Example of Row Reduced Echelon Form
Consider solving the equation






3 −2
0 2
0
6 −5 −2 4 −3
0
5 10 0 15
6
0
8 4 18
1
2
0
2









x1
x2
x2
x4
x5
x6














=
0
−1
5
6





.
To begin with, augmenting the matrix with the right hand side vector. Then apply a
combination of elimination matrices and row swap matrices to transform the problem to row
reduced echelon form. The elimination matrices will always have 1 on the main diagonal.
Such a procdure is more consistent with how the system would be solved on a computer.
Since the number of elimination matrices is not known ahead of time, the subscript on the
outside elimination matrix will have to be determined.



E? · · · E1 A = E? · · · E2 





= E? · · · E3 




= E? · · · E4 


= E? · · · E5 




= E6 






= 
1
0
0
0
1
−2
0
−2
0
1
0
0
0
0
1
0
0
0
0
1





1
0
0
0
0
1
5
4
0
0
1
0
0
0
0
1

1
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0









1
0
0 0
0 −1
0 0
0
0 1/6 0
0
0
0 1
1
0
0
0
0
0 0
1 −3 0
0
1 0
0
0 1
2
1
0
0
0
0
1
0
0
0
0
1










1
0
0
0
1
0
0
0
3 −2
0 2
0
0
6 −5 −2 4 −3 −1 


0
5 10 0 15
5 
6
0
8 4 18
6
1
2
0
2

(1)
1
0
0
0
3 −2
0 2
0
0
0 −1 −2 0 −3 −1 


0
5 10 0 15
5 
0
4
8 0 18
6
(2)
1
0
0
0
3 −2
0 2
0
0
0 −1 −2 0 −3 −1 


0
0
0 0
0
0 
0
0
0 0
6
2
(3)







1
0
0
0
3 −2
0 2
0
0
0 −1 −2 0 −3 −1 


0
0
0 0
6
2 
0
0
0 0
0
0

3 −2 0 2 0
0
0
1 2 0 3
1
0
0 0 0 1 1/3
0
0 0 0 0
0
3 −2 0 2 0
0
0
1 2 0 0
0
0
0 0 0 1 1/3
0
0 0 0 0
0
(4)





(5)





(6)




= 
1
0
0
0
3
0
0
0
0
1
0
0
4
2
0
0
2
0
0
0
0
0
0
0
1 1/3
0
0





(7)
What did multiplication by each elimination matrix do? For an explanation, understand
that the values stored in the A matrix are changing after each operation.
1. E1 eliminated a21 , a31 , a41 in A.
2. E2 eliminated a33 , a43 in E1 A.
3. E3 swapped rows 3 and 4 of E2 E1 A.
4. E4 made the pivot positions of E3 E2 E1 A be 1.
5. E5 eliminated the a26 postition in E4 E3 E2 E1 A.
6. E6 eliminated the a13 postion in E5 E4 E3 E2 E1 A.
The resulting matrix transformation has privots in the a11 , a23 and a36 positions. This
leads to free variables x2 , x4 and x5 . With respect to notating the solution for A~x = ~b in
vector format, start with
x1
x2
x3
x4
x5
x6
= −3x2 −4x4 −2x5
=
x2
=
−2x4
=
x4
=
x5
=
1/3
and then rewrite the equations in vector format as

~x =




x2 




−3
1
0
0
0
0










 + x4 








−4
0
−2
1
0
0










 + x5 








−2
0
0
0
1
0


 
 
 
 
+
 
 
 
 
0
0
0
0
0
1/3










If the variable x2 , x4 and x5 on the right hand side were replaced by r, s and t then the
equation would automatically imply that x2 = r, x4 = s, and x5 = t. Most often the letters
r, s and t would be called free variables. Furthmore, if ~xH and ~xP were define as

~xH =




x2 




−3
1
0
0
0
0










 + x4 








−4
0
−2
1
0
0










 + x5 








−2
0
0
0
1
0




















~xP =
0
0
0
0
0
1/3










then the solution would be define as having two components, one called the homogeneous
solution and the other the particular solution. Note that the format of these solutions could
change if operations such as row swaps at the beginning were performed.