A First Course in Linear Algebra, 2017a

20 Systems of Equations
2. Step two involves creating zeros in the entries below the first pivot position. The first entry of the
second row is already a zero. All we need to do is subtract 5 times the first row from the third row.
The resulting matrix is
⎡
⎤
1 4 3
⎣ 0 −5 −4 ⎦
0 −10 −8
3. Now ignore the top row. Apply steps 1 and 2 to the smaller matrix
[ ]
−5 −4
−10
−8
In this matrix, the first column is a pivot column, and −5 is in the first pivot position. Therefore, we
need to create a zero below it. To do this, add −2 times the first row (of this matrix) to the second.
The resulting matrix is [ −5
] −4
0 0
Our original matrix now looks like
⎡
⎣
1 4 3
0 −5 −4
0 0 0
We can see that there are no more rows to modify.
4. Now, we need to create leading 1s in each row. The first row already has a leading 1 so no work is
needed here. Divide the second row by −5 to create a leading 1. The resulting matrix is
⎡
1 4
⎤
3
⎣ 0 1 4 5
⎦
0 0 0
This matrix is now in row-echelon form.
5. Now create zeros in the entries above pivot positions in each column, in order to carry this matrix
all the way to reduced row-echelon form. Notice that there is no pivot position in the third column
so we do not need to create any zeros in this column! The column in which we need to create zeros
is the second. To do so, subtract 4 times the second row from the first row. The resulting matrix is
⎡ ⎤
⎢
⎣
1 0 − 1 5
0 1
4
5
0 0 0
⎤
⎦
⎥
⎦
This matrix is now in reduced row-echelon form.
♠
The above algorithm gives you a simple way to obtain the row-echelon form and reduced row-echelon
form of a matrix. The main idea is to do row operations in such a way as to end up with a matrix in
row-echelon form or reduced row-echelon form. This process is important because the resulting matrix
will allow you to describe the solutions to the corresponding linear system of equations in a meaningful
way.