I'm asked if the matrices
$\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}$, $\begin{pmatrix} 0 & 2\\ 3 & 0 \end{pmatrix}$, $\begin{pmatrix} 2 & 1\\ -3 & -2 \end{pmatrix}$, $\begin{pmatrix} -1 & 4\\ 5 & 1 \end{pmatrix}$
span $M_2 (\mathbb{R})$.
I know that the traces of each of them are $0$ so they can't possibly span $M_2 (\mathbb{R})$ since you can't write them as a linear combination with matrices that have non-zero traces.
However, I also attempted to do this with a system of equations. If I make a coefficient matrix from this, I get:
$\begin{pmatrix} 1 & 0 & 2 & -1 \\ 1 & 2 & 1 & 4 \\ 1 & 3 & -3 & 5 \\ -1 & 0 & -2 & 1 \\ \end{pmatrix}$.
After row reducing, I get:
$\begin{pmatrix} 1 & 0 & 0 & -13/7 \\ 0& 1 & 0& 19/7 \\ 0 & 0 & 1 & 3/7 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$
this means the last variable is free and I have a consistent solution and hence, should span $M_2 (\mathbb{R})$ but it doesn't.
Why is my reasoning wrong?
As the last row is equal to zero, you won't be able to generate a vector having the last coordinate not equal to zero.
Hence those four matrices can't span $M_2(\mathbb R)$.