Matrices & Determinants
Determinants
Grade 12
Question:
<p>If \(\alpha, \beta \neq 0\), and \(f(n) = \alpha^n + \beta^n\) and \[\begin{vmatrix} 3 & 1+f(1) & 1+f(2) \\ 1+f(1) & 1+f(2) & 1+f(3) \\ 1+f(2) & 1+f(3) & 1+f(4) \end{vmatrix} = K(1-\alpha)^2(1-\beta)^2(\alpha-\beta)^2,\] then \(K\) is equal to</p>
<p>\(1\)</p>
<p>\(-1\)</p>
<p>\(\alpha\beta\)</p>
<p>\(\dfrac{1}{\alpha\beta}\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(n) = α^n + β^n satisfies a recurrence relation, and the determinant's entries form a pattern allowing factorization through the characteristic structure of linear recurrences.
<p><strong>Step 1:</strong> Express f(n) = α^n + β^n. Note that f(n) satisfies the recurrence: f(n) = (α+β)f(n-1) - αβ·f(n-2) with f(0)=2, f(1)=α+β.</p><p><strong>Step 2:</strong> Denote the determinant matrix entries. The key observation is that each entry 1+f(k) = 1 + α^k + β^k can be factored. Notice: 1 + α^n + β^n relates to products involving (1-α) and (1-β).</p><p><strong>Step 3:</strong> Expand: 1 + f(n) = 1 + α^n + β^n. For systematic evaluation, apply row/column operations. Subtract row 1 from rows 2,3 to expose the recurrence structure, which creates factors of the form (α^i - α^j) terms.</p><p><strong>Step 4:</strong> The determinant evaluates through Vandermonde-like structure. Each difference operation introduces factors related to (α-β). The pattern yields three independent factors of (1-α), (1-β), and (α-β) appearing twice each.</p><p><strong>Step 5:</strong> Direct calculation or using the systematic structure shows the determinant equals (1-α)²(1-β)²(α-β)² × K. Computing the leading coefficient (using limiting behavior or explicit substitution): K = <strong>1</strong>.</p><p>∴ Answer: A (K = 1)</p>
Correct Answer: A