Let `A=[(0,alpha),(0,0)]` and `(A+I)^(50)-50A=[(a,b),(c,d)]`, find abc+abd+bcd+acd

Options

A. 0
B. -1
C. 1
D. Noneof these

Answer

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 find the values of \( a, b, c, \) and \( d \) from the resulting matrix and then compute \( abc + abd + bcd + acd \).

### Step-by-Step Solution

1. **Define the Matrices**:
Let \( A = \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix} \) and \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

2. **Calculate \( A^2 \)**:
\[ A^2 = A \cdot A = \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & \alpha \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]
Since \( A^2 = 0 \), we can conclude that \( A^n = 0 \) for all \( n \geq 2 \).

3. **Calculate \( (A + I)^{50} \)**:
Using the binomial expansion:
\[ (A + I)^{50} = \sum_{k=0}^{50} \binom{50}{k} A^k I^{50-k} \]
Since \( A^k = 0 \) for \( k \geq 2 \), the only terms that contribute are for \( k = 0 \) and \( k = 1 \):
\[ (A + I)^{50} = \binom{50}{0} A^0 I^{50} + \binom{50}{1} A^1 I^{49} = I + 50A \]
Thus, we have:
\[ (A + I)^{50} = I + 50A \]

4. **Substitute into the Expression**:
Now we substitute this back into our expression:
\[ (A + I)^{50} - 50A = (I + 50A) - 50A = I \]

5. **Identify the Resulting Matrix**:
The resulting matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
Therefore, we can identify:
- \( a = 1 \)
- \( b = 0 \)
- \( c = 0 \)
- \( d = 1 \)

6. **Calculate the Expression \( abc + abd + bcd + acd \)**:
Now we compute:
\[ abc + abd + bcd + acd = (1)(0)(0) + (1)(0)(1) + (0)(0)(1) + (1)(0)(1) = 0 + 0 + 0 + 0 = 0 \]

### Final Answer
The value of \( abc + abd + bcd + acd \) is \( \boxed{0} \).