Cfrgtkky

Let $A,B,C,D in mathbb{R}^{n×n}$. Show that if $A, C, B−AC^{−1}D,$ and $D−CA^{−1}B$ are nonsingular then

$left[ begin{smallmatrix} A&B\ C&D end{smallmatrix} right]^{-1} = left[ begin{smallmatrix} A^{-1} - A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\ (B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1} end{smallmatrix} right]$

So i think that probably i have to use The Determinant and the adjacency matrix but then i get determinant =
$AD - BC $

Adjacency matrix $= left[ begin{smallmatrix} D&-B\ -C&A end{smallmatrix} right]$

But i cant divide the determinant by the adjacency matrix because AD - BC doesnt have an inverse so what do i do?emphasized text

To prove that
$$left[ begin{smallmatrix} A&B\ C&D end{smallmatrix} right]^{-1} = left[ begin{smallmatrix} A^{-1} - A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\ (B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1} end{smallmatrix} right]$$

I would first try to calculate

$$left[ begin{smallmatrix} A&B\ C&D end{smallmatrix} right]cdot left[ begin{smallmatrix} A^{-1} - A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\ (B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1} end{smallmatrix} right]$$

and show that this equals the identity matrix.

Let $I$ denote the $n$-by-$n$ identity matrix. For simplicity, let $$X:=begin{bmatrix}A&B\C&Dend{bmatrix},.$$
As $A$ is invertible,
$$X=begin{bmatrix}I&0\CA^{-1}&Iend{bmatrix},begin{bmatrix}A&B\0&D-CA^{-1}Bend{bmatrix},.$$
Since $D-CA^{-1}B$ is invertible,
$$X=begin{bmatrix}I&0\CA^{-1}&Iend{bmatrix},begin{bmatrix}A&0\0&D-CA^{-1}Bend{bmatrix},begin{bmatrix}I&A^{-1}B\0&Iend{bmatrix},.$$
Consequently,
$$X^{-1}=begin{bmatrix}I&-A^{-1}B\0&Iend{bmatrix},begin{bmatrix}A^{-1}&0\0&(D-CA^{-1}B)^{-1}end{bmatrix},begin{bmatrix}I&0\-CA^{-1}&Iend{bmatrix},.$$
That is,
$$X^{-1}=begin{bmatrix}A^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\0&(D-CA^{-1}B)^{-1}end{bmatrix},begin{bmatrix}I&0\-CA^{-1}&Iend{bmatrix},.$$
Hence,
$$X^{-1}=begin{bmatrix}A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}end{bmatrix},.tag{*}$$

Similarly,
$$X=begin{bmatrix}I&AC^{-1}\0&Iend{bmatrix},begin{bmatrix}0&B-AC^{-1}D\C&Dend{bmatrix},.$$
Thus,
$$X=begin{bmatrix}I&AC^{-1}\0&Iend{bmatrix},begin{bmatrix}0&B-AC^{-1}D\C&0end{bmatrix},begin{bmatrix}I&C^{-1}D\0&Iend{bmatrix},.$$
Because $B-AC^{-1}D$ is invertible,
$$X^{-1}=begin{bmatrix}I&-C^{-1}D\0&Iend{bmatrix},begin{bmatrix}0&C^{-1}\(B-AC^{-1}D)^{-1}&0end{bmatrix},begin{bmatrix}I&-AC^{-1}\0&Iend{bmatrix},.$$
Ergo,
$$X^{-1}=begin{bmatrix}-C^{-1}D(B-AC^{-1}D)^{-1}&C^{-1}\(B-AC^{-1}D)^{-1}&0end{bmatrix},begin{bmatrix}I&-AC^{-1}\0&Iend{bmatrix},.$$
Thence,
$$X^{-1}=begin{bmatrix}-C^{-1}D(B-AC^{-1}D)^{-1} & C^{-1}+C^{-1}D(B-AC^{-1}D)^{-1}AC^{-1}\ (B-AC^{-1}D)^{-1}& -(B-AC^{-1}D)^{-1}AC^{-1}end{bmatrix},.tag{#}$$

From (*) and (#), we conclude that
$$ (D-CA^{-1}B)^{-1}=-(B-AC^{-1}D)^{-1}AC^{-1},.$$
That is,
$$(D-CA^{-1}B)^{-1}CA^{-1}=-(B-AC^{-1}D)^{-1},.$$
Hence,
$$A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}=A^{-1}-A^{-1}B(B-AC^{-1}D)^{-1},.$$
This shows that
$$X=begin{bmatrix}A^{-1}-A^{-1}B(B-AC^{-1}D)^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\(B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1}end{bmatrix},.$$

If $D$ is invertible, then
$$X=begin{bmatrix}I&BD^{-1}\0&Iend{bmatrix},begin{bmatrix}A-BD^{-1}C&0\C &Dend{bmatrix}=begin{bmatrix}I&BD^{-1}\0&Iend{bmatrix},begin{bmatrix}A-BD^{-1}C&0\0 &Dend{bmatrix},begin{bmatrix}I&0\D^{-1}C&Iend{bmatrix},.$$
Ergo,
$$X^{-1}=begin{bmatrix}I&0\-D^{-1}C&Iend{bmatrix},begin{bmatrix}(A-BD^{-1}C)^{-1}&0\0 &D^{-1}end{bmatrix},begin{bmatrix}I&-BD^{-1}\0&Iend{bmatrix},,$$
provided that $B-D^{-1}C$ is nonsingular. Thus,
$$X^{-1}=begin{bmatrix}(A-BD^{-1}C)^{-1} & 0\-D^{-1}C(A-BD^{-1}C)^{-1} & D^{-1}end{bmatrix},begin{bmatrix}I&-BD^{-1}\0&Iend{bmatrix},,$$
or
$$X=begin{bmatrix} (A-BD^{-1}C)^{-1} & -(A-BD^{-1}C)^{-1} BD^{-1}\ -D^{-1}C(A-BD^{-1}C)^{-1}& D^{-1}+D^{-1}C(A-BD^{-1}C)^{-1}BD^{-1}end{bmatrix},.tag{@}$$

