Question

(c) Prove that \[ \begin{vmatrix} 1 \omega & \omega^2 \\ \omega & \omega^2 1 \\ \omega^2 1 \omega \end{vmatrix} = 0 \], where \( \omega \) is a cube root of unity.

Hint:

Use properties of roots of unity such as \( 1 + \omega + \omega^2 = 0 \) to simplify expressions.

Answer 1

The properties of cube roots of unity are:

\[ \omega^3 = 1, \quad 1 + \omega + \omega^2 = 0. \]

Expanding the determinant:

\[ \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} \]

Using cofactor expansion along the first row:

\[ 1 \cdot \begin{vmatrix} \omega^2 & 1 \\ 1 & \omega \end{vmatrix} - \omega \cdot \begin{vmatrix} \omega & 1 \\ \omega^2 & \omega \end{vmatrix} + \omega^2 \cdot \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & 1 \end{vmatrix} \]

Evaluating each minor and simplifying using properties of \( \omega \):

\[ = 0. \]