Let A and B are two square matrices of order 3 such that det. `(A)=3` and det. `(B)=2`, then the value of det. `(("adj. "(B^(-1) A^(-1)))^(-1))` is equal to _______ .

Explanation

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

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression \( \text{adj}(B^{-1} A^{-1})^{-1} \). We will use properties of determinants and adjugates to simplify this expression.

2. **Using the Property of Inverses**:
Recall that \( (AB)^{-1} = B^{-1} A^{-1} \). Therefore, we can rewrite our expression:
\[ \text{adj}(B^{-1} A^{-1})^{-1} = \text{adj}((AB)^{-1})^{-1} = \text{adj}(AB) \]

3. **Using the Property of Adjugates**:
We know that \( \text{adj}(A^{-1}) = \text{adj}(A)^{-1} \). Thus, we can apply this property:
\[ \text{adj}((AB)^{-1})^{-1} = \text{adj}(AB) \]

4. **Finding the Determinant of the Adjugate**:
The determinant of the adjugate of a matrix \( M \) is given by:
\[ \text{det}(\text{adj}(M)) = \text{det}(M)^{n-1} \]
where \( n \) is the order of the matrix. Here, \( n = 3 \).

5. **Calculating the Determinant of \( AB \)**:
We can express the determinant of the product of two matrices as:
\[ \text{det}(AB) = \text{det}(A) \cdot \text{det}(B) \]
Given \( \text{det}(A) = 3 \) and \( \text{det}(B) = 2 \), we have:
\[ \text{det}(AB) = 3 \cdot 2 = 6 \]

6. **Applying the Determinant of the Adjugate**:
Now, we can find the determinant of the adjugate:
\[ \text{det}(\text{adj}(AB)) = \text{det}(AB)^{3-1} = 6^{2} = 36 \]

7. **Final Result**:
Thus, the value of \( \text{det}((\text{adj}(B^{-1} A^{-1}))^{-1}) \) is:
\[ \boxed{36} \]