Complex Numbers And Quadratic Equations question 263
Question: Let $ \Delta =| \begin{matrix} 1 & \omega & 2{{\omega }^{2}} \\ 2 & 2{{\omega }^{2}} & 4{{\omega }^{3}} \\ 3 & 3{{\omega }^{3}} & 6{{\omega }^{4}} \\ \end{matrix} | $ where $ \omega $ is the cube root of unity, then
Options:
A) $ \Delta =0 $
B) $ \Delta =1 $
C) $ \Delta =2 $
D) $ \Delta =3 $
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Answer:
Correct Answer: A
Solution:
We have $ \Delta = \begin{vmatrix} 1 & \omega & 2{{\omega }^{2}} \\ 2 & 2{{\omega }^{2}} & 4{{\omega }^{3}} \\ 3 & 3{{\omega }^{3}} & 6{{\omega }^{4}} \\ \end{vmatrix} =2\omega \begin{vmatrix} 1 & \omega & \omega \\ 2 & 2{{\omega }^{2}} & 2{{\omega }^{2}} \\ 3 & 3{{\omega }^{3}} & 3{{\omega }^{3}} \\ \end{vmatrix} =0 $
BETA
