\(A^{2}=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]\)
$$^{2}=\left[\begin{array}{ll}
1 \times 1+(-1 \times 2) & (1 \times-1)+(-1 \times 3) \\
(2 \times 1+3 \times 2) & (2 \times-1+3 \times 3)
\end{array}\right]
$$
\(A ^{2}=\left[\begin{array}{cc}-1 & -4 \\ 8 & 7\end{array}\right]\)
using the steps we use to get \(A^{2}\)
\(A B=\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 4 & 3\end{array}\right]=\left[\begin{array}{cc}-4 & -1 \\ 12 & 13\end{array}\right]\)
\(2.A=2\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]=\left[\begin{array}{cc}2 & -2 \\ 4 & 6\end{array}\right]\)
So that.
$$^{2}+A B-2 . A=\left[\begin{array}{cc}
-1 & -4 \\
8 & 7
\end{array}\right]+\left[\begin{array}{cc}
-4 & -1 \\
12 & 13
\end{array}\right]-\left[\begin{array}{cc}
2 & -2 \\
4 & 6
\end{array}\right]
$$
$$
=\left[\begin{array}{cc}
-7 & -3 \\
16 & 14
\end{array}\right]
$$