If A is a square matrix of order 3 such that `det(A)=(1)/(sqrt(3))` then `det(Adj(-2A^(-2)))` is equal to

Options

A. 24
B. 16
C. 576
D. 144

Explanation

To solve the problem, we need to find the value of \( \det(\text{Adj}(-2A^{-2})) \) given that \( \det(A) = \frac{1}{\sqrt{3}} \) and \( A \) is a square matrix of order 3.

### Step-by-Step Solution:

1. **Understanding the Determinant of the Adjoint**:
The determinant of the adjoint of a matrix \( B \) is given by the formula:
\[ \det(\text{Adj}(B)) = (\det(B))^{n-1} \]
where \( n \) is the order of the matrix. Since \( A \) is a \( 3 \times 3 \) matrix, we have \( n = 3 \).

2. **Finding the Determinant of \( -2A^{-2} \)**:
We first need to find \( \det(-2A^{-2}) \):
\[ \det(-2A^{-2}) = \det(-2I) \cdot \det(A^{-2}) \]
where \( I \) is the identity matrix. The determinant of \( -2I \) for a \( 3 \times 3 \) matrix is:
\[ \det(-2I) = (-2)^3 = -8 \]

3. **Calculating \( \det(A^{-2}) \)**:
We know that:
\[ \det(A^{-2}) = \det(A^{-1})^2 = \left(\frac{1}{\det(A)}\right)^2 \]
Given \( \det(A) = \frac{1}{\sqrt{3}} \), we find:
\[ \det(A^{-1}) = \sqrt{3} \]
Therefore:
\[ \det(A^{-2}) = (\sqrt{3})^2 = 3 \]

4. **Combining the Determinants**:
Now we can combine the results:
\[ \det(-2A^{-2}) = -8 \cdot 3 = -24 \]

5. **Finding the Determinant of the Adjoint**:
Using the formula for the determinant of the adjoint:
\[ \det(\text{Adj}(-2A^{-2})) = (\det(-2A^{-2}))^{3-1} = (-24)^{2} \]
Calculating this gives:
\[ (-24)^2 = 576 \]

### Final Answer:
Thus, the value of \( \det(\text{Adj}(-2A^{-2})) \) is \( \boxed{576} \).