Matrices & Determinants
Sum of matrix elements
Grade 12
Question:
<p><strong>17.</strong> Let \(M_n = (a_{ij})\) where \(i, j = 1, 2, 3, \ldots, n\). We first find out \(a_{11}\) for the \(n^{\text{th}}\) matrix, which is the \(n^{\text{th}}\) term in the series: \(1, 2, 6, 15, \ldots\). The diagonal elements of the \(n^{\text{th}}\) matrix form an arithmetic progression with first term \(1 + \dfrac{n(n-1)(2n-1)}{6}\) and common difference \(n+1\). Find the required sum \(M_n\).</p>
<p>\(\dfrac{n}{6}(2n^3 + n + 3)\)</p>
<p>\(\dfrac{n}{6}(2n^3 + n - 3)\)</p>
<p>\(\dfrac{n}{3}(n^3 + n + 3)\)</p>
<p>\(\dfrac{n}{6}(n^3 + 2n + 3)\)</p>
Step-by-Step Solution
Key Concept: Recognize that a₁₁ follows the pattern n(n+1)(n-1)/3 (differences of cubes), then use the sum formula for diagonal elements in AP: S = (number of terms/2)(first + last term), where the diagonal has n elements.
<p><strong>Step 1:</strong> Identify the pattern for a₁₁. The sequence 1, 2, 6, 15, ... represents:</p><ul><li>n=1: 1 = 1(0)(1)/3</li><li>n=2: 2 = 2(1)(3)/3</li><li>n=3: 6 = 3(2)(5)/3</li><li>n=4: 15 = 4(3)(7)/3</li></ul><p>Pattern: a₁₁ = n(n-1)(2n-1)/3 for the nth term.</p><p><strong>Step 2:</strong> The diagonal elements form an AP with:</p><ul><li>First term: d₁ = 1 + n(n-1)(2n-1)/6</li><li>Common difference: D = n+1</li><li>Number of terms: n</li></ul><p><strong>Step 3:</strong> The last diagonal element is:</p><p>d_n = d₁ + (n-1)D = 1 + n(n-1)(2n-1)/6 + (n-1)(n+1)</p><p>= 1 + n(n-1)(2n-1)/6 + n² - 1</p><p>= n² + n(n-1)(2n-1)/6</p><p><strong>Step 4:</strong> Sum of diagonal elements (Trace):</p><p>S = (n/2)(d₁ + d_n)</p><p>= (n/2)[1 + n(n-1)(2n-1)/6 + n² + n(n-1)(2n-1)/6]</p><p>= (n/2)[1 + n² + n(n-1)(2n-1)/3]</p><p>= (n/2)[1 + n² + n(2n² - 3n + 1)/3]</p><p>= (n/2)[(3 + 3n² + 2n³ - 3n² + n)/3]</p><p>= (n/2) · (2n³ + n + 3)/3</p><p>= <strong>n(2n³ + n + 3)/6</strong></p><p>∴ Answer: A</p>
Correct Answer: A