Question

If $1,\omega ,{\omega }^2$ are cube roots of unity then $\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^n& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right|$ equals

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

Answer 1

Answer: (D)

$0$

Since, $1,\omega ,{\omega }^2$ are the cube roots of unity.
${\omega }^3=1$
$\Rightarrow {\omega }^{3n}=1$
$\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^n& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right|$
$1({\omega }^{3n}-1)-{\omega }^n({\omega }^{2n}-{\omega }^{2n})+{\omega }^{2n}({\omega }^n-{\omega }^{4n})=0$