If $A=\left[\begin{array}{rr}-2 & 6 \\ -5 & -1\end{array}\right]$ then $A^{-1}$ is:

$\frac{1}{28}\left[\begin{array}{ll}2 & -5 \\ 6 & -1\end{array}\right]$ $\frac{1}{32}\left[\begin{array}{rr}-1 & 6 \\ -5 & -2\end{array}\right]$ $\frac{1}{32}\left[\begin{array}{rr}-1 & -6 \\ 5 & -2\end{array}\right]$ $\frac{-1}{28}\left[\begin{array}{ll}-1 & -6 \\ -5 & -2\end{array}\right]$

$\frac{1}{32}\left[\begin{array}{rr}-1 & -6 \\ 5 & -2\end{array}\right]$

The correct answer is Option (3) - $\frac{1}{32}\left[\begin{array}{rr}-1 & -6 \\ 5 & -2\end{array}\right]$