Question

If $\omega$ is one of the imaginary cube roots of unity, find the value of
$\left|\begin{array}{ccc}1& {\omega }^3& {\omega }^2\\ {\omega }^3& 1& \omega \\ {\omega }^2& \omega & 1\end{array}\right|$.

Answer 1

$\Delta =\left|\begin{array}{ccc}1& {\omega }^3& {\omega }^2\\ {\omega }^3& 1& \omega \\ {\omega }^2& \omega & 1\end{array}\right|$

we know ${\omega }^3=1$

$\Rightarrow \Delta =\left|\begin{array}{ccc}1& 1& {\omega }^2\\ 1& 1& \omega \\ {\omega }^2& \omega & 1\end{array}\right|R_1=R_1-R_2\Rightarrow \Delta =\left|\begin{array}{ccc}0& 0& {\omega }^2-\omega \\ 1& 1& \omega \\ {\omega }^2& \omega & 1\end{array}\right|$

Expanding along $R_1$

$\Rightarrow \Delta =0+0+({\omega }^2-\omega )\{\omega -{\omega }^2\}\Rightarrow \Delta =-{({\omega }^2-\omega )}^2\Rightarrow \Delta =-({\omega }^4+{\omega }^2-2{\omega }^3)\Rightarrow \Delta =-(\omega +{\omega }^2-2)$

As $1+\omega +{\omega }^2=0$

$\Rightarrow \Delta =-(-1-2)\Rightarrow \Delta =3$