Complex Numbers And Quadratic Equations question 292
Question: If $ 1,\omega ,{{\omega }^{2}} $ are the cube roots of unity, then $ \Delta = \begin{vmatrix} 1 & {{\omega }^{n}} & {{\omega }^{2n}} \\ {{\omega }^{n}} & {{\omega }^{2n}} & 1 \\ {{\omega }^{2n}} & 1 & {{\omega }^{n}} \\ \end{vmatrix} $ = [AIEEE 2003]
Options:
A) 0
B) 1
C) $ \omega $
D) $ {{\omega }^{2}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \Delta =({{\omega }^{3n}}-1)+{{\omega }^{n}}({{\omega }^{2n}}-{{\omega }^{2n}})+{{\omega }^{2n}}({{\omega }^{n}}-{{\omega }^{4n}}) $ $ \Delta =(1-1)+0+{{\omega }^{2n}}[{{\omega }^{n}}-{{({{\omega }^{3}})}^{n}}{{\omega }^{n}}] $ $ \Delta =0+0+0=0 $ .
BETA
