Answer
See explanation
Work Step by Step
We get the right side of the equation:
\[
\left[\begin{array}{cc}
I & 0 \\
X & I
\end{array}\right]\left[\begin{array}{cc}
A_{11} & 0 \\
0 & S
\end{array}\right]\left[\begin{array}{cc}
I & Y \\
0 & I
\end{array}\right]=\left[\begin{array}{cc}
A_{11} & 0 \\
X A_{11} & S
\end{array}\right]\left[\begin{array}{cc}
I & Y \\
0 & I
\end{array}\right]=\left[\begin{array}{cc}
A_{11} & A_{11} Y \\
X A_{11} & X A_{11} Y+S
\end{array}\right]
\]
Set this equal to the left side of the equation:
\[
\begin{array}{l}
\left.\qquad \begin{array}{rl}
\quad \begin{array}{c}
A_{11} \\
X A_{11} \quad X A_{11}+S
\end{array}
\end{array}\right]=\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right] \\
\text { So, } \\
\qquad \begin{array}{rl}
A_{11}=A_{11} & A_{11} Y=A_{12} \\
X A_{11}=A_{21} & X A_{11}+S=A_{22}
\end{array}
\end{array}
\]
Because the (1,2) blocks are equal, $A_{11} Y=A_{12} .$ since $A_{11}$ is invertible, left multiplication by $A_{11}^{-1}$ gives $Y=A_{11}^{-1} A_{12}$. Likewise since the (2,1) block are equal, $X A_{11}=A_{21}$. since $A_{11}$ is invertible, right multiplication by $A_{11}^{-11}$ gives that $X=A_{21} A_{11}^{-1}$. One can check that the matrix $S$ as given in the exercise satisfies the equation $X A_{11} Y+S=A_{22}$ with the calculated values of $X$ and $Y$ given above,
