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

Answer: (C)

0

Answer 2

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