Answer
See solution
Work Step by Step
Get the left side of the equation:
\[
\left[\begin{array}{lll}
I & 0 & 0 \\
X & I & 0 \\
Y & 0 & I
\end{array}\right]\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22} \\
A_{31} & A_{32}
\end{array}\right]=\left[\begin{array}{cc}
I A_{11}+0 A_{21}+0 A_{31} & I A_{12}+0 A_{22}+0 A_{32} \\
X A_{11}+I A_{21}+0 A_{31} & X A_{12}+I A_{22}+0 A_{32} \\
Y A_{11}+0 A_{21}+I A_{31} & Y A_{12}+0 A_{22}+I A_{32}
\end{array}\right]
\]
Set this equal to the right side of the equation:
\[
\left[\begin{array}{cc}
I A_{11} & I A_{12} \\
X A_{11}+I A_{21} & X A_{12}+I A_{22} \\
Y A_{11}+I A_{31} & Y A_{12}+I A_{32}
\end{array}\right]=\left[\begin{array}{cc}
B_{11} & B_{12} \\
0 & B_{22} \\
0 & B_{32}
\end{array}\right]
\]
Equating both sides
\[
\begin{array}{cc}
A_{11}=B_{11} & A_{12}=B_{12} \\
X A_{11}+A_{21}=0 & X A_{12}+A_{22}=B_{22} \\
Y A_{11}+I A_{31}=0 & Y A_{12}+I A_{32}=B_{32}
\end{array}
\]
Because the (2,1) blocks are equal, $X A_{11}+A_{21}=0$ and $X A_{11}=-A_{21}$ because $A_{11}$ is invertible, right multiplication by $A_{11}^{-1}$ gives $X=-A_{21} A_{11}^{-11}$ Likewise since the (3,1) blocks are equal, $Y A_{11}+I A_{31}=0$ and $.$ since $A_{11} 11$ is invertible, right multiplication by $A_{11}^{-1}$ gives . $Y=-A_{31} A_{11}^{-11}$ Finally, from the (2,2) entries, $X A_{12}+A_{22}=B_{22}$. Because $X=-A_{21} A_{11}^{-1}, B_{22}=$
\[
-A_{21} A_{11}^{-1} A_{12}+A l 22
\]
