If omega is a complex cube root of unity,then | vedclass

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(A) Let $\Delta = \left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \ 1 & 1 & 1 \ 1 & -1 & 0 \end{array} ight|$.Expanding along the third row:$\De

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(A) Let $\Delta = \left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \ 1 & 1 & 1 \ 1 & -1 & 0 \end{array} ight|$.Expanding along the third row:$\Delta = 1 \cdot \left| \begin{array}{cc} 2\omega & -\omega^2 \ 1 & 1 \end{array} ight| - (-1) \cdot \left| \begin{array}{cc} 2 & -\omega^2 \ 1 & 1 \end{array} ight| + 0 \cdot \left| \begin{array}{cc} 2 & 2\omega \ 1 & 1 \end{array} ight|$$\Delta = 1(2\omega - (-\omega^2)) + 1(2 - (-\omega^2))$$\Delta = 2\omega + \omega^2 + 2 + \omega^2$$\Delta = 2\omega + 2\omega^2 + 2$$\Delta = 2(1 + \omega + \omega^2)$Since $\omega$ is a complex cube root of unity,$1 + \omega + \omega^2 = 0$.Therefore,$\Delta = 2(0) = 0$.

If $\omega$ is a complex cube root of unity,then the value of the determinant $\left| \begin{array}{ccc} 2 & 2\omega & -\omega^2 \\ 1 & 1 & 1 \\ 1 & -1 & 0 \end{array} \right|$ is:

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