if `Aa n dB` are squares matrices such that `A^(2006)=Oa n dA B=A+B ,t h e n"det"(B)` equals `0` b. `1` c. `-1` d. none of these

Options

A. 0
B. 1
C. `-1`
D. none of these

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the required result.

### Step 1: Understand the given conditions
We have two square matrices \( A \) and \( B \) such that:
1. \( A^{2006} = 0 \) (the null matrix)
2. \( AB = A + B \)

### Step 2: Determine the implications of \( A^{2006} = 0 \)
Since \( A^{2006} = 0 \), this implies that the matrix \( A \) is a singular matrix. Therefore, the determinant of \( A \) must be zero:
\[ \text{det}(A) = 0 \]

### Step 3: Analyze the equation \( AB = A + B \)
Rearranging the equation \( AB = A + B \) gives us:
\[ AB - A - B = 0 \]
This can be factored as:
\[ A(B - I) = 0 \]
where \( I \) is the identity matrix.

### Step 4: Take the determinant of both sides
Taking the determinant of both sides of the equation \( A(B - I) = 0 \):
\[ \text{det}(A(B - I)) = \text{det}(0) \]
Since the determinant of the null matrix is 0, we have:
\[ \text{det}(A) \cdot \text{det}(B - I) = 0 \]

### Step 5: Use the known value of \( \text{det}(A) \)
From Step 2, we know that \( \text{det}(A) = 0 \). Therefore, we can conclude:
\[ 0 \cdot \text{det}(B - I) = 0 \]
This equation holds true for any value of \( \text{det}(B - I) \).

### Step 6: Analyze the implications for \( \text{det}(B) \)
Since \( A \) is singular, we cannot directly conclude anything about \( \text{det}(B) \) from the previous step. However, we can analyze the equation \( AB = A + B \) further.

### Step 7: Consider the case when \( B = I \)
If we assume \( B = I \), then:
\[ AB = A + I \]
This would imply:
\[ A \cdot I = A + I \implies A = A + I \]
which is a contradiction unless \( A \) is the null matrix.

### Step 8: Conclude about \( \text{det}(B) \)
Since \( A \) is singular and \( A(B - I) = 0 \) holds, it suggests that \( B \) must also be singular for the equation to hold true. Therefore, we can conclude:
\[ \text{det}(B) = 0 \]

### Final Answer
Thus, the determinant of \( B \) is:
\[ \text{det}(B) = 0 \]