Gaussian Elimination Calculator
This Gaussian elimination calculator solves a system of linear equations from its augmented matrix, with steps: every row operation, then every finished row read back as the equation it actually stands for, then back substitution worked upwards.
Take 2x + y − z = 8, −3x − y + 2z = −11, −2x + y + 2z = −3. Seven steps of elimination bring the grid to echelon form, the bottom row says z = −1, and working upwards gives y = 3 and x = 2. The calculator then puts those three numbers back into the equations exactly as you typed them, because that is the only way to be sure a sign did not go missing on the way.
Writing the System as an Augmented Matrix
The letters never change while you work, so they can be left out. One row per equation, one column per unknown, and the numbers from the right-hand side in a last column separated by a bar. The system above becomes:
2 1 −1 | 8
−3 −1 2 | −11
−2 1 2 | −3
That bar matters. Everything left of it is a coefficient; the single column to its right is what each equation adds up to. A term missing from an equation is a zero, not a gap — if your second equation has no y in it, type 0 in that cell.
How Gaussian Elimination Works
Gaussian elimination runs in two halves, and the first allows three moves and no others. Swap two rows, multiply a row through by a non-zero number, or add a multiple of one row to another. These are the elementary row operations, and each is something you could do to the equations themselves without changing what solves them — which is the whole reason the method is trustworthy.
Forward elimination comes first, and it is the half that gives the method its name. Working left to right, each column gets a pivot, and everything below that pivot is cleared to zero. Our example needs seven operations, including a row swap when a pivot position turns up empty. What comes out is echelon form, a staircase of zeros. If the exercise in front of you stops there, so can you — the row echelon form calculator does that half on its own and shows nothing more.
Back substitution finishes the job. The bottom row now holds a single unknown, so it hands over a value immediately. Put that into the row above, which then has one unknown left, and keep climbing. There is a second route, which keeps clearing until the answer can be read straight off the grid without any climbing at all, and the reduced row echelon form calculator takes it. Neither is more correct; they simply divide the work differently.
Reading the Final Rows: One Answer, None, or Infinitely Many
Elimination is only half the job. This is the step most calculators skip, and it is where the three outcomes announce themselves. Every row is still an equation. Translate each one back and the answer tells you what it is.
A row that still holds letters ties one unknown to the ones after it. That is what back substitution feeds on.
A row reading 0 = something. Try x + y = 2 with 2x + 2y = 5. One operation, R2 − 2R1, turns the second row into 0 = 1. No values of x and y can make that true, so the equations contradict each other and the system is inconsistent — there is no solution and nothing to substitute back. That row is the proof, not a breakdown of the method.
A row reading 0 = 0. Change that 5 to a 4 and the same operation gives 0 = 0 instead. True whatever x and y are, so it constrains nothing: two equations, but only one real condition. One letter stays free, and the answer is a family — here x = 2 − y, with y taking any value you like. Every member of that family satisfies both equations.
One elimination therefore settles all three cases; you only have to be willing to read the rows afterwards. If you would rather work with the equations as they are printed, without turning them into a grid at all, the elimination method calculator does exactly that, and the system of equations calculator covers the wider family of approaches.
How to Use This Gaussian Elimination Calculator
The screen has three parts: the answer band at the top, the grid you fill in below it, and the keypad underneath.
1. Set the size. Two rows of buttons choose how many equations and how many unknowns you have, from 1 to 4 each. They are separate, so four equations in two unknowns is perfectly allowed.
2. Fill in the grid. Each row is one equation. The cells before the bar are the coefficients of x, y, z and w in that order, and the cell after the bar is the number on the right. Blanks count as zero, and fractions such as 3/4 or decimals such as 0.25 are both accepted.
3. Press =. The answer appears at the top — one value per letter, a whole family when letters are free, or a plain statement that there is no solution, with the row that proves it named.
4. Open the working. Show the elimination lays out the whole elimination: every row operation with the grid after it, then each finished row written out as an equation, then the values found one at a time, and finally the check.
Every key and control, in one place:
| equations / unknowns | Sets the shape of the system, 1 to 4 each, chosen separately |
| the cells | Coefficients before the bar, the right-hand side after it; blank counts as zero |
| ← → | Move to the previous or next cell, wrapping round the whole grid |
| ↑ ↓ | Move up or down a row, for filling a column at a time |
| ± | Flips the sign of the cell you are in, so a minus cannot be typed twice by accident |
| ⁄ | The fraction bar. 3⁄4 stays a fraction all the way through |
| . | Decimal point. 0.25 is read as one quarter, exactly |
| ⌫ | Deletes one character from the cell you are in |
| AC | Clears the whole grid and starts again |
| = | Solves the system |
| My keyboard | Switches that cell to your own keyboard, for anything the pad does not carry |
| copy | The icon in the band copies the answer |
Nothing is rounded at any stage, so a coefficient of one third stays one third rather than becoming 0.333. That matters more here than it looks: a rounded value carried through the elimination can turn a row that should read 0 = 0 into one that reads 0 = 0.0000001, which changes the verdict from infinitely many solutions to none at all.
Gaussian Elimination FAQ
What is an augmented matrix?
What are the steps of Gaussian elimination?
How do you know a system has no solution?
What does a row of all zeros mean?
What are the three elementary row operations?
What is back substitution?
Write the grid, eliminate below each pivot, read the rows, climb back up — that is Gaussian elimination in four moves. The reading is the part worth slowing down for — get that right and the system tells you itself whether it has one answer, none, or a whole family of them.