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

Options

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

Answer

To solve the problem of finding the value of \(|A^{-1} \, \text{adj}(B^{-1}) \, \text{adj}(3A^{-1})|\), we will use properties of determinants. Here are the steps:

### Step 1: Write the expression using properties of determinants
We start with the expression:
\[ |A^{-1} \, \text{adj}(B^{-1}) \, \text{adj}(3A^{-1})| \]
Using the property of determinants, we can separate this into:
\[ |A^{-1}| \cdot |\text{adj}(B^{-1})| \cdot |\text{adj}(3A^{-1})| \]

### Step 2: Calculate \(|A^{-1}|\)
Using the property of determinants:
\[ |A^{-1}| = \frac{1}{|A|} \]
Given that \(|A| = 3\), we have:
\[ |A^{-1}| = \frac{1}{3} \]

### Step 3: Calculate \(|\text{adj}(B^{-1})|\)
Using the property of the adjoint:
\[ |\text{adj}(B^{-1})| = |B^{-1}|^{n-1} \]
where \(n\) is the order of the matrix (which is 3 here). We know:
\[ |B^{-1}| = \frac{1}{|B|} = \frac{1}{2} \]
Thus,
\[ |\text{adj}(B^{-1})| = \left(\frac{1}{2}\right)^{3-1} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \]

### Step 4: Calculate \(|\text{adj}(3A^{-1})|\)
Using the property of the adjoint again:
\[ |\text{adj}(3A^{-1})| = |3A^{-1}|^{n-1} \]
Now, we calculate \(|3A^{-1}|\):
\[ |3A^{-1}| = 3^n |A^{-1}| = 3^3 \cdot \frac{1}{3} = 27 \]
Thus,
\[ |\text{adj}(3A^{-1})| = |3A^{-1}|^{2} = 27^{2} = 729 \]

### Step 5: Combine all parts
Now we can combine all parts to find the determinant:
\[ |A^{-1} \, \text{adj}(B^{-1}) \, \text{adj}(3A^{-1})| = |A^{-1}| \cdot |\text{adj}(B^{-1})| \cdot |\text{adj}(3A^{-1})| \]
Substituting the values we calculated:
\[ = \frac{1}{3} \cdot \frac{1}{4} \cdot 729 \]
Calculating this gives:
\[ = \frac{729}{12} = \frac{243}{4} \]

### Final Answer
Thus, the value of \(|A^{-1} \, \text{adj}(B^{-1}) \, \text{adj}(3A^{-1})|\) is:
\[ \frac{243}{4} \]