Question

If $\omega$ is imaginary cube root of unity then prove that the value of
$\left|\begin{array}{ccc}1+\omega & {\omega }^2& -\omega \\ 1+{\omega }^2& \omega & -{\omega }^2\\ {\omega }^2+\omega & \omega & -{\omega }^2\end{array}\right|$ is equal to $-3{\omega }^2$.

Answer 1

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

We know that $1+\omega +{\omega }^2=0,{\omega }^3=1,1+\omega =-{\omega }^2$

$1+{\omega }^2=-\omega$

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

$\omega (1+\omega )=-1$

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

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

$=0-{\omega }^2(1-{\omega }^2)+{\omega }^3-{\omega }^2$

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

$=1+\omega -2{\omega }^2$

$=-3{\omega }^2$