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If $\omega$ is the cube root of unity, then what is one root of the equation
$\left|\begin{array}{ccc}x^{2} & -2 x & -2 \omega^{2} \\ 2 & \omega & -\omega \\ 0 & \omega & 1\end{array}\right|=0 ?$
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$\left|\begin{array}{ccc}x^{2} & -2 x & -2 \omega^{2} \\ 2 & \omega & -\omega \\ 0 & \omega & 1\end{array}\right|=0 ?$
Solution:
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Verified Answer
The correct answer is:
$-2$
Given matrix is :
$\left|\begin{array}{ccc}x^{2} & -2 x & -2 \omega^{2} \\ 2 & \omega & -\omega \\ 0 & \omega & 1\end{array}\right|=0$
$\mathrm{By} \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}$, we get
$\Rightarrow\left|\begin{array}{ccc}\mathrm{x}^{2} & -2 \mathrm{x}-2 \omega^{2} & -2 \omega^{2} \\ 2 & 0 & -\omega \\ 0 & 1+\omega & 1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}\mathrm{x}^{2} & -2 \mathrm{x}-2 \omega^{2} & -2 \omega^{2} \\ 2 & 0 & -\omega \\ 0 & -\omega^{2} & 1\end{array}\right|=0$
$\quad\left[\because \quad 1+\omega=-\omega^{2}\right]$
$\Rightarrow \omega^{2}\left|\begin{array}{cc}x^{2} & -2 \omega^{2} \\ 2 & -\omega\end{array}\right|+1\left|\begin{array}{cc}x^{2} & -2 x-2 \omega^{2} \\ 2 & -0\end{array}\right|=0$
$\Rightarrow \omega^{2}\left(-\omega x^{2}+4 \omega^{2}\right)-\left(-4 x-4 \omega^{2}\right)=0$
$\Rightarrow-x^{2}+4 \omega+4 x+4 \omega^{2}=0$
$\Rightarrow-x^{2}+4 \omega-4 x-4-4 \omega=0 \Rightarrow-x^{2}-4 x-4=0$
$\Rightarrow(x+2)^{2}=0 \Rightarrow x=-2$
$\left|\begin{array}{ccc}x^{2} & -2 x & -2 \omega^{2} \\ 2 & \omega & -\omega \\ 0 & \omega & 1\end{array}\right|=0$
$\mathrm{By} \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3}$, we get
$\Rightarrow\left|\begin{array}{ccc}\mathrm{x}^{2} & -2 \mathrm{x}-2 \omega^{2} & -2 \omega^{2} \\ 2 & 0 & -\omega \\ 0 & 1+\omega & 1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}\mathrm{x}^{2} & -2 \mathrm{x}-2 \omega^{2} & -2 \omega^{2} \\ 2 & 0 & -\omega \\ 0 & -\omega^{2} & 1\end{array}\right|=0$
$\quad\left[\because \quad 1+\omega=-\omega^{2}\right]$
$\Rightarrow \omega^{2}\left|\begin{array}{cc}x^{2} & -2 \omega^{2} \\ 2 & -\omega\end{array}\right|+1\left|\begin{array}{cc}x^{2} & -2 x-2 \omega^{2} \\ 2 & -0\end{array}\right|=0$
$\Rightarrow \omega^{2}\left(-\omega x^{2}+4 \omega^{2}\right)-\left(-4 x-4 \omega^{2}\right)=0$
$\Rightarrow-x^{2}+4 \omega+4 x+4 \omega^{2}=0$
$\Rightarrow-x^{2}+4 \omega-4 x-4-4 \omega=0 \Rightarrow-x^{2}-4 x-4=0$
$\Rightarrow(x+2)^{2}=0 \Rightarrow x=-2$
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