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# If $1,\omega,\omega^2$ are the cube roots of unity ,then $\Delta=\begin{vmatrix}1&\omega^n&\omega^{2n}\\\omega^{2n}&1&\omega^n\\\omega^{2n}&1&\omega^n\end{vmatrix}$ is equal to

$(a)\;\omega^2\qquad(b)\;0\qquad(c)\;1\qquad(d)\;\omega$

Can you answer this question?

$\Rightarrow 1(\omega^{3n}-1)-\omega^n(\omega^{2n}-\omega^{2n})+\omega^{2n}(\omega^n-\omega^{4n})$
$\Rightarrow \omega^{3n}-1-0+\omega^{3n}-\omega^{6n}$
$\omega^{3n}=1$
$\Rightarrow 1-1+1-1=0$
Hence (b) is the correct answer.
answered Nov 21, 2013