<p>The number of solutions of the matrix equation \(X^2 = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}\) is</p>
Step-by-Step Solution
Key Concept: Finding all matrices X such that X² = A requires analyzing the minimal polynomial and characteristic polynomial of A. The number of square roots of a 2×2 matrix depends on whether it has repeated eigenvalues and the geometric multiplicity of those eigenvalues.
<p><strong>Step 1:</strong> Find the eigenvalues of A = [[1,1],[2,3]].</p><p>Characteristic polynomial: det(A - λI) = (1-λ)(3-λ) - 2 = λ² - 4λ + 1 = 0</p><p>λ = (4 ± √(16-4))/2 = (4 ± 2√3)/2 = 2 ± √3</p><p><strong>Step 2:</strong> Since A has two distinct positive eigenvalues λ₁ = 2+√3 and λ₂ = 2-√3, each eigenvalue √λᵢ can be taken with ± sign.</p><p><strong>Step 3:</strong> For each eigenvalue, we can choose positive or negative square root independently. This gives 2 × 2 = 4 combinations.</p><p><strong>Step 4:</strong> Each combination yields a distinct matrix X satisfying X² = A. We can verify: if X = PDP⁻¹ where D is diagonal with square roots of eigenvalues, then X² = PD²P⁻¹ = A.</p><p>∴ The number of solutions is <strong>4</strong> (Answer: B)</p>
Correct Answer: B