NDA 1 2025 Maths, question 12

Topic: Matrices and determinants

Q12

If $\omega$ is a non-real cube root of unity, then what is a root of the following equation? $\left| \begin{matrix} x + 1 & \omega & \omega^{2} \\ \omega & x + \omega^{2} & 1 \\ \omega^{2} & 1 & x + \omega \end{matrix} \right| = 0$

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

Solution.

Answer: (a) $x = 0$

Try $x = 0$ .
The first row becomes $\left( 1,\omega,\omega^{2} \right)$ .
Using $\omega^{3} = 1$ , the second row is $\omega$ times the first, and the third is $\omega^{2}$ times the first.
The rows are proportional, so the determinant is zero.
Therefore $x = 0$ is a root.