Answer
$C=\left[\begin{array}{rrr}
1 & 1 & -1\\
-1 & 1 & 0
\end{array}\right]$
$AC= \left[\begin{array}{lll}
-1 & 3 & -1\\
-2 & 4 & -1\\
-4 & 6 & -1
\end{array}\right]$
Work Step by Step
After trial and error,
$C=\left[\begin{array}{lll}
1 & 1 & -1\\
-1 & 1 & 0
\end{array}\right]$
$CA=I_{2}$
$AC$ is a 3$\times$3 matrix, so it can not equal $I_{3}.$
$AC= \left[\begin{array}{lll}
-1 & 3 & -1\\
-2 & 4 & -1\\
-4 & 6 & -1
\end{array}\right]$
