Complex Numbers 5 Question 7
7. Let $\omega=-\frac{1}{2}+i \frac{\sqrt{3}}{2}$, then value of the determinant $\left|\begin{array}{ccc}1 & 1 & 1 \ 1 & -1-\omega^{2} & \omega^{2} \ 1 & \omega^{2} & \omega\end{array}\right|$ is
(2002, 1M)
(a) $3 \omega$
(b) $3 \omega(\omega-1)$
(c) $3 \omega^{2}$
(d) $3 \omega(1-\omega)$
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Answer:
Correct Answer: 7. (b)
Solution:
- $L e t \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \ 1 & -1-\omega^{2} & \omega^{2} \ 1 & \omega^{2} & \omega\end{array}\right|$
Applying $R _2 \rightarrow R _2-R _1 ; R _3 \rightarrow R _3-R _1$
$$ \begin{aligned} & \begin{array}{llll} 1 & 1 & 1 \end{array} \\ & =0 \quad-2-\omega^{2} \quad \omega^{2}-1 \\ & 0 \quad \omega^{2}-1 \quad \omega-1 \\ & =\left(-2-\omega^{2}\right)(\omega-1)-\left(\omega^{2}-1\right)^{2} \\ & =-2 \omega+2-\omega^{3}+\omega^{2}-\left(\omega^{4}-2 \omega^{2}+1\right) \\ & =3 \omega^{2}-3 \omega=3 \omega(\omega-1) \quad\left[\because \omega^{4}=\omega\right] \end{aligned} $$
