A and B are square matrices such that det. `(A)=1, B B^(T)=I`, det `(B) gt 0`, and A( adj. A + adj. B)=B.
The value of det `(A+B)` is

Options

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

Explanation

To solve the problem, we need to find the value of \( \text{det}(A + B) \) given the conditions on matrices \( A \) and \( B \).

### Step-by-Step Solution:

1. **Given Information**:
- \( \text{det}(A) = 1 \)
- \( B B^T = I \) (which implies \( B \) is an orthogonal matrix)
- \( \text{det}(B) > 0 \)
- \( A (\text{adj}(A) + \text{adj}(B)) = B \)

2. **Finding \( \text{det}(B) \)**:
Since \( B B^T = I \), we can take the determinant on both sides:
\[ \text{det}(B) \cdot \text{det}(B^T) = \text{det}(I) \]
Since \( \text{det}(B^T) = \text{det}(B) \), we have:
\[ \text{det}(B)^2 = 1 \]
Given that \( \text{det}(B) > 0 \), we conclude:
\[ \text{det}(B) = 1 \]

3. **Using the property of adjugate**:
We know that:
\[ A \cdot \text{adj}(A) = \text{det}(A) \cdot I = 1 \cdot I = I \]
Therefore, \( \text{adj}(A) = A^{-1} \).

4. **Finding \( \text{adj}(B) \)**:
Similarly, since \( \text{det}(B) = 1 \):
\[ \text{adj}(B) = \text{det}(B) \cdot B^{-1} = 1 \cdot B^{-1} = B^{-1} \]

5. **Substituting into the equation**:
Now substituting \( \text{adj}(A) \) and \( \text{adj}(B) \) into the equation:
\[ A (A^{-1} + B^{-1}) = B \]
This simplifies to:
\[ A A^{-1} + A B^{-1} = B \]
Which simplifies to:
\[ I + A B^{-1} = B \]
Rearranging gives:
\[ A B^{-1} = B - I \]

6. **Taking determinants**:
Now we take the determinant of both sides:
\[ \text{det}(A B^{-1}) = \text{det}(B - I) \]
Using the property of determinants:
\[ \text{det}(A) \cdot \text{det}(B^{-1}) = \text{det}(B - I) \]
Since \( \text{det}(A) = 1 \) and \( \text{det}(B^{-1}) = \frac{1}{\text{det}(B)} = 1 \):
\[ 1 \cdot 1 = \text{det}(B - I) \]
Thus:
\[ \text{det}(B - I) = 1 \]

7. **Finding \( \text{det}(A + B) \)**:
We can use the property of determinants that states:
\[ \text{det}(A + B) = \text{det}(A) + \text{det}(B) + \text{det}(A B) \]
Since \( \text{det}(A) = 1 \) and \( \text{det}(B) = 1 \):
\[ \text{det}(A + B) = 1 + 1 + \text{det}(A B) \]
We know \( \text{det}(A B) = \text{det}(A) \cdot \text{det}(B) = 1 \cdot 1 = 1 \):
\[ \text{det}(A + B) = 1 + 1 + 1 = 3 \]

### Final Result:
Thus, the value of \( \text{det}(A + B) \) is \( \boxed{3} \).