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UPSC NDA II 2024 Mathematics Paper

Question : 12 of 120

Marks: +1, -0

\rm \omega=-\frac{1}{2}+i\frac{\sqrt3}{2}

\rm \begin{vmatrix}1+\omega &1+\omega^2&\omega+\omega^2\\\ 1&\omega\ & \omega^2\\\ \frac{1}{\omega}&\frac{1}{\omega^2}&1\end{vmatrix}

equal to?

Solution:

Concept:

The cube root of unity

\omega

satisfies

1+\omega+\omega^2 = 0

If any column of a determinant becomes all zeros after an elementary operation, the determinant is zero.

Explanation:

Let

\Delta = \begin{vmatrix} 1+\omega & 1+\omega^2 & \omega+\omega^2 \\ 1 & \omega & \omega^2 \\ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \end{vmatrix}

Apply the column operation

C_1 \rightarrow C_1 + C_2 + C_3

The new first column entries are:

Row 1:

(1+\omega) + (1+\omega^2) + (\omega+\omega^2) = 2 + \omega + \omega + \omega^2 + \omega^2 = 2 + 2\omega + 2\omega^2 = 2(1+\omega+\omega^2) = 0

Row 2:

1 + \omega + \omega^2 = 0

Row 3:

\frac{1}{\omega} + \frac{1}{\omega^2} + 1 = \frac{\omega^2 + \omega + 1}{\omega^2} = \frac{0}{\omega^2} = 0

Thus

C_1

becomes

\begin{bmatrix}0 & 0 & 0\end{bmatrix}^T

A determinant with an all‑zero column equals

0

Answer:

0

(Option A).

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