1. Once the pivot is selected, focus on the "pivot column". This is the column that contains the pivot you have already designated:

\mat2{\[ 2 , 3 , 4 , 1 \]! 0 , 2 , 4 , 1! 0 , -6 , 1 , 2}

\mat2{\[ 2 , 4 , 0 , 4 \]! 0 , 5 , 2 , 1! 0 , 0 , 1 , 2}

2. To clear the pivot column, you will need to get zeros in place of all the blue-colored entries by changing their rows. Next to each row you want to change -- that is, next to each row with a blue entry that is not already zero -- write the name of that row on the left, and the pivot row (the row that contains the pivot) on the right:

\mat2{\[ 3 , 4 , 1 \]! , , ! -8 , 1 , 2, R_2 R_1}

\mat2{\[ 2 , 3 , 4 , 1 \], R_1 R_2! 0 , 2 , 4 , 1! 0 , -6 , 1 , 2 , R_3 R_2}

\mat2{\[ 2 , 4 , 0 , 4 \]! 0 , 5 , 2 , 1 , R_2 R_3! 0 , 0 , 1 , 2}

3. Next to the blue row label, write the absolute value of the red number, and next to the red row label, write the absolute value of the blue number:

\mat2{\[ 3 , 4 , 1 \]! , , ! -8 , 1 , 2, 3R_2 8R_1}

\mat2{\[ 2 , 3 , 4 , 1 \], 2R_1 3R_2! 0 , 2 , 4 , 1! 0 , -6 , 1 , 2 , 2R_3 6R_2}

\mat2{\[ 2 , 4 , 0 , 4 \]! 0 , 5 , 2 , 1 , 1R_2 2R_3! 0 , 0 , 1 , 2}

4. If the blue and red entries have the same sign, insert a minus (−), but if they have different signs, insert a plus (+):

\mat2{\[ 3 , 4 , 1 \]! , , ! -8 , 1 , 2, 3R_2 + 8R_1}

\mat2{\[ 2 , 3 , 4 , 1 \], 2R_1 - 3R_2! 0 , 2 , 4 , 1! 0 , -6 , 1 , 2 , 2R_3 + 6R_2^*}

\mat2{\[ 2 , 4 , 0 , 4 \]! 0 , 5 , 2 , 1 , R_2 - 2R_3! 0 , 0 , 1 , 2}

The result of doing these operations will be to clear the pivot column. (Try them!)

^*NOTE We can (and should!) use smaller numbers when the coefficients have a common factor. For instance, instead of using 2R_3 + 6R_2 as we did in the second matrix, it is better to use R_3 + 3R_2 (divide the original coefficients 2 and 6 by the common factor 2). Similarly, 8R_3 - 12R_2 could be replaced by 2R_3 - 3R_2 (divide by the common factor 4). The result of using the smaller coefficients will be a matrix with smaller entries.