Let Delta = left| begin{array}{ccc} 1 & omega | vedclass

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(A) Given $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \ 2 & 2\omega^2 & 4\omega^3 \ 3 & 3\omega^3 & 6\omega^4 \end{array} ight|$.We

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$\Delta = 0$

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(A) Given $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \ 2 & 2\omega^2 & 4\omega^3 \ 3 & 3\omega^3 & 6\omega^4 \end{array} ight|$.We know that $\omega^3 = 1$ and $\omega^4 = \omega$.Substituting these values,we get $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \ 2 & 2\omega^2 & 4 \ 3 & 3 & 6\omega \end{array} ight|$.Alternatively,observe the columns of the original determinant:Column $2$ is $\omega$ times Column $1$ (if we consider the ratio of elements).More simply,factor out constants from the rows/columns:$\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \ 2 & 2\omega^2 & 4\omega^3 \ 3 & 3\omega^3 & 6\omega^4 \end{array} ight| = 1 \cdot 2 \cdot 3 \left| \begin{array}{ccc} 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega^3 \ 1 & \omega^3 & \omega^4 \end{array} ight| 	imes 2 = 12 \left| \begin{array}{ccc} 1 & \omega & \omega^2 \ 1 & \omega^2 & 1 \ 1 & 1 & \omega \end{array} ight|$.Since Column $2$ and Column $3$ are proportional or by direct expansion,we see that the determinant is $0$ because the rows are linearly dependent (e.g.,$R_2 = 2R_1$ if we simplify the terms using $\omega^3=1$).Thus,$\Delta = 0$.

Let $\Delta = \left| \begin{array}{ccc} 1 & \omega & 2\omega^2 \\ 2 & 2\omega^2 & 4\omega^3 \\ 3 & 3\omega^3 & 6\omega^4 \end{array} \right|$ where $\omega$ is the cube root of unity,then

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