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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): 2R2 + R1
3×4 | x | y | z | = c |
R1: | 1 | 0 | 9 | 19 |
R2: | 0 | 1 | 3 | 5 |
R3: | 0 | 0 | 1 | 2 |
Matrix (Iteration 2):
−9R3 + R1 then −3R3 + R2
3×4 | x | y | z | = c |
R1: | 1 | 0 | 0 | 1 |
R2: | 0 | 1 | 0 | −1 |
R3: | 0 | 0 | 1 | 2 |
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 = 1 − 2(−1) + 3(2) = 9
−x + 3y = −1 + 3(−1) = −4
2x − 5y + 5z = 2(1) − 5(−1) + 5(2) = 17
The system of equations is now solved by Gauss-Jordan Elimination.
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