(2) Here, $\left| {\begin{array}{*{20}{c}}
{x – 4}&{2x}&{2x}\\
{2x}&{x – 4}&{2x}\\
{2x}&{2x}&{x – 4}
\end{array}} \right| = \left( {A + Bx} \right){\left( {x – A} \right)^2}$

Put $x = 0 \Rightarrow \left| {\begin{array}{*{20}{c}}
{ – 4}&0&0\\
0&{ – 4}&0\\
0&0&{ – 4}
\end{array}} \right| = {A^3} \Rightarrow {A^3} = {\left( { – 4} \right)^3}$

$ \Rightarrow A =  – 4$

$ \Rightarrow \left| {\begin{array}{*{20}{c}}
{x – 4}&{2x}&{2x}\\
{2x}&{x – 4}&{2x}\\
{2x}&{2x}&{x – 4}
\end{array}} \right| = \left( {Bx – 4} \right){\left( {x – 4} \right)^2}$

Now take $x$ common from both the sides

$\therefore \left| {\begin{array}{*{20}{c}}
{1 – \frac{4}{x}}&{2x}&{2x}\\
{2x}&{1 – \frac{4}{x}}&{1 – \frac{4}{x}}\\
{2x}&{2x}&{2x}
\end{array}} \right| = \left( {B – \frac{4}{x}} \right){\left( {1 + \frac{4}{x}} \right)^2}$