= Block matrix inverse
{title2=$M^{-1}$}
A <block matrix inverse> expresses the <matrix inverse> in terms of smaller <matrices>. Partition a <square matrix> over a <field> as
$$
M=\begin{pmatrix}A&B\\C&D\end{pmatrix},\qquad
A\in\mathbb F^{p\times p},\quad D\in\mathbb F^{q\times q}.
$$
If $A$ is an <invertible matrix>, define its <Schur complement> $S=D-CA^{-1}B$. Then $M$ is an <invertible matrix> exactly when $S$ is an <invertible matrix>, and
$$
\boxed{
M^{-1}=\begin{pmatrix}
A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\
-S^{-1}CA^{-1}&S^{-1}
\end{pmatrix}.
}
$$
The order of the <matrix products> matters. In particular, the lower-right block is $S^{-1}$, rather than generally $D^{-1}$: the coupling through $B$ and $C$ changes the response.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[Appendix A.5.5, printed p. 650: inverse of a symmetric block matrix; §C.4.1, printed p. 673: elimination for general blocks]{ref}
For an original example, take
$$
M=\begin{pmatrix}2&0&1\\0&3&2\\1&2&5\end{pmatrix},\quad
A=\begin{pmatrix}2&0\\0&3\end{pmatrix},\quad
B=C^T=\begin{pmatrix}1\\2\end{pmatrix},\quad D=5.
$$
The <Schur complement> is $S=5-1/2-4/3=19/6$. Substitution gives the concise answer
$$
\boxed{M^{-1}=\frac1{19}\begin{pmatrix}11&2&-3\\2&9&-4\\-3&-4&6\end{pmatrix}}.
$$
Multiplying in either order gives the <identity matrix>; the lower-right entry is $6/19$, whereas $D^{-1}=1/5$. The difference records the feedback through the two eliminated variables.
\Image[/block-matrix-schur-inverse-example.png]
{title=A <block matrix inverse> computed from the smaller <Schur complement>}
{description=Original exact example. Colours mark the two-by-two block $A$, the coupling blocks $B,C$, and the scalar block $D$. The same partition is retained in the <matrix inverse>; its blocks are not the entrywise inverses of the original blocks.}
Often the desired output is a solution of a <linear system>, rather than every entry of a <matrix inverse>. <Block elimination> solves with $A$ and then with $S$, recovering the eliminated variables by <back substitution>. Reusing an <LU decomposition> of $A$ avoids explicitly forming $A^{-1}$. A favourable <block diagonal matrix> or <tridiagonal matrix> block can make this useful; merely partitioning an unstructured dense <matrix> does not guarantee a faster solve.<Convex Optimization by Stephen Boyd and Lieven Vandenberghe>[§C.4.1, printed pp. 672–675: Algorithm C.4, factor-solve interpretation and dense complexity comparison]{ref}
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