Question

If $\omega$ is a complex 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)

Let us denote the given determinant by $\Delta$.
Applying $C_1 \rightarrow C_1+C_2+C_3$, we get
$\Delta=\begin{vmatrix}x+1+\omega+\omega^2 & \omega & \omega^2 \\x+1+\omega+\omega^2 & x+\omega^2 & 1 \\x+1+\omega+\omega^2 & 1 & x+\omega \end{vmatrix}=\begin{vmatrix}x & \omega & \omega^2 \\x & x+\omega^2 & 1 \\x & 1 & x+\omega\end{vmatrix}$
Clearly $\Delta=0$ for $x=0$.