Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, find the pivot positions and pivot columns.

Step 1 ：Step 1: We start with the leftmost non-zero column, which is the first column in this case. The pivot position in this column is the first entry, i.e., 1.

Step 2 ：Step 2: To create zeros below this pivot, we can subtract 4 times the first row from the second row and 7 times the first row from the third row, giving us: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix}$

Step 3 ：Step 3: The next pivot position is the second entry in the second column (since the first entry in this column is now zero). To create zeros below this pivot, we can add 2 times the second row to the third row: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & 0 & 0 \end{bmatrix}$

Step 4 ：Step 4: The final pivot position is the third entry in the third column. So, the pivot positions are (1, 1), (2, 2), and (3, 3).

Step 5 ：Step 5: The corresponding pivot columns are the first, second, and third columns of the original matrix.