Answer
See explanation
Work Step by Step
Because the two matrices are inverses
\[
\left[\begin{array}{lll}
I & 0 & 0 \\
C & I & 0 \\
A & B & I
\end{array}\right]\left[\begin{array}{lll}
I & 0 & 0 \\
Z & I & 0 \\
X & Y & I
\end{array}\right]=\left[\begin{array}{lll}
I & 0 & 0 \\
0 & I & 0 \\
0 & 0 & I
\end{array}\right]
\]
2
Get the left side of the equation:
\[
\left[\begin{array}{lll}
I & 0 & 0 \\
C & I & 0 \\
A & B & I
\end{array}\right]\left[\begin{array}{lll}
I & 0 & 0 \\
Z & I & 0 \\
X & Y & I
\end{array}\right]=\left[\begin{array}{ccc}
I I+0 Z+0 X & I 0+0 I+0 Y & I 0+00+0 I \\
C I+I Z+0 X & C 0+I I+0 Y & C 0+I 0+0 I \\
A I+B Z+I X & A 0+B I+I Y & A 0+B 0+I I
\end{array}\right]
\]
3
Set this equal to the right side of the equation:
\[
\left[\begin{array}{ccc}
I & 0 & 0 \\
C+Z & I & 0 \\
A+B Z+X & B+Y & I
\end{array}\right]=\left[\begin{array}{ccc}
I & 0 & 0 \\
0 & I & 0 \\
0 & 0 & I
\end{array}\right]
\]
4
$\mathrm{So}$
\[
\begin{array}{ccc}
I=I & 0=0 & 0=0 \\
C+Z=0 & I=I & 0=0 \\
A+B Z+X=0 & B+Y=0 & I=I
\end{array}
\]
5
Because the (2,1) blocks are equal, $C+Z=0$ and $C=-Z$. Likewise since the (3,2) blocks are equal, $B+Y=0$ and $Y=-B .$ Finally, from the (3,1) entries, $A+B Z+X=0$ and $. X=-A-B Z$ since
\[
Z=-C, X=-A-B(-C)=-A+B C
\]
