Lesson 3: Gauss-Jordan and Reduced Row Echelon Form
Estimated time: 30-40 minutes
Learning Objectives
By the end of this lesson, you will be able to:
- Define reduced row echelon form (RREF) and its properties
- Continue from REF to RREF using back-elimination
- Read solutions directly from RREF without back-substitution
- Understand why RREF is unique for any given matrix
From REF to RREF
In Lesson 2, Gaussian elimination produced row echelon form (REF), and we used back-substitution to finish. Gauss-Jordan elimination takes it further: we continue using row operations to also eliminate entries above each pivot, and we scale each pivot to 1. The result is reduced row echelon form (RREF).
Reduced Row Echelon Form (RREF): A matrix is in RREF if it satisfies all REF conditions plus:
- Every pivot is equal to 1.
- Every pivot is the only nonzero entry in its column (zeros above and below).
Example: REF vs. RREF
REF (not reduced):
[ 0 1 1 | 4 ]
[ 0 0 2 | 6 ]
Pivots are 1, 1, and 2. The last pivot is not 1, and there are nonzero entries above pivots.
RREF:
[ 0 1 0 | 1 ]
[ 0 0 1 | 3 ]
Every pivot is 1, and each pivot column has zeros everywhere else. You can read off: x_1 = 1, x_2 = 1, x_3 = 3.
Gauss-Jordan Elimination Algorithm
Steps (continuing from REF)
- Scale each pivot to 1: Divide each pivot row by its pivot value.
- Eliminate above each pivot: Working from the rightmost pivot to the left, use row replacement to create zeros above each pivot.
Worked Example: Full Gauss-Jordan Elimination
Take the REF from above:
[ 0 1 1 | 4 ]
[ 0 0 2 | 6 ]
Step 1: Scale row 3. (1/2)R_3 → R_3:
[ 0 1 1 | 4 ]
[ 0 0 1 | 3 ]
Step 2: Eliminate above pivot in column 3.
R_2 - R_3 → R_2: [0, 1, 1-1 | 4-3] = [0, 1, 0 | 1]
R_1 - 3R_3 → R_1: [1, 2, 3-3 | 10-9] = [1, 2, 0 | 1]
[ 0 1 0 | 1 ]
[ 0 0 1 | 3 ]
Step 3: Eliminate above pivot in column 2.
R_1 - 2R_2 → R_1: [1, 2-2, 0 | 1-2] = [1, 0, 0 | -1]
[ 0 1 0 | 1 ]
[ 0 0 1 | 3 ]
Read the solution: x_1 = -1, x_2 = 1, x_3 = 3.
Uniqueness of RREF
Uniqueness Theorem: Every matrix has exactly one reduced row echelon form. Different sequences of row operations may be used, but they all lead to the same RREF.
This is important: while there are many possible REFs for a matrix (depending on which operations you choose), the RREF is unique. This makes RREF the canonical or standard form for any matrix.
Pivot Columns and Free Variables
In RREF, each pivot occupies a column. We call these pivot columns. Columns without a pivot correspond to free variables -- variables that can take any value. This is the key to understanding infinite solution sets, which we explore fully in Lesson 4.
Example: Identifying Pivot and Free Variables
RREF:
[ 0 1 -2 | 1 ]
[ 0 0 0 | 0 ]
Pivot columns: Columns 1 and 2 (where the 1s are).
Free variable: x_3 (column 3 has no pivot).
Reading the solution: x_1 + 3x_3 = 5 and x_2 - 2x_3 = 1. Let x_3 = t (any real number). Then x_1 = 5 - 3t, x_2 = 1 + 2t, x_3 = t.
Check Your Understanding
1. Is this matrix in RREF? [ 1 0 2 | 3 ] / [ 0 2 1 | 4 ]
2. Reduce to RREF: [ 1 3 | 7 ] / [ 0 2 | 4 ]
3. What is the advantage of RREF over REF?
4. In the RREF [ 1 0 -1 | 2 ] / [ 0 1 3 | -1 ] / [ 0 0 0 | 0 ], identify the pivot variables and free variables.
Key Takeaways
- RREF requires all pivots equal to 1 and zeros both above and below each pivot
- Gauss-Jordan elimination = forward elimination (to REF) + scaling pivots to 1 + back-elimination (clearing above pivots)
- Every matrix has a unique RREF
- Columns with pivots give pivot variables; columns without pivots give free variables
- RREF lets you read solutions directly without back-substitution
Next Lesson
Lesson 4 explores solution sets in depth -- unique, infinitely many, and no solution -- and how to express them in parametric vector form.
Start Lesson 4