Matrices & Determinants
Determinant Equations
Grade 12
Question:
<p>For the equation \(\begin{vmatrix} 1 & x & x^2 \\ x^2 & 1 & x \\ x & x^2 & 1 \end{vmatrix} = 0\),</p>
<p>(1) There are exactly two distinct roots</p>
<p>(2) There is one pair of equation real roots</p>
<p>(3) There are three pairs of equal roots</p>
<p>(4) Modulus of each root is 2</p>
Step-by-Step Solution
Key Concept: This is a circulant matrix where each row is a cyclic shift of the previous row. Use the property that for circulant matrices with first row [1, x, x²], the determinant factors as (1 + x + x²)(1 + ωx + ω²x²)(1 + ω²x + ωx²) where ω is a cube root of unity, or directly expand and factor recognizing the symmetric structure.
<p><strong>Step 1:</strong> Recognize this as a circulant matrix. Expand the determinant directly:</p><p>Δ = 1(1 - x⁴) - x(x - x³) + x²(x³ - x)</p><p>= 1 - x⁴ - x² + x⁴ + x⁵ - x³</p><p>= 1 - x² - x³ + x⁵</p><p><strong>Step 2:</strong> Factor by recognizing the structure. Rewrite as:</p><p>Δ = 1 - x² - x³(1 - x²) = (1 - x²)(1 - x³)</p><p>= (1 - x)(1 + x)(1 - x)(1 + x + x²)</p><p>= (1 - x)²(1 + x)(1 + x + x²)</p><p><strong>Step 3:</strong> Set Δ = 0:</p><p>(1 - x)² = 0 → x = 1 (double root)</p><p>(1 + x) = 0 → x = -1</p><p>(1 + x + x²) = 0 → x = ω, ω² (complex cube roots of unity)</p><p><strong>Step 4:</strong> The real solutions are x ∈ {1, -1} and complex solutions are the non-real cube roots of unity.</p><p>∴ Answer: A</p>
Correct Answer: A