1.2.2 Gauss-Jordan Elimination
In the method of Gaussian elimination, starting from a system A x = b of the general
form
a11
a21

:

:
an1
a12 a13 ...
a22 a23 ...
:
:
:
:
an2 an3 ...
a1n 
a2n 
: 

: 
ann 
 x1   b1 
 x2   b2 
   
:  = : 
   
:  : 
 xN  bn 
is converted to an equivalent system A’ x = b ’ after
(1.2.2-1)
2 3
N FLOP’s that is of upper
3
triangular form
a





'
11
a
'
12
a '22
'
13
...
a '23
...
'
33
...
a
a


a '2N 

'
a 3N

'
a NN 
a
'
1N
'
 x1   b 1 
 x2] b ' 
   2
:  = b 3' 
   
:  : 
 xN  b ' 
 N 
(1.2.2-2)
At this point, it is possible, through backward substitution, to solve for the unknowns in
N2
the order xN, xN-1, xN-2, … in
steps.
2
In the method of Gauss-Jordan elimination, one continues the work of elimination,
placing zeros above the diagonal.
To “zero” the element at (N-1, N), we write the last two equations of (1.2.2-2)
a 'N −1, N −1 x N −1 + a 'N −1, N x N = b 'N −1 

a 'N, N x N = b 'N

(1.2.2-3)
We then define λ N −1, N =
a 'N −1, N
(1.2.2-4)
a 'NN
And replace the N-1st row with the equation obtained after performing the row operation
(a 'N −1, N −1 x N −1 + a 'N −1, N x N = b 'N −1 )
− λ N −1, N (a 'NN x N = b 'N )
(1.2.2-5)
a 'N −1, N −1 x N −1 + (a 'N −1, N − λ N −1, N a 'NN )x N = b 'N −1 − b 'N λ N −1, N
Defining
b 'N' −1 = b 'N −1 − b 'N λ N −1, N
(1.2.2-6)
and noting
a
''
N −1
=a
'
N −1, N
− λ N −1, N a
'
NN
=a
'
N -1, N
−(
a 'N −1, N
a
'
NN
)a 'NN = 0
(1.2.2-7)
After this row operation the set of equations becomes
'
a 11









'
a 12
'
a 13
...
a 1,' N-1
a '22
a '23
...
a '2, N-1
'
a 33
...
a 3,' N-1
:
a 'N' -1, N-1
'
'
 x1
a 1.N

  b1 



a '2, N   x2  b '2 


a 3,' N   x 3  b 3' 

=
:  :  : 


  '' 
0   x N -1  b N -1 
 x
 ' 
a 'NN   N  b N 
(1.2.2-2)
We can continue this process until the set of equations is in diagonal form

a


'''

 a 22


'''
a 33


''' 

a NN 
'''
11
'''
 x1   b 1 
 x2   b ' ' ' 
   2
:  = b 3' ' ' 
   
:  : 
 xN  b ' ' ' 
 N 
(1.2.2-9)
Dividing each equation by the value of its single coefficient yields
 b1' ' '
 '''
 a 11
1  x 1   b '2' '
1  x   a ' ' '
   2   22
1  x 3  =  b 3' ' '
    '''
:  :   a 33
1  x N  :

 b'''
 ' 'N'
 a NN














(1.2.2-10)
The matrix on the left that has a one everywhere along the principal diagonal and zeros
everywhere else is called the identity matrix, and has the property that for any vector v,
Iv = v
(1.2.2-11)
The form (1.2.2-10) therefore immediately gives the solution to the problem.
In practice, we use Gaussian Elimination, stopping at (1.2.2-2) to begin backward
substitution rather than continue the elimination process because backward substitution is
so fast, N2 << 2N3/3 for all but small problems.
We therefore do not consider the method of Gauss-Jordan Elimination further.