If A is a square matrix of order 3 such that `abs(A)=2,` then
`abs((adjA^(-1))^(-1))` is

Options

A. 1
B. 2
C. 4
D. 8

Explanation

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

### Step-by-step Solution:

1. **Understanding the Problem**:
We know that \( A \) is a square matrix of order 3, which means \( n = 3 \). We are given that \( \text{abs}(A) = 2 \).

2. **Using the Properties of Determinants**:
We need to find \( \text{abs}((\text{adj} A^{-1})^{-1}) \). We can use the property of determinants:
\[ \text{abs}(A^{-1}) = \frac{1}{\text{abs}(A)} \]
Therefore,
\[ \text{abs}(A^{-1}) = \frac{1}{2} \]

3. **Finding the Determinant of the Adjoint**:
The determinant of the adjoint of a matrix \( A \) is given by:
\[ \text{abs}(\text{adj} A) = \text{abs}(A)^{n-1} \]
For our matrix \( A \) of order 3, this becomes:
\[ \text{abs}(\text{adj} A) = \text{abs}(A)^{3-1} = \text{abs}(A)^2 = 2^2 = 4 \]

4. **Finding the Determinant of the Adjoint of the Inverse**:
Now, we need to find \( \text{abs}(\text{adj} A^{-1}) \). Using the property of the adjoint:
\[ \text{abs}(\text{adj} A^{-1}) = \text{abs}(A^{-1})^{n-1} \]
Substituting \( \text{abs}(A^{-1}) = \frac{1}{2} \):
\[ \text{abs}(\text{adj} A^{-1}) = \left(\frac{1}{2}\right)^{3-1} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

5. **Finding the Determinant of the Inverse of the Adjoint**:
We now need \( \text{abs}((\text{adj} A^{-1})^{-1}) \):
\[ \text{abs}((\text{adj} A^{-1})^{-1}) = \frac{1}{\text{abs}(\text{adj} A^{-1})} \]
Substituting \( \text{abs}(\text{adj} A^{-1}) = \frac{1}{4} \):
\[ \text{abs}((\text{adj} A^{-1})^{-1}) = \frac{1}{\frac{1}{4}} = 4 \]

### Final Answer:
Thus, the value of \( \text{abs}((\text{adj} A^{-1})^{-1}) \) is \( 4 \).