Matrices and Determinants
Gaussian Elimination and Row-Echelon Form
Detaching the coefficients into an augmented matrix, the three legitimate row operations, and reducing to row-echelon form.
What this page covers
- Write a system as an augmented matrix
- State the three permitted row operations
- Reduce a matrix to row-echelon form
- Recover the solution by back-substitution
The augmented matrix
During elimination the variable names never change and never affect the arithmetic. Dropping them leaves the coefficients in a grid, with the constants separated by a rule.
Forming the matrix — the source's example
The system x - 3y = 4, 2x + y = 1 becomes
| 1 | -3 | 4 |
| 2 | 1 | 1 |
Each row is an equation, each column a variable, and the column after the rule holds the constants. The typed notes describe the whole method as row operations followed by back substitution.
The system must be in standard form, with variables in the same order in every equation and constants on the right. A missing variable needs an explicit zero — omitting it misaligns every subsequent column.
The three row operations
The source lists exactly three, and the typed notes give the same list:
| Operation | Effect on the system | Why it is safe |
|---|---|---|
| Multiply a row by a non-zero constant | Multiplies an equation through | An equation multiplied by a non-zero number has the same solutions |
| Interchange two rows | Reorders the equations | Order never affected the solution set |
| Add a multiple of one row to another | Replaces an equation by a combination | This is exactly elimination |
The constant in the first operation must be non-zero. Multiplying a row by zero destroys the equation entirely and changes the solution set.
Row-echelon form
Each row's leading entry is 1 and sits strictly to the right of the leading entry of the row above, with zeros beneath every leading 1. The result is a triangular arrangement.
- Choose the pivot. The leading entry of the current row.
- Make it 1 by dividing the row through.
- Clear below it by subtracting multiples of the pivot row from the rows underneath.
- Move down and right and repeat.
- Back-substitute from the bottom row upward.
Worked example — the source's 2×2 case
Reduce the matrix for x - 3y = 4, 2x + y = 1.
| 1 | -3 | 4 |
| 2 | 1 | 1 |
Subtract twice the first row from the second:
| 1 | -3 | 4 |
| 0 | 7 | -7 |
Divide the second row by 7:
| 1 | -3 | 4 |
| 0 | 1 | -1 |
Back-substitution: the second row says y = -1; the first says x - 3y = 4, so x = 4 + 3(-1) = 1.
1 - 3(-1) = 4 ✓ and 2(1) + (-1) = 1 ✓.
Worked example: three variables
The source's 3×3 system
For x + y + z = 5, x - 4y - 3z = 11, -2x + 2y + 5z = -30:
| 1 | 1 | 1 | 5 |
| 1 | -4 | -3 | 11 |
| -2 | 2 | 5 | -30 |
Clear the first column: subtract row 1 from row 2, and add twice row 1 to row 3.
| 1 | 1 | 1 | 5 |
| 0 | -5 | -4 | 6 |
| 0 | 4 | 7 | -20 |
Divide row 2 by -5 to make the pivot 1:
| 1 | 1 | 1 | 5 |
| 0 | 1 | 4/5 | -6/5 |
| 0 | 4 | 7 | -20 |
Subtract 4 times row 2 from row 3:
| 1 | 1 | 1 | 5 |
| 0 | 1 | 4/5 | -6/5 |
| 0 | 0 | 19/5 | -76/5 |
Divide row 3 by 195:
| 1 | 1 | 1 | 5 |
| 0 | 1 | 4/5 | -6/5 |
| 0 | 0 | 1 | -4 |
Back-substituting: z = -4; then y + 45(-4) = -65 gives y = 2; then x + 2 - 4 = 5 gives x = 7.
In the third original equation: -2(7) + 2(2) + 5(-4) = -30 ✓.
Recognising the outcome
The reduced matrix announces which of the three cases has occurred, without any back-substitution.
| Bottom row | Reads as | Outcome |
|---|---|---|
| [0 0 1 | k] | z = k | Unique solution |
| [0 0 0 | k], k ≠ 0 | 0 = k, false | Inconsistent — no solution |
| [0 0 0 | 0] | 0 = 0, true | Dependent — infinitely many |
The source's inconsistent example ends with a row reading [0 0 0 | 4] and is annotated inconsistent, no solution. Its consistent example ends with a zero row and is marked consistent system with solution y = 0, x = 4.
If the pivot position holds a zero, swap with a lower row that has a non-zero entry there. The typed notes do exactly this in one example, interchanging the second and third rows to bring a non-zero pivot into position.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Omitting a zero for a missing variable | Write the zero explicitly | Columns must align |
| Operating on columns | Only row operations are permitted | A column operation mixes different variables |
| Multiplying a row by zero | The constant must be non-zero | It destroys the equation |
| Clearing above the pivot too soon | Row-echelon form clears only below | That is Gauss-Jordan |
| Forgetting the augmented column | Every operation applies to the whole row | The constants must travel with the coefficients |
| Back-substituting from the top | Start at the bottom | The last row has one unknown |
Frequently asked questions
Why detach the coefficients?
Because the variable names carry no information during the elimination. Writing only the numbers is faster, less error-prone and is what a computer does.
What is a pivot?
The leading entry in a row, the one used to clear the entries below it. The typed notes circle each pivot as it is chosen.
Must the pivot be 1?
Not for row-echelon form in the loosest sense, but making it 1 simplifies every subsequent step. The typed notes reduce each pivot to 1 before clearing the column.
What if a pivot position holds a zero?
Swap that row with a lower one having a non-zero entry there. Row interchange is one of the three permitted operations for exactly this reason.
Source. Handwritten teaching notes, Week 11, pages 3-6, with the typed supplementary notes on Gaussian elimination.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.