Checking if a matrix is in the span of other matrices Problem: Expand the following set matrices \begin{align*} \left\{ \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}, \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}, \begin{pmatrix} 0 & -3 \\ 5 & 4 \end{pmatrix}\right\} \end{align*} to a basis of $\mathbb{R}^{2 \times 2}$.
What is the best way to do this? Should I just pick an arbitrary matrix and check for linear dependence? 
I picked the matrix $\begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix}$. So now we need to know if there are scalars $a,b$ and $c$ such that \begin{align*} \begin{pmatrix} 3 & 1 \\ 2 & 1 \end{pmatrix} = a \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} + b \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix} + c \begin{pmatrix} 0 & -3 \\ 5 & 4 \end{pmatrix}. \end{align*} This gives rise to the following system of equations: \begin{align*} \begin{cases} a + 2b &= 3 \\ -a + b -c &= 1 \\ 2a - b + 5c &= 2 \\ 3a + 4b + 4c &= 1 \end{cases} \end{align*} I put this in augmented matrix-form as follows: \begin{align*}\left(\begin{array}{ccc|c}  1 & 2 & 0 & 3 \\ -1 & 1 & -3 & 1 \\ 2 & -1 & 5 & 2 \\ 3 & 4 & 4 & 1 \end{array}\right) \end{align*} I eventually row-reduced it to \begin{align*}\left(\begin{array}{ccc|c}  1 & 2 & 0 & 3 \\ 0 & 1 & -1 & 4/3 \\ 0 & 0 & 0 & 8/3 \\ 0 & 0 & 2 & -16/3 \end{array}\right) \end{align*} Since the third row implies a contradiction, can I conclude on the basis of this that my four matrices are linearly independent?

Instead of a blind guess you could check which standard basis element is not in the span of your set. (You should also check that your set is linearly independent if it is not already assumed so).
i.e. check that these systems have solutions
\begin{align*} \begin{cases} a + 2b &= 1 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 0 \\ 3a + 4b + 4c &= 0 \end{cases} \end{align*}
\begin{align*} \begin{cases} a + 2b &= 0 \\ -a + b -c &= 1 \\ 2a - b + 5c &= 0 \\ 3a + 4b + 4c &= 0 \end{cases} \end{align*}
\begin{align*} \begin{cases} a + 2b &= 0 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 1 \\ 3a + 4b + 4c &= 0 \end{cases} \end{align*}
\begin{align*} \begin{cases} a + 2b &= 0 \\ -a + b -c &= 0 \\ 2a - b + 5c &= 0 \\ 3a + 4b + 4c &= 1 \end{cases} \end{align*}
I think only the last one doesn't work. (You should check) So just add in the matrix
\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}
