Find the pivot positions and pivot columns of the following matrix, \[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]

Step 1 ：First we will reduce the given matrix to row-echelon form. The first entry is already 1, so we use it to clear out the entries below it by subtracting \(4\times\) the first row from the second row, and subtracting \(7\times\) the first row from the third row. This gives us \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \\ 0 & -6 & -12 \end{bmatrix} \]

Step 2 ：Next, we divide the second row by -3 to make the leading entry 1, giving us \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -6 & -12 \end{bmatrix} \]. Now we can clear out the entry below in the second column by adding \(6\times\) the second row to the third row. This gives us \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} \]

Step 3 ：This is now in row-echelon form. The pivot positions are the first non-zero entries in each row, so they are the (1,1) and (2,2) positions. These correspond to the first and second columns.