Let A be a matrix of order ` 3 xx 3 `such that det ( A)= 2 , `B = 2A^(-1)` and `C = (( adjA))/(root(3)(16))`,then the value of `det(A^(3) B^(2) C^(3))` is

Answer

To find the value of \( \det(A^3 B^2 C^3) \), we will use the properties of determinants and the definitions of matrices \( B \) and \( C \) in terms of matrix \( A \).

Given:
- \( \det(A) = 2 \)
- \( B = 2A^{-1} \)
- \( C = \frac{\text{adj}(A)}{\sqrt[3]{16}} \)

### Step 1: Calculate \( \det(B) \)

Using the property of determinants, we have:
\[ \det(B) = \det(2A^{-1}) = 2^3 \det(A^{-1}) = 8 \cdot \frac{1}{\det(A)} = 8 \cdot \frac{1}{2} = 4 \]

### Step 2: Calculate \( \det(C) \)

Using the property of determinants for the adjugate matrix:
\[ \det(\text{adj}(A)) = \det(A)^{n-1} \quad \text{where } n \text{ is the order of the matrix} \]
Since \( A \) is a \( 3 \times 3 \) matrix, we have:
\[ \det(\text{adj}(A)) = \det(A)^{3-1} = \det(A)^2 = 2^2 = 4 \]
Thus, we can calculate \( \det(C) \):
\[ \det(C) = \det\left(\frac{\text{adj}(A)}{\sqrt[3]{16}}\right) = \frac{1}{\sqrt[3]{16^3}} \det(\text{adj}(A)) = \frac{1}{16} \cdot 4 = \frac{4}{16} = \frac{1}{4} \]

### Step 3: Calculate \( \det(A^3) \)

Using the property of determinants:
\[ \det(A^3) = (\det(A))^3 = (2)^3 = 8 \]

### Step 4: Combine the determinants

Now we can compute \( \det(A^3 B^2 C^3) \):
\[ \det(A^3 B^2 C^3) = \det(A^3) \cdot \det(B^2) \cdot \det(C^3) \]

Calculating \( \det(B^2) \):
\[ \det(B^2) = (\det(B))^2 = 4^2 = 16 \]

Calculating \( \det(C^3) \):
\[ \det(C^3) = (\det(C))^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \]

### Final Calculation

Now substituting back:
\[ \det(A^3 B^2 C^3) = 8 \cdot 16 \cdot \frac{1}{64} \]
Calculating this:
\[ = \frac{8 \cdot 16}{64} = \frac{128}{64} = 2 \]

Thus, the value of \( \det(A^3 B^2 C^3) \) is \( \boxed{2} \).