Let $A$ be a matrix such that $A\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right]$ is a scalar matrix and $\left| {3A} \right| = 108$ . Then ${A^2}$ equals:
A) $\left[ {\begin{array}{*{20}{c}}
  4&{ - 32} \\
  0&{36}
\end{array}} \right]$
B) $\left[ {\begin{array}{*{20}{c}}
  4&0 \\
  { - 32}&{36}
\end{array}} \right]$
C) $\left[ {\begin{array}{*{20}{c}}
  {36}&0 \\
  { - 32}&4
\end{array}} \right]$
D) $\left[ {\begin{array}{*{20}{c}}
  {36}&{ - 32} \\
  0&4
\end{array}} \right]$

Hint: We have given that the product of the matrices is a scalar matrix, we will use this fact and multiply both sides by the inverse of the given matrix. Then we will find the required matrix in general terms. Later we will use the other given condition to find the unknown.

Complete step-by-step answer:
It is given that the product $A\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right]$ is a scalar matrix.
Let us assume that the product is $A\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  x&0 \\
  0&x
\end{array}} \right]$ where $x$ is a real number.
If $\left[ {\begin{array}{*{20}{c}}
  a&d \\
  b&c
\end{array}} \right]$ is a $2 \times 2$ matrix then the inverse of the matrix is given by $\dfrac{1}{{ac - bd}}\left[ {\begin{array}{*{20}{c}}
  c&{ - d} \\
  { - b}&a
\end{array}} \right]$ .
Let us assume that the matrix $B = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right]$ then ${B^{ - 1}} = \dfrac{1}{3}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  0&1
\end{array}} \right]$.
Now we already have assumed the following:
$\Rightarrow$$A\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  x&0 \\
  0&x
\end{array}} \right]$
Post multiply by ${B^{ - 1}}$ on both sides:
$\Rightarrow$$A\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  0&3
\end{array}} \right]\left( {\dfrac{1}{3}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  0&1
\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}
  x&0 \\
  0&x
\end{array}} \right]\left( {\dfrac{1}{3}\left[ {\begin{array}{*{20}{c}}
  3&{ - 2} \\
  0&1
\end{array}} \right]} \right)$
Therefore, the matrix $A$ is given as follows:
$\Rightarrow$$A = \left[ {\begin{array}{*{20}{c}}
  x&0 \\
  0&x
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&{\dfrac{{ - 2}}{3}} \\
  0&{\dfrac{1}{3}}
\end{array}} \right]$
On completing the multiplication, we get,
$\Rightarrow$\[A = \left[ {\begin{array}{*{20}{c}}
  x&{\dfrac{{ - 2x}}{3}} \\
  0&{\dfrac{x}{3}}
\end{array}} \right]\]
If we take determinant of the above matrix then we get,
$\left| A \right| = \dfrac{{{x^2}}}{3}$ ………….… (1)
Another condition given is that $\left| {3A} \right| = 108$.
Note that for any matrix $A$, $\left| {kA} \right| = {k^n}\left| A \right|$.
We will put $k = 3$ and $n = 2$ then we get,
$\Rightarrow$$\left| {3A} \right| = 9\left| A \right|$
Substituting equation (1) we get,
$\Rightarrow$$\left| {3A} \right| = 9\left( {\dfrac{{{x^2}}}{3}} \right)$
From the given condition we can write:
$\Rightarrow$$3{x^2} = 108$
Therefore, the value of $x$ is given by $x = \pm 6$ .
Hence, the matrix $A$ is given by:
$\Rightarrow$$A = \left[ {\begin{array}{*{20}{c}}
  6&{ - 4} \\
  0&2
\end{array}} \right]$
Now the product ${A^2}$ is given by:
$\Rightarrow$${A^2} = \left[ {\begin{array}{*{20}{c}}
  6&{ - 4} \\
  0&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  6&{ - 4} \\
  0&2
\end{array}} \right]$
Multiplying we get,
$\Rightarrow$${A^2} = \left[ {\begin{array}{*{20}{c}}
  {36}&{ - 32} \\
  0&4
\end{array}} \right]$

Hence, the correct option is D.

Note: Identity matrix of a particular size is the matrix with 1 on the main diagonal and 0 in the remaining places. Determinant of an Identity matrix is always equal to 1. If any two rows or columns of a determinant are equal then the determinant value will be equal to 0.