JEE Main Solved Paper 2017 Question 18

Question: Let $ \omega $ be a complex number such that $ 2\omega +1=z $ where $ z=\sqrt{-3}. $ If $ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -{{\omega }^{2}}-1 & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & {{\omega }^{7}} \\ \end{vmatrix} =3k, $ then k is equal to:- JEE Main Solved Paper-2017

Options:

A) 1

B) $ -z $

C) z

D) -1

Show Answer

Answer:

Correct Answer: B

Solution:

  • Here $ \omega $ is complex cube root of unity $ R _1\to R _1+R _2+R _3 $ $ = \begin{vmatrix} 3 & 0 & 0 \\ 1 & -{{\omega }^{2}}-1 & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & \omega \\ \end{vmatrix} $ $ =3(-1-\omega -\omega )=-3z\Rightarrow k=-z $