Hint:
A matrix $$ A= \begin{bmatrix} a&b\\c&d \end{bmatrix} $$ is represented in the canonical basis as: $$ A=a\begin{bmatrix} 1&0\\0&0 \end{bmatrix} +b\begin{bmatrix} 0&b\\0&0 \end{bmatrix} +c\begin{bmatrix} 0&0\\c&0 \end{bmatrix} +d\begin{bmatrix} 0&0\\0&d \end{bmatrix} $$$$ A=a\begin{bmatrix} 1&0\\0&0 \end{bmatrix} +b\begin{bmatrix} 0&1\\0&0 \end{bmatrix} +c\begin{bmatrix} 0&0\\1&0 \end{bmatrix} +d\begin{bmatrix} 0&0\\0&1 \end{bmatrix} $$ so, in this basis, you can represent it as a vector: $$ \vec A= \begin{bmatrix} a\\b\\c\\d \end{bmatrix} $$ So your matrix $M$ acts as: $$ M\vec A= \begin{bmatrix} 0 & 0 & 0 & 0 \\ -1 & -3 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a\\b\\c\\d \end{bmatrix} $$ and you have to solve : $$ \begin{bmatrix} 0 & 0 & 0 & 0 \\ -1 & -3 & 3 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a\\b\\c\\d \end{bmatrix}=0 $$