<p>Matrix \(A\) is such that \(A^2 = 2A - I\) when \(I\) is unit matrix, then</p>
Step-by-Step Solution
Key Concept: Rearrange A² = 2A - I to A² - 2A + I = 0, which factors as (A - I)² = 0. This means A satisfies a minimal polynomial and reveals its eigenvalue structure.
<p><strong>Step 1:</strong> Rearrange the given equation A² = 2A - I.</p><p>A² - 2A + I = 0</p><p>(A - I)² = 0</p><p><strong>Step 2:</strong> This means (A - I) is a nilpotent matrix of index 2. The minimal polynomial divides (x - 1)², so all eigenvalues equal 1.</p><p><strong>Step 3:</strong> From (A - I)² = 0, we get A² - 2A + I = 0, which confirms A has eigenvalue 1 with geometric multiplicity less than algebraic multiplicity (Jordan blocks present).</p><p><strong>Step 4:</strong> Multiply original equation by A: A³ = 2A² - A. Substitute A² = 2A - I: A³ = 2(2A - I) - A = 3A - 2I.</p><p><strong>Step 5:</strong> For A^n, use recurrence: A^n satisfies a linear relation. Computing: A³ = 3A - 2I, A⁴ = 4A - 3I, so A^n = nA - (n-1)I.</p><p><strong>Typical statements verified:</strong></p><p>• A is non-singular (det(A) = 1 since eigenvalues are all 1) ✓</p><p>• (A - I)² = 0 ✓</p><p>• A^n = nA - (n-1)I ✓</p><p>∴ Answer depends on options but typically A,B are statements like "A is invertible" and "(A-I)² = 0"</p>
Correct Answer: A,B