Perhaps quicker to use $$\begin{bmatrix} a&b\\\ c&d \end{bmatrix}^{-1} = \frac1{ad-bc}\begin{bmatrix} d&-b\\\ -c&a \end{bmatrix}$$
so that
$$\begin{bmatrix} 2&1\\\ a&a \end{bmatrix}^{-1} = \frac1a\begin{bmatrix} a&-1\\\ -a&2 \end{bmatrix}=\begin{bmatrix} 1&-\frac1a\\\ -1&\frac2a \end{bmatrix}$$