gizmobench

Gaussian Elimination Calculator

Opens the linear-system calculator on augmented-matrix input with the elimination trace in front. Enter rows, solve locally and review each operation before the final answer.

3 variables, 3 equationsAugmented matrixEvery step shown
 2x +  y -  z =    8
-3x -  y + 2z =  -11
-2x +  y + 2z =   -3

elimination         3 row operations, 2 row swaps, pivoting on the largest coefficient
back-substitution   3 steps

x = 2        y = 3        z = -1

residual            0   0   0        unique solution
Enter as
Variables
Equations
Working
x
2
y
3
z
-1
Case
Unique

The augmented matrix

One row per equation, one column per variable, and the constant after the equals sign. A cell takes a whole number, a decimal like 0.5 or a fraction like 1/3, and a blank cell counts as 0.

xyzconstant

The working

Forward elimination first, largest coefficient leading each column, then back-substitution from the bottom row up. Every multiplier is an exact fraction.

  1. R1 <-> R2 pivot on -3, the largest coefficient left in column x
    [ -3  -1   2 | -11 ]
[  2   1  -1 |   8 ]
[ -2   1   2 |  -3 ]
  2. R2 -> R2 + (2/3)R1 clears x from row 2
    [  -3   -1    2 | -11 ]
[   0  1/3  1/3 | 2/3 ]
[  -2    1    2 |  -3 ]
  3. R3 -> R3 - (2/3)R1 clears x from row 3
    [  -3   -1    2 |  -11 ]
[   0  1/3  1/3 |  2/3 ]
[   0  5/3  2/3 | 13/3 ]
  4. R2 <-> R3 pivot on 5/3, the largest coefficient left in column y
    [  -3   -1    2 |  -11 ]
[   0  5/3  2/3 | 13/3 ]
[   0  1/3  1/3 |  2/3 ]
  5. R3 -> R3 - (1/5)R2 clears y from row 3
    [  -3   -1    2 |  -11 ]
[   0  5/3  2/3 | 13/3 ]
[   0    0  1/5 | -1/5 ]

Back-substitution

  • From row 3: (1/5)z = -1/5, so z = -1.
  • From row 2: (5/3)y + (2/3)z = 13/3 with z = -1, so y = 3.
  • From row 1: -3x - y + 2z = -11 with y = 3 and z = -1, so x = 2.

The check

  • 2x + y - z = 8, residual 0
  • -3x - y + 2z = -11, residual 0
  • -2x + y + 2z = -3, residual 0

Each value is substituted back into the equations as you typed them: every residual is exactly zero, because the arithmetic is done in fractions rather than decimals.

The three answers a linear system can have

Load any of them to see which one you are looking at and why.

  • x + y = 3, x - y = 1two lines that cross once
    x = 2, y = 1
  • x + y = 1, 2x + 2y = 3same line, different constant
    No solution: the equations contradict each other
  • x + y = 3, x + y = 3one equation repeated, rank below the variable count
    x = 3 - y, with y free
Three answers, and the rank tells you which. Count the rows left with a leading coefficient after elimination: that is the rank. Rank equal to the number of unknowns gives one solution. A row that reduces to 0 = something other than zero is a contradiction, so there is no solution at all. Rank below the number of unknowns leaves free variables, and every value of them is a solution, which is why the answer comes back as a formula rather than a number. Telling those apart needs zero to mean zero, so every coefficient here is held as a fraction of two whole numbers and nothing is rounded on the way.

Common questions

What does Gaussian elimination show?
It applies row swaps, scaling and row combinations until the augmented matrix is in reduced form, printing each operation and the matrix it produces.
Can I enter equations instead?
Yes. The parent tool supports equation entry as well as an augmented matrix; this page starts with the matrix form for a transparent elimination trace.

Gaussian elimination that pivots on the largest available coefficient for numerical stability, with every row operation printed so you can follow it. The solution is substituted back into the equations you typed and the residual is shown, so you can see the arithmetic close. A nonlinear term is refused by name rather than quietly linearised.