Question

If $\omega \ne 1$ is a cube root of unity and $\left|\begin{array}{ccc}x+{\omega }^2& \omega & 1\\ \omega & {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0$, then value of x is

• A. 0
• B. 1
• C. -1
• D. none of these

Answer 1

The correct option is A

0

$Let\Delta =\left|\begin{array}{ccc}x+{\omega }^2& \omega & 1\\ \omega & {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|$
Applying $C_1\to C_1+C_2+C_3$ we get
$\Delta =\left|\begin{array}{ccc}x+{\omega }^2+\omega +1& \omega & 1\\ \omega +{\omega }^2+1+x& {\omega }^2& 1+x\\ 1+x+\omega +{\omega }^2& x+\omega & {\omega }^2\end{array}\right|=\left|\begin{array}{ccc}x& \omega & 1\\ x& {\omega }^2& 1+x\\ x& x+\omega & {\omega }^2\end{array}\right|(\because 1+\omega +{\omega }^2=0)$
$\Delta$ is clearly equal to 0 for x = 0.