Let A be a square matrix of order n.
Statement - 1 : `abs(adj(adj A))=absA^(n-1)^2`
Statement -2 : `adj(adj A)=absA^(n-2)A`

Options

A. Statement -1 is True, Statement -2 is true, Statement -2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement -2 is True, Statement -2 is not a correct explanation for Statement -1.
C. Statement -1 is True, Statement -2 is False.
D. Statement -1 is False, Statement -2 is True.

Explanation

To solve the problem, we need to verify both statements regarding the adjoint of a square matrix \( A \) of order \( n \).

### Step 1: Understanding the adjoint of a matrix
The adjoint of a matrix \( A \), denoted as \( \text{adj}(A) \), is defined as the transpose of the cofactor matrix of \( A \). For a square matrix \( A \) of order \( n \), the following properties hold:
- \( \text{adj}(A) \cdot A = \det(A) I_n \), where \( I_n \) is the identity matrix of order \( n \).
- \( \det(\text{adj}(A)) = \det(A)^{n-1} \).

### Step 2: Analyzing Statement 1
We need to verify if \( |\text{adj}(\text{adj}(A))| = |\det(A)|^{(n-1)^2} \).

Using the property of determinants:
\[ \det(\text{adj}(A)) = \det(A)^{n-1} \]
Now, applying the adjoint operation again:
\[ \det(\text{adj}(\text{adj}(A))) = \det(\text{adj}(A))^{n-1} = (\det(A)^{n-1})^{n-1} = \det(A)^{(n-1)^2} \]
Thus, we have:
\[ |\text{adj}(\text{adj}(A))| = |\det(A)|^{(n-1)^2} \]
This confirms that Statement 1 is true.

### Step 3: Analyzing Statement 2
Now we need to verify if \( \text{adj}(\text{adj}(A)) = |\det(A)|^{(n-2)} A \).

From the property of adjoints, we know:
\[ \text{adj}(\text{adj}(A)) = \det(A)^{n-2} A \]
This is derived from the fact that the adjoint of the adjoint of a matrix is proportional to the original matrix scaled by the determinant raised to the power of \( n-2 \).

Thus, Statement 2 is also true.

### Conclusion
Both statements are true:
1. \( |\text{adj}(\text{adj}(A))| = |\det(A)|^{(n-1)^2} \)
2. \( \text{adj}(\text{adj}(A)) = |\det(A)|^{(n-2)} A \)