Question

If $1, \omega, \omega^2$ are the cube roots of unity, then $\Delta= \begin{vmatrix}1 & \omega^n & \omega^{2 n} \\ \omega^n & \omega^{2 n} & 1 \\ \omega^{2 n} & 1 & \omega^n\end{vmatrix}$ is equal to-

• A. 0
• B. 1
• C. $\omega$
• D. $\omega^2$

Answer 1

Answer: (A)

Given $\Delta=\begin{vmatrix}1 & \omega^n & \omega^{2 n} \\ \omega^n & \omega^{2 n} & 1 \\ \omega^{2 n} & 1 & \omega^n\end{vmatrix}=1\left(\omega^{3 n}-1\right)-\omega^n\left(\omega^{2 n}-\omega^{2 n}\right)+\omega^{2 n}\left(\omega^n-\omega^{4 n}\right)$
$=1(1-1)-0+\omega^{2 n}\left(\omega^n-\omega^n\right)=0$