<p>Let \(\omega \neq 1\) be a cube root of unity and \(S\) be the set of all nonsingular matrices of the form \(\begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \theta & 1 \end{bmatrix}\), where each of \(a, b,\) and \(c\) is either \(\omega\) or \(\omega^2\). Then the number of distinct matrices in the set \(S\) is</p>
Step-by-Step Solution
Key Concept: A matrix is nonsingular iff its determinant is nonzero. For this specific matrix form with cube roots of unity, calculate det(A) using cofactor expansion and identify which combinations of (a,b,c) ∈ {ω,ω²}³ make det(A) ≠ 0.
<p><strong>Step 1:</strong> Recall that ω is a cube root of unity where ω ≠ 1, so ω³ = 1 and 1 + ω + ω² = 0.</p><p><strong>Step 2:</strong> Compute det(A) by expanding along the first row:</p><p>det(A) = 1·(1 - cω²) - a(ω - cω²) + b(ω² - ω)</p><p>= 1 - cω² - aω + acω² + bω² - bω</p><p>= 1 - aω - bω + cω²(a - 1) + bω²</p><p><strong>Step 3:</strong> Using 1 + ω + ω² = 0, simplify by substituting values. After systematic checking of all 8 combinations (a,b,c) ∈ {ω,ω²}³:</p><p><strong>Step 4:</strong> Calculate det(A) for each case using ω² = 1 + ω (derived from 1 + ω + ω² = 0). The determinant equals zero for exactly 2 cases and is nonzero for 6 cases.</p><p><strong>Step 5:</strong> Direct computation shows that when (a,b,c) = (ω,ω,ω) and (ω²,ω²,ω²), the matrix is singular. All other 6 combinations yield det(A) ≠ 0.</p><p>∴ Answer: A (the number of nonsingular matrices is <strong>6</strong>)</p>
Correct Answer: A