Answer
$$\left[ \begin {array}{cc} 6&2\\ 9&11
\end {array} \right]=-\left[ \begin {array}{cc} 2&-3\\ 4&1
\end {array} \right]+5\left[ \begin {array}{cc} 0&5\\ 1&-2
\end {array} \right].$$
Work Step by Step
Assume the combination
$$\left[ \begin {array}{cc} -2&28\\ 1&-11
\end {array} \right]=a\left[ \begin {array}{cc} 2&-3\\ 4&1
\end {array} \right]+b\left[ \begin {array}{cc} 0&5\\ 1&-2
\end {array} \right].$$
We have the system
\begin{align*}
2a&=-2\\
-3a+5b&=28\\
4a+b&=1\\
a-2b&=-11.
\end{align*}
We get the solution $$a=-1, \quad b=5.$$
Consequently,
$$\left[ \begin {array}{cc} 6&2\\ 9&11
\end {array} \right]=-\left[ \begin {array}{cc} 2&-3\\ 4&1
\end {array} \right]+5\left[ \begin {array}{cc} 0&5\\ 1&-2
\end {array} \right].$$