Answer
$S$ is not a basis for $M_{2,2}$.
Work Step by Step
The set
$S=\left\{\left[\begin{array}{rr}{1} & {2} \\ {-5} & {4}\end{array}\right],\left[\begin{array}{rr}{2} & {-7} \\ {6} & {2}\end{array}\right],\left[\begin{array}{rr}{4} & {-9} \\ {11} & {12}\end{array}\right],\left[\begin{array}{rr}{12} & {-16} \\ {17} & {42}\end{array}\right]\right\}$
is not a basis for $M_{2,2}$. Indeed, suppose the following linear combination
$$a\left[\begin{array}{rr}{1} & {2} \\ {-5} & {4}\end{array}\right]+b\left[\begin{array}{rr}{2} & {-7} \\ {6} & {2}\end{array}\right]+c\left[\begin{array}{rr}{4} & {-9} \\ {11} & {12}\end{array}\right]+d\left[\begin{array}{rr}{12} & {-16} \\ {17} & {42}\end{array}\right]=\left[\begin{array}{cc}{0} & {0} \\ {0} & {0}\end{array}\right].$$
Comparing the entries of both sides of the above equation, we have
\begin{align*}
a+2b+4c+12d&=0\\
2a-7b-9c-16d&=0\\
-5a+6b+11c+17d&=0\\
4a+2b+12c+42d&=0.
\end{align*}
The augmented matrix of the above system is given by
$$\left[\begin{array}{cccc}
1&2&4&12\\
2&-7&-9&-16
\\
-5&6&11&17\\
4&2&12&42
\\
\end{array}\right]$$
and its determinant is $0$. Therefore, the above system has infinite number of solutions, that is, there is lots of non trivial solution. Then $S$ is linearly dependent set of vectors and then $S$ is not a basis for $M_{2,2}$.
