NDA 2 2024 Maths, question 12

Topic: Matrices and determinants

Q12

$\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$

then what is

$\left| \begin{matrix} 1+\omega & 1+\omega^{2} & \omega+\omega^{2} \\ 1 & \omega & \omega^{2} \\ \frac{1}{\omega} & \frac{1}{\omega^{2}} & 1 \end{matrix} \right|$

equal to?

(a) $0$ Correct answer
(b) $\omega$
(c) $\omega^{2}$
(d) $1-\omega^{2}$

Solution.

Answer: (a) $0$

Since ω³=1, 1/ω=ω² and 1/ω²=ω.
The first row is (−ω²,−ω,−1), the negative of the third row.
Two proportional rows make the determinant zero.