Gauss Jordan Elimination Method

Gauss Jordan Elimination Method is a method to solve large linear equation numerically. It is done by manipulating the given matrix using elementary row operations. It puts zero both above and below each pivot element as it goes from top row of the matrix to the bottom. This method is also known as Back Substitution.

 

Gauss Jordan Elimination Method

Gauss Jordan elimination is an algorithm for getting matrices in reduced row echelon form using elementary row operations. Gauss Jordan is well suited for solving linear systems.
Consider n linear equations to be solved
a₁₁x₁ + a₁₂ x₂ + . ......+ a_1nx_n = b₁
a₂₁x₁ + a₂₂ x₂ + . ......+ a_2nx_n = b_{2
........................................................}
a_n1x₁ + a_n2 x₂ + . ......+ a_nnx_n = b_n
If ⎡⎣⎢⎢⎢⎢⎢⎢a11a21..an1a12a22..an2..........a1na2n..ann∣∣∣∣∣∣∣b1b2..bn⎤⎦⎥⎥⎥⎥⎥⎥ is augmented matrix associated with a linear system, then elementary row operations are used to reduced this matrix to ⎡⎣⎢⎢⎢⎢⎢⎢10..001..0..........0...1∣∣∣∣∣∣∣m1m2..mn⎤⎦⎥⎥⎥⎥⎥⎥

The solution appears in last column, i.e. xi = mi.

Gauss Jordan Elimination Step by Step

In order to solve a system of three linear equations by Gauss Jordan elimination method, elementary row operation has to be performed on the augmented matrix as follow:

Step 1:

• Transform the element at a₁₁ position to 1, by a suitable elementary row transformation using the element at a₂₁ or a₃₁ position or otherwise.
• Transform the non-zero elements , if any , at a₂₁, a₃₁ position as zeros (other elements of the first column) by using the element 1 at a₁₁ position. If at the end of step 1, there is a non-zero element at a₂₂ or a₃₂ position, go to step 2. Otherwise skip it.

Step 2:

• Transform the element at a₂₂ position as 1 by a suitable elementary row transformation using the element at a₃₂ position or otherwise.
• Transform the other non-zero elements, if any, of the second column (i.e., the non-zero elements, if any, at a₁₂ and a₃₂ positions) as zeros, by using the element 1 at a₂₂ position. At the end of step 2 or after skipping it for reasons specified above, examine the element at a₃₃ position.If it is non-zero, go to step3.

Step 3:

• Transform the element at a₃₃ position as 1 by dividing R₃ with a suitable number.
• Transform the other non-zero elements if any of the third column (i.e., the non-zero elements, if any, at a₁₃ , a₂₃ position) as zeros by using the 1 present at a₃₃ position.

Gauss Jordan Elimination Example

Below you could see examples on Gauss Jordan Elimination methods.

Solved Examples

Question 1: Solve the following equations by Gauss Jordan Elimination method.
x + 2y = 3
-x - 2z = -5
-3x - 5y + z = -4
Solution:

Given system of linear equations is x + 2y = 3
-x - 2z = -5
-3x - 5y + z = -4
Augmented matrix is
⎡⎣⎢1−1−320−50−21∣∣∣∣3−5−4⎤⎦⎥

Using operations R₂ ∼ R₂ + R₁ and R₃ ∼ R₃ + 3R₁
⎡⎣⎢1002210−21∣∣∣∣3−25⎤⎦⎥

R₁ ~ R₁ - R₂ and R₂ = R22

⎡⎣⎢1000112−11∣∣∣∣5−15⎤⎦⎥

R₃ ~ R₃ - R₂
⎡⎣⎢1000102−12∣∣∣∣5−16⎤⎦⎥
R₃ ~ R32
⎡⎣⎢1000102−11∣∣∣∣5−13⎤⎦⎥
R₁ ~ R₁ - 2R₃ and R₂ ~ R₂ + R₃
⎡⎣⎢100010001∣∣∣∣−123⎤⎦⎥
Therefore, the solution of the system is x = -1, y = 2 and z = 3.

Question 2: Solve the following equations by Gauss Jordan Elimination method. 3x + 4y + 5z = 18
2x - y + 8z = 13
5x - 2y + 7z = 20

Solution: The augmented Matrix is
⎡⎣⎢3254−1−2587181320⎤⎦⎥
On applying operation R₁ --> R₁ - R₂, we get
⎡⎣⎢1255−1−2−38751320⎤⎦⎥
On applying R₂ -->R₂ - 2R₁,R₃ --> R₃ - 5R₁ ,we get
⎡⎣⎢1005−11−27−3142253−5⎤⎦⎥
On applying R₂ --> -5R₂ + 2R₃, we get
⎡⎣⎢10051−27−3−26225−25−5⎤⎦⎥
On applying R₁ --> R₁ - 5R₂, R₃ --> R₃ + 27R₂ ,we obtain
⎡⎣⎢100010127−26−680130−25−680⎤⎦⎥
On applying R₃ -->R₃/(-680), we get
⎡⎣⎢100010127−261130−251⎤⎦⎥
On applying R₁ --> R₁ - 127R₃, R₂ --> R₂ + 26R_{3 ,}we obtain
⎡⎣⎢100010001311⎤⎦⎥
Hence, the solution is x = 3, y = 1, z = 1.