If alpha = cos frac{pi}{3} + i sin frac{pi}{3 | vedclass

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(D) Given $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \pi / 3}$.Note that $\alpha^3 = e^{i \pi} = -1$ and $\alpha^6 = 1$.The determinan

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$4$

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(D) Given $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = e^{i \pi / 3}$.Note that $\alpha^3 = e^{i \pi} = -1$ and $\alpha^6 = 1$.The determinant is $D = \left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \ \alpha^2 & 1 & \alpha \ \alpha & \alpha^2 & 1 \end{array} ight|$.Applying the operation $C_1 	o C_1 + C_2 + C_3$,we get:$D = \left| \begin{array}{ccc} 1 + \alpha + \alpha^2 & \alpha & \alpha^2 \ 1 + \alpha + \alpha^2 & 1 & \alpha \ 1 + \alpha + \alpha^2 & \alpha^2 & 1 \end{array} ight| = (1 + \alpha + \alpha^2) \left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \ 1 & 1 & \alpha \ 1 & \alpha^2 & 1 \end{array} ight|$.Expanding the determinant:$D = (1 + \alpha + \alpha^2) [1(1 - \alpha^3) - \alpha(1 - \alpha) + \alpha^2(\alpha^2 - 1)]$.Since $\alpha^3 = -1$,$1 - \alpha^3 = 2$.$D = (1 + \alpha + \alpha^2) [2 - \alpha + \alpha^2 + \alpha^4 - \alpha^2] = (1 + \alpha + \alpha^2) [2 - \alpha + \alpha^4]$.Since $\alpha = \frac{1}{2} + i \frac{\sqrt{3}}{2}$,$\alpha^2 = \alpha - 1$ is not true here,but $1 + \alpha + \alpha^2 = \frac{1 - \alpha^3}{1 - \alpha} = \frac{2}{1 - \alpha}$.Alternatively,calculating directly: $D = 1(1 - \alpha^3) - \alpha(\alpha^2 - \alpha^2) + \alpha^2(\alpha^4 - \alpha) = 1(1 - (-1)) - 0 + \alpha^2(\alpha^4 - \alpha) = 2 + \alpha^6 - \alpha^3 = 2 + 1 - (-1) = 4$.

If $\alpha = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}$,then the value of the determinant $\left| \begin{array}{ccc} 1 & \alpha & \alpha^2 \\ \alpha^2 & 1 & \alpha \\ \alpha & \alpha^2 & 1 \end{array} \right|$ is

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