Given $A\in \mathbb{N}^{n\times n}$, then $A(\mathcal{I})$ is defined by first deleting the those columns with index in $\mathcal{I}$ and then extracting the first $n-|\mathcal{I}|$ rows.
Note that the definition of $A(\mathcal{I})$ is similar to principle submatrix, but they are not the same.
For example, if $$A=\begin{bmatrix} 11 & 12 & 13 & 14 & 15\\ 21 & 22 & 23 & 24 & 25\\ 31 & 32 & 33 & 34 & 35\\ 41 & 42 & 43 & 44 & 45\\ 51 & 52 & 53 & 54 & 55\end{bmatrix},$$ then $$A(\{1,3\})=\begin{bmatrix} 12 & 14 & 15\\ 22 & 24 & 25\\ 32 & 34 & 35\end{bmatrix}, A(\{2,3,5\})=\begin{bmatrix} 11 & 14 \\ 21 & 24 \end{bmatrix}.$$
The inverse of $A(\mathcal{I})$ can be computed in some ordinary ways. However, my question is: How can we calculate the inverse of $A(\mathcal{I})$ faster if we are given arbitrary $A(\mathcal{I})$ everytime, given that the inverse of $A$ is known by us?
If the knowledge of $A$ cannot help, which kind of prior knowledge can help us calculate the inverse of $A(\mathcal{I})$ faster?
If there is no such prior knowledge that can be helpful, I also want to ask whether there is some kind of special form of $A$ such that there exists some prior knowledge which can enable us to calculate $A(\mathcal{I})$ much faster?
