How to determine if set of linear mappings are linearly independent or dependent? I missed a couple of my Linear algebra classes, so I'm a little lost on this question...
Given $S_1$, $S_2$, $S_3 : \mathbf{R}^2\to \mathbf{R}^2$ are linear mappings defined by:
$S_1(x_1, x_2) = (x_1-x_2, -2x_1+x_2)$
$S_2(x_1,x_2) = (2x_1- x_2, -4x_1+ x_2)$
$S_3(x_1,x_2) = (-x_1+ x_2 , 2x_1+x_2)$
And I need to figure out whether the set $\{S_1, S_2, S_3\}$ is linearly independent or dependent.

I began to try to use constants $c_1$, $c_2$,$c_3$ and made each mapping a matrix, but that did not seem to work. Any help is appreciated. 
 A: Saying $c_1S_1+c_2S_2+c_3S_3=0$ means that, for all $(x_1,x_2)\in\mathbf{R}^2$,
$$
c_1S_1(x_1,x_2)+c_2S_2(x_1,x_2)+c_3S_3(x_1,x_2)=0
$$
that is,
$$
c_1(x_1-x_2, -2x_1+x_2)+c_2(2x_1−x_2,−4x_1+x_2)+c_3(−x_1+x_2,2x_1+x_2)=0
$$
In particular, if $x_1=1$ and $x_2=0$, you get
$$
c_1(1,-2)+c_2(2,-4)+c_3(-1,2)=0
$$
You also get, for $x_1=0$ and $x_2=1$,
$$
c_2(-1,1)+c_2(-1,1)+c_3(1,1)=0
$$
and similarly for $x_1=2$, $x_2=1$.
However, there is a simpler method. The matrices associated to $S_1$, $S_2$ and $S_3$ are
$$
\begin{bmatrix}
1 & -1\\
-2 & 1
\end{bmatrix},
\qquad
\begin{bmatrix}
2 & -1\\
-4 & 1
\end{bmatrix},
\qquad
\begin{bmatrix}
1 & 1\\
-2 & 1
\end{bmatrix}.
$$
Take a linear combination which is $0$:
$$
a\begin{bmatrix}
1 & -1\\
-2 & 1
\end{bmatrix}+
b\begin{bmatrix}
2 & -1\\
-4 & 1
\end{bmatrix}+
c\begin{bmatrix}
1 & 1\\
-2 & 1
\end{bmatrix}=
\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix}
$$
which gives
$$
\begin{cases}
a+2b+c=0\\
-a-b+c=0\\
-4a-2b-2c=0\\
a+b+c=0
\end{cases}
$$
Does this system have non trivial solutions?
A: Hint:
Consider the isomorphism $\;\begin{aligned}\mathcal M_{2\times 2}(\mathbf R)&\longrightarrow \mathbf R^4\\\begin{pmatrix}a&b\\c&d\end{pmatrix} &\longmapsto (a,b,c,d)\end{aligned},\enspace$
and use row reduction.
