Question

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

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

Answer 1

Answer: (A)

$0$

$\left|\begin{array}{ccc}x+{\omega }^2& \omega & 1\\ \omega & {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0$

Applying $C1=C1+C2+C3$

$\left|\begin{array}{ccc}x+1+\omega +{\omega }^2& \omega & 1\\ x+1+\omega +{\omega }^2& {\omega }^2& 1+x\\ x+1+\omega +{\omega }^2& x+\omega & {\omega }^2\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x& \omega & 1\\ x& {\omega }^2& 1+x\\ x& x+\omega & {\omega }^2\end{array}\right|=0$ $\because 1+\omega +{\omega }^2=0$

$\Rightarrow x\left|\begin{array}{ccc}1& \omega & 1\\ 1& {\omega }^2& 1+x\\ 1& x+\omega & {\omega }^2\end{array}\right|=0$

Applying $R2=R2-R1,R3=R3-R2$

$\Rightarrow x\left|\begin{array}{ccc}1& \omega & 1\\ 0& {\omega }^2-\omega & x\\ 0& x+\omega -{\omega }^2& {\omega }^2-1-x\end{array}\right|=0$

$\Rightarrow x\left|\begin{array}{cc}{\omega }^2-\omega & x\\ x+\omega -{\omega }^2& {\omega }^2-1-x\end{array}\right|=0$

$\Rightarrow x({\omega }^4-{\omega }^2-x{\omega }^2-{\omega }^3+\omega +\omega x-x^2-\omega x+x{\omega }^2)=0$

$\Rightarrow x(\omega -{\omega }^2-{\omega }^3+\omega -x^2)=0$

$\Rightarrow x(-{\omega }^2+2\omega -x^2-1)=0$

$\therefore x=0$ or $-{\omega }^2+2\omega -x^2-1=0$

$\therefore x=0$ or $-{\omega }^2+2\omega -1=x^2$

$\therefore x=0$ or $x^2=-(\omega -1{)}^2$

Hence, $x=0$