Let A be a square matrix of order 3 such that transpose of inverse of A is A itself then |adj (adjA)| is equal to

Options

A. 9
B. 27
C. 4
D. 1

Answer

To solve the problem, we need to find the value of |adj(adj A)| given that A is a square matrix of order 3 and that the transpose of the inverse of A is equal to A itself.

### Step-by-step Solution:

1. **Understanding the Condition**:
We are given that \( A^{-T} = A \). This implies that \( A \) is an orthogonal matrix because the transpose of the inverse equals the matrix itself.

2. **Finding the Determinant**:
For an orthogonal matrix, the determinant is either \( 1 \) or \( -1 \). Therefore, we can write:
\[ |A|^2 = 1 \implies |A| = 1 \text{ or } -1 \]
Since we are interested in the absolute value, we have \( |A| = 1 \).

3. **Using the Property of Adjoint**:
The determinant of the adjoint of a matrix \( A \) is given by:
\[ |adj(A)| = |A|^{n-1} \]
where \( n \) is the order of the matrix. Here, \( n = 3 \), so:
\[ |adj(A)| = |A|^{3-1} = |A|^2 \]
Since \( |A| = 1 \):
\[ |adj(A)| = 1^2 = 1 \]

4. **Finding the Determinant of the Adjoint of the Adjoint**:
We need to find \( |adj(adj(A))| \). Using the property again:
\[ |adj(adj(A))| = |adj(A)|^{n-1} \]
Here, \( n = 3 \) again, so:
\[ |adj(adj(A))| = |adj(A)|^{3-1} = |adj(A)|^2 \]
Since we found \( |adj(A)| = 1 \):
\[ |adj(adj(A))| = 1^2 = 1 \]

5. **Final Result**:
Therefore, the value of \( |adj(adj A)| \) is:
\[ \boxed{1} \]