Answer
$S$ is a basis for $M_{2,2}$.
Work Step by Step
The set
$S=\left\{\left[\begin{array}{ll}{2} & {0} \\ {0} & {3}\end{array}\right],\left[\begin{array}{ll}{1} & {4} \\ {0} & {1}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {3} & {2}\end{array}\right],\left[\begin{array}{ll}{0} & {1} \\ {2} & {0}\end{array}\right]\right\}$
is a basis for $M_{2,2}$. Indeed, suppose the following linear combination
$$a\left[\begin{array}{ll}{2} & {0} \\ {0} & {3}\end{array}\right]+b\left[\begin{array}{ll}{1} & {4} \\ {0} & {1}\end{array}\right]+c\left[\begin{array}{ll}{0} & {1} \\ {3} & {2}\end{array}\right]+d\left[\begin{array}{ll}{0} & {1} \\ {2} & {0}\end{array}\right]=\left[\begin{array}{cc}{0} & {0} \\ {0} & {0}\end{array}\right].$$
Comparing the entries of both sides of the above equation, we have
\begin{align*}
2a+b&=0\\
4b+ c+d&=0\\
3c+2d&=0\\
3a+b+2c&=0.
\end{align*}
The augmented matrix of the above system is given by
$$\left[\begin{array}{cccc}
{2}&{1}&{0}&{0}\\
{0}&{4}&{1}&{1}\\
{0}&{0}&{3}&{2}\\
{3}&{1}&{2}&{0}\\
\end{array}\right]$$
and its determinant is $-31$. Therefore, the above system has unique solution, that is, the trivial solution. Then $S$ is linearly independent set of vectors and since $M_{2,2}$ has dimension $4$, then by Theorem 4.12, $S$ is a basis for $M_{2,2}$.
