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

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 $A$?

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 $A$  ?

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

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 $A^{-1}$.

How can I efficiently compute the following $3 \times 3 $ matrix inverse $((A^{-1})_{\cal I \times \cal I})^{-1}$ with no need to inverse the large matrix $A$ ?

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 $A$ ?