Matrices and Determinants 2 Question 10
12. If
$ \begin{aligned} \begin{vmatrix} & x-4 & 2 x & 2 x \\ & 2 x & x-4 & 2 x \\ & 2 x & 2 x & x-4 \\ \end{vmatrix} =(A+B x)(x-A)^{2} \end{aligned} $ , then the ordered pair $(A, B)$ is equal to
(2018 Main)
(a) $(-4,-5)$
(b) $(-4,3)$
(c) $(-4,5)$
(d) $(4,5)$
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Answer:
Correct Answer: 12. (c)
Solution:
- Given,
$ \begin{aligned} \begin{vmatrix} & x-4 & 2 x & 2 x \\ & 2 x & x-4 & 2 x \\ & 2 x & 2 x & x-4 \\ \end{vmatrix} =(A+B x)(x-A)^{2} \end{aligned} $
$\Rightarrow \text { Apply } C_{1} \rightarrow C_{1}+C_{2}+C_{3} $
$ \begin{aligned} \begin{vmatrix} & 5 x-4 & 2 x & 2 x \\ & 5 x-4 & x-4 & 2 x\\ & 5 x-4 & 2 x & x-4 \end{vmatrix} \end{aligned} $
$=(A+B x)(x-A)^{2} $
Taking common $(5 x-4)$ from $C_{1}$, we get
(5x-4) $ \begin{vmatrix} 1 & 2 x & 2 x \\ & x-4 & 2 x \\ 1 & 2 x & x-4 \end{vmatrix}=(A+B x)(x-A)^{2} $
Apply $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$
$ \therefore(5 x-4) \begin{vmatrix} 1 & 2 x & 0 \\ 0 & -x-4 & 0 \\ 0 & 0 & -x-4 \end{vmatrix}=(A+B x)(x-A)^{2} $
Expanding along $C_{1}$, we get
$ (5 x-4)(x+4)^{2}=(A+B x)(x-A)^{2} $
Equating, we get, $A=-4$ and $B=5$
