A. $\left[\begin{array}{ll}1 & 2 \\ 0 & 3\end{array}\right]$ is a scalar matrix and $|3 \mathrm{~A}|=108$ suppose the scalar matrix is $\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$
$$
\begin{aligned}
&\therefore A \cdot\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right] \\
&\Rightarrow A=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]^{-1} \\
&{\left[\therefore A B=C \Rightarrow A B B^{-1}=C B^{-1} \Rightarrow A=C B^{-1}\right]} \\
&\Rightarrow A=\frac{1}{3}\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{cc}
3 & -2 \\
0 & 1
\end{array}\right] \\
&\Rightarrow A=\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]\left[\begin{array}{cc}
1 & -\frac{2}{3} \\
0 & \frac{1}{3}
\end{array}\right]
\end{aligned}
$$
$\Rightarrow A=\left[\begin{array}{cc}k & -\frac{2}{3} k \\ 0 & \frac{k}{3}\end{array}\right]$
$\because|3 \mathrm{~A}|=108$
$\Rightarrow 108=\left|\begin{array}{cc}3 k & -2 k \\ 0 & k\end{array}\right|$
$\Rightarrow 3 k^2=108 \Rightarrow k^2=36 \Rightarrow k=\pm 6$
For $k=6$
$A=\left[\begin{array}{cc}6 & -4 \\ 0 & 2\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$
For $k=-6$
$\Rightarrow A=\left[\begin{array}{cc}-6 & 4 \\ 0 & -2\end{array}\right]$
$\Rightarrow A^2=\left[\begin{array}{cc}36 & -32 \\ 0 & 4\end{array}\right]$