Matrices & Determinants
Powers of matrices
Grade 12
Question:
<p>Matrix \(A\) such that \(A^2 = 2A - I\), where \(I\) is the identity matrix, then for \(n \geq 2\), \(A^n\) is equal to</p>
<p>\(2^{n-1}A - (n-1)I\)</p>
<p>\(2^{n-1}A - I\)</p>
<p>\(nA - (n-1)I\)</p>
<p>\(nA - I\)</p>
Step-by-Step Solution
Key Concept: Rearrange A² = 2A - I to get A² - 2A + I = 0, then recognize this as (A - I)² = 0. Use this recurrence relation to find a pattern for A^n by expressing it in terms of A and I.
<p><strong>Step 1:</strong> Start with the given equation A² = 2A - I</p><p>Rearrange: A² - 2A + I = 0, which gives (A - I)² = 0</p><p><strong>Step 2:</strong> This means A - I is nilpotent. Write A = I + N where N = A - I and N² = 0</p><p><strong>Step 3:</strong> For n ≥ 2, expand A^n = (I + N)^n using binomial theorem:</p><p>A^n = I^n + nI^(n-1)N + (higher terms with N²)</p><p>Since N² = 0, all terms with N² and higher vanish:</p><p>A^n = I + nN = I + n(A - I)</p><p><strong>Step 4:</strong> Simplify: A^n = I + nA - nI = nA - (n-1)I</p><p><strong>Verification:</strong> For n = 2: A² = 2A - I ✓</p><p>∴ Answer: A^n = nA - (n-1)I for n ≥ 2</p>
Correct Answer: C