<p>If \(A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\) and \(I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), then which one of the following holds for all \(n \geq 1\), by the principle of mathematical induction?</p>
<p>\(A^n = nA - (n-1)I\)</p>
<p>\(A^n = 2^{n-1}A - (n-1)I\)</p>
<p>\(A^n = nA + (n-1)I\)</p>
<p>\(A^n = 2^{n-1}A + (n-1)I\)</p>
Step-by-Step Solution
Key Concept: When proving matrix power formulas by induction, the inductive step requires verifying that A^(k+1) = A · A^k holds the predicted form, which demands careful matrix multiplication and algebraic simplification to match the proposed pattern.
<p><strong>Step 1: Identify the pattern</strong></p><p>For A = [1, 0; 1, 1], compute the first few powers:</p><p>A¹ = [1, 0; 1, 1]</p><p>A² = [1, 0; 2, 1]</p><p>A³ = [1, 0; 3, 1]</p><p><strong>Conjecture:</strong> A^n = [1, 0; n, 1] for all n ≥ 1</p><p><strong>Step 2: Base case (n=1)</strong></p><p>A¹ = [1, 0; 1, 1] = [1, 0; 1·1, 1] ✓</p><p><strong>Step 3: Inductive step</strong></p><p>Assume A^k = [1, 0; k, 1]</p><p>Then A^(k+1) = A · A^k = [1, 0; 1, 1] · [1, 0; k, 1]</p><p>= [1·1 + 0·k, 1·0 + 0·1; 1·1 + 1·k, 1·0 + 1·1]</p><p>= [1, 0; k+1, 1] ✓</p><p><strong>Step 4: Conclusion</strong></p><p>By mathematical induction, A^n = [1, 0; n, 1] holds for all n ≥ 1</p><p>∴ Answer: A</p>
Correct Answer: A