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Gauss-Jordan Elimination - Solving any systems of equation - calculator


You can calculate with explanations any system of linear equations, both homogeneous and heterogeneous with any number of unknowns by Gauss-Jordan elimination method. As a result, apart from the solution, you will also receive a complete analysis and a step-by-step calculation.
To get the result, enter the coefficients in empty cells or by clicking on , change to the text field and enter the coefficients separated by a space. You can also enter full equations with unknowns in the text field (click the example below).

You can add or subtract another equation and coefficients using the buttons or position the cursor over a cell in the last row or column and press the arrow in the appropriate direction on the keyboard.

You can move around cells using tabs, spaces or arrows on the keyboard.

In each field of the calculator you can enter decimal numbers, simple fractions, basic mathematical operation, for example: 1.5; 1/2; 5 + 2; 5-2; 5 * 2; 1/2 + 5; 5 * (8 + 2); (5-2) / 2; 1.5e-3; 1.5 (3); (2/5) * (1/2); 5 + 2 * (5 / 3-2); etc.

You can also enter data into the calculator from the result obtained, just grab any matrix from the results and drop it on cells or a text field.

The results are stored in cookies, which means that the result will be saved until you click .
Thanks to the stored results, you can compare the actions using other linear equation calculators available on our website, eg.
  • Solving System of linear equations by Cramer's method
  • Solving Systems with Gaussian Elimination calculator
  • System of equations using the inverse matrix method

  • Example 1:
    2x-2y+z=-3
    x+3y-2z=1
    3x-y-z=2
    Example 2:
    2x-0y+z=2
    x-3y-0z=1
    x+y-2z=0
    Example 3:
    2x-2y+z-w=7
    -x+4y-5z+3w=1
    x+y-z+w=4
    -4x+2y+z-2w=6


    System of linear equations by Gauss-Jordan method













    Result

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