NIMCET 2025 Paper

The cube roots of 27 are $3,3\mathit{\omega },3{\mathit{\omega }}^2$, where $\mathit{\omega }$ is a cube root of unity.
A circulant matrix like this has a determinant equal to:
$\left(\mathit{x}+\mathit{y}+\mathit{z}\right)\left(\mathit{x}+\mathit{\omega }\mathit{y}+{\mathit{\omega }}^2\mathit{z}\right)\left(\mathit{x}+{\mathit{\omega }}^2\mathit{y}+\mathit{\omega }\mathit{z}\right)$
For cube roots of unity, one linear factor always becomes zero.
Cube roots of 27:
$\mathit{x}=3,\mathit{y}=3\mathit{\omega },\mathit{z}=3{\mathit{\omega }}^2$
Sum of cube roots of unity is zero:
$1+\mathit{\omega }+{\mathit{\omega }}^2=0$
Thus,
$\mathit{x}+\mathit{y}+\mathit{z}=3\left(1+\mathit{\omega }+{\mathit{\omega }}^2\right)=0$
Since one factor in the determinant formula becomes zero:
$\det =\left(\mathit{x}+\mathit{y}+\mathit{z}\right)(\cdots )(\cdots )=0$