Question

If ω is a non-real cube root of unity and n is not a multiple of 3, then

$∆=\left|\begin{array}{ccc}1& {\omega }^n& {\omega }^{2n}\\ {\omega }^{2n}& 1& {\omega }^n\\ {\omega }^n& {\omega }^{2n}& 1\end{array}\right|$ is equal to

(a) 0
(b) ω
(c) ω²
(d) 1

Answer 1

(a) 0
$$\begin{gathered} \mathrm{\Delta }=\left|1w^n{w}^{2n} \\ {w}^{2n}1w^n \\ {w}^n{w}^{2n}1\right| \\ =\left|1+w^n+w^{2n}{w}^n{w}^{2n} \\ {w}^{2n}+1+w^{n}1w^n \\ {w}^n+w^{2n}+1w^{2n}1\right|\left[\mathrm{Appplying}{C}_1\to C_1+C_2+C_3\right] \\ \mathrm{Now}, \\ 1+w+w^2=0\left[\because w\mathrm{is}\mathrm{a}\mathrm{complex}\mathrm{cube}\mathrm{root}\mathrm{of}\mathrm{unity}\right] \\ \Rightarrow 1+w^n+w^{2n}=0\left[\because n\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{multiple}\mathrm{of}3\right] \\ \Rightarrow \mathrm{\Delta }=\left|0w^n{w}^{2n} \\ 01w^n \\ 0w^{2n}1\right|=0 \end{gathered}$$