Swap two rows
Exchange the position of two equations when a better pivot is available.
Linear Systems Guide
To solve a linear system with matrices, place the coefficients in A, the variables in x, and the constants in b so the system becomes Ax = b. Form the augmented matrix [A | b], use row operations to reduce it to RREF, and then read the solution from the final rows.
Start with the same variable order in every equation:
x + y = 5
2x - y = 1
The coefficients become A, the variables become x, and the constants become b:
The matrix form keeps the coefficients, variables, and constants in separate roles.
Solve with Equations or A and b →Place b as the final column beside A. Each equation becomes one row:
[1 1 | 5]
[2 -1 | 1]The separator is a visual reminder that the last column contains constants, not another variable coefficient.
Open the Augmented Matrix Calculator →Gauss-Jordan elimination uses only operations that keep the solution set equivalent:
Exchange the position of two equations when a better pivot is available.
Scale a row to create a pivot of 1 without changing the solution set.
Clear entries above or below a pivot while preserving equivalent equations.
The goal is RREF, where pivot columns make the solution easy to read.
For the system x + y = 5 and 2x - y = 1, start with:
[1 1 | 5]
[2 -1 | 1]R2 ← R2 − 2R1
[1 1 | 5]
[0 -3 | -9]R2 ← −⅓R2
[1 1 | 5]
[0 1 | 3]R1 ← R1 − R2
[1 0 | 2]
[0 1 | 3]The final RREF gives x = 2 and y = 3.
[1 0 | 2]
[0 1 | 3]Every variable column has a pivot, so each variable has one value.
[1 0 | 2]
[0 0 | 1]The second row says 0 = 1, so the system is inconsistent.
[1 2 | 5]
[0 0 | 0]The system is consistent with a free variable, so the solutions use a parameter.
Use the same route for a larger system:
x + y + z = 6
2x - y + z = 3
x + 2y - z = 2The augmented matrix is:
[1 1 1 | 6]
[2 -1 1 | 3]
[1 2 -1 | 2]One efficient reduction route is R2 ← R2 − 2R1, R3 ← R3 − R1, scale the second and third pivot rows, then clear the entries above the pivots. The final RREF is:
[1 0 0 | 1]
[0 1 0 | 2]
[0 0 1 | 3]Therefore x = 1, y = 2, and z = 3.
You can use x = A⁻¹b only when A is square and invertible. Many systems are rectangular or singular, so RREF is the main method for solving systems with matrices.
Fractions preserve exact values during row reduction. Decimals can be useful for estimates, but rounding too early can hide exact relationships. Keep fractions such as 1/2 and -3/4 exact when possible.
Use the calculator to enter equations, A and b, or an augmented matrix and verify the RREF result.
A is the coefficient matrix, x is the column vector of unknown variables, and b is the constants vector. Together they represent the same linear system in matrix form.
An augmented matrix places the constants beside the coefficient matrix so each row operation updates the equations and constants together.
RREF shows no solution when a row becomes a contradiction, such as [0 0 | 1]. That row means 0 = 1, which cannot be true.
Infinitely many solutions occur when the system is consistent but at least one variable column has no pivot. That variable is free and can be written as a parameter.
No. The inverse method x = A⁻¹b works only when A is square and invertible. RREF is the more general method for linear systems.
Valid row operations do not change the solution set. They rewrite the system into an equivalent form that is easier to read.