Question

If $\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix}=\left(A + B x\right) \, \left(x - A\right)^{2},$ then the ordered pair $\left(\right.A, \, B\left.\right)$ is equal to

• A. $\left(4, 5\right)$
• B. $\left(- 4 , - 5\right)$
• C. $\left(- 4 , \, 3\right)$
• D. $\left(- 4 , \, 5\right)$

Answer 1

Answer: (D)

$\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix}=\left(A + B x\right) \, \left(x - A\right)^{2}$
Put $x=0\Rightarrow \begin{vmatrix} -4 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & -4 \end{vmatrix}=A^{3}\Rightarrow A=-4$
$\begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \end{vmatrix}=\left(B x - 4\right)\left(x + 4\right)^{2}$
$\begin{vmatrix} 1-\frac{4}{x} & 2 & 2 \\ 2 & 1-\frac{4}{x} & 2 \\ 2 & 2 & 1-\frac{4}{x} \end{vmatrix}=\left(B - \frac{4}{x}\right)\left(1 + \frac{4}{x}\right)^{2}$
Put $x \rightarrow \in fty\Rightarrow \begin{vmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{vmatrix}=B$
On expanding the determinant along the first row, we get $B=5$ .