JEE Main On 08 April 2018 Question 14
Question: If $ \begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{vmatrix} =(A+Bx){{(x-A)}^{2}} $ , then the ordered pair (A, B) is equal to: [JEE Main Online 08-04-2018]
Options:
A) $ (-4,5) $
B) $ (4,5) $
C) $ (-4,-5) $
D) $ (-4,3) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \begin{vmatrix} x-4 & 2x & 2x \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{vmatrix} $ $ {R_1}\to {R_1}\text{+}{R_2}\text{+}{R_3} $
$ \begin{vmatrix} 5x-4 & 5x-4 & 5x-4 \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{vmatrix} $ $ (5x-4) \begin{vmatrix} 1 & 1 & 1 \\ 2x & x-4 & 2x \\ 2x & 2x & x-4 \\ \end{vmatrix} $
$ C _1\to C _1-C _2,C _2\to C _2-C _3 $
$ (5x-4) \begin{vmatrix} 0 & 0 & 0 \\ x+4 & -x-4 & 2x \\ 0 & x+4 & x-4 \\ \end{vmatrix} $
$ ={{(x+4)}^{2}}(5x-4)=(A+Bx){{(x-A)}^{2}} $ $ A=-4,B=5\Rightarrow (-4,5) $
