Sec 3.3 Reduced Row-Echelon Matrices
Def:
A matrix A in reduced-row-echelon form if
1) A is row-echelon form
2) All leading entries = 1
3) A column containing a leading entry 1 has 0’s everywhere else
1 5 0 2 
A  0 1 0 1 
0 0 0 0 
1
0

0

0
0
0 0 0 1 0
1 0 0 2 0

0 0 0 0 0

0 0 1 0 0
0 0 0 0 0
1 0 5 2 
B  0 1 0 1 
0 0 0 0 
1
0

0

0
0
0 0 0 1 0
1 0 0 1 0

0 0 1 0 0

0 0 0 0 0
0 0 0 0 0
Gauss-Jordan Elimination
* * * *
* * * *


* * * *
* * * *
* * * *


* * * *
Gaussian
Elimination
Echelon
Gauss-Jordan
Elimination
Re duced 
 Row



 Echelon 
Echelon Matrix  Reduced Echelon Matrix
1) A 
row-echelon form
2) Make All leading entries = 1 (by division)
3) Use each leading 1 to clear out any nonzero elements in its
column
1 5 0 2 
A  0 1 0 1 
0 0 0 0 
1
0

0

0
0
0 0 0 1 0
1 0 0 2 0
0 0 0 0 1

0 0 1 0 0
0 0 0 0 0
1 0 5 2 
B  0 1 0 1 
0 0 0 0 
1 1 2 1 1 3 4
0 0 3 3 0 6 9


0 0 0 0 0 2 4
Leading variables and Free variables
1 1 0 1 1 0 4 
0 0 1 3 0 0 9 


0 0 0 0 0 1 0 
x1
x2
x3
x4
x5
x6
x2  t
Free Variables
x4  s
x5  w
Leading variables and Free variables
Example 3: Use Gauss-Jordan elimination to solve the linear system
x1  x2  x3  x4  12
x1  2 x2
 5 x4  17
3x1  2 x2  4 x3  x4  31
Solution:
1 1 1 1 12
1 2 0 5 17 


3 2 4  1 31
Gauss-Jordan
1 0 2  3 7
0 1  1 4 5


0 0 0
0 0
Reduced Echelon is
Unique
Theorem 1 : Every matrix is row equivalent
to one and only one reduced echelon matrix
NOTE: Every matrix is row equivalent to one
and only one echelon matrix
1 1 1 1 12
1 2 0 5 17 


3 2 4  1 31
Row-equivalent
Row-equivalent
1 0 2  3 7
0 1  1 4 5


0 0 0
0 0
1 1 1 1 12  1 2  6 5 17 
0 1  7 4 5  0 1  7 4 5 


 
0 0 0 0 0  0 0 0 0 0 
1 2  6 5 17 
0 3  13 9 2 


0 0 0 0 0 
What is common
The Three Possibilities
#unknowns =#equs
Example


nn


 *


 *
3x  2 y  5
x  9y 1
#unknowns > #equs
Square systm
Example


nm

unique




3x  2 y  z  w  5
x  9 y  2z  w  1
No sol.

No sol.
Homogeneous System
a11x1  a12 x2    a1n xn  0
Homogeneous
System
a21x1  a22 x2    a2 n xn  0


am1 x1  am 2 x2    amn xn  0
Example
x  2y  z  0
3x  8 y  7 z  0
2x  7 y  9z  0
* *  * 0
     


* *  * 0
NOTE:
Every homog system has at least the trivial solution
x1  0, x2  0, , xn  0
Homogeneous System
NOTE: Every homog system either has only the trivial solution or has
infinitely many solutions
Homog System
1
Unique
Solution
2
Infinitely
many
solutions
3
No
Solution
Special case ( more variables than equations
Theorem: Every homog system with more variables than equations has
Homogeneous System
Theorem: Every homog system with more variables than equations has
infinitely many solutions
Homog
#unknowns =#equs Example


nn


 *


 *
3x  2 y  0
x  9y  0
Homog
#unknowns > #equs
Square systm
Example


nm

unique




3x  2 y  z  w  0
x  9 y  2z  w  0
No sol.

No sol.