Question

If $\omega$ is one of the imaginary cube roots of unity, then the value of the determinant $\begin{vmatrix}1 & \omega^3 & \omega^2 \\ \omega^3 & 1 & \omega \\ \omega^2 & \omega & 1\end{vmatrix}=$

• A. 1
• B. 2
• C. 3
• D. none

Answer 1

Answer: (C)

Put $\omega^3=1\begin{vmatrix}1 & 1 & \omega^2 \\ 1 & 1 & \omega \\ \omega^2 & \omega & 1\end{vmatrix}$ and open by $R_1$ to get $\left(1-\omega^2\right)+(1-\omega)=3$