Linear Equation System Solver

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matrix row reduction

Matrix row reduction with every row operation visible

Row-reducing a matrix means applying a sequence of elementary row operations (swapping two rows, multiplying a row by a nonzero number, or adding a multiple of one row to another) until you reach a simpler form to solve, the reduced row echelon form (RREF). This page opens a sample 3x3 system, large enough to show a row swap from partial pivoting.

Why the row order changes during elimination

This calculator uses partial pivoting: in each column, before eliminating, it picks as the pivot the row with the largest absolute value in that column among the rows still available, and swaps that row into the pivot position if needed. That avoids two practical problems: dividing by a pivot equal to zero (which would halt the algorithm) and dividing by a pivot very close to zero, which amplifies rounding error in floating point. Since this tool uses exact fractions instead of floating point, the second problem does not exist here, but pivoting is still needed for the first: without it, a column with a zero in the pivot's natural position would halt the elimination.
Every row of the reduced row echelon form has a pivot equal to 1, and that pivot is the only nonzero value in its column, both above and below it. Once the augmented matrix reaches that form, the system's solution can be read directly off the last column, with no further calculation, which is exactly why the reduced row echelon form is the destination of the process, not just an intermediate step.

Row-reducing a 3x3 system

Initial augmented matrix

Input: [1 1 1 | 6] [0 2 5 | -4] [2 5 -1 | 27]
Expected output: [1 0 0 | 5] [0 1 0 | 3] [0 0 1 | -2]

After full row reduction, the solution (x=5, y=3, z=-2) shows up directly in the last column.

Row swap from partial pivoting

Input: [1 1 1 | 6] [0 2 5 | -4] [2 5 -1 | 27]
Expected output: L1 <-> L3

The third row has the largest absolute value in the first column (2), so it becomes the new first row before eliminating the rest, avoiding a small pivot.

Full tool FAQ

How should I type the equations?

One equation per line, with an equals sign separating the two sides, for example "2x + 3y = 8" or "x - y = -1". The calculator accepts an implicit coefficient ("x" means the same as "1x"), a negative sign, unknowns with any name, a variable appearing on both sides of the equals sign ("2x + 1 = x + 4"), and a variable that simply doesn't appear in one of the equations, it is treated as a zero coefficient automatically.

What's the difference between the Gaussian elimination and Cramer's rule shown here?

What does a "consistent and dependent" system mean?

How do I know a system is inconsistent, rather than that I typed something wrong?

Why does the answer show up as a fraction instead of a decimal number?

How many unknowns can a system have?

What does the "ill-conditioned" system warning mean?

Is my data saved on any server?

Frequently asked questions

What is the difference between row echelon form and reduced row echelon form?

In row echelon form, each pivot just needs to be nonzero and the rows below it are zeroed out beneath it. In reduced row echelon form (RREF), on top of that, each pivot is exactly 1 and is the only nonzero value in its entire column, including the rows above it. This calculator always produces the reduced form, because it lets you read the solution off directly.

Why do rows sometimes swap position during elimination?

Because of partial pivoting: the row with the largest absolute value in that column is moved into the pivot position, which avoids halting the algorithm when the natural value in that spot is zero.

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matrix row reduction

Matrix row reduction with every row operation visible

Why the row order changes during elimination

This calculator uses partial pivoting: in each column, before eliminating, it picks as the pivot the row with the largest absolute value in that column among the rows still available, and swaps that row into the pivot position if needed. That avoids two practical problems: dividing by a pivot equal to zero (which would halt the algorithm) and dividing by a pivot very close to zero, which amplifies rounding error in floating point. Since this tool uses exact fractions instead of floating point, the second problem does not exist here, but pivoting is still needed for the first: without it, a column with a zero in the pivot's natural position would halt the elimination.

Every row of the reduced row echelon form has a pivot equal to 1, and that pivot is the only nonzero value in its column, both above and below it. Once the augmented matrix reaches that form, the system's solution can be read directly off the last column, with no further calculation, which is exactly why the reduced row echelon form is the destination of the process, not just an intermediate step.

Row-reducing a 3x3 system

Initial augmented matrix

Input: [1 1 1 | 6] [0 2 5 | -4] [2 5 -1 | 27]
Expected output: [1 0 0 | 5] [0 1 0 | 3] [0 0 1 | -2]

After full row reduction, the solution (x=5, y=3, z=-2) shows up directly in the last column.

Row swap from partial pivoting

Input: [1 1 1 | 6] [0 2 5 | -4] [2 5 -1 | 27]
Expected output: L1 <-> L3

The third row has the largest absolute value in the first column (2), so it becomes the new first row before eliminating the rest, avoiding a small pivot.

Full tool FAQ

How should I type the equations?

What's the difference between the Gaussian elimination and Cramer's rule shown here?

What does a "consistent and dependent" system mean?

How do I know a system is inconsistent, rather than that I typed something wrong?

Why does the answer show up as a fraction instead of a decimal number?

How many unknowns can a system have?

What does the "ill-conditioned" system warning mean?

Is my data saved on any server?

Frequently asked questions

What is the difference between row echelon form and reduced row echelon form?

Why do rows sometimes swap position during elimination?

Because of partial pivoting: the row with the largest absolute value in that column is moved into the pivot position, which avoids halting the algorithm when the natural value in that spot is zero.