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COMEDK · Maths · 3. Complex Number

If \(\omega\) is a cube root of unity, then the value of \(\operatorname{determinant}\left|\begin{array}{ccc}1+\omega & \omega^{2} & \omega \\ \omega^{2}+\omega & -\omega & \omega^{2} \\ 1+\omega^{2} & \omega & \omega^{2}\end{array}\right|\) is equal to

  1. A \(1+\omega\)
  2. B \(1-\omega\)
  3. C 0
  4. D \(\omega^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1-\omega\)

Step-by-step Solution

Detailed explanation

We have, \(\left|\begin{array}{ccc}1+\omega & \omega^{2} & \omega \\ \omega^{2}+\omega & -\omega & \omega^{2} \\ 1+\omega^{2} & \omega & \omega^{2}\end{array}\right|\)…