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 the value of $x$ is

• A. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is A $0$

$△=\left|\begin{array}{ccc}x+{\omega }^2& \omega & 1\\ \omega & {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0,$
$C_1\to C_1+C_2+C_3$
$△=\left|\begin{array}{ccc}x+{\omega }^2+\omega +1& \omega & 1\\ x+{\omega }^2+\omega +1& {\omega }^2& 1+x\\ x+{\omega }^2+\omega +1& x+\omega & {\omega }^2\end{array}\right|=0,$
$△=(x+{\omega }^2+\omega +1)\left|\begin{array}{ccc}1& \omega & 1\\ 1& {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0,$
$\because {\omega }^2+\omega +1=0$
$△=x\left[1\right({\omega }^4-(1+x)(x+\omega ))-\omega ({\omega }^2-1-x)+1(x+\omega -{\omega }^2\left)\right]=0$
$\Rightarrow x[x^2-2\omega +1+{\omega }^2]=0$
$\Rightarrow x=0,x^2=3\omega$