Correct option: D.$\left[ {\begin{array}{*{20}{c}}
{36}&{ - 32}\\
0&4
\end{array}} \right]$
d
$(d)$ Since
$A.\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]$ is a scalar matrix and $\left| {3A} \right| = 108$
Suppose the scalar matrix is $\left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]$
$\therefore A.\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]$
$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]^{ - 1}}$
$\left[ {\therefore AB = C \Rightarrow AB{B^{ - 1}} = C{B^{ - 1}} \Rightarrow A = C{B^{ - 1}}} \right]$
$ \Rightarrow A = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
3&{ - 2}\\
0&1
\end{array}} \right]$
$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ - \frac{2}{3}}\\
0&{\frac{1}{3}}
\end{array}} \right]$
$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
k&{ - \frac{2}{3}k}\\
0&{\frac{k}{3}}
\end{array}} \right]\,\,\,\,\,\,\,\,\,.......\left( 1 \right)$
$\because$ $\left| {3A} \right| = 108$
$ \Rightarrow 108 = \left| {\begin{array}{*{20}{c}}
{3k}&{ - 2k}\\
0&k
\end{array}} \right|$
$ \Rightarrow 3{k^2} = 108 \Rightarrow {k^2} = 36 \Rightarrow k = \pm 6$
For $k=6$
$A = \left[ {\begin{array}{*{20}{c}}
6&{ - 4}\\
0&2
\end{array}} \right]$ ....From $(1)$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{36}&{ - 32}\\
0&4
\end{array}} \right]$
For $k=-6$
$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
{ - 6}&4\\
0&{ - 2}
\end{array}} \right]$ ....From$(1)$
$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{36}&{ - 32}\\
0&4
\end{array}} \right]$