Answer
See answer below
Work Step by Step
Set the values to 0,
$av_1+bv_2+cv_3+dv_4=0$
we have:
$\begin{bmatrix}
1 & 2 & 1 & 2 | 0\\
1 & -1 & 1 & -1 | 0\\
-1 & 3 & 2 &2 |0\\
1 & 1 & 1 & 1 | 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 2 & 1 & 2 | 0\\
0 & 3 & 0 & 3 | 0\\
0 & 5 & 3 &4 |0\\
0 & 1 & 0 & 1| 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 2 & 1 & 2 | 0\\
0 & 3 & 0 & 3 | 0\\
0 & 0 & 9 &3 |0\\
0 & 0 & 0 & 0| 0
\end{bmatrix}$
$rank (A)=3$
Hence, the set $\{ v_1, v_2, v_3 \}$ spans the same subspace of V as that spanned by the original set of vectors $\{ v_1, v_2, v_3, v_4 \}$
