Question

If $\omega$ is imaginary cube root of unity, then a root of the equation $\begin{vmatrix}x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega\end{vmatrix}=0$ is

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

Answer 1

Answer: (D)

Put $x=0(x=0) \Rightarrow \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega\end{vmatrix}=0$