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 need to find the value of \( \det(BA^{-1}B^T) \) given the conditions \( \det(ABA^T) = 8 \) and \( \det(AB^{-1}) = 8 \).

### Step-by-Step Solution:

1. **Use the property of determinants**:
We know that for any square matrix \( M \), \( \det(M^T) = \det(M) \). Therefore, we can express \( \det(ABA^T) \) as:
\[ \det(ABA^T) = \det(A) \cdot \det(B) \cdot \det(A^T) = \det(A)^2 \cdot \det(B) \]
Given that \( \det(ABA^T) = 8 \), we have:
\[ \det(A)^2 \cdot \det(B) = 8 \quad \text{(1)} \]

2. **Analyze the second condition**:
For the second condition, we have:
\[ \det(AB^{-1}) = \det(A) \cdot \det(B^{-1}) = \det(A) \cdot \frac{1}{\det(B)} \]
Given that \( \det(AB^{-1}) = 8 \), we can write:
\[ \det(A) \cdot \frac{1}{\det(B)} = 8 \quad \text{(2)} \]

3. **From equations (1) and (2)**:
From equation (1), we have:
\[ \det(B) = \frac{8}{\det(A)^2} \]
Substituting this into equation (2):
\[ \det(A) \cdot \frac{1}{\frac{8}{\det(A)^2}} = 8 \]
Simplifying this gives:
\[ \det(A) \cdot \frac{\det(A)^2}{8} = 8 \]
\[ \frac{\det(A)^3}{8} = 8 \]
\[ \det(A)^3 = 64 \quad \Rightarrow \quad \det(A) = 4 \]

4. **Finding \( \det(B) \)**:
Now substitute \( \det(A) = 4 \) back into equation (1):
\[ \det(4)^2 \cdot \det(B) = 8 \]
\[ 16 \cdot \det(B) = 8 \quad \Rightarrow \quad \det(B) = \frac{8}{16} = \frac{1}{2} \]

5. **Calculate \( \det(BA^{-1}B^T) \)**:
Now we can find \( \det(BA^{-1}B^T) \):
\[ \det(BA^{-1}B^T) = \det(B) \cdot \det(A^{-1}) \cdot \det(B^T) = \det(B) \cdot \frac{1}{\det(A)} \cdot \det(B) \]
Since \( \det(B^T) = \det(B) \):
\[ \det(BA^{-1}B^T) = \det(B)^2 \cdot \frac{1}{\det(A)} \]
Substituting the values we found:
\[ \det(BA^{-1}B^T) = \left(\frac{1}{2}\right)^2 \cdot \frac{1}{4} = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \]

### Final Answer:
Thus, \( \det(BA^{-1}B^T) = \frac{1}{16} \).