If omega is a cube root of unity,then a root | vedclass

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Question

(D) Given the determinant equation: $\left| \begin{array}{ccc} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{arra

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$x = 0$

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(D) Given the determinant equation: $\left| \begin{array}{ccc} x+1 & \omega & \omega^2 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{array} ight| = 0$ Applying the transformation $R_1 	o R_1 + R_2 + R_3$,the sum of elements in the first row becomes: $(x+1+\omega+\omega^2) = (x+0) = x$. Thus,we get: $x \left| \begin{array}{ccc} 1 & 1 & 1 \ \omega & x+\omega^2 & 1 \ \omega^2 & 1 & x+\omega \end{array} ight| = 0$ This implies $x = 0$ is a root. Alternatively,by substituting $x = 0$,the determinant becomes $\left| \begin{array}{ccc} 1 & \omega & \omega^2 \ \omega & \omega^2 & 1 \ \omega^2 & 1 & \omega \end{array} ight|$. Since $1+\omega+\omega^2 = 0$,the rows are linearly dependent,making the determinant $0$.

If $\omega$ is a cube root of unity,then a root of the equation $\left| \begin{array}{ccc} x+1 & \omega & \omega^2 \\ \omega & x+\omega^2 & 1 \\ \omega^2 & 1 & x+\omega \end{array} \right| = 0$ is

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