Let A and B be two invertible matrices of order `3xx3`. If det. `(ABA^(T))` = 8 and det. `(AB^(-1))` = 8, then det. `(BA^(-1)B^(T))` is equal to

Options

A. 16
B. `(1)/(16)`
C. `(1)/(4)`
D. 1

Answer

To solve the problem, we will use properties of determinants and the given information about the matrices A and B.

### Step-by-Step Solution:

1. **Given Information**:
- We have two invertible matrices \( A \) and \( B \) of order \( 3 \times 3 \).
- \( \det(ABA^T) = 8 \)
- \( \det(AB^{-1}) = 8 \)

2. **Using the Determinant Properties**:
- We know that \( \det(ABA^T) = \det(A) \det(B) \det(A^T) \).
- Since \( \det(A^T) = \det(A) \), we can rewrite this as:
\[ \det(ABA^T) = \det(A)^2 \det(B) \]
- Therefore, we have:
\[ \det(A)^2 \det(B) = 8 \quad \text{(1)} \]

3. **For the Second Determinant**:
- We also know that \( \det(AB^{-1}) = \det(A) \det(B^{-1}) \).
- Since \( \det(B^{-1}) = \frac{1}{\det(B)} \), we can rewrite this as:
\[ \det(AB^{-1}) = \det(A) \cdot \frac{1}{\det(B)} \]
- Therefore, we have:
\[ \det(A) \cdot \frac{1}{\det(B)} = 8 \quad \text{(2)} \]

4. **Solving the Equations**:
- From equation (1):
\[ \det(A)^2 \det(B) = 8 \]
- From equation (2):
\[ \det(A) = 8 \det(B) \]
- Substitute \( \det(A) \) from (2) into (1):
\[ (8 \det(B))^2 \det(B) = 8 \]
\[ 64 \det(B)^2 \det(B) = 8 \]
\[ 64 \det(B)^3 = 8 \]
\[ \det(B)^3 = \frac{8}{64} = \frac{1}{8} \]
\[ \det(B) = \frac{1}{2} \]

5. **Finding \( \det(A) \)**:
- Substitute \( \det(B) \) back into equation (2):
\[ \det(A) = 8 \cdot \frac{1}{2} = 4 \]

6. **Finding \( \det(BA^{-1}B^T) \)**:
- We need to find \( \det(BA^{-1}B^T) \).
- Using the properties of determinants:
\[ \det(BA^{-1}B^T) = \det(B) \det(A^{-1}) \det(B^T) \]
- Since \( \det(B^T) = \det(B) \), we have:
\[ \det(BA^{-1}B^T) = \det(B)^2 \det(A^{-1}) \]
- We know \( \det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{4} \):
\[ \det(BA^{-1}B^T) = \left(\frac{1}{2}\right)^2 \cdot \frac{1}{4} \]
\[ = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

### Final Answer:
\[ \det(BA^{-1}B^T) = \frac{1}{16} \]