Gauss elimination & Gauss–Jordan method

Introduction

The Gauss elimination and Gauss–Jordan methods are systematic procedures for solving systems of linear equations.
Both methods convert a system into a simpler equivalent form using elementary row operations, but they differ in how far the simplification goes.

  • Gauss elimination reduces the matrix to Row Echelon Form (REF) and then uses back-substitution.
  • Gauss–Jordan reduces the matrix all the way to Reduced Row Echelon Form (RREF), giving the solution directly.

These techniques are essential for understanding matrix rank, solution types, and numerical linear algebra.


Elementary Row Operations

These operations do not change the solution set of the system:

  1. Swap two rows:

R_i \leftrightarrow R_j

Multiply a row by a non-zero constant:

R_i \to k R_i

Add a multiple of one row to another:

R_i \to R_i + k R_j

All matrix reduction methods rely on these operations.


Gauss Elimination Method (Forward Elimination)

Objective

Convert the augmented matrix into upper triangular form (REF), where all elements below the diagonal are zero.

Example system:

 \begin{aligned} x + y + z &= 6 \\ 2x + 3y + 7z &= 20 \\ x - y + z &= 4 \end{aligned}

Step 1: Form Augmented Matrix

 \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 3 & 7 & | & 20 \\ 1 & -1 & 1 & | & 4 \end{bmatrix}

Step 2: Eliminate entries below the first pivot

Pivot = 1 (row 1, column 1).

Perform:

R_2 \to R_2 - 2R_1

R_3 \to R_3 - R_1

Result:

 \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 5 & | & 8 \\ 0 & -2 & 0 & | & -2 \end{bmatrix}

Step 3: Eliminate entries below second pivot

Pivot = 1 (row 2, column 2).

R_3 \to R_3 + 2R_2

 \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 5 & | & 8 \\ 0 & 0 & 10 & | & 14 \end{bmatrix}

Step 4: Back-Substitution (Gauss Method)

From bottom:

 10z = 14 \Rightarrow z = 1.4

 y + 5z = 8 \Rightarrow y = 8 - 7 = 1

 x + y + z = 6 \Rightarrow x = 6 - 1 - 1.4 = 3.6

Solution:

 (x, y, z) = (3.6,, 1,, 1.4)


Row Echelon Form (REF)

A matrix is in REF if:

  • All-zero rows are at bottom
  • Each leading entry (pivot) is to the right of the pivot above it
  • All entries below pivots are zero

Example:

 \begin{bmatrix} 1 & 2 & -1 \\ 0 & 3 & 5 \\ 0 & 0 & 7 \end{bmatrix}

Gauss elimination always produces REF.


Gauss–Jordan Method (Full Elimination)

Objective

Convert matrix all the way to Reduced Row Echelon Form (RREF):

  • Pivots are 1
  • Each pivot is the only non-zero entry in its column
  • No back-substitution required

Procedure

Same steps as Gauss elimination (forward elimination), plus backward elimination.

Continuing the same example, after Gauss elimination we had:

 \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 5 & | & 8 \\ 0 & 0 & 10 & | & 14 \end{bmatrix}

Step 1: Make last pivot = 1

R_3 \to \frac{1}{10}R_3

 \begin{bmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 5 & | & 8 \\ 0 & 0 & 1 & | & 1.4 \end{bmatrix}

Step 2: Eliminate above the third pivot

R_2 \to R_2 - 5R_3

R_1 \to R_1 - R_3

 \begin{bmatrix} 1 & 1 & 0 & | & 4.6 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & 1.4 \end{bmatrix}

Step 3: Eliminate above the second pivot

R_1 \to R_1 - R_2

Final RREF:

 \begin{bmatrix} 1 & 0 & 0 & | & 3.6 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & 1.4 \end{bmatrix}

Directly giving:

 (x, y, z) = (3.6,, 1,, 1.4)


Comparison: Gauss vs Gauss–Jordan

AspectGauss EliminationGauss–Jordan
Final formREFRREF
Requires back-substitutionYesNo
Computation costLowerHigher
Preferred for hand calculationsYesNo
Common in numerical algorithmsYesSometimes

When to Use Which?

Use Gauss Elimination when:

  • Solving systems by hand
  • Only need one solution vector
  • Speed is important

Use Gauss–Jordan when:

  • Finding matrix inverse
  • Solving multiple systems with same coefficient matrix
  • Want direct solutions with no back-substitution

Key Points for GATE

  • Understand pivot positions
  • Know how row operations affect rank
  • Identify inconsistent systems from REF/RREF
  • Quickly reduce 2×2 or 3×3 systems
  • Recall that Gauss–Jordan is used to compute A^{-1}
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