Gauss-Jordan Elimination Calculator

Enter the coefficients and constants for a system of three linear equations to reduce the augmented matrix to reduced row echelon form (RREF) and solve for x, y, and z.

Quick Facts

Goal
Reduce [A | b] to [I | x]
Row-reduce the augmented matrix to reduced row echelon form (RREF) so the solution can be read off directly.
Allowed row operations
Swap, scale, add multiple
Swap two rows, multiply a row by a nonzero scalar, or add a multiple of one row to another.
vs. Gaussian elimination
No back-substitution needed
Gauss-Jordan clears entries above each pivot too, so it skips the back-substitution step.

Your Results

Calculated
x
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Solution for x
y
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Solution for y
z
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Solution for z
System type
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Unique, dependent, or inconsistent

Ready

Enter the coefficients for three equations, then press Calculate.

How Gauss-Jordan Elimination Works

Gauss-Jordan elimination solves a system of linear equations by writing the coefficients and constants as an augmented matrix [A | b] and applying row operations until the left side becomes the identity matrix. For a 3×3 system a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃, the calculator builds the augmented matrix, row-reduces it to reduced row echelon form (RREF), and reads the solution x, y, z directly from the last column.

Formula and method

Three elementary row operations are allowed, each of which preserves the solution set: (1) swap two rows, (2) multiply a row by a nonzero scalar, and (3) add a multiple of one row to another. Starting from column 1, the calculator selects the row with the largest available entry in that column as the pivot (partial pivoting, which reduces rounding error), scales that row so the pivot equals 1, then uses row operation (3) to zero out every other entry — above and below — in that column. Repeating this for each column leaves an identity matrix on the left and the solution on the right: [I | x, y, z]. This is what distinguishes Gauss-Jordan elimination from plain Gaussian elimination, which only zeroes out entries below the pivot (row echelon form) and then requires a separate back-substitution pass to recover each variable.

Common sources of error

  • Zero pivot: if the entry you would pivot on is zero, you must swap with a row below it first — dividing by zero corrupts the whole row.
  • Sign errors when eliminating: when subtracting a multiple of the pivot row from another row, double-check the sign of the multiplier — this is the most common hand-calculation mistake.
  • Rounding mid-calculation: keep full-precision fractions or decimals through every row operation; rounding early compounds errors by the final step.

Checking your result

Once you have x, y, and z, substitute them back into all three original equations — each one must balance. If a row of the reduced matrix reads all zeros with a nonzero constant (like [0 0 0 | 5]), the system is inconsistent and has no solution. If a row reduces entirely to zeros (including the constant), the equations are dependent and the system has infinitely many solutions along a line or plane, rather than a single point.

Applications

Beyond solving linear systems, Gauss-Jordan elimination is the standard way to compute a matrix inverse (row-reduce [A | I] to get [I | A⁻¹]) and to determine a matrix's rank. It underlies circuit analysis (Kirchhoff's laws), balancing chemical equations, linear regression, and the simplex method in linear programming.

Frequently Asked Questions

What is Gauss-Jordan elimination?
Gauss-Jordan elimination is a method for solving a system of linear equations by writing the coefficients and constants as an augmented matrix and applying row operations (swapping rows, scaling a row, adding a multiple of one row to another) until the matrix reaches reduced row echelon form (RREF) — an identity matrix on the left with the solution values on the right.
How is Gauss-Jordan elimination different from Gaussian elimination?
Gaussian elimination stops once the matrix reaches row echelon form (zeros only below the diagonal) and then requires back-substitution to find each variable. Gauss-Jordan elimination continues clearing entries above the pivots too, producing a fully reduced identity matrix so the solution can be read directly off the augmented column with no back-substitution step.
What does it mean if the system has no solution or infinitely many solutions?
If row reduction produces a row like [0 0 0 | k] with k not equal to 0, the equations contradict each other and the system has no solution (it is inconsistent). If instead a row reduces entirely to zeros (including the constant), the equations are dependent and the system has infinitely many solutions along a line or plane.
Can Gauss-Jordan elimination find a matrix inverse?
Yes. Augmenting a square matrix A with the identity matrix and row-reducing [A | I] until the left side becomes the identity turns the right side into A⁻¹, provided A is invertible (its determinant is nonzero, which is equivalent to full rank).