Question

If $1,\omega ,{\omega }^{2n}$ are the cube roots of unity, then $△=\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$ has the value.

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

Answer 1

The correct option is A $0$

$\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$
$=1(1-{\omega }^{3n})-{\omega }^n({\omega }^{2n}-{\omega }^{2n})+{\omega }^{2n}({\omega }^{4n}-{\omega }^n)$
$=(1-1)-{\omega }^n\left(0\right)+{\omega }^{2n}({\omega }^n-{\omega }^n)=0+0=0$.