Answer
$\left[\begin{array}{ll}
25 & -9\\
4 & 20
\end{array}\right]$
Work Step by Step
We found $AC$ is problem 15:
$ AC =\begin{bmatrix}
28 & -9\\
4 & 23
\end{bmatrix}$
Next, calculate $3$ times the identity matrix:
$3I=3 \begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}=\begin{bmatrix}
3 & 0\\
0 & 3
\end{bmatrix}$
Now, find the difference:
$AC-3I=\begin{bmatrix}
28 & -9\\
4 & 23
\end{bmatrix} \begin{bmatrix}
3 & 0\\
0 & 3
\end{bmatrix} \\=\left[\begin{array}{ll}
28-3 & -9-0\\
4-0 & 23-3
\end{array}\right] \\=\left[\begin{array}{ll}
25 & -9\\
4 & 20
\end{array}\right]$
