Let omega be a complex number such that 2omeg | vedclass

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(B) Given $2\omega + 1 = z$ and $z = \sqrt{-3} = i\sqrt{3}$.Thus,$\omega = \frac{i\sqrt{3} - 1}{2}$,which is a complex cube root of unity.We know that

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$-z$

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(B) Given $2\omega + 1 = z$ and $z = \sqrt{-3} = i\sqrt{3}$.Thus,$\omega = \frac{i\sqrt{3} - 1}{2}$,which is a complex cube root of unity.We know that $1 + \omega + \omega^2 = 0$,so $-\omega^2 - 1 = \omega$.Also,$\omega^3 = 1$,so $\omega^7 = \omega^{3 	imes 2 + 1} = \omega$.The determinant is $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight|$.Applying $R_1 	o R_1 + R_2 + R_3$:$\Delta = \left| \begin{array}{ccc} 3 & 1+\omega+\omega^2 & 1+\omega^2+\omega \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight| = \left| \begin{array}{ccc} 3 & 0 & 0 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{array} ight|$.Expanding along $R_1$: $\Delta = 3(\omega^2 - \omega^4) = 3(\omega^2 - \omega)$.Since $\omega = \frac{-1 + i\sqrt{3}}{2}$,$\omega^2 = \frac{-1 - i\sqrt{3}}{2}$.$\Delta = 3\left( \frac{-1 - i\sqrt{3}}{2} - \frac{-1 + i\sqrt{3}}{2} ight) = 3\left( \frac{-2i\sqrt{3}}{2} ight) = -3i\sqrt{3} = -3z$.Given $\Delta = 3k$,we have $3k = -3z$,so $k = -z$.

Let $\omega$ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt{-3}$. If $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{array} \right| = 3k$,then $k$ is equal to:

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