Gaussjordan elimination method

With the concept of matrices one can solve the system of linear equations apart from usual algebraic methods. Even in matrices topic, a system there are different methods to solve and one such method is called as Gauss Jordan elimination method named after the two mathematicians who introduced this elimination method. It is also called in short as Gauss elimination method or as Jordan elimination.
Let us first study the concept behind the Gauss elimination method. A system of equations is expressed in the matrix form as, AX = B, where A is the matrix of the coefficients, X is the matrix of the variables and B is the matrix of the constants on the right side of the system of equations. If the augmented matrix [A¦B] can be expressed in reduced row echelon form, by the method of Gauss-Jordan, that is, in the form [I¦C] (I being the identity matrix), then X = C, which gives us the solution to the system.
Now let us do Gauss-Jordan elimination step by step, with an example for clear understanding. Let a system of equations is,


-3x + 4y + z = -5
x +3y -2z = -6
2x –y + 3z = 9

   Then,A is,

-3	4	1
1	3	-2
2	-1	3

x

y

z

X is, 

-5

-6

9

B is

The following are the steps of elimination.

-3	4	1	-5
1	3	-2	-6
2	-1	3	9

[A│B]

1	-4/3	-1/3	5/3
1	3	-2	-6
2	-1	3	9

[R₁│-3]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
2	-1	3	9

[R₂│-R₁]

1	-4/3	-1/3	5/3
0	13/3	-5/3	-23/3
0	5/3	11/3	17/3

[R₃-2R₁]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	5/3	11/3	17/3

[R₂/(13/3]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	168/39	336/39

[R₃│-(5/3)R₂]

1	-4/3	-1/3	5/3
0	1	-5/13	-23/13
0	0	1	2

[R₃/(168/39)]

1	-4/3	0	7/3
0	1	-5/13	-23/13
0	0	1	2

[R₁+(R₃/3)]

1	-4/3	0	7/3
0	1	0	-1
0	0	1	2

[R₂+(5R₃/13)]

1	0	0	1
0	1	0	-1
0	0	1	2

[R₁+(4R₂/3)]

Now the finally reduced matrix is in the form [I│C], where,

1

-1

2

           C and therefore X = 

Hence, the solution to the given system of equations is, x = 1, y = -1 and z = 2. 

Thus, we had seen the used of the elimination method introduced by Gauss and Jordon.