Let $\omega=-\frac{1}{2}+\mathrm{i} \frac{\sqrt{3}}{2}, \mathrm{i}=\sqrt{-1}$, then the value of…

Let $\omega=-\frac{1}{2}+\mathrm{i} \frac{\sqrt{3}}{2}, \mathrm{i}=\sqrt{-1}$, then the value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4\end{array}\right|$ is
  1. $3 \omega$
  2. $3 \omega^2$
  3. $3 \omega(\omega-1)$
  4. $3 \omega(1-\omega)$

Solution

$\begin{aligned} & \quad \omega^2=\left(-\frac{1}{2}\right)^2+\left(\mathrm{i} \frac{\sqrt{3}}{2}\right)^2-2 \mathrm{i}\left(\frac{\sqrt{3}}{4}\right)=-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i} \\ & \quad \omega^3=\frac{1}{4}+\frac{3}{4}=1 \\ & \therefore \quad 1+\omega+\omega^2=0\end{aligned}$ $\begin{aligned} & \therefore \quad\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega^4\end{array}\right|=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega\end{array}\right| \\ & \quad=1\left(\omega^2-\omega^4\right)-1\left(\omega-\omega^2\right)+1\left(\omega^2-\omega\right) \\ & \quad=\omega^2-\omega^4-\omega+\omega^2+\omega^2-\omega \\ & \quad=-\omega^4+3 \omega^2-2 \omega \\ & \quad=-\omega+3 \omega^2-2 \omega \\ & =3 \omega^2-3 \omega \\ & \quad=3 \omega(\omega-1)\end{aligned}$

Asked in: MHT CET 2024 (11 May Shift 2)

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