Answer
$\begin{bmatrix}
25&-9\\
4&20
\end{bmatrix}$
Work Step by Step
Refer to sum $\#15$ for determination of $AC$.
$\implies I_{2} = \begin{bmatrix}
1&0\\
0&1\end{bmatrix} \implies 3I_{2} = \begin{bmatrix}3&0\\0&3\end{bmatrix}$
$\implies \begin{bmatrix}28&-9\\4&23\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix} = AC-3I_{2}$
We subtract matrices by subtracting the respective individual elements.
$\therefore AC-3I_{2} = \begin{bmatrix}
28-3&-9-0\\
4-0&23-3
\end{bmatrix} = \begin{bmatrix}
25&-9\\
4&20
\end{bmatrix}$
