Question

If $\omega$ is a complex cube-root of unity then, $\begin{vmatrix} {1}&{\omega} &{\omega^2}\\ {\omega}&{\omega^2}& {1} \\ {\omega^2}&{1}&{\omega}\\ \end{vmatrix} $ is equal to

• A. -1
• B. 1
• C. 0
• D. $\omega$

Answer 1

The Correct Option is C

Let $ A =\begin{vmatrix}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{vmatrix}$ $=\begin{vmatrix}1+\omega+\omega^{2} & \omega & \omega^{2} \\ 1+\omega+\omega^{2} & \omega^{2} & 1 \\ 1+\omega+\omega^{2} & 1 & \omega\end{vmatrix}$ $ C_{1} \rightarrow C_{1}+C_{2}+C_{3} $ $=\begin{vmatrix}0 & \omega & \omega^{2} \\ 0 & \omega^{2} & 1 \\ 0 & 1 & \omega\end{vmatrix}$ $\left[\because 1+\omega+\omega^{2}=0\right]$ $=0$