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Question: If \(n \neq 3k\) and 1, \(\omega\), \(\omega^{2}\) are the cube roots of unity, then \(\Delta = \lef...

If n3kn \neq 3k and 1, ω\omega, ω2\omega^{2} are the cube roots of unity, then Δ=1ωnω2nω2n1ωnωnω2n1\Delta = \left| \begin{matrix} 1 & \omega^{n} & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^{n} & \omega^{2n} & 1 \end{matrix} \right| has the value

A

0

B

ω\omega

C

ω2\omega^{2}

D

1

Answer

0

Explanation

Solution

Applying C1C1+C2+C3C_{1} \rightarrow C_{1} + C_{2} + C_{3}, we get

Δ=1+ωn+ω2nωnω2n1+ωn+ω2n1ωn1+ωn+ω2nω2n1\Delta = \left| \begin{matrix} 1 + \omega^{n} + \omega^{2n} & \omega^{n} & \omega^{2n} \\ 1 + \omega^{n} + \omega^{2n} & 1 & \omega^{n} \\ 1 + \omega^{n} + \omega^{2n} & \omega^{2n} & 1 \end{matrix} \right| =0ωnω2n01ωn0ω2n1=0= \left| \begin{matrix} 0 & \omega^{n} & \omega^{2n} \\ 0 & 1 & \omega^{n} \\ 0 & \omega^{2n} & 1 \end{matrix} \right| = 0

(1+ωn+ω2n=0ifnisnotmultipleof3)(\because 1 + \omega^{n} + \omega^{2n} = 0\text{if}nis\text{notmultipleof}3)