9.6Solving Systems by Row Reduction
Write augmented matrices, apply row operations (swap, scale, add), and reduce to row-echelon form (Gaussian) or reduced row-echelon form (Gauss-Jordan) to solve linear systems.
Gaussian elimination is the systematic algorithm for solving any linear system. It is the foundation of numerical linear algebra and is used in every scientific computing application.
Essential Question
How does transforming an augmented matrix to row echelon form through systematic row operations give you the solution to a linear system, and what does it mean when you reach a row of all zeros?
Lesson Overview
Augmented Matrix [A|b]
The augmented matrix is formed by appending the constant vector b to the coefficient matrix A. A vertical bar separates them: [A|b].
The Three Row Operations (preserve the solution set)
Rᵢ ↔ Rⱼ — swap two rows
kRᵢ → Rᵢ — multiply a row by a nonzero constant k
Rᵢ + kRⱼ → Rᵢ — add k times row j to row i
Row Echelon Form (REF)
- • All zero rows are at the bottom
- • Each pivot is to the right of the pivot above it
- • All entries below a pivot are 0
Reduced Row Echelon Form (RREF)
- • All REF conditions, PLUS
- • Each pivot equals 1
- • All entries above a pivot are also 0
Two Algorithms
Gaussian Elimination: Reduce to REF, then back-substitute from the bottom row up.
Gauss-Jordan Elimination: Reduce all the way to RREF — the solution can be read directly.
Interpreting Results
✅ Unique solution: each variable has a pivot column.
❌ No solution: a row of the form [0 0 0 | k] where k ≠ 0 (contradiction).
∞ Infinitely many solutions: a row of all zeros (free variable exists).
Worked Examples
Solve using Gaussian elimination: 2x + y = 5, x + 2y = 4.
Augmented matrix: [[2, 1 | 5], [1, 2 | 4]]
Swap R₁ ↔ R₂: [[1, 2 | 4], [2, 1 | 5]]
R₂ → R₂ − 2R₁: [[1, 2 | 4], [0, −3 | −3]]
R₂ → R₂ ÷ (−3): [[1, 2 | 4], [0, 1 | 1]]
Back-substitute: y = 1; x + 2(1) = 4 → x = 2
Solve: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
Augmented: [[1,1,1|6],[2,−1,1|3],[1,2,−1|2]]
R₂ → R₂ − 2R₁: [[1,1,1|6],[0,−3,−1|−9],[1,2,−1|2]]
R₃ → R₃ − R₁: [[1,1,1|6],[0,−3,−1|−9],[0,1,−2|−4]]
Swap R₂ ↔ R₃: [[1,1,1|6],[0,1,−2|−4],[0,−3,−1|−9]]
R₃ → R₃ + 3R₂: [[1,1,1|6],[0,1,−2|−4],[0,0,−7|−21]]
R₃ → R₃ ÷ (−7): [[1,1,1|6],[0,1,−2|−4],[0,0,1|3]]
Back-substitute: z = 3; y − 2(3) = −4 → y = 2; x + 2 + 3 = 6 → x = 1
Identify the solution type: x + 2y = 3, 2x + 4y = 6.
Augmented: [[1, 2 | 3], [2, 4 | 6]]
R₂ → R₂ − 2R₁: [[1, 2 | 3], [0, 0 | 0]]
Row of all zeros → dependent system (free variable)
Identify the solution type: x + y = 3, x + y = 5.
Augmented: [[1, 1 | 3], [1, 1 | 5]]
R₂ → R₂ − R₁: [[1, 1 | 3], [0, 0 | 2]]
[0 0 | 2] means 0 = 2 → contradiction → inconsistent
Solve using Gauss-Jordan (RREF): x + 2y = 5, 3x + 5y = 13.
Augmented: [[1, 2 | 5], [3, 5 | 13]]
R₂ → R₂ − 3R₁: [[1, 2 | 5], [0, −1 | −2]]
R₂ → −R₂: [[1, 2 | 5], [0, 1 | 2]]
R₁ → R₁ − 2R₂: [[1, 0 | 1], [0, 1 | 2]]
RREF reached — read solution directly
Guided Practice
Write the augmented matrix for: 3x − y = 7, 2x + 4y = 10. Then reduce to REF.
Hint: Swap rows if needed to get a leading 1 in position (1,1).
Solve using Gaussian elimination: x + 2y + z = 8, 2x + y − z = 3, x − y + 2z = 5.
Hint: Use R₂ → R₂ − 2R₁ and R₃ → R₃ − R₁ to eliminate x from rows 2 and 3.
Solve using Gauss-Jordan (RREF): 2x + y = 7, x − 3y = −4.
Hint: Reduce all the way to RREF so the solution can be read directly.
Determine the solution type: 2x − 4y + 6z = 4, x − 2y + 3z = 2, 3x − 6y + 9z = 6.
Hint: What happens when you eliminate x from rows 2 and 3?
Solve: x + y − z = 2, 2x + 3y + z = 11, x + 2y + 2z = 9.
Hint: Systematically eliminate x, then y.
Key Vocabulary
Augmented matrix
The coefficient matrix with the constant column appended, written [A|b].
Example: 2x + y = 5 → [[2, 1 | 5]]
Row operations
Swap rows, multiply a row by a nonzero constant, or add a multiple of one row to another. All three preserve the solution set.
Row Echelon Form (REF)
Each pivot is to the right of the one above it; all entries below each pivot are 0.
Reduced Row Echelon Form (RREF)
REF with each pivot equal to 1 and all other entries in pivot columns equal to 0.
Pivot
The leading nonzero entry in a row; used to eliminate entries below (and above in RREF).
Gaussian elimination
Reducing to REF then back-substituting from the bottom row up to find all variable values.
Gauss-Jordan elimination
Reducing all the way to RREF; the solution can be read directly without back-substitution.
Free variable
A variable without a pivot column; its presence indicates infinitely many solutions.
Check Your Understanding
Interactive Practice — 5 Questions
The augmented matrix for 2x − y = 3, x + 4y = 7 is:
A row of [0 0 0 | 5] in a reduced matrix means:
Which is NOT a valid row operation?
A row of [0 0 0 | 0] in a reduced matrix means:
Gauss-Jordan elimination produces:
Common Mistakes
Applying a row operation to only part of the row
A row operation applies to the ENTIRE row, including the augmented column. Never forget to update the constant column.
Multiplying a row by 0 to create zeros
Multiplying by 0 destroys the row. Use Rᵢ + kRⱼ → Rᵢ to create zeros by adding a multiple of another row.
Stopping at REF and forgetting to back-substitute
After reaching REF, use back-substitution from the bottom row up to find all variable values.
Confusing a row of zeros [0 0 0 | 0] with [0 0 0 | k] (k ≠ 0)
[0 0 0 | 0] means a free variable (infinitely many solutions). [0 0 0 | k] means no solution.
Math Tips
Always aim to get a leading 1 in each pivot position. Swap rows or divide a row to create a 1 before eliminating below.
Work column by column, left to right. Eliminate all entries below the current pivot before moving to the next column.
Label each row operation clearly (e.g., R₂ → R₂ − 2R₁). This makes it easy to check your work and find errors.
For RREF, after getting REF, work back up: use each pivot to eliminate entries ABOVE it as well.
The number of pivots = the number of independent equations = the number of variables with unique values.