# How to Solve a System of Linear Equations by Gauss-Jordan Elimination

Gauss-Jordan Elimination applies elementary row operations to a matrix until a row-echelon form of the matrix is obtained; the same matrix process used for Gaussian Elimination. The reduction process of Gauss-Jordan continues the reduction process beyond Gaussian Elimination until a reduced row-echelon form is attained.

The advantage of using matrices to solve systems of linear equations is that it is a procedural and rule-based process. When the rules are followed exactly, the solving of a complex system of linear equations, equations that each has multiple terms with coefficients, is simplified.

To solve the following system of linear equations using Gauss-Jordan Elimination:

First write the linear equations system to be solved.

Linear Equations System to be Solved

x − 2y + 3z = 9

−x + 3y = −4

2x − 5y + 5z = 17

Next use Gaussian Elimination to obtain the row-echelon form of the linear system.

(Review Gaussian Elimination if you need to see how to transform a system of linear equations to row-echelon form)

Row-Echelon Form Matrix

 3×4 x y z = c R1: 1 −2 3 9 R2: 0 1 3 5 R3: 0 0 1 2

Now apply elementary row operations to obtain zeros above each of the leading 1’s.

Matrix (Iteration 1)

 3×4 x y z = c R1: 1 0 9 19 R2: 0 1 3 5 R3: 0 0 1 2

2R2 + R1

Matrix (Iteration 2)

 3×4 x y z = c R1: 1 0 0 1 R2: 0 1 0 −1 R3: 0 0 1 2

−9R3 + R1

−3R3 + R2

Reduced Row-Echelon Matrix

 3×4 x y z = c R1: 1 0 0 1 R2: 0 1 0 −1 R3: 0 0 1 2

This provides us the values to each variable x, y, and z of the system of linear equations.

1x = 1

1y = −1

1z = 2

x − 2y + 3z = 9

−x + 3y = −4

2x − 5y + 5z = 17

1 − 2(−1) + 3(2) = 9

−1 + 3(−1) = −4

2(1) − 5(−1) + 5(2) = 17

The system of equations is now solved by Gauss-Jordan Elimination.

Top of Page