Question

If $1,\omega$ and ${\omega }^2$ are the cube roots of unity, then $\Delta =\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^n& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right|$ is equal to

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

Answer 1

The correct option is A $0$

Expanding given determinant
$\Delta =1\cdot \left|\begin{array}{cc}{\omega }^{2n}& 1\\ 1& {\omega }^n\end{array}\right|-{\omega }^n\left|\begin{array}{cc}{\omega }^n& 1\\ {\omega }^{2n}& {\omega }^n\end{array}\right|+{\omega }^{2n}\left|\begin{array}{cc}{\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1\end{array}\right|\Rightarrow {\omega }^{3n}-1-{\omega }^{3n}+{\omega }^{3n}+{\omega }^{3n}-{\omega }^{6n}=0(\because {\omega }^{3k}=1,k\in I)$