Question

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

• A. 0
• B. $2\omega$
• C. $2{\omega }^2$
• D. $-3{\omega }^2$

Answer 1

The correct option is D $-3{\omega }^2$

$\left|\begin{array}{ccc}1+\omega & {\omega }^2& -\omega \\ 1+{\omega }^2& \omega & -{\omega }^2\\ {\omega }^2+\omega & \omega & -{\omega }^2\end{array}\right|$
$={\omega }^2\left|\begin{array}{ccc}1+\omega & \omega & -1\\ 1+{\omega }^2& 1& -\omega \\ {\omega }^2+\omega & 1& -\omega \end{array}\right|$
$R_2\to R_2-R_3$
$={\omega }^2\left|\begin{array}{ccc}1+\omega +& \omega & -1\\ 1-\omega & 0& 0\\ {\omega }^2+\omega & 1& -\omega \end{array}\right|$
$={\omega }^2\left[\right(\omega -1\left)\right(-{\omega }^2+1\left)\right]$
$=-3{\omega }^2$