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If $\omega$ is a cube root of unity, then the value of $\operatorname{determinant}\left|\begin{array}{ccc}1+\omega & \omega^{2} & \omega \\ \omega^{2}+\omega & -\omega & \omega^{2} \\ 1+\omega^{2} & \omega & \omega^{2}\end{array}\right|$ is equal to
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Verified Answer
The correct answer is:
$1-\omega$
We have, $\left|\begin{array}{ccc}1+\omega & \omega^{2} & \omega \\ \omega^{2}+\omega & -\omega & \omega^{2} \\ 1+\omega^{2} & \omega & \omega^{2}\end{array}\right|$
$$
=\left|\begin{array}{ccc}
1+\omega+\omega^{2} & \omega^{2} & \omega \\
\omega^{2} & -\omega & \omega^{2} \\
1+\omega+\omega^{2} & \omega & \omega^{2}
\end{array}\right| \quad\left(C_{1} \rightarrow C_{1}+C_{2}\right)
$$
$$
\begin{aligned}
&=\left|\begin{array}{ccc}
0 & \omega^{2} & \omega \\
\omega^{2} & -\omega & \omega^{2} \\
0 & \omega & \omega^{2}
\end{array}\right| \\
&\left.=0-\omega^{2}\left(\omega^{4}-0\right)+\omega(\omega)^{3}-0\right) \\
&=-\omega^{6}+\omega^{4}
\end{aligned}
$$
$=-1+\omega$
$\left[\because \omega^{3}=1\right]$
$=\omega-1$
$$
=\left|\begin{array}{ccc}
1+\omega+\omega^{2} & \omega^{2} & \omega \\
\omega^{2} & -\omega & \omega^{2} \\
1+\omega+\omega^{2} & \omega & \omega^{2}
\end{array}\right| \quad\left(C_{1} \rightarrow C_{1}+C_{2}\right)
$$
$$
\begin{aligned}
&=\left|\begin{array}{ccc}
0 & \omega^{2} & \omega \\
\omega^{2} & -\omega & \omega^{2} \\
0 & \omega & \omega^{2}
\end{array}\right| \\
&\left.=0-\omega^{2}\left(\omega^{4}-0\right)+\omega(\omega)^{3}-0\right) \\
&=-\omega^{6}+\omega^{4}
\end{aligned}
$$
$=-1+\omega$
$\left[\because \omega^{3}=1\right]$
$=\omega-1$
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