Question

If $\omega$ is an imaginary cube root of unity, then the value of the determinant

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

• A. $-2\omega$
• B. $-3{\omega }^2$
• C. $-1$
• D. $0$ (Zero)

Answer 1

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

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

Apply $C_1\to C_1+C_2$

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

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

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

Using determinant expansion, we get
$\Delta =(\omega -1)(-{\omega }^4+{\omega }^2)\Rightarrow \Delta =(\omega -1)(-\omega +{\omega }^2)\Rightarrow \Delta =-{\omega }^2+{\omega }^3+\omega -{\omega }^2\Rightarrow \Delta =1-2{\omega }^2+\omega \Rightarrow \Delta =-3{\omega }^2$