At what value of x will left| begin{array}{cc | vedclass

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(A) Given the determinant: $\Delta = \left| \begin{array}{ccc} x + \omega^2 & \omega & 1 \ \omega & \omega^2 & 1 + x \ 1 & x + \omega & \omega^2 \en

Vedclass · Accepted Answer

$x = 0$

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(A) Given the determinant: $\Delta = \left| \begin{array}{ccc} x + \omega^2 & \omega & 1 \ \omega & \omega^2 & 1 + x \ 1 & x + \omega & \omega^2 \end{array} ight| = 0$. Applying the column operation $C_1 	o C_1 + C_2 + C_3$: $\Delta = \left| \begin{array}{ccc} x + \omega^2 + \omega + 1 & \omega & 1 \ \omega + \omega^2 + 1 + x & \omega^2 & 1 + x \ 1 + x + \omega + \omega^2 & x + \omega & \omega^2 \end{array} ight|$. Since $1 + \omega + \omega^2 = 0$,the first column becomes: $\Delta = \left| \begin{array}{ccc} x & \omega & 1 \ x & \omega^2 & 1 + x \ x & x + \omega & \omega^2 \end{array} ight| = x \left| \begin{array}{ccc} 1 & \omega & 1 \ 1 & \omega^2 & 1 + x \ 1 & x + \omega & \omega^2 \end{array} ight|$. For the determinant to be $0$,either $x = 0$ or the remaining determinant must be $0$. Checking $x = 0$ in the original determinant: $\left| \begin{array}{ccc} \omega^2 & \omega & 1 \ \omega & \omega^2 & 1 \ 1 & \omega & \omega^2 \end{array} ight| = \omega^2(\omega^4 - \omega) - \omega(\omega^3 - 1) + 1(\omega^2 - \omega^2) = \omega^2(\omega - \omega) - \omega(1 - 1) + 0 = 0$. Thus,$x = 0$ is the correct value.

At what value of $x$ will $\left| \begin{array}{ccc} x + \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 + x \\ 1 & x + \omega & \omega^2 \end{array} \right| = 0$?

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