Question

If $\omega$ is complex (non real) cube root of $1$ then show that $\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|=0$.

Answer 1

$\left|\begin{array}{ccc}1& \omega & {\omega }^2\\ \omega & {\omega }^2& 1\\ {\omega }^2& 1& \omega \end{array}\right|$

$=1({\omega }^3-1)-\omega ({\omega }^2-{\omega }^2)+{\omega }^2(\omega -{\omega }^4)$

$=1(1-1)-\omega \left(0\right)+{\omega }^2(\omega -{\omega }^4)$

$=0-0+0$

$=0$