Let $A$ be a matrix of large size (say, $1000 \times 1000$), and $\cal I=\{2,3,5\}$ be the column/row index number. The notation $(A^{-1})_{\cal I \times \cal I}$ is the submatrix of $A^{-1}$ that consists of the $\{2,3,5\}$ rows of $A^{-1}$ and the $\{2,3,5\}$ columns of $A^{-1}$.

Is there any efficient way of computing the following $3 \times 3 $ matrix inverse $((A^{-1})_{\cal I \times \cal I})^{-1}$ with no need to inverse the large matrix $A$ ?