Gaussian Elimination Formula – Row Reduction and Back Substitution Reference

Complete Gaussian elimination formula reference: row operation notation, elimination formula, back substitution, and worked examples for 2×2 and 3×3 systems.

Formula
Each elimination step zeros out entry (i, j) below the pivot at row p by subtracting the appropriate multiple of the pivot row from row i. After reaching upper triangular form, back substitution recovers each variable from the last equation upward.
Variables
SymbolNameDescriptionUnit
RᵢRow iThe row being modified (below the pivot row)—
RₚPivot rowThe current pivot row used for elimination—
aᵢⱼEntry to zeroThe element in row i, pivot column j that is being eliminated—
aₚⱼPivot elementThe leading non-zero element of the pivot row in column j—
kMultiplierk = aᵢⱼ / aₚⱼ, the factor subtracted to produce a zero—
xₙLast variableFound directly from the last equation in back substitution—
How to Use
  1. Write the augmented matrix [A|b].
  2. For each pivot column j (left to right): find the row with the largest |aᵢⱼ| below and including the pivot row (partial pivoting) and swap it to the pivot position.
  3. For each row i below the pivot row: compute k = aᵢⱼ / aₚⱼ and apply Rᵢ → Rᵢ − k·Rₚ.
  4. Move to the next pivot (row p+1, column j+1) and repeat.
  5. Check for special rows: [0…0 | c≠0] → no solution; [0…0 | 0] → free variable.
  6. Back substitution: xₙ = bₙ / aₙₙ, then for i = n−1 down to 1: xᵢ = (bᵢ − Σⱼ>ᵢ aᵢⱼxⱼ) / aᵢᵢ.
Examples
1. 2×2 system: x₁ + 3x₂ = 9, 2x₁ + x₂ = 8
ℹSolution: x₁ = 3, x₂ = 2. Verify: 3+3(2)=9 ✓, 2(3)+2=8 ✓
2. 3×3 system: x₁ + x₂ + x₃ = 6, 2x₁ + 3x₂ + x₃ = 10, x₁ + 2x₂ + 3x₃ = 14

R₂ → R₂ − 2R₁, R₃ → R₃ − R₁:

R₃ → R₃ − R₂:

Back substitution: x₃ = 10/3, x₂ = −2 + 10/3 = 4/3, x₁ = 6 − 4/3 − 10/3 = 4/3

ℹSolution: x₁ = 4/3, x₂ = 4/3, x₃ = 10/3. Verify: 4/3 + 4/3 + 10/3 = 18/3 = 6 ✓
3. Inconsistent system: x₁ + x₂ = 2, 3x₁ + 3x₂ = 9
⚠Row 2: 0 = 3, a contradiction. No solution exists.
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