Question

If $\omega$ is a complex cube root of unity, then a root of the equation $\left|\begin{array}{ccc}x+1& \omega & {\omega }^2\\ \omega & x+{\omega }^2& 1\\ {\omega }^2& 1& x+\omega \end{array}\right|=0$ 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 $C_1\to C_1+C_2+C_3$, we get
$\Delta =\left|\begin{array}{ccc}x+1+\omega +{\omega }^2& \omega & {\omega }^2\\ x+1+\omega +{\omega }^2& x+{\omega }^2& 1\\ x+1+\omega +{\omega }^2& 1& x+\omega \end{array}\right|=\left|\begin{array}{ccc}x& \omega & {\omega }^2\\ x& x+{\omega }^2& 1\\ x& 1& x+\omega \end{array}\right|$
Clearly $\Delta =0$ for $x=0$.