Theorem LIVRN suggests we analyze a matrix whose columns are the vectors from the set,
\begin{equation*}
A=\begin{bmatrix}
1 & 3 & 4 & -1\\
2 & 2 & 4 & 2\\
-1 & -1 & -2 & -1\\
0 & 2 & 2 & -2\\
1 & 2 & 3 & 0
\end{bmatrix}
\end{equation*}
Row-reducing the matrix $A$ yields,
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 1 & 2\\
0 & \leading{1} & 1 & -1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
We see that $r=2\neq 4=n$, where $r$ is the number of nonzero rows and $n$ is the number of columns. By Theorem LIVRN, the set $S$ is linearly dependent.