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Previous Year Question (PYQs)



If $\omega \ne 1$ is a cube root of unity and $i = \sqrt{-1}$, the value of the determinant $\left|\begin{matrix} 1 & 1+i+\omega^2 & \omega \\ 1-i & -1 & \omega^2 - 1 \\ -i & -i+\omega-1 & -\omega^3 \end{matrix}\right|$ is





Solution

Solution: Using $\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0,$ simplify the entries. After row/column reduction and applying cube root identities, the determinant becomes $\omega^2.$


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