Answer
See solution
Work Step by Step
We will verify that outside factors on the right side of (7) are invertible.
Note that:
$\left[\begin{array}{ll}I & 0 \\ X & I\end{array}\right] \cdot\left[\begin{array}{cc}I & 0 \\ -X & I\end{array}\right]=\left[\begin{array}{ll}I & 0 \\ 0 & I\end{array}\right]$
$\left[\begin{array}{cc}I & Y \\ 0 & I\end{array}\right] \cdot\left[\begin{array}{cc}I & -Y \\ 0 & I\end{array}\right]=\left[\begin{array}{ll}I & 0 \\ 0 & I\end{array}\right]$
So, both $\left[\begin{array}{ll}I & 0 \\ X & I\end{array}\right]$ and $\left[\begin{array}{ll}I & Y \\ 0 & I\end{array}\right]$ are invertible.
We can multiply equation (7) from left with $\left[\begin{array}{ll}I & 0 \\ X & I\end{array}\right]^{-1}$ and from right with $\left[\begin{array}{ll}I & Y \\ 0 & I\end{array}\right]^{-1}$ to obtain
$\left[\begin{array}{cc}A_{11} & 0 \\ 0 & S\end{array}\right]=\left[\begin{array}{cc}I & 0 \\ X & I\end{array}\right]^{-1} \cdot A \cdot\left[\begin{array}{cc}I & Y \\ 0 & I\end{array}\right]^{-1}$
From Theorem 6 it follows that $\left[\begin{array}{cc}A_{11} & 0 \\ 0 & S\end{array}\right]$ is invertible as product of invertible matrices $(A$ is given as invertible matrix
By exercise $13,$ we know that $S$ is invertible.
