Question

If $1,\omega,\omega^2$ are the cube roots of unity, then $\Delta=\begin{vmatrix} 1&\omega^n &\omega^{2n} \\[0.3em] \omega^n &\omega^{2n} & 1 \\[0.3em] \omega^{2n} &1& \omega^n \end{vmatrix}$ 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 $\Delta= \omega^n \begin{vmatrix} \omega^{2n} &1&\omega^n \\[0.3em] \omega^n &\omega^{2n} & 1 \\[0.3em] \omega^{2n} &1& \omega^n \end{vmatrix}$ = 0 [$\therefore \, R_1 , R_3 $ are identical]