<p>Let \(A = \begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix}\) and \((A+I)^{50} - 50A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\). Then the value of \(a + b + c + d\) is</p>
Step-by-Step Solution
Key Concept: Recognize that A is nilpotent (A² = 0), so (A+I)⁵⁰ expands using binomial theorem with only two non-zero terms. The constraint that (A+I)⁵⁰ - 50A equals a specific matrix form allows you to extract the values directly.
<p><strong>Step 1:</strong> Verify that A is nilpotent: A² = <math><mrow><begin matrix>0 & α<\matrix><\begin matrix>0 & 0<\matrix></mrow></math> × <math><mrow><begin matrix>0 & α<\matrix><\begin matrix>0 & 0<\matrix></mrow></math> = <math><mrow><begin matrix>0 & 0<\matrix><\begin matrix>0 & 0<\matrix></mrow></math> = 0</p><p><strong>Step 2:</strong> Since A² = 0, use binomial expansion: (A+I)⁵⁰ = I + 50A + ⁵⁰C₂A² + ... = I + 50A (all higher powers vanish)</p><p><strong>Step 3:</strong> Calculate (A+I)⁵⁰ - 50A = I + 50A - 50A = I = <math><mrow><begin matrix>1 & 0<\matrix><\begin matrix>0 & 1<\matrix></mrow></math></p><p><strong>Step 4:</strong> Therefore a = 1, b = 0, c = 0, d = 1</p><p>∴ a + b + c + d = 1 + 0 + 0 + 1 = <strong>2</strong></p>
Correct Answer: A