How to determine if a set of five $2\times2$ matrices is independent $$S=\bigg\{\left[\begin{matrix}1&2\\2&1\end{matrix}\right], \left[\begin{matrix}2&1\\-1&2\end{matrix}\right], \left[\begin{matrix}0&1\\1&2\end{matrix}\right],\left[\begin{matrix}1&0\\1&1\end{matrix}\right], 
\left[\begin{matrix}1&4\\0&3\end{matrix}\right]\bigg\}$$
How can I determine if a set of five $2\times2$ matrices are independent? 
 A: As has been pointed out, four matrices form a basis for the $2\times2$ matrices (the easiest would be
$$
\left[\begin{matrix}1&0\\0&0\end{matrix}\right], \left[\begin{matrix}0&1\\0&0\end{matrix}\right], \left[\begin{matrix}0&0\\1&0\end{matrix}\right], \left[\begin{matrix}0&0\\0&1\end{matrix}\right]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
\left[\begin{matrix}1&2\\2&1\end{matrix}\right] + \left[\begin{matrix}2&1\\-1&2\end{matrix}\right] + \left[\begin{matrix}0&1\\1&2\end{matrix}\right] - 2\left[\begin{matrix}1&0\\1&1\end{matrix}\right] - 
\left[\begin{matrix}1&4\\0&3\end{matrix}\right] = 
\left[\begin{matrix}0&0\\0&0\end{matrix}\right].
$$
A: Since the space of all $2\times2$ matrices is $4$-dimensional, every set of $5$  such matrices is linearly dependent.
A: As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2\times 2$ matrices is $4$.  
More generally, how do you show that a set of vectors is linearly dependent or independent?  Create a linear combination of the vectors, set it equal to $0$, and try to solve it.  
$$
a_1X_1 + a_2X_2 + \dotsb+ a_nX_n = 0
$$
If the only possible solution is $a_1 = a_2 = \dotsb = a_n = 0$ then the set is independent.  If a different solution exists then the set is dependent.
A: Stretch out the matrices to complete the rows of the following matrix
$$\newcommand{\adj}{\operatorname{adj}}
M(v)=\begin{bmatrix}
v_1&1&2&2&1\\
v_2&2&1&-1&2\\
v_3&0&1&1&2\\
v_4&1&0&1&1\\
v_5&1&4&0&3
\end{bmatrix}\tag1
$$
Note that the top row of the adjugate of $M(v)$
$$
\begin{bmatrix}
-14&-14&-14&28&14
\end{bmatrix}\tag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $\adj M(v)$.  By Laplace's Formula,
$$
\det M(v)=u\cdot v\tag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $\det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-\frac1{14}u$ with the fixed columns of $M(v)$ as
$$
1\ \overbrace{\begin{bmatrix}
1&2\\2&1
\end{bmatrix}}^\text{row $1$}
+
1\ \overbrace{\begin{bmatrix}
2&1\\-1&2
\end{bmatrix}}^\text{row $2$}
+
1\ \overbrace{\begin{bmatrix}
0&1\\1&2
\end{bmatrix}}^\text{row $3$}
-2\ \overbrace{\begin{bmatrix}
1&0\\1&1
\end{bmatrix}}^\text{row $4$}
-
1\ \overbrace{\begin{bmatrix}
1&4\\0&3
\end{bmatrix}}^\text{row $5$}
=
\begin{bmatrix}
0&0\\0&0
\end{bmatrix}\tag4
$$
