<p>If \(A = [a_{ij}]_{n \times n}\) and \(a_{ij} = (i^2 + j^2 - ij)(j - i)\), where \(n\) is odd, then the value of \(tr.(A)\) is equal to:</p>
Step-by-Step Solution
Key Concept: The trace only involves diagonal elements where i=j, making a_ii = (i² + i² - i²)(i - i) = i²·0 = 0 for every diagonal entry. For odd n, all diagonal elements vanish, so tr(A) = 0.
<p><strong>Step 1:</strong> Recall that the trace of a matrix is the sum of its diagonal elements: tr(A) = Σ a_ii where i goes from 1 to n.</p><p><strong>Step 2:</strong> For diagonal elements, i = j, so:</p><p>a_ii = (i² + i² - i·i)(i - i) = (2i² - i²)(0) = i²·0 = 0</p><p><strong>Step 3:</strong> Since every diagonal element a_ii = 0, regardless of the value of i or whether n is odd:</p><p>tr(A) = a₁₁ + a₂₂ + ... + a_nn = 0 + 0 + ... + 0 = 0</p><p><strong>Step 4:</strong> The condition that n is odd is a distractor; it doesn't affect the result since each diagonal term vanishes independently.</p><p>∴ Answer: <strong>0</strong></p>
Correct Answer: A