A and B are square matrices such that det. `(A)=1, B B^(T)=I`, det `(B) gt 0`,B + A= B^(2) and A( adj. A + adj. B)=B.
`AB^(-1)=`

Options

A. `B^(-1)A`
B. `AB^(-1)`
C. `A^(T)B^(-1)`
D. `B^(T)A^(-1)`

Explanation

To solve the problem step-by-step, we will analyze the given equations and use properties of matrices.

### Step 1: Analyze the given equations
We are given the following:
1. \( \det(A) = 1 \)
2. \( BB^T = I \) (which means \( B \) is an orthogonal matrix)
3. \( \det(B) > 0 \)
4. \( B + A = B^2 \)
5. \( A(\text{adj}(A) + \text{adj}(B)) = B \)

### Step 2: Rearranging the equation \( B + A = B^2 \)
From the equation \( B + A = B^2 \), we can rearrange it to find \( A \):
\[ A = B^2 - B \]

### Step 3: Pre-multiply by \( B^T \)
We will pre-multiply both sides of the equation \( B + A = B^2 \) by \( B^T \):
\[ B^T(B + A) = B^TB^2 \]
Since \( BB^T = I \), we have:
\[ B^T B + B^T A = B^T B^2 \]
This simplifies to:
\[ I + B^T A = B \]

### Step 4: Isolate \( B^T A \)
Now, we can isolate \( B^T A \):
\[ B^T A = B - I \]

### Step 5: Substitute \( A \) into the equation
We already found that \( A = B^2 - B \). We can substitute this into the equation \( A(\text{adj}(A) + \text{adj}(B)) = B \):
\[ (B^2 - B)(\text{adj}(A) + \text{adj}(B)) = B \]

### Step 6: Find \( AB^{-1} \)
We know from our previous steps that:
\[ B^T A = B - I \]
Now, we can express \( A \) in terms of \( B \):
\[ A = B^2 - B \]
Next, we can multiply both sides of the equation \( B^T A = B - I \) by \( B^{-1} \):
\[ B^{-1} B^T A = B^{-1}(B - I) \]
This simplifies to:
\[ A = B^{-1}(B - I) \]

### Step 7: Final expression for \( AB^{-1} \)
Now, we can find \( AB^{-1} \):
\[ AB^{-1} = (B^2 - B)B^{-1} = B - I \]

### Conclusion
Thus, the final result is:
\[ AB^{-1} = B - I \]