Matrices & Determinants
Determinant and Roots of Equations
Grade 12
Question:
<p>If \(\alpha, \beta, \gamma\) are the roots of \(ax^3 + bx^2 + cx + d = 0\) and \[\begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0,\] \(\alpha \neq \beta \neq \gamma\), then find the equation whose roots are \(\alpha+\beta-\gamma\), \(\beta+\gamma-\alpha\), and \(\gamma+\alpha-\beta\).</p>
Step-by-Step Solution
Key Concept: The determinant equals zero only when α + β + γ = 0 (using the circulant matrix property), which means the new roots can be expressed as 2α, 2β, 2γ respectively. Apply Vieta's formulas with the substitution y = 2x to get the transformed equation.
<p><strong>Step 1:</strong> Evaluate the circulant determinant. For the given circulant matrix, the determinant equals (α + β + γ)[(α - β)² + (β - γ)² + (γ - α)²].</p><p><strong>Step 2:</strong> Since the determinant = 0 and α ≠ β ≠ γ ≠ α, we must have α + β + γ = 0.</p><p><strong>Step 3:</strong> With α + β + γ = 0, the new roots become:</p><ul><li>α + β - γ = -2γ</li><li>β + γ - α = -2α</li><li>γ + α - β = -2β</li></ul><p><strong>Step 4:</strong> If the roots of the new equation are -2α, -2β, -2γ, substitute y = -2x (or x = -y/2) into ax³ + bx² + cx + d = 0:</p><p>a(-y/2)³ + b(-y/2)² + c(-y/2) + d = 0</p><p>-ay³/8 + by²/4 - cy/2 + d = 0</p><p><strong>Step 5:</strong> Multiply through by -8 to clear denominators:</p><p>∴ <strong>ay³ - 2by² + 4cy - 8d = 0</strong></p><p>Replacing y with x: <strong>ax³ - 2bx² + 4cx - 8d = 0</strong></p>
Correct Answer: ax^3 - 2bx^2 + 4cx - 8d = 0