<p>Let \(A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\). Then \(A^n\) equals</p>
<p>\(\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 0 \\ n^2 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Use mathematical induction or pattern recognition to find that A^n follows a specific structure where the (1,2) entry becomes n times the (1,1) entry, while diagonal remains 1.
<p><strong>Step 1:</strong> Compute the first few powers to identify a pattern.</p><p>A = [1, 0; 1, 1]</p><p>A² = [1, 0; 1, 1]·[1, 0; 1, 1] = [1, 0; 2, 1]</p><p>A³ = A²·A = [1, 0; 2, 1]·[1, 0; 1, 1] = [1, 0; 3, 1]</p><p><strong>Step 2:</strong> Observe the pattern: A^n = [1, 0; n, 1]</p><p><strong>Step 3:</strong> Verify by induction. If A^n = [1, 0; n, 1], then A^(n+1) = A^n·A = [1, 0; n, 1]·[1, 0; 1, 1] = [1, 0; n+1, 1] ✓</p><p><strong>Step 4:</strong> The formula holds for all positive integers n.</p><p>∴ Answer: A^n = <strong>[1, 0; n, 1]</strong></p>
Correct Answer: A