Gaussian elimination
Gaussian elimination turns the system into an equivalent one in row echelon form, where each equation has one unknown fewer than the one above it. Then you solve it from the bottom up. You work with the augmented matrix (coefficients | constants) and only three row operations are allowed, none of which changes the solutions:
- Swap two rows.
- Multiply a row by a non-zero number.
- Add a multiple of one row to another.
Example · 3 equations in 3 unknowns
2x + y − z = 8, −3x − y + 2z = −11, −2x + y + 2z = −3. With R₂ ← R₂ + (3/2)·R₁ and R₃ ← R₃ + R₁ the x is eliminated from rows 2 and 3; with R₃ ← R₃ − 4·R₂ the y is eliminated from row 3, leaving −z = 1. Back-substituting: x = 2, y = 3, z = −1.
How to classify a linear system (Rouché–Capelli theorem)
Compare the rank of the coefficient matrix A with the rank of the augmented matrix A|b and with the number of unknowns n:
| Condition | Type | Solutions |
|---|---|---|
| rank(A) = rank(A|b) = n | Consistent and independent | Exactly one |
| rank(A) = rank(A|b) < n | Consistent and dependent | Infinitely many, depending on parameters |
| rank(A) < rank(A|b) | Inconsistent | None |
In practice, while reducing: if a row 0 = k with k ≠ 0 appears the system is inconsistent; if fewer useful equations remain than unknowns, it has infinitely many solutions.
Frequently asked questions
Why does the calculator use fractions?
To avoid rounding errors. Fractions appear constantly during elimination; with decimals, a badly rounded 0.333… can make an inconsistent system look consistent.
My teacher uses different row operations and gets different matrices. Is that wrong?
No, that is normal: there are many ways to reach row echelon form (for example, scaling rows to avoid fractions). The intermediate matrices change, but the final solution and the classification must match.
What is the difference with Gauss-Jordan elimination?
Gauss-Jordan keeps eliminating above the pivots too, until the matrix is in reduced row echelon form (a diagonal of ones), so the solutions can be read directly without back substitution.
Can I solve a 2×2 system with substitution instead?
Yes, substitution and elimination give the same answer. For 2 equations either is quick; for 3 or more, Gaussian elimination is far more systematic.