3.2. SYSTEMS OF EQUATIONS, ALGEBRAIC PROCEDURES for the next step
0 0 0 0 0
1 0 0 0 0
1 2 0 0 0
4 3 −1 0 0
3 2 −2 0 −3
35
.
Now the algorithm will be applied to the matrix, µ ¶ 0 −3 There is only one column and it is nonzero so this single column is the pivot column. Therefore, the algorithm yields the following matrix for the echelon form. 0 1 1 4 3 0 0 2 3 2 0 0 0 −1 −2 . 0 0 0 0 −3 0 0 0 0 0 To complete placing the matrix in reduced echelon form, multiply the third row by 3 and add −2 times the fourth row to it. This yields 0 1 1 4 3 0 0 2 3 2 0 0 0 −3 0 0 0 0 0 −3 0 0 0 0 0 Next multiply the second row by 3 and take 2 times the fourth row and add to it. Then add the fourth row to the first. 0 1 1 4 0 0 0 6 9 0 0 0 0 −3 0 . 0 0 0 0 −3 0 0 0 0 0 Next work on the fourth column in the same way. 0 3 3 0 0 0 0 6 0 0 0 0 0 −3 0 0 0 0 0 −3 0 0 0 0 0 Take −1/2 times the second row and 0 0 0 0 0
add to the first. 3 0 0 0 0
0 6 0 0 0
0 0 0 0 −3 0 0 −3 0 0
.