`AB =I , detA=1/8` then find the value of `det(adj(B adj 2A))` when `A` is `3xx3` matrix

Options

A. 128
B. 32
C. 64
D. 16

Answer

To solve the problem, we need to find the value of \( \text{det}(\text{adj}(B \cdot \text{adj}(2A))) \) given that \( AB = I \) and \( \text{det}(A) = \frac{1}{8} \), where \( A \) is a \( 3 \times 3 \) matrix.

### Step-by-Step Solution:

1. **Understanding the properties of determinants**:
- We know that if \( A \) is an \( n \times n \) matrix, then:
\[ \text{det}(\text{adj}(A)) = \text{det}(A)^{n-1} \]
- For a scalar \( k \) and a matrix \( A \):
\[ \text{det}(kA) = k^n \cdot \text{det}(A) \]

2. **Finding \( \text{det}(B) \)**:
- Since \( AB = I \), we have:
\[ \text{det}(A) \cdot \text{det}(B) = \text{det}(I) = 1 \]
- Given \( \text{det}(A) = \frac{1}{8} \), we can find \( \text{det}(B) \):
\[ \text{det}(B) = \frac{1}{\text{det}(A)} = 8 \]

3. **Finding \( \text{det}(2A) \)**:
- Using the property of determinants:
\[ \text{det}(2A) = 2^3 \cdot \text{det}(A) = 8 \cdot \frac{1}{8} = 1 \]

4. **Finding \( \text{det}(\text{adj}(2A)) \)**:
- Now applying the adjoint property:
\[ \text{det}(\text{adj}(2A)) = \text{det}(2A)^{3-1} = \text{det}(2A)^2 = 1^2 = 1 \]

5. **Finding \( \text{det}(B \cdot \text{adj}(2A)) \)**:
- We can find \( \text{det}(B \cdot \text{adj}(2A)) \) using the property of determinants:
\[ \text{det}(B \cdot \text{adj}(2A)) = \text{det}(B) \cdot \text{det}(\text{adj}(2A)) = 8 \cdot 1 = 8 \]

6. **Finding \( \text{det}(\text{adj}(B \cdot \text{adj}(2A))) \)**:
- Finally, we apply the adjoint property again:
\[ \text{det}(\text{adj}(B \cdot \text{adj}(2A))) = \text{det}(B \cdot \text{adj}(2A))^{3-1} = \text{det}(B \cdot \text{adj}(2A))^2 = 8^2 = 64 \]

### Final Answer:
\[ \text{det}(\text{adj}(B \cdot \text{adj}(2A))) = 64 \]