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 solve the problem, we need to find the value of \( \det(A^3 B^2 C^3) \) given that \( \det(A) = 2 \), \( B = 2A^{-1} \), and \( C = \frac{\text{adj}(A)}{\sqrt[3]{16}} \).

### Step-by-Step Solution:

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

2. **Calculate \( \det(C) \)**:
\[ C = \frac{\text{adj}(A)}{\sqrt[3]{16}} \]
The determinant of the adjugate of a matrix is given by:
\[ \det(\text{adj}(A)) = \det(A)^{n-1} \quad \text{(where \( n \) is the order of the matrix)} \]
For a \( 3 \times 3 \) matrix:
\[ \det(\text{adj}(A)) = \det(A)^{3-1} = \det(A)^2 = 2^2 = 4 \]
Therefore,
\[ \det(C) = \frac{\det(\text{adj}(A))}{\sqrt[3]{16}} = \frac{4}{\sqrt[3]{16}} = \frac{4}{2^{4/3}} = 4 \cdot 2^{-4/3} = 2^{2} \cdot 2^{-4/3} = 2^{2 - 4/3} = 2^{6/3 - 4/3} = 2^{2/3} \]

3. **Calculate \( \det(A^3) \)**:
\[ \det(A^3) = (\det(A))^3 = 2^3 = 8 \]

4. **Combine the determinants**:
Now we can find \( \det(A^3 B^2 C^3) \):
\[ \det(A^3 B^2 C^3) = \det(A^3) \cdot \det(B^2) \cdot \det(C^3) \]
First, calculate \( \det(B^2) \):
\[ \det(B^2) = (\det(B))^2 = 4^2 = 16 \]
Next, calculate \( \det(C^3) \):
\[ \det(C^3) = (\det(C))^3 = (2^{2/3})^3 = 2^{2} = 4 \]

5. **Final calculation**:
Now, substituting back into the determinant product:
\[ \det(A^3 B^2 C^3) = 8 \cdot 16 \cdot 4 \]
Calculate:
\[ 8 \cdot 16 = 128 \]
Then,
\[ 128 \cdot 4 = 512 \]

Thus, the final answer is:
\[ \boxed{512} \]