Question

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

• A. 2
• B. 1
• C. 4
• D. None of these

Hint:

For nilpotent matrices, binomial expansion truncates.

Answer 1

The Correct Option is A

To solve the problem, we need to evaluate \( (A + I)^{50} - 50A \) and find the value of \( a + b + c + d \) from the resulting matrix.

Given the matrices: 

\( A = \begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix} \)

\( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

First, let's calculate \( A + I \):

\( A + I = \begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix} \)

We need to find \( (A+I)^{50} \). The matrix \( \begin{bmatrix} 1 & \alpha \\ 0 & 1 \end{bmatrix} \) is a special kind of matrix known as a Jordan block of eigenvalue 1. The general exponential formula for this kind of matrix is:

\( (A + I)^n = \begin{bmatrix} 1 & n\alpha \\ 0 & 1 \end{bmatrix} \)

Therefore,

\( (A + I)^{50} = \begin{bmatrix} 1 & 50\alpha \\ 0 & 1 \end{bmatrix} \)

Next, we compute \( 50A \):

\( 50A = 50 \times \begin{bmatrix} 0 & \alpha \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 50\alpha \\ 0 & 0 \end{bmatrix} \)

Now, calculate \( (A + I)^{50} - 50A \):

\( (A + I)^{50} - 50A = \begin{bmatrix} 1 & 50\alpha \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 50\alpha \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

The resulting matrix is simply the identity matrix:

\( \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

The sum of all elements is:

\( a + b + c + d = 1 + 0 + 0 + 1 = 2 \)

Therefore, the value of \( a + b + c + d \) is 2.