Let A and B be two square matrices of order 3 such that `|A|=3 and |B|=2`, then the value of
`|A^(-1).adj(B^(-1)).adj(2A^(-1))|` is equal to (where adj(M) represents the adjoint matrix of M)

Options

A. 72
B. `(64)/(27)`
C. `(8)/(9)`
D. `(16)/(27)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Determinants**:
We know that for any square matrix \( M \) of order \( n \):
\[ |\text{adj}(M)| = |M|^{n-1} \]
Here, \( n = 3 \) since \( A \) and \( B \) are \( 3 \times 3 \) matrices.

2. **Calculating \( |A^{-1}| \)**:
The determinant of the inverse of a matrix is given by:
\[ |A^{-1}| = \frac{1}{|A|} \]
Given \( |A| = 3 \), we have:
\[ |A^{-1}| = \frac{1}{3} \]

3. **Calculating \( |\text{adj}(B^{-1})| \)**:
First, we find \( |B^{-1}| \):
\[ |B^{-1}| = \frac{1}{|B|} = \frac{1}{2} \]
Now, applying the adjoint property:
\[ |\text{adj}(B^{-1})| = |B^{-1}|^{3-1} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \]

4. **Calculating \( |\text{adj}(2A^{-1})| \)**:
First, we find \( |2A^{-1}| \):
\[ |2A^{-1}| = 2^3 |A^{-1}| = 8 \cdot \frac{1}{3} = \frac{8}{3} \]
Now, applying the adjoint property:
\[ |\text{adj}(2A^{-1})| = |2A^{-1}|^{3-1} = \left(\frac{8}{3}\right)^{2} = \frac{64}{9} \]

5. **Combining the Determinants**:
Now we can combine these results:
\[ |A^{-1} \cdot \text{adj}(B^{-1}) \cdot \text{adj}(2A^{-1})| = |A^{-1}| \cdot |\text{adj}(B^{-1})| \cdot |\text{adj}(2A^{-1})| \]
Substituting the values we calculated:
\[ = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{64}{9} \]

6. **Calculating the Final Value**:
\[ = \frac{64}{3 \cdot 4 \cdot 9} = \frac{64}{108} = \frac{16}{27} \]

### Final Answer:
Thus, the value of \( |A^{-1} \cdot \text{adj}(B^{-1}) \cdot \text{adj}(2A^{-1})| \) is \( \frac{16}{27} \).