Reduced Row Echelon (RREF)
Introduction
Reduced Row Echelon Form (RREF) is a powerful way to simplify a system of linear equations so that the solutions become obvious.
Since you already know Row Echelon Form (REF), you can think of RREF as “REF, but cleaner and more decisive.”
RREF helps you:
- Identify unique solutions quickly
- Detect when a system has no solution
- Detect when a system has infinitely many solutions
- Read solutions directly from the matrix without back‑substitution
What is RREF?
A matrix is in Reduced Row Echelon Form (RREF) when:
- Every pivot is 1
- Each pivot is the only nonzero entry in its column
- Pivots move to the right as you go down
- Any zero rows are at the bottom
- The matrix is simplified enough that the solution can be read directly, without back‑substitution
To see what this means, let’s compare a matrix in REF with its fully simplified RREF version.
Example: REF vs RREF
Matrix in REF (Row Echelon Form)
$$\left[ \begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right]$$ Why this is REF:
- Pivots move right as you go down (columns 1 → 2).
- All entries below pivots are zero.
- Pivots do not need to be the only nonzero entries in their columns.
- Back‑substitution is still required to read the solution.
For example, the first row still has a “2” above the pivot in column 2.
Matrix in RREF (Reduced Row Echelon Form)
$$\left[ \begin{array}{cc|c} 1 & 0 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right]$$ Why this is RREF:
This is the “cleanest possible” form of the system.
Summary of the contrast
| Feature | REF | RREF |
|---|
| Pivots move right as you go down | ✔️ | ✔️ |
| Zeros below pivots | ✔️ | ✔️ |
| Pivots must be 1 | ❌ | ✔️ |
| Pivots must be the only nonzero entry in their column | ❌ | ✔️ |
| Back‑substitution required | ✔️ | ❌ |
| Solutions readable immediately | ❌ | ✔️ |
Why RREF is Useful
RREF makes solutions easy to read:
- If each variable has a pivot → unique solution
- If a row becomes $[0\ 0\ 0\ |\ b]$ with $b \neq 0$ → no solution
- If some variables have no pivot → infinitely many solutions
RREF removes the need for back‑substitution because the matrix is already fully simplified.
How to Compute RREF
Start from REF and continue simplifying:
1. Make each pivot equal to 1
If a pivot is $k$, divide the entire row by $k$.
2. Clear the column above each pivot
Use row operations to make all entries above the pivot equal to 0.
3. Clear the column below each pivot
(You already did this when forming REF.)
4. Move to the next pivot
Repeat until all pivots are clean and isolated.
A Simple Example
Consider the system: $$\left[ \begin{array}{cc|c} 1 & 2 & 5 \\ 2 & 4 & 10 \end{array} \right]$$ Steps:
- Row 2 → Row 2 $- 2 \cdot$ Row 1
Gives: $$\left[ \begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 0 & 0 \end{array} \right]$$
This is already in RREF because:
- Pivot in column 1 is 1
- Pivot column has zeros everywhere else
- No other pivots exist
This tells us:
- $x + 2y = 5$
- $y$ is free
- Infinitely many solutions
Example: From REF to RREF (pivots need scaling)
Start with this REF augmented matrix: $$\left[ \begin{array}{cc|c} 2 & 4 & 6 \\ 0 & 3 & 9 \end{array} \right]$$ This corresponds to:
Why this is REF:
- Pivot in row 1 is in column 1 (value $2$)
- Pivot in row 2 is in column 2 (value $3$)
- Zeros below each pivot
- But pivots are not 1, and pivot columns are not yet “clean”
Step 1: Scale the pivots to 1
- Row 1 → Row 1 ÷ $2$
- Row 2 → Row 2 ÷ $3$
We get: $$\left[ \begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 1 & 3 \end{array} \right]$$ Now both pivots are 1.
Step 2: Clear above the pivot in column 2
Pivot: the $1$ in row 2, column 2.
Row 1 has a 2 above it.
- Row 1 → Row 1 $- 2$·Row 2
Compute:
- New column 1: $1 - 2(0) = 1$
- New column 2: $2 - 2(1) = 0$
- New augmented entry: $3 - 2(3) = -3$
So the matrix becomes: $$\left[ \begin{array}{cc|c} 1 & 0 & -3 \\ 0 & 1 & 3 \end{array} \right]$$
Final RREF
$$\left[ \begin{array}{cc|c} 1 & 0 & -3 \\ 0 & 1 & 3 \end{array} \right]$$ This is RREF:
- Pivots are 1
- Each pivot is the only nonzero entry in its column
- Pivots move right as you go down
Interpreting the solution
Let the variables be $x$ and $y$.
From the RREF:
So the system has a single unique solution: $$(x, y) = (-3,\ 3)$$
Example (No solution)
REF matrix
Start with: $$\left[ \begin{array}{cc|c} 1 & 2 & 3 \\ 2 & 4 & 10 \end{array} \right]$$ This corresponds to:
- $x + 2y = 3$
- $2x + 4y = 10$
Step 1: Put into REF
Eliminate the $2$ below the first pivot:
- Row 2 → Row 2 $- 2$·Row 1
Compute Row 2:
- New column 1: $2 - 2(1) = 0$
- New column 2: $4 - 2(2) = 0$
- New augmented entry: $10 - 2(3) = 4$
So we get: $$\left[ \begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 4 \end{array} \right]$$ This is REF.
Step 2: Interpret the RREF (it’s already “reduced enough”)
The second row says: $$0x + 0y = 4$$ which simplifies to: $$0 = 4$$ This is impossible, so the system has no solution.
In RREF language:
- A row of the form $[0\ 0\ |\ b]$ with $b \neq 0$ means inconsistent system
- Therefore, no solution exists for this system.
Calculator
Calculating the RREF
- To calculate the reduced row echelon form we use the $\operatorname{rref}()$ function
rref([2, 4, 6; 0, 3, 9]) rref([3, 6, 9; 0, 2, 4])
Exercises
- Reduce the REF matrix $$\left[ \begin{array}{cc|c} 2 & 4 & 6 \\ 0 & 3 & 9 \end{array} \right]$$ to RREF and state the solution.
- Reduce the REF matrix $$\left[ \begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 0 & 0 \end{array} \right]$$ to RREF and describe the solution set.
- Reduce the REF matrix $$\left[ \begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 4 \end{array} \right]$$ to RREF and determine whether a solution exists.
- Convert the REF matrix $$\left[ \begin{array}{cc|c} 3 & 6 & 9 \\ 0 & 2 & 4 \end{array} \right]$$ into RREF.
- Reduce the REF matrix $$\left[ \begin{array}{cc|c} 1 & 1 & 2 \\ 0 & 1 & 1 \end{array} \right]$$ to RREF and state the solution.