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NIMCET 2025 Paper

Section: Mathematics

Question : 9 of 120

Marks: +1, -0

x, y

z

\begin{array}{ccc} x & y & z \\ y & z & x \\ z & x & y \end{array}

Solution:

The cube roots of 27 are

3, 3\omega, 3\omega^{2}

, where

\omega

is a cube root of unity.

A circulant matrix like this has a determinant equal to:

(x+y+z)(x+\omega y+\omega^{2} z)(x+\omega^{2} y+\omega z)

For cube roots of unity, one linear factor always becomes zero.

Cube roots of 27:

x=3, \;\; y=3\omega, \;\; z=3\omega^{2}

Sum of cube roots of unity is zero:

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

Thus,

x+y+z=3(1+\omega+\omega^{2})=0

Since one factor in the determinant formula becomes zero:

\det=(x+y+z)(\cdots)(\cdots)=0

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