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Home Linear Systems Gauss_Jordan_Elimination

Gauss-Jordan Elimination Calculator - RREF Solver

Gauss-Jordan elimination transforms any augmented matrix into Reduced Row Echelon Form (RREF). Every pivot becomes 1 and the only nonzero entry in its column, making the solution directly readable.

Calculator

Enter your matrix below and click "Calculate" to see the step-by-step solution.

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Learn About Gauss_Jordan_Elimination

Understanding the concepts behind the calculations.

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What is Gauss-Jordan Elimination?

Gauss-Jordan elimination is an extension of Gaussian elimination that continues the row reduction process until the matrix is in Reduced Row Echelon Form (RREF). In RREF:

  1. The matrix is in Row Echelon Form (REF).
  2. Each leading coefficient (pivot) is 1.
  3. Each pivot is the only non-zero entry in its column.
Key Benefit: The solution can be read directly from the augmented matrix without back-substitution!

Why Use Gauss-Jordan Elimination?

While Gaussian elimination stops at REF and requires back-substitution, Gauss-Jordan elimination:

  • Gives you the solution directly - Each variable's value appears explicitly.
  • Creates the inverse matrix - When applied to $[A|I]$, it produces $[I|A^{-1}]$.
  • Shows the complete reduction - Students see both forward and backward elimination steps.

The Two Phases of Gauss-Jordan Elimination

Phase 1: Forward Elimination (to REF)

Same as Gaussian elimination - create zeros below each pivot:

$$ \left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\\\ -3 & -1 & 2 & -11 \\\\ -2 & 1 & 2 & -3 \end{array}\right] \rightarrow \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & 3 \\\\ 0 & 0 & 1 & -1 \end{array}\right] $$

Phase 2: Backward Elimination (to RREF)

Continue eliminating above each pivot to get zeros above as well:

  • Create zeros above the pivot in row 2.
  • Create zeros above the pivot in row 1.
  • Result: Each pivot is the only non-zero entry in its column.

Complete Example

Solve the system using Gauss-Jordan elimination:

$$ \begin{cases} 2x + y - z = 8 \\\\ -3x - y + 2z = -11 \\\\ -2x + y + 2z = -3 \end{cases} $$

Step 1: Write the augmented matrix

$$ \left[\begin{array}{ccc|c} 2 & 1 & -1 & 8 \\\\ -3 & -1 & 2 & -11 \\\\ -2 & 1 & 2 & -3 \end{array}\right] $$

Step 2: Create pivot in row 1, column 1 (divide row 1 by 2)

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & -1/2 & 4 \\\\ -3 & -1 & 2 & -11 \\\\ -2 & 1 & 2 & -3 \end{array}\right] $$

Step 3: Eliminate below pivot (rows 2 and 3)
R₂ ← R₂ + 3R₁, R₃ ← R₃ + 2R₁

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & -1/2 & 4 \\\\ 0 & 1/2 & 1/2 & 1 \\\\ 0 & 2 & 1 & 5 \end{array}\right] $$

Step 4: Create pivot in row 2, column 2 (multiply row 2 by 2)

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & -1/2 & 4 \\\\ 0 & 1 & 1 & 2 \\\\ 0 & 2 & 1 & 5 \end{array}\right] $$

Step 5: Eliminate below pivot (row 3)
R₃ ← R₃ - 2R₂

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & -1/2 & 4 \\\\ 0 & 1 & 1 & 2 \\\\ 0 & 0 & -1 & 1 \end{array}\right] $$

Step 6: Create pivot in row 3, column 3 (multiply by -1)

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & -1/2 & 4 \\\\ 0 & 1 & 1 & 2 \\\\ 0 & 0 & 1 & -1 \end{array}\right] $$

Step 7: Eliminate above pivot (back substitution phase)

R₂ ← R₂ - R₃:

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & -1/2 & 4 \\\\ 0 & 1 & 0 & 3 \\\\ 0 & 0 & 1 & -1 \end{array}\right] $$

R₁ ← R₁ + \frac{1}{2}R₃:

$$ \left[\begin{array}{ccc|c} 1 & 1/2 & 0 & 7/2 \\\\ 0 & 1 & 0 & 3 \\\\ 0 & 0 & 1 & -1 \end{array}\right] $$

R₁ ← R₁ - \frac{1}{2}R₂:

$$ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & 3 \\\\ 0 & 0 & 1 & -1 \end{array}\right] $$
Solution: $x = 2$, $y = 3$, $z = -1$

Gauss-Jordan vs Gaussian Elimination

Gaussian Elimination

  • Stops at: Row Echelon Form (REF)
  • Back-substitution? Yes
  • Readability: Requires solving from bottom up
  • Cost: Lower (fewer operations)
  • Best for: Solving systems efficiently

Gauss-Jordan Elimination

  • Stops at: Reduced Row Echelon Form (RREF)
  • Back-substitution? No
  • Readability: Directly readable
  • Cost: Higher (more operations)
  • Best for: Finding inverses, theoretical work

Common Questions

Q: When should I use Gauss-Jordan instead of Gaussian elimination?

A: Use Gauss-Jordan when you need the solution directly without back-substitution, or when finding matrix inverses.

Q: Does Gauss-Jordan always work?

A: Yes, for any system with a unique solution. For inconsistent systems, you'll get a row like $[0\ 0\ ...\ 0\ |\ c]$ with $c \neq 0$.

Q: What if the system has infinite solutions?

A: The RREF will have rows of zeros and free variables. The solution can be expressed with parameters.

Practice Problems

Try these systems with the calculator above:

  1. 2×2 System:

    $$ \begin{cases} 2x + 3y = 8 \\\\ x - y = -1 \end{cases} $$

  2. 3×3 System:

    $$ \begin{cases} x + y + z = 6 \\\\ 2x - y + z = 3 \\\\ x + 2y - z = 2 \end{cases} $$

  3. System with fractions:

    $$ \begin{cases} \frac{1}{2}x + \frac{1}{3}y = \frac{5}{6} \\\\ \frac{1}{3}x + \frac{1}{4}y = \frac{7}{12} \end{cases} $$

Applications

  • Finding matrix inverses: Apply Gauss-Jordan to $[A|I]$.
  • Solving multiple systems: With same coefficient matrix but different constants.
  • Linear algebra theory: Understanding fundamental subspaces.
  • Computer graphics: Transforming coordinates.

Master Gauss-Jordan elimination by practicing with the calculator above. Watch how the matrix transforms to RREF with each operation!