Question

If $1,\omega ,{\omega }^2$ are cube roots of unity and $n\ne 3p,p\in Z$, then $\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$ is equal to

• A. $0$
• B. $\omega$
• C. ${\omega }^2$
• D. $1$

Answer 1

The correct option is A $0$

$\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$

$\because n\ne 3p$, i.e., not a multiple of $3\left[\text{Given}\right]$

$\therefore 1+{\omega }^n+{\omega }^{2n}=0$

Now solving the given determent, we have

$\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$

Applying $C_1\to C_1+C_2+C_3$

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

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

$=0$

Hence $\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|=0$.