If 1, omega, omega^2 are the cube roots of un | vedclass

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(A) We are given the determinant $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \ \omega^n & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^n \end

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(A) We are given the determinant $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \ \omega^n & \omega^{2n} & 1 \ \omega^{2n} & 1 & \omega^n \end{vmatrix}$.Expanding the determinant along the first row:$\Delta = 1(\omega^n \cdot \omega^n - 1 \cdot 1) - \omega^n(\omega^n \cdot \omega^n - \omega^{2n} \cdot 1) + \omega^{2n}(\omega^n \cdot 1 - \omega^{2n} \cdot \omega^{2n})$$\Delta = 1(\omega^{2n} - 1) - \omega^n(\omega^{2n} - \omega^{2n}) + \omega^{2n}(\omega^n - \omega^{4n})$Since $\omega^3 = 1$,we have $\omega^{3n} = 1$ and $\omega^{4n} = \omega^n$.$\Delta = (\omega^{2n} - 1) - 0 + \omega^{2n}(\omega^n - \omega^n)$$\Delta = \omega^{2n} - 1 + 0 = \omega^{2n} - 1$.However,if we look at the properties of the cube roots of unity,the sum of the columns $C_1 + C_2 + C_3$ gives:$1 + \omega^n + \omega^{2n}$.If $n$ is not a multiple of $3$,$1 + \omega^n + \omega^{2n} = 0$,making the determinant $0$.If $n$ is a multiple of $3$,then $\omega^n = 1$,so the determinant becomes $\begin{vmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{vmatrix} = 0$.Thus,in all cases,$\Delta = 0$.

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix}$ is equal to

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