Matrices and Determinants 2 Question 10
«««< HEAD
12. If $2 x \quad x-4 \quad 2 x=(A+B x)(x-A)^{2}$, then the $2 x \quad 2 x \quad x-4$
======= ####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
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed ordered pair $(A, B)$ is equal to
(2018 Main)
(a) $(-4,-5)$
(b) $(-4,3)$
(c) $(-4,5)$
(d) $(4,5)$
Show Answer
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 $B=5$
