If $\omega$ is a root of the equation $x+\frac{1}{x}+1=0$, then $\left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array} ight|$ is equal to

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If $\omega$ is a root of the equation $x+\frac{1}{x}+1=0$, then $\left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array}ight|$ is equal to

Aicharya · Accepted Answer

$\omega$ is root of $x+\frac{1}{x}+1=0$, i.e. $x^2+x+1=0$ This gives roots $\frac{-1 \pm \sqrt{1-4}}{2}=\frac{-1 \pm i \sqrt{3}}{2}$ $	herefore \omega=\frac{-1+\sqrt{-3}}{2}$ and $\omega^2=\frac{-1-\sqrt{-3}}{2}$ are roots of $x^2+x+1=0$ Also, $\omega^3=\omega \cdot \omega^2=1$ and $1+\omega+\omega^2=0$ Then, $A=\left|\begin{array}{ccc}1 & 1+\omega & 1+\omega+\omega^2 \ 3 & 4+3 \omega & 5+4 \omega+3 \omega^2 \ 6 & 9+6 \omega & 11+9 \omega+6 \omega^2\end{array}ight|$ is written as, $ \begin{aligned} & |A|=\left|\begin{array}{ccc} 1 & 1+\omega & 1+\omega+\omega^2 \ 3 & 4+3 \omega & 3\left(1+\omega+\omega^2ight)+2+\omega \ 6 & 9+6 \omega & 6\left(1+\omega+\omega^2ight)+5+3 \omega \end{array}ight| \ & |A|=\left|\begin{array}{ccc} 1 & 1+\omega & 0 \ 3 & 4+3 \omega & 2+\omega \ 6 & 9+6 \omega & 5+3 \omega \end{array}ight| \end{aligned} $ Expand along row 1, $ \begin{aligned} & |A|=(4+3 \omega)(5+3 \omega)-(2+\omega)(9+6 \omega) \ & -(1+\omega)\{15+9 \omega-12-6 \omega\} \ & =20+27 \omega+9 \omega^2-18-21 \omega-6 \omega^2 \ & -3-6 \omega-3 \omega^2=-1 \end{aligned} $

If $\omega$ is a root of the equation $x+\frac{1}{x}+1=0$, then $\left|\begin{array}{ccc}1 & 1+\omega &…

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