Question

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

Answer 1

Answer: (B)

Write 1 as ${\omega }^{3n}$ in $R_1$ and take out ${\omega }^n$ from $R_1$,
we get $\Delta ={\omega }^n\left|\begin{array}{ccc}{\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\\ {\omega }^{2n}& 1& {\omega }^n\end{array}\right|$ = 0 [$\therefore R_1,R_3$ are identical]