Let A and B are two matrices of order `3xx3`, where `|A|=-2 and |B|=2`, then `|A^(-1)adj(B^(-1))adj(2A^(-1))|` is equal to

Options

A. 2
B. `-2`
C. 4
D. 8

Answer

To solve the problem, we need to find the value of \(|A^{-1} \text{adj}(B^{-1}) \text{adj}(2A^{-1})|\).

### Step-by-Step Solution:

1. **Understanding the Determinant Properties**:
We know that for any square matrix \( M \):
\[ |M^{-1}| = \frac{1}{|M|} \]
and for the adjugate of a matrix \( M \):
\[ |\text{adj}(M)| = |M|^{n-1} \]
where \( n \) is the order of the matrix.

2. **Applying the Properties**:
Given that \( |A| = -2 \) and \( |B| = 2 \), we can find:
- \( |A^{-1}| = \frac{1}{|A|} = \frac{1}{-2} = -\frac{1}{2} \)
- \( |B^{-1}| = \frac{1}{|B|} = \frac{1}{2} \)

3. **Calculating the Determinant of the Adjugate**:
Since both \( A \) and \( B \) are \( 3 \times 3 \) matrices (\( n = 3 \)):
- For \( B^{-1} \):
\[ |\text{adj}(B^{-1})| = |B^{-1}|^{3-1} = |B^{-1}|^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
- For \( 2A^{-1} \):
\[ |\text{adj}(2A^{-1})| = |2A^{-1}|^{3-1} = |2|^2 |A^{-1}|^2 = 4 \cdot \left(-\frac{1}{2}\right)^2 = 4 \cdot \frac{1}{4} = 1 \]

4. **Putting it All Together**:
Now we can compute the determinant:
\[ |A^{-1} \text{adj}(B^{-1}) \text{adj}(2A^{-1})| = |A^{-1}| \cdot |\text{adj}(B^{-1})| \cdot |\text{adj}(2A^{-1})| \]
Substituting the values we calculated:
\[ = \left(-\frac{1}{2}\right) \cdot \left(\frac{1}{4}\right) \cdot 1 = -\frac{1}{8} \]

5. **Final Calculation**:
The final value is:
\[ |A^{-1} \text{adj}(B^{-1}) \text{adj}(2A^{-1})| = -\frac{1}{8} \]

### Conclusion:
Thus, the value of \(|A^{-1} \text{adj}(B^{-1}) \text{adj}(2A^{-1})|\) is \(-\frac{1}{8}\).