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 $
