Complex Numbers And Quadratic Equations question 247
Question: If $ \omega $ is a cube root of unity, then a root of the equation $ \begin{vmatrix} x+1 & \omega & {{\omega }^{2}} \\ \omega & x+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & x+\omega \\ \end{vmatrix} =0 $ is [MNR 1990; MP PET 1999]
Options:
A) $ x=1 $
B) $ x=\omega $
C) $ x={{\omega }^{2}} $
D) $ x=0 $
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Answer:
Correct Answer: D
Solution:
Given that $ \begin{vmatrix} x+1 & \omega & {{\omega }^{2}} \\ \omega & x+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & x+\omega \\ \end{vmatrix} =0 $ Applying transformation $ R_1\to R_1+R_2+R_3 $ , we get $ x \begin{vmatrix} 1 & 1 & 1 \\ \omega & x+{{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & x+\omega \\ \end{vmatrix} =0 $
Þ $ (x+{{\omega }^{2}})(x+\omega )-1+{{\omega }^{2}}-\omega (x+\omega )+\omega $ $ -{{\omega }^{2}}(x+{{\omega }^{2}})=0 $
Þ $ x^{2}=0 $
Þ $ x=0 $ Trick: Putting $ x=0, $ we get $ \begin{vmatrix} 1 & \omega & {{\omega }^{2}} \\ \omega & {{\omega }^{2}} & 1 \\ {{\omega }^{2}} & 1 & \omega \\ \end{vmatrix} =0 $
BETA
