<p>If \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\) (where \(bc \neq 0\)) satisfies the equations \(x^2 + k = 0\), then</p>
Step-by-Step Solution
Key Concept: A matrix satisfying a polynomial equation x² + k = 0 means A² = -kI, which directly constrains the eigenvalues and trace of the matrix. Use the characteristic equation and properties of matrix powers to establish relationships between a, b, c, d.
<p><strong>Step 1:</strong> If matrix A satisfies x² + k = 0, then A² + kI = 0, which gives A² = -kI.</p><p><strong>Step 2:</strong> Calculate A²: A² = [[a² + bc, ab + bd], [ac + cd, bc + d²]] = [[a² + bc, b(a+d)], [c(a+d), d² + bc]]</p><p><strong>Step 3:</strong> Equating A² = -kI: <br>• a² + bc = -k<br>• b(a+d) = 0 (since k ≠ 0, bc ≠ 0)<br>• c(a+d) = 0<br>• d² + bc = -k</p><p><strong>Step 4:</strong> From b(a+d) = 0 with b ≠ 0, we get a + d = 0, so d = -a. From c(a+d) = 0 with c ≠ 0, we again get d = -a.</p><p><strong>Step 5:</strong> Substituting d = -a into a² + bc = -k and d² + bc = -k: both give a² + bc = -k. Also trace(A) = a + d = 0 and a² + bc = -k.</p><p>∴ Answer: C</p>
Correct Answer: C