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 that the following matrix has the matrix: 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
 A: 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.
A: This is what you get from performing a blockwise Gaussian elimination on colums, assuming everything is invertible when you need it:
Starting with $\left[\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right]$ you want to multiply the first block of colums by $A^{-1}$, you obtain
$$
\left[\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right]
\left[\begin{smallmatrix}A^{-1}&0\\0&I\end{smallmatrix}\right]
=
\left[\begin{smallmatrix}I&B\\CA^{-1}&D\end{smallmatrix}\right].
$$
Now subtract the first block column $B$-times from the second:
$$
\left[\begin{smallmatrix}I&B\\CA^{-1}&D\end{smallmatrix}\right]
\left[\begin{smallmatrix}I&-B\\0&I\end{smallmatrix}\right]
=
\left[\begin{smallmatrix}I&0\\CA^{-1}&D-CA^{-1}B\end{smallmatrix}\right].
$$
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
$$
\left[\begin{smallmatrix}I&0\\CA^{-1}&D-CA^{-1}B\end{smallmatrix}\right]
\left[\begin{smallmatrix}I&0\\0&(D-CA^{-1}B)^{-1}\end{smallmatrix}\right]
=
\left[\begin{smallmatrix}I&0\\CA^{-1}&I\end{smallmatrix}\right].
$$
Finally subtract the second block column $CA^{-1}$-times from the first block column
$$
\left[\begin{smallmatrix}I&0\\CA^{-1}&I\end{smallmatrix}\right]
\left[\begin{smallmatrix}I&0\\-CA^{-1}&I\end{smallmatrix}\right]
=
\left[\begin{smallmatrix}I&0\\0&I\end{smallmatrix}\right].
$$
In summary, you obtained
\begin{align*}
\left[\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right]^{-1}
&=
\left[\begin{smallmatrix}A^{-1}&0\\0&I\end{smallmatrix}\right]
\left[\begin{smallmatrix}I&-B\\0&I\end{smallmatrix}\right]
\left[\begin{smallmatrix}I&0\\0&(D-CA^{-1}B)^{-1}\end{smallmatrix}\right]
\left[\begin{smallmatrix}I&0\\-CA^{-1}&I\end{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.
