Question

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

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

Answer 1

Answer: (B)

$3\omega(\omega - 1)$

Answer 2

$\Delta = \left| \begin{matrix} 3 & 1 & 1 \\ 0 & - 1 - \omega^{2} & \omega^{2} \\ 0 & \omega^{2} & \omega \end{matrix} \right|$ $(C_{1} \rightarrow C_{1} + C_{2} + C_{3})$

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

$= 3\lbrack\omega.\omega - \omega^{4}\rbrack = 3(\omega^{2} - \omega)$ $= 3\omega(\omega - 1)$.