TITLE: Relation between inverse of truncated and original matrix
QUESTION [0 upvotes]: I've got matrix A:
\begin{bmatrix}0&0&0.9070&0.4724&0\\0.2740&0.8045&0&0&0.7579\\0.7292&    0.6490&0&0&0.5162\\0&0&0.3600&0.0949&0\\0.4633&0.3413&0&0&0.7306\end{bmatrix}
with inv(A):
\begin{bmatrix}0&-1.5372&1.6981&0&0.3949\\0&1.5148&0.7787&0&-2.1214\\-1.1309&0&0&5.6272&0\\4.2879&0&0&-10.8036&0\\0&0.2674&-1.4406&0&2.1091\end{bmatrix}
If I remove rows(1,4) and cols(3,4) from A, it gives matrix B:
\begin{bmatrix}0.2740&0.8045&0.7579\\0.7292&0.6490&0.5162\\0.4633&0.3413&0.7306\end{bmatrix}
with inv(B):
\begin{bmatrix}-1.5372&1.6981&0.3949\\1.5148&0.7787&-2.1214\\0.2674&-1.4406&2.1091\end{bmatrix}
It is obvious that I can get inv(B) from inv(A) by:
inv(B) = inv(A) (unremovedCols,unremovedRows) = inv(A)([1,2,5],[2,3,5])
What principle/theorem is behind this observation?

REPLY [1 votes]: This is a consequence of the following two facts:

For a block-diagonal matrix whose diagonal blocks are square and invertible, we have
$$
\pmatrix{A&0\\0 & B}^{-1} = \pmatrix{A^{-1}&0\\0 & B^{-1}}.
$$
For permutation matrices $P,Q$ and an invertible matrix $M$, we have
$$
(PMQ)^{-1} = Q^TM^{-1}P^T.
$$

Now, try to find two permutation matrices $P,Q$ and a block diagonal matrix $M$ such that your matrix can be expressed as $A = PMQ$.  In particular, we can take
$$
P = \left[\begin{matrix}0 & 0 & 0 & 0 & 1\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0\\1 & 0 & 0 & 0 & 0\end{matrix}\right], \quad 
Q = \left[\begin{matrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 1 & 0\\0 & 0 & 1 & 0 & 0\end{matrix}\right],
$$
${}$
$$
M = \left[\begin{array}{ccc|cc}
0.4633 & 0.3413 & 0.7306 & 0 & 0\\
0.2740 & 0.8045 & 0.7579 & 0 & 0\\
0.7292 & 0.6490 & 0.5162 & 0 & 0\\
\hline 
0 & 0 & 0 & 0.0949 & 0.3600\\
0 & 0 & 0 & 0.4724 & 0.9070\end{array}\right].
$$
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