For
\[\mathbf{A} = \begin{pmatrix} 2 & 7 \\ 13 & -2 \end{pmatrix},\]compute $(\mathbf{A}^{-1})^{-1}.$
We can check that $\mathbf{A}$ is invertible, and so
\[\mathbf{A} \mathbf{A}^{-1} = \mathbf{I}.\]By the same token,
\[\mathbf{A}^{-1} (\mathbf{A}^{-1})^{-1} = \mathbf{I}.\]Therefore, $(\mathbf{A}^{-1})^{-1} = \mathbf{A} = \boxed{\begin{pmatrix} 2 & 7 \\ 13 & -2 \end{pmatrix}}.$