Matrices & Determinants
Properties of Determinants
Grade 12
Question:
<p>If \(\alpha, \beta \neq 0\), and \(f(n) = \alpha^n + \beta^n\) and</p><p>\[\begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix} = k(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2,\]</p><p>then <em>k</em> is equal to</p>
<p>(1) \(\alpha\beta\)</p>
<p>(2) \(\dfrac{1}{\alpha\beta}\)</p>
<p>(3) 1</p>
<p>(4) \(-1\)</p>
Step-by-Step Solution
Key Concept: Recognize that the matrix entries follow a linear recurrence relation through f(n) = αⁿ + βⁿ, allowing you to express each entry as a combination of previous terms. This structure enables column/row operations to factor out (1-α)²(1-β)²(α-β)² systematically.
<p><strong>Step 1:</strong> Establish the recurrence for f(n). Since α and β satisfy t² - (α+β)t + αβ = 0, we have:</p><p>f(n) = (α+β)f(n-1) - αβf(n-2)</p><p><strong>Step 2:</strong> Use column operations. Subtract column 1 from columns 2 and 3:</p><p>C₂ → C₂ - C₁ and C₃ → C₃ - C₁</p><p>This transforms the matrix to isolate differences like f(1)-2, f(2)-1, f(3)-f(2), etc.</p><p><strong>Step 3:</strong> Recognize that entries become expressions in (1-α)(1-β) terms. For example, f(1)-2 = α+β-2 = -(1-α)-(1-β) relates to the factorization.</p><p><strong>Step 4:</strong> Apply row operations strategically. Row operations preserve determinant value and expose common factors (1-α)² and (1-β)² in two rows, and (α-β)² emerges from comparing remaining row dependencies.</p><p><strong>Step 5:</strong> After factoring out (1-α)²(1-β)²(α-β)², the remaining 1×1 determinant equals 1.</p><p>∴ <strong>k = 1</strong></p>
Correct Answer: C