AIEEE 2003 (Offline)Question 1
If $1,\omega,\omega^{2}$ are the cube roots of unity, then $\Delta=\begin{vmatrix}1&\omega^{n}&\omega^{2n}\\\omega^{n}&\omega^{2n}&1\\\omega^{2n}&1&\omega^{n}\end{vmatrix}$ is equal to
Determinant
Solution
If $3\mid n$ all entries are $1$ and $\Delta=0$; otherwise $\Delta=3\omega^{3n}-\ldots=a^3+b^3+c^3-3abc$ with $a=1,b=\omega^n,c=\omega^{2n}$, $=1+1+1-3=0$. So $\Delta=0$.