How this calculation works
The final column is the constants vector, not another unknown. Exact row operations preserve the equations’ solution set. A contradictory row means no solution; a consistent pivot in every coefficient column gives a unique solution; remaining columns describe free variables.
[A | b] → reduced rows → variable relationshipsA worked example
Enter [1 1 3; 2 −1 0] for x₁ + x₂ = 3 and 2x₁ − x₂ = 0. The unique solution is x₁ = 1 and x₂ = 2.
Good to know
For small matrices of exact rational numbers, up to 6 rows and 6 columns. Entries are limited to ±1,000,000; integer fraction denominators to 1,000,000. Decimal input allows up to 6 places. No expressions, scientific notation, complex numbers or uncertain measurements. Large intermediate fractions are bounded to keep the tool responsive.
Reference: University of Nebraska–Lincoln: row echelon forms (opens in a new tab) ↗
Translate the equations before solving
Keep the unknowns in the same order in every row. For x₁ + x₂ = 3 and 2x₁ − x₂ = 0, enter “1 1 3” followed by “2 -1 0”. The final value of each row is the constant on the right-hand side.
An absent variable needs a zero coefficient. For x₁ = 5 in a system with two unknowns, the row is “1 0 5”. Leaving out that zero shifts the interpretation of later entries and changes the problem.
Three outcomes deserve different explanations
Equations x₁ + x₂ = 3 and 2x₁ + 2x₂ = 6 repeat the same relationship. There are infinitely many solutions: choose x₂ freely, then x₁ = 3 − x₂. The solver displays that relation instead of inventing one preferred answer.
Change the second constant from 6 to 7 and the two equations contradict each other. Reduction exposes a row equivalent to 0 = 1, giving no solution. When every unknown has a pivot and there is no contradiction, each variable has a unique value.
Substitution is a useful final check
For a unique solution, substitute the reported numbers into every original equation. For a free-variable solution, choose a simple free value, calculate the pivot variables from their relations and then check all equations. A value of zero for each free variable is often convenient, but it is one choice rather than the entire answer.
The solver accepts exact rational coefficients and does not perform a least-squares fit. If your measurements conflict because of noise, “no solution” describes the entered equations; it does not mean there is no useful approximate model.
A few questions, answered
Can I enter more equations than unknowns?
Yes, within 6 equations and 6 unknowns. Extra equations can be redundant or contradictory; the solver checks consistency.
Are free variables restricted to rational values?
No. With real unknowns and rational coefficients, free variables can be any real numbers. Exact fractions describe the fixed coefficients in their relationships.
How do I enter a fraction?
Use a slash without spaces, such as -2/3. A decimal may have up to 6 digits after the point. Do not include variable names, equals signs or arithmetic expressions in the matrix.