If $B$ is invertible, then
$$X=begin{bmatrix}I&0\DB^{-1}&Iend{bmatrix},begin{bmatrix}A&B\C -DB^{-1}A&0end{bmatrix}=begin{bmatrix}I&0\DB^{-1}&Iend{bmatrix},begin{bmatrix}0&B\C-DB^{-1}A &0end{bmatrix},begin{bmatrix}I&0\B^{-1}A&Iend{bmatrix},.$$
Ergo,
$$X^{-1}=begin{bmatrix}I&0\-B^{-1}A&Iend{bmatrix},begin{bmatrix}0&(C-DB^{-1}A )^{-1}\B^{-1} &0end{bmatrix},begin{bmatrix}I&0\-DB^{-1}&Iend{bmatrix},,$$
provided that $C-DB^{-1}A$ is nonsingular. Thus,
$$X^{-1}=begin{bmatrix}0 & (C-DB^{-1}A )^{-1}\B^{-1} & -B^{-1}A(C-DB^{-1}A )^{-1}end{bmatrix},begin{bmatrix}I&0\-DB^{-1}&Iend{bmatrix},,$$
or
$$X=begin{bmatrix} -(C-DB^{-1}A )^{-1}DB^{-1} & (C-DB^{-1}A )^{-1}\B^{-1}+B^{-1}A(C-DB^{-1}A )^{-1}DB^{-1}& -B^{-1}A(C-DB^{-1}A )^{-1}end{bmatrix},.tag{$}$$

In the special case where $A,B,C,D$ are invertible matrices such that $A-BD^{-1}C$, $B-AC^{-1}D$, $C-DB^{-1}A$, and $D-CA^{-1}B$ are all nonsingular, we have the following compact form of $X^{-1}$:
$$X^{-1}=begin{bmatrix}(A-BD^{-1}C)^{-1}&(C-DB^{-1}A)^{-1}\(B-AC^{-1}D)^{-1}&(D-CA^{-1}B)^{-1}end{bmatrix},.$$
While this looks nice, I think it is computationally more expensive than using (*), (#), (@), or ($) alone.

This is what you get from performing a blockwise Gaussian elimination on colums, assuming everything is invertible when you need it:

Starting with $left[begin{smallmatrix}A&B\C&Dend{smallmatrix}right]$ you want to multiply the first block of colums by $A^{-1}$, you obtain
$$
left[begin{smallmatrix}A&B\C&Dend{smallmatrix}right]
left[begin{smallmatrix}A^{-1}&0\0&Iend{smallmatrix}right]
=
left[begin{smallmatrix}I&B\CA^{-1}&Dend{smallmatrix}right].
$$
Now subtract the first block column $B$-times from the second:
$$
left[begin{smallmatrix}I&B\CA^{-1}&Dend{smallmatrix}right]
left[begin{smallmatrix}I&-B\0&Iend{smallmatrix}right]
=
left[begin{smallmatrix}I&0\CA^{-1}&D-CA^{-1}Bend{smallmatrix}right].
$$
Now you want to get an $I$ in the lower right block, so you multiply the second column block by $(D-CA^{-1}B)^{-1}$ to obtain
$$
left[begin{smallmatrix}I&0\CA^{-1}&D-CA^{-1}Bend{smallmatrix}right]
left[begin{smallmatrix}I&0\0&(D-CA^{-1}B)^{-1}end{smallmatrix}right]
=
left[begin{smallmatrix}I&0\CA^{-1}&Iend{smallmatrix}right].
$$
Finally subtract the second block column $CA^{-1}$-times from the first block column
$$
left[begin{smallmatrix}I&0\CA^{-1}&Iend{smallmatrix}right]
left[begin{smallmatrix}I&0\-CA^{-1}&Iend{smallmatrix}right]
=
left[begin{smallmatrix}I&0\0&Iend{smallmatrix}right].
$$
In summary, you obtained
begin{align*}
left[begin{smallmatrix}A&B\C&Dend{smallmatrix}right]^{-1}
&=
left[begin{smallmatrix}A^{-1}&0\0&Iend{smallmatrix}right]
left[begin{smallmatrix}I&-B\0&Iend{smallmatrix}right]
left[begin{smallmatrix}I&0\0&(D-CA^{-1}B)^{-1}end{smallmatrix}right]
left[begin{smallmatrix}I&0\-CA^{-1}&Iend{smallmatrix}right] \
&=
left[begin{smallmatrix}A^{-1}+A^{-1}B(D-CA^{-1}B)^{-1}CA^{-1}&-A^{-1}B(D-CA^{-1}B)^{-1}\-(D-CA^{-1}B)^{-1}CA^{-1}&(D-CA^{-1}B)^{-1}end{smallmatrix}right]
end{align*}
using only the assumptions that $A$ and $D-CA^{-1}B$ are invertible.

Assuming that $C$ and $B-AC^{-1}D$ are also invertible, you can simplify this to the matrix in your question by using $Y^{-1} Z = (Z^{-1} Y)^{-1}$ a few times.