If n ne 3k and 1, omega, omega^2 are the cube | vedclass

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(A) Applying the column operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \begin{vmatrix} 1 + \omega^n + \omega^{2n} & \omega^n & \omega^{2n} \ 1

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(A) Applying the column operation $C_1 	o C_1 + C_2 + C_3$,we get:$\Delta = \begin{vmatrix} 1 + \omega^n + \omega^{2n} & \omega^n & \omega^{2n} \ 1 + \omega^n + \omega^{2n} & 1 & \omega^n \ 1 + \omega^n + \omega^{2n} & \omega^{2n} & 1 \end{vmatrix}$Since $n 
e 3k$,$n$ is not a multiple of $3$. Therefore,$1 + \omega^n + \omega^{2n} = 0$.Substituting this into the determinant:$\Delta = \begin{vmatrix} 0 & \omega^n & \omega^{2n} \ 0 & 1 & \omega^n \ 0 & \omega^{2n} & 1 \end{vmatrix}$Since all elements in the first column are $0$,the value of the determinant is $0$.

If $n \ne 3k$ and $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^n \\ \omega^n & \omega^{2n} & 1 \end{vmatrix}$ has the value

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