Question

If $1,\omega ,{\omega }^2$ are two cube roots of unity, then
$\Delta =\left|\begin{array}{c}1\\ {\omega }^{2n}\\ {\omega }^n\end{array}\begin{array}{c}{\omega }^n\\ 1\\ {\omega }^{2n}\end{array}\begin{array}{c}{\omega }^{2n}\\ {\omega }^n\\ 1\end{array}\right|$ has the value

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

Answer 1

The correct option is A $0$

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

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

${\omega }^4=\omega$

$1-{\omega }^{3n}-{\omega }^n({\omega }^{2n}-{\omega }^{2n})+{\omega }^{2n}({\omega }^{4n}-{\omega }^n)$

$1-{\left(1\right)}^n-{\omega }^n\left(0\right)+{\omega }^{2n}({\left(\omega \right)}^n-{\omega }^n)$

$=0$

Answer option A