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If $ω \neq 1$ is a cube root of unity, then $|\begin{array}{ccc} 1 & 1 + i + ω^{2} & ω^{2} \\ 1 - i & - 1 & ω^{2} - 1 \\ - i & - i + ω - 1 & - 1 \end{array}|$ equal

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detailed solution

Correct option is A

Use R2→R2−R1−R3, then the first row of the determinant elements are all zero's, hence the determinant value=0

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If ω ≠ 1 is a cube root of unity, then 11+i+ω2ω21−i−1ω2−1−i−i+ω−1−1 equal

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If $ω \neq 1$ is a cube root of unity, then $|\begin{array}{ccc} 1 & 1 + i + ω^{2} & ω^{2} \\ 1 - i & - 1 & ω^{2} - 1 \\ - i & - i + ω - 1 & - 1 \end{array}|$ equal