Answer
X=I/A
Y=0
Z=-B/A
Work Step by Step
To solve the system of equations, we first find that the product of \[
\left[ {\begin{array}{cc}
A&B\\
0&I\\
\end{array} } \right]
\] and
\[
\left[ {\begin{array}{cc}
X&Y&Z\\
0&0&1
\end{array} } \right]
\]
is
\[
\left[ {\begin{array}{cc}
AX&AY&AZ+B\\
0&0&I
\end{array} } \right]
\]
Then, we set up a system of equations based on the given fact that this product is also equal to \[
\left[ {\begin{array}{cc}
I&0 &0\\
0&0&1\\
\end{array} } \right]
\]
in order to isolate equations for X, Y, and Z using A, B, and C.
$AX=1\\
AY=0\\
AZ+B=0$.
The fastest thing to deal with is the second equation, because it means that at least one of A or Y must be equal to 0. Because A cannot be equal to 0, Y=0.
The first equation gives us that X=I/A based on simple division, and now we use the third equation.
$AZ+B=0\\
AZ=-B\\
AZ/A=-B/A\\
Z=-B/A$,
which means:
X=I/A
Y=0
Z=-B/A
