b \\ c
\end{bmatrix}=\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}\\
\rightarrow \begin{bmatrix}
1 & -1 & 3 \\ 0 & -4 & 4\\ 0 & -8 & 8\\ 0 & -3 & 3
\end{bmatrix}\begin{bmatrix}
a \\ b \\ c
\end{bmatrix}=\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}\\
\rightarrow \begin{bmatrix}
1 & -1 & 3 \\ 0 & -4 & 4\\ 0 & 0 & 0\\0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
a \\ b \\ c
\end{bmatrix}=\begin{bmatrix}
0 \\ 0 \\ 0
\end{bmatrix}$
then we have $-4b+4c=0\rightarrow b=c\\
a-b+3c=0\rightarrow a=b-3c=-2c$
Suppose $a=-2\\
b=c=1$
We can see that $-2v_1+v_2+v_3=0$
Thus, we can span the same subspace with just two matrices.
We can choose between: $v_1,v_2\\
v_2,v_3\\
v_1,v_3$
