Let A and b are two square idempotent matrices such that `ABpm BA` is a null matrix, the value of det (A - B)
cann vbe equal

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem, we need to analyze the given conditions about the idempotent matrices \( A \) and \( B \). An idempotent matrix is defined as a matrix \( M \) such that \( M^2 = M \).

### Step-by-Step Solution:

1. **Understanding the Idempotent Property**:
Since \( A \) and \( B \) are idempotent matrices, we have:
\[ A^2 = A \quad \text{and} \quad B^2 = B \]

2. **Using the Given Conditions**:
We are given that:
\[ AB + BA = 0 \quad \text{and} \quad AB - BA = 0 \]
From \( AB - BA = 0 \), we can deduce that:
\[ AB = BA \]
Thus, we can rewrite the first condition as:
\[ 2AB = 0 \implies AB = 0 \]

3. **Finding \( A - B \)**:
We need to find the determinant of \( A - B \). We can express \( (A - B)^2 \):
\[ (A - B)^2 = A^2 - AB - BA + B^2 \]
Substituting the idempotent properties:
\[ = A - AB - AB + B = A - 2AB + B \]
Since \( AB = 0 \):
\[ = A + B \]

4. **Finding the Determinant**:
Now we have:
\[ (A - B)^2 = A + B \]
Taking the determinant on both sides:
\[ \det((A - B)^2) = \det(A + B) \]
Using the property of determinants:
\[ \det(A - B)^2 = \det(A + B) \]
This implies:
\[ \det(A - B)^2 = \det(A + B) \]

5. **Possible Values of Determinants**:
Since \( A \) and \( B \) are idempotent, their determinants can be either 0 or 1. Thus, \( \det(A + B) \) can take values depending on the eigenvalues of \( A \) and \( B \):
- If both \( A \) and \( B \) are zero matrices, \( \det(A + B) = 0 \).
- If both are identity matrices, \( \det(A + B) = 2 \).
- If one is a zero matrix and the other is an identity matrix, \( \det(A + B) = 1 \).

6. **Conclusion**:
Therefore, the determinant \( \det(A - B) \) can be:
\[ \det(A - B) = 0 \quad \text{or} \quad \det(A - B) = \pm 1 \]

### Final Answer:
The value of \( \det(A - B) \) can be equal to \( 0 \) or \( \pm 1 \).