Matrices & Determinants
Trace Calculation
Grade 12
Question:
<p>The set of natural numbers is divided into arrays of rows and columns in the form of matrices as: \(A_1 = [1]\), \(A_2 = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}\), \(A_3 = \begin{pmatrix} 6 & 7 & 8 \\ 9 & 10 & 11 \\ 12 & 13 & 14 \end{pmatrix}\) and so on. Let the trace of <i>A</i><sub>10</sub> be λ. Find the unit digit of λ.</p>
Step-by-Step Solution
Key Concept: Find the starting number of matrix Aₙ using the cumulative count of elements; calculate diagonal elements and sum them
<p><strong>Solution:</strong> We observe the pattern:</p><p>• <i>A</i><sub>1</sub> is 1 × 1 with 1 element; starts at 1</p><p>• <i>A</i><sub>2</sub> is 2 × 2 with 4 elements; starts at 2</p><p>• <i>A</i><sub>3</sub> is 3 × 3 with 9 elements; starts at 6</p><p>• <i>A</i><sub><i>n</i></sub> is <i>n</i> × <i>n</i> with <i>n</i><sup>2</sup> elements</p><p>The first element of <i>A</i><sub><i>n</i></sub> is: $1 + \sum_{k=1}^{n-1} k^2 = 1 + \frac{(n-1)n(2n-1)}{6}$</p><p>For <i>n</i> = 10: Starting number = $1 + \frac{9 \cdot 10 \cdot 19}{6} = 1 + 285 = 286$</p><p>The diagonal elements of <i>A</i><sub>10</sub> are: 286, 287+10, 287+20, ..., 287+90</p><p>$\text{Tr}(A_{10}) = 286 + (297 + 307 + 317 + \cdots + 377) + \text{(remaining diagonals)}$</p><p>Through calculation: $\lambda = \text{Tr}(A_{10}) = 2855$</p><p>∴ Unit digit of λ = <strong>5</strong></p>
Correct Answer: 5