Matrices & Determinants
Trace of a Matrix
Grade 12
Question:
<p><strong>Paragraph for Question nos. 644 and 645</strong><br>If \(A = [a_{ij}]_{n \times n}\), where \(a_{ij} = i^2 + j^2\), \(\forall\, i\) and \(j\), then:<br><br>\(\lim_{n \to \infty} \dfrac{\text{tr.}(A)}{n^3}\) is equal to:</p>
<p>\(\dfrac{1}{6}\)</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{2}{3}\)</p>
<p>\(\dfrac{4}{3}\)</p>
Step-by-Step Solution
Key Concept: The trace of A is the sum of diagonal elements where a_ii = i² + i². We need to find tr(A) = Σ(2i²) for i=1 to n, then divide by n³ and take the limit. Use the formula Σi² = n(n+1)(2n+1)/6 to convert the sum into a polynomial in n.
<p><strong>Step 1:</strong> Find the trace of A. Since A is n×n with a_ij = i² + j², the trace includes only diagonal elements where i = j.</p><p>tr(A) = Σ(i=1 to n) a_ii = Σ(i=1 to n) (i² + i²) = 2Σ(i=1 to n) i²</p><p><strong>Step 2:</strong> Use the standard formula for sum of squares: Σ(i=1 to n) i² = n(n+1)(2n+1)/6</p><p>tr(A) = 2 · n(n+1)(2n+1)/6 = n(n+1)(2n+1)/3</p><p><strong>Step 3:</strong> Expand the numerator: n(n+1)(2n+1) = n(2n² + 3n + 1) = 2n³ + 3n² + n</p><p>tr(A) = (2n³ + 3n² + n)/3</p><p><strong>Step 4:</strong> Calculate the limit:</p><p>lim(n→∞) tr(A)/n³ = lim(n→∞) (2n³ + 3n² + n)/(3n³) = lim(n→∞) (2/3 + 1/n + 1/(3n²)) = <strong>2/3</strong></p><p>∴ Answer: C</p>
Correct Answer: C