If A is a square matrix of order `n xx n` then adj(adj A) is equal to

Options

A. `|A|^(n)A`
B. `|A|^(n-1)A`
C. `|A|^(n-2)A`
D. `|A|^(n-3)A`

Explanation

To solve the problem, we need to find the expression for \( \text{adj}(\text{adj} A) \) when \( A \) is a square matrix of order \( n \times n \).

### Step-by-Step Solution:

1. **Understanding the Adjoint**:
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 adjoint has the property that:
\[ A \cdot \text{adj} A = \det(A) I_n \]
where \( I_n \) is the identity matrix of order \( n \).

2. **Applying the Adjoint to the Adjoint**:
We need to find \( \text{adj}(\text{adj} A) \). Using the property of adjoints, we have:
\[ \text{adj}(\text{adj} A) = \det(\text{adj} A) I_n \]

3. **Finding \( \det(\text{adj} A) \)**:
For any \( n \times n \) matrix \( A \):
\[ \det(\text{adj} A) = \det(A)^{n-1} \]
This is a key property of the adjoint.

4. **Substituting Back**:
Now substituting this result back into our expression for \( \text{adj}(\text{adj} A) \):
\[ \text{adj}(\text{adj} A) = \det(\text{adj} A) I_n = \det(A)^{n-1} I_n \]

5. **Final Result**:
Therefore, we conclude that:
\[ \text{adj}(\text{adj} A) = \det(A)^{n-1} I_n \]

### Summary:
The final answer is:
\[ \text{adj}(\text{adj} A) = \det(A)^{n-1} I_n \]