Question

$\omega$ is an imaginary 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 one of the value of x is

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is B $0$

Given, $\left|\begin{array}{ccc}x+{\omega }^2& \omega & 1\\ \omega & {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0$
Applying $C_1\to C_1+C_2+C_3$
$=\left|\begin{array}{ccc}x+{\omega }^2+\omega +1& \omega & 1\\ \omega +{\omega }^2+1+x& {\omega }^2& 1+x\\ 1+\omega +{\omega }^2+x& x+\omega & {\omega }^2\end{array}\right|=0$
$=\left|\begin{array}{ccc}x& \omega & 1\\ x& {\omega }^2& 1+x\\ x& x+\omega & {\omega }^2\end{array}\right|=0[\because 1+\omega +{\omega }^2=0]$
$\Rightarrow =\left|\begin{array}{ccc}1& \omega & 1\\ 1& {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Rightarrow x\left|\begin{array}{ccc}1& \omega & 1\\ 0& {\omega }^2-\omega & x\\ 0& x& {\omega }^2-1\end{array}\right|=0$
$\Rightarrow x\left\{\begin{array}{c}({\omega }^2-\omega )({\omega }^2-1)-x^2\end{array}\right\}=0$
$\Rightarrow x=0$ i.e., one value of $x=0$.