Question

Let $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} i$. Then the value of the determinant $\Delta=\begin{vmatrix}1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4\end{vmatrix}$ is

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

Answer 1

Answer: (B)

Using $1+\omega^2=-\omega, \omega^4=\omega$ and applying $C_2 \rightarrow C_2-C_1, C_3 \rightarrow C_3-C_1$ we get
$\Delta =\begin{vmatrix}1 & 0 & 0 \\1 & \omega-1 & \omega^2-1 \\1 & \omega^2-1 & \omega-1\end{vmatrix}$
$ =(\omega-1)^2-\left(\omega^2-1\right)^2$
$ =\left(\omega+\omega^2-2\right)\left(\omega-1-\omega^2+1\right) $
$ =(-3)\left(\omega-\omega^2\right)=3 \omega(\omega-1)$