If A and B are square matrices of order 3 such that `|A| = 3 and |B| = 2`, then the value of `|A^(-1) adj (3A^(-1))|` is equal to

Options

A. `27`
B. `27/4`
C. `1/108`
D. `1/4`

Answer

To solve the problem, we need to find the value of the determinant \( |A^{-1} \, \text{adj}(3A^{-1})| \) given that \( |A| = 3 \) and \( |B| = 2 \).

### Step-by-step Solution:

1. **Use the Property of Determinants**:
The determinant of the product of matrices is the product of their determinants:
\[ |A^{-1} \, \text{adj}(3A^{-1})| = |A^{-1}| \cdot |\text{adj}(3A^{-1})| \]

2. **Calculate \( |A^{-1}| \)**:
Using the property of determinants:
\[ |A^{-1}| = \frac{1}{|A|} = \frac{1}{3} \]

3. **Calculate \( |\text{adj}(3A^{-1})| \)**:
We use the property of the adjoint:
\[ |\text{adj}(kA)| = k^{n-1} |A|^n \]
where \( n \) is the order of the matrix. Here, \( n = 3 \) and \( k = 3 \):
\[ |\text{adj}(3A^{-1})| = 3^{3-1} |A^{-1}|^3 = 3^2 \cdot |A^{-1}|^3 \]

4. **Calculate \( |A^{-1}|^3 \)**:
We already found \( |A^{-1}| = \frac{1}{3} \):
\[ |A^{-1}|^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]

5. **Substitute Back**:
Now substituting back into the equation for \( |\text{adj}(3A^{-1})| \):
\[ |\text{adj}(3A^{-1})| = 3^2 \cdot \frac{1}{27} = 9 \cdot \frac{1}{27} = \frac{9}{27} = \frac{1}{3} \]

6. **Combine Results**:
Now we can combine the results:
\[ |A^{-1} \, \text{adj}(3A^{-1})| = |A^{-1}| \cdot |\text{adj}(3A^{-1})| = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} \]

### Final Answer:
Thus, the value of \( |A^{-1} \, \text{adj}(3A^{-1})| \) is \( \frac{1}{9} \).