Answer
$\left[\begin{array}{ll}
25 & -9\\
4 & 20
\end{array}\right]$
Work Step by Step
We found in exercise 15:
$ AC =\left[\begin{array}{ll}
28 & -9\\
4 & 23
\end{array}\right],\qquad$ a 2 by 2 matrix.
$3I_{2}=3\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
3 & 0\\
0 & 3
\end{array}\right]\qquad $also a 2 by 2 matrix,
so the difference is defined.
$AC-3I=\left[\begin{array}{ll}
28-3 & -9-0\\
4-0 & 23-3
\end{array}\right]=\left[\begin{array}{ll}
25 & -9\\
4 & 20
\end{array}\right]$