$\omega$ is an imaginary cube root of unity and $\left| \begin{matrix} x + \omega^{2} & \omega & 1 \ \omega & \omega^{2} & 1 + x \ 1 & x + \omega & \omega^{2} \end{matrix} ight| = 0$, then one of the value of $x$ is

Question

$\omega$ is an imaginary cube root of unity and $\left| \begin{matrix} x + \omega^{2} & \omega & 1 \ \omega & \omega^{2} & 1 + x \ 1 & x + \omega & \omega^{2} \end{matrix} ight| = 0$, then one of the value of $x$ is

Dr. Abubacker · Accepted Answer

(b)
$\left| \begin{matrix}
x + \omega^{2} & \omega & 1 \
\omega & \omega^{2} & 1 + x \
1 & x + \omega & \omega^{2}
\end{matrix} ight| = 0$
Applying $C_{1} ightarrow C_{1} + C_{2} + C_{3}$
$\left| \begin{matrix}
x & \omega & 1 \
x & \omega^{2} & 1 + x \
x & x + \omega & \omega^{2}
\end{matrix} ight| = 0\ \ \ \ \ \ \left( 	herefore 1 + \omega + \omega^{2} = 0 ight)$
$\Rightarrow \ \ \ x = 0\$is one of the values of $x$ which satisfy the above determinant equation.

Q. $\omega$ is an imaginary cube root of unity and $\left| \begin{matrix} x + \omega^{2} & \omega & 1 \\ \omega & \omega^{2} & 1 + x \\ 1 & x + \omega & \omega^{2} \end{matrix} \right| = 0$, then one of the value of $x$ is

Explanation

Reviewed by Dr. Abubacker

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