Question 10

Mathematics Determinants Hard

If&nbsp;&omega; is a cube root of unity, then find the value of the determinant&nbsp;<span class="math-tex">\(\rm \begin{vmatrix} 1 + \omega &amp; \omega^2 &amp; -\omega \\\ 1 + \omega^2 &amp; \omega &amp; -\omega^2 \\\ \omega^2 + \omega &amp; \omega &amp; -\omega^2 \end{vmatrix}\)</span>&nbsp;is

(A) 3&omega;&nbsp;
(B) -3&omega;
(C) 3&omega;<sup>2</sup>
(D) -3&omega;<sup>2</sup>
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We are given the determinant: \[ \left| \begin{array}{ccc} -\omega^2 & \omega^2 & \omega \\ \omega & \omega & -\omega^2 \\ -1 & \omega & -\omega^2 \end{array} \right| \] First, factor out \( \omega^2 \) from the first row: \[ = \omega^2 \left| \begin{array}{ccc} -\omega & \omega & -1 \\ -1 & 1 & -\omega \\ -1 & \omega & \omega^2 \end{array} \right| \] Next, simplify the determinant: \[ = \omega^2 \left| \begin{array}{ccc} 0 & \omega & -1 \\ 0 & 1 & -1 \\ \omega-1 & \omega & \omega^2 \end{array} \right| \] Finally, evaluating the determinant gives: \[ = -3 \omega^2 \]

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