Linear Algebra Tool

Gaussian Elimination Calculator - REF & RREF Linear Equation Solver

Free Gaussian elimination calculator. Solve linear systems, calculate row echelon form (REF) & reduced row echelon form (RREF) with step-by-step solutions. Our calculator uses elementary row operations including forward elimination and back substitution to transform matrices and solve systems of linear equations.

Last updated: December 15, 2024

Step-by-step row operations
Forward elimination and back substitution
REF and RREF calculations

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Gaussian Elimination Calculator
Solve systems of linear equations using Gaussian elimination
1:|
2:|
3:|

Enter coefficients for the system. The augmented column contains the constants.

Solution

✓ Unique Solution
x1 =2.000
x2 =3.000
x3 =-1.000

Row Echelon Form (REF):

[2.001.00-1.00|8.00]
[0.000.500.50|1.00]
[0.000.00-1.00|1.00]

Reduced Row Echelon Form (RREF):

[1.000.000.00|2.00]
[0.001.000.00|3.00]
[0.000.001.00|-1.00]

Step-by-Step Process:

  1. Forward elimination complete
  2. Back substitution complete
  3. Solution type: unique

Gaussian Elimination Steps:

  • Forward Elimination: Create zeros below pivots (REF)
  • Back Substitution: Create zeros above pivots (RREF)
  • Row Operations: Swap, scale, and add/subtract rows
  • Pivot: First non-zero element in each row

Gaussian Elimination Calculator Features

Row Echelon Form (REF)
Forward elimination process

Process

Zeros Below Pivots

Create triangular form with row operations

Reduced Row Echelon Form (RREF)
Complete row reduction

Process

Identity Matrix Form

Solutions directly visible in final column

Elementary Row Operations
Three fundamental transformations

Operations

Swap, Scale, Add

Preserve solution set while simplifying

Forward Elimination
First phase of Gaussian elimination

Goal

Upper Triangular

Eliminate entries below main diagonal

Back Substitution
Second phase for RREF

Goal

Diagonal Form

Eliminate entries above pivots

Solution Classification
Unique, infinite, or no solution

Types

1, ∞, or 0

Determined by pivot positions

Example System Solution

System: 2x + y - z = 8, -3x - y + 2z = -11, -2x + y + 2z = -3

x =

2

y =

3

z =

-1

How Gaussian Elimination Works

Gaussian elimination solves linear systems by systematically transforming the augmented matrix using elementary row operations. The algorithm proceeds in two phases: forward elimination to reach REF, and back substitution to reach RREF.

The Gaussian Elimination Algorithm

Phase 1: Forward Elimination (REF)

  1. Identify pivot in first column, swap rows if needed
  2. Eliminate all entries below the pivot
  3. Move to next column and repeat
  4. Result: upper triangular matrix (REF)

Phase 2: Back Substitution (RREF)

  1. Scale each row to make pivots equal to 1
  2. Starting from bottom, eliminate entries above pivots
  3. Result: identity matrix on left (RREF)
  4. Solution appears directly in augmented column

Mathematical Foundation

Elementary row operations preserve the solution set because they represent equivalent systems. The three operations (row swapping, row scaling, row addition) correspond to algebraic manipulations that don't change the equations' meaning, only their representation.

  • Row operations preserve the solution set of the system
  • REF reveals the rank of the coefficient matrix
  • RREF provides solutions directly without back calculation
  • Pivot positions determine free and basic variables
  • Algorithm complexity: O(n³) for n×n matrices
  • Forms the basis for matrix inversion and determinant calculation

Sources & References

  • Linear Algebra and Its Applications - David C. Lay (6th Edition)Standard reference for Gaussian elimination and row reduction
  • Introduction to Linear Algebra - Gilbert Strang (5th Edition)Comprehensive treatment of elimination methods
  • Khan Academy - Linear Algebra CourseFree educational resources for row operations

Gaussian Elimination Example

Step-by-Step Example
Solve: 2x + y - z = 8, -3x - y + 2z = -11, -2x + y + 2z = -3

Initial Augmented Matrix:

[ 2 1 -1 | 8 ]

[ -3 -1 2 | -11]

[ -2 1 2 | -3 ]

After Forward Elimination (REF):

[ 2 1 -1 | 8 ]

[ 0 * * | * ]

[ 0 0 * | * ]

Solution: x = 2, y = 3, z = -1

System has a unique solution (consistent and independent).

Row Operations Used

Multiple row additions and scalings to create zeros

Verification

Substitute back to verify: 2(2) + 3 - (-1) = 8 ✓

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