If \omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2} then what is \begin{vmatrix} 1+\omega & 1+\omega^{2} & \omega+\omega^{2} \\ 1 & \omega & \omega^{2} \\ \frac{1}{\omega} & \frac{1}{\omega^{2}} & 1 \end{vmatrix} equal to?

Explanation

Using the properties of cube roots of unity, 1+\omega+\omega^2 = 0. The first row of the determinant becomes -\omega^2, -\omega, -1. The third row simplifies to \omega^2, \omega, 1 since 1/\omega = \omega^2 and 1/\omega^2 = \omega. The first row is the exact negative of the third row (R_1 = -R_3), making the rows proportional. Thus, the determinant evaluates to 0.

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