To solve this problem, we need to determine the value of the determinant of the transpose of the given matrix \( A \) (i.e., \( |A^T| \)). First, let's define and construct the matrix \( A \) using the provided rules:
| Condition | \( a_{ij} \) Value |
| \( i = j \) | \( i - 2j \) |
| \( i > j \) | 0 |
| \( i < j \) | 1 |
Let's construct the matrix \( A \) for order \( 3 \times 3 \):
| \[ A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \] | \[ = \begin{bmatrix} 1 - 2 \times 1 & 1 & 1 \\ 0 & 2 - 2 \times 2 & 1 \\ 0 & 0 & 3 - 2 \times 3 \end{bmatrix} \] | \[ = \begin{bmatrix} -1 & 1 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & -3 \end{bmatrix} \] |
Next, let's transpose the matrix \( A \), denoted as \( A^T \):
| \[ A^T = \begin{bmatrix} -1 & 0 & 0 \\ 1 & -2 & 0 \\ 1 & 1 & -3 \end{bmatrix} \] |
Now, we find the determinant of \( A^T \), denoted as \(|A^T|\):
Using cofactor expansion along the first row, we have:
| \[ |A^T| = \left(-1\right)\begin{vmatrix} -2 & 0 \\ 1 & -3 \end{vmatrix} - 0 + 0 \] |
Calculate the determinant of the 2x2 matrix:
| \[ \begin{vmatrix} -2 & 0 \\ 1 & -3 \end{vmatrix} = (-2)(-3) - (0)(1) = 6 - 0 = 6 \] |
Substituting back, we get:
| \[ |A^T| = -1 \times 6 = -6 \] |
Thus, the determinant of \( A^T \) is \( -6 \). Therefore, the correct option is \( -6 \).