Let $A$ be a large matrix (say, $1000 \times 1000$), and let $\mathcal I = \{2,3,5\}$ be a set of row/column indices. Let $(A^{-1})_{\cal I \times I}$ denote the submatrix of $A^{-1}$ that consists of the $\{2,3,5\}$ rows and columns of $A^{-1}$.


Is there an efficient way of computing the following $3 \times 3$ matrix inverse $((A^{-1})_{\cal I \times \cal I})^{-1}$ without inverting the large matrix $A$?