Let `A=[(0, alpha),(0,0)]` and `(A+I)^(50) -50A=[(a,b),(c,d)]`. Then the value of `a+b+c+d` is

Options

A. 2
B. 1
C. 4
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \((A + I)^{50} - 50A\) where \(A = \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix}\) and \(I\) is the identity matrix. We will then find the values of \(a\), \(b\), \(c\), and \(d\) from the resulting matrix and compute \(a + b + c + d\).

### Step-by-Step Solution:

1. **Define the matrices**:
\[ A = \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

2. **Calculate \(A + I\)**:
\[ A + I = \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & \alpha \\ 0 & 1 \end{pmatrix} \]

3. **Compute \((A + I)^{50}\)**:
To compute \((A + I)^{50}\), we can use the property of upper triangular matrices. The matrix \(\begin{pmatrix} 1 & \alpha \\ 0 & 1 \end{pmatrix}\) raised to any power \(n\) is given by:
\[ \begin{pmatrix} 1 & n\alpha \\ 0 & 1 \end{pmatrix} \]
Therefore,
\[ (A + I)^{50} = \begin{pmatrix} 1 & 50\alpha \\ 0 & 1 \end{pmatrix} \]

4. **Calculate \(50A\)**:
\[ 50A = 50 \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 50\alpha \\ 0 & 0 \end{pmatrix} \]

5. **Evaluate \((A + I)^{50} - 50A\)**:
\[ (A + I)^{50} - 50A = \begin{pmatrix} 1 & 50\alpha \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & 50\alpha \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

6. **Identify the values of \(a\), \(b\), \(c\), and \(d\)**:
From the resulting matrix:
\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
We have:
- \(a = 1\)
- \(b = 0\)
- \(c = 0\)
- \(d = 1\)

7. **Calculate \(a + b + c + d\)**:
\[ a + b + c + d = 1 + 0 + 0 + 1 = 2 \]

### Final Answer:
The value of \(a + b + c + d\) is \(2\).