1.2 Row Reduction & Echelon Forms
Definitions
►A matrix is said to be in row echelon
form if it satisfies the following:
1. All non-zero rows are above all rows
of all zeros.
2. Each leading entry of a row is in a
column to the right of the leading entry of
the row above it.
3. All entries in a column below the
leading entry are zeros.
►A matrix is said to be in reduced
row echelon form if it satisfies the
following additional properties:
4. The leading entry in each non zero
row is a 1.
5. Each leading 1 is the only non zero
entry in its column.
To put a matrix into echelon form,
start with the first leading entry, and
use it to clear the column. This entry
is called a pivot.
 0 3 6 4 9
 1  2 1 3 1 


 2  3 0 3 - 1 


 1 4 5  9  7
R1 ↔ R4
 1 4 5 - 1 - 7
  1  2  1 3 1


 2  3 0 3 - 1 


 0  3  6 4 9
R2+R1→R2
R3+2R1→R3






5 -9 -7 
4 - 6 - 6 
10 - 15 - 15

0 3 6 4
9
1
0
0
4
2
5
½R2 (optional)






5 -9 -7 

2 -3 -3 
10 - 15 - 15

0 3 6 4
9
1
0
0
4
1
5
R3+(-5R2) →R3
R4+3R2→R4






1
0
0
4
1
0
0
0
5 - 9 - 7

2 - 3 - 3
0 0 0

0 - 5 0
R3 ↔ R4






1
0
0
4
1
0
0
0
5 - 9 - 7

2 - 3 - 3
0 -5 0

0 0 0
Note: There are only 3 leading entries.
The pivot columns are 1, 2, and 4.
Note the difference between this and
the Example on p16-17.






1
0
0
4
2
0
0
0
5 - 9 - 7
4 - 6 - 6
0 -5 0

0 0 0
►Theorem 1.1: Each matrix is
equivalent to one and only one reduced
row echelon matrix.
Find the reduced row echelon matrix:
(-1/5)R3






1
0
0
4
1
0
0
0
5 - 9 - 7

2 - 3 - 3
0 1 0

0 0 0
R1 + 9R3 →R1
R2 + 3R3 →R2






1
0
0
4
1
0
0
0
5 0 - 7

2 0 - 3
0 1 0

0 0 0
Clear Column 2:
R1 + (-4R2) →R1






1
0
0
0
0 - 3 0 - 7

1 2 0 - 3
0 0 1 0

0 0 0 0
The original system was:
 3x2  6 x3  4 x4  9
 x1  2 x2  x3  3x4  1
 2 x1  3x2
 3x4  1
x1  4 x2  5 x3  9 x4  7
The solution is
x1
 x3
 7
x2  2 x3
 3
x4  0
x3 is called a free variable.
x1, x2, and x4 are called basic variables.
Definitioin: Any variable that
corresponds to a pivot column is a
basic variable.
Note: Use only the reduced echelon
form to solve a system.
Theorem 1.2
1. A linear system is consistent if and
only if the rightmost column of the
augmented matrix is not a pivot
column, i.e. an echelon form of the
augmented matrix has no row of the
form
[ 0 … 0 b]
2. If a linear system is consistent,
then the solution contains either
a) a unique solution (when there are no
free variables)
b) infinitely many solutions (when
there is at least one free variable).