MCQ
If $\left| {\begin{array}{*{20}{c}}{x - 4}&{2x}&{2x}\\{2x}&{x - 4}&{2x}\\{2x}&{2x}&{x - 4}\end{array}} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2},$ then the ordered pair $\left( {A,B} \right) = $. . . . .
  • A
    $\left( { - 4,3} \right)$
  • $\left( { - 4,5} \right)$
  • C
    $\left( {4,5} \right)$
  • D
    $\left( { - 4, - 5} \right)$

Answer

Correct option: B.
$\left( { - 4,5} \right)$
b
(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

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

 

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