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Reduced Row-Echelon Form of a Matrix — Definition & Examples

Reduced Row-Echelon Form of a Matrix

A matrix form used in solving linear systems of equations.

 

Properties and example of a reduced row-echelon matrix with three rules listed and a 3×4 matrix showing...

 

See also

Gauss-Jordan elimination, row-echelon form

Key Formula

A matrix is in RREF if:\text{A matrix is in RREF if:} 1. All rows of all zeros are at the bottom.2. The leading (first nonzero) entry of each nonzero row is 1 (called a pivot).3. Each pivot is to the right of the pivot in the row above it.4. Each pivot is the only nonzero entry in its column.\begin{aligned} &1.\ \text{All rows of all zeros are at the bottom.} \\ &2.\ \text{The leading (first nonzero) entry of each nonzero row is } 1 \text{ (called a pivot).} \\ &3.\ \text{Each pivot is to the right of the pivot in the row above it.} \\ &4.\ \text{Each pivot is the only nonzero entry in its column.} \end{aligned}
Where:
  • pivot\text{pivot} = The leading 1 in each nonzero row; it is the first nonzero entry reading left to right.
  • pivot column\text{pivot column} = A column that contains a pivot. In RREF, every other entry in a pivot column is 0.

Worked Example

Problem: Use row operations to reduce the following augmented matrix to reduced row-echelon form and solve the system: [12−1325080122]\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 2 & 5 & 0 & 8 \\ 0 & 1 & 2 & 2 \end{array}\right]
Step 1: Eliminate the 2 in row 2, column 1 by replacing R₂ with R₂ − 2R₁.
R2←R2−2R1:[12−1301220122]R_2 \leftarrow R_2 - 2R_1: \quad \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 1 & 2 & 2 \end{array}\right]
Step 2: Eliminate the 1 in row 3, column 2 by replacing R₃ with R₃ − R₂.
R3←R3−R2:[12−1301220000]R_3 \leftarrow R_3 - R_2: \quad \left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]
Step 3: The matrix is now in row-echelon form. To reach reduced row-echelon form, eliminate the 2 in row 1, column 2 by replacing R₁ with R₁ − 2R₂.
R1←R1−2R2:[10−5−101220000]R_1 \leftarrow R_1 - 2R_2: \quad \left[\begin{array}{ccc|c} 1 & 0 & -5 & -1 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right]
Step 4: The matrix is in RREF. Each pivot column (columns 1 and 2) has a leading 1 with all other entries 0. Column 3 has no pivot, so x₃ is a free variable. Read off the solution: x₁ = −1 + 5x₃ and x₂ = 2 − 2x₃, where x₃ = t for any real number t.
x1=−1+5t,x2=2−2t,x3=tx_1 = -1 + 5t, \quad x_2 = 2 - 2t, \quad x_3 = t
Answer: The RREF reveals infinitely many solutions: (x1,x2,x3)=(−1+5t,  2−2t,  t)(x_1, x_2, x_3) = (-1 + 5t,\; 2 - 2t,\; t) for any real number tt.

Another Example

This example shows a system with a unique solution (every variable column is a pivot column) and requires scaling rows to create leading 1s, unlike the first example where the leading entries were already 1.

Problem: Reduce the following 2×3 augmented matrix to RREF and solve: [24103511]\left[\begin{array}{cc|c} 2 & 4 & 10 \\ 3 & 5 & 11 \end{array}\right]
Step 1: Scale R₁ by 1/2 so the leading entry becomes 1.
R1←12R1:[1253511]R_1 \leftarrow \tfrac{1}{2}R_1: \quad \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 3 & 5 & 11 \end{array}\right]
Step 2: Eliminate the 3 in row 2, column 1 by replacing R₂ with R₂ − 3R₁.
R2←R2−3R1:[1250−1−4]R_2 \leftarrow R_2 - 3R_1: \quad \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & -1 & -4 \end{array}\right]
Step 3: Scale R₂ by −1 so the leading entry becomes 1.
R2←−R2:[125014]R_2 \leftarrow -R_2: \quad \left[\begin{array}{cc|c} 1 & 2 & 5 \\ 0 & 1 & 4 \end{array}\right]
Step 4: Eliminate the 2 in row 1, column 2 by replacing R₁ with R₁ − 2R₂. This makes the pivot in column 2 the only nonzero entry in that column.
R1←R1−2R2:[10−3014]R_1 \leftarrow R_1 - 2R_2: \quad \left[\begin{array}{cc|c} 1 & 0 & -3 \\ 0 & 1 & 4 \end{array}\right]
Answer: The RREF gives the unique solution x1=−3x_1 = -3 and x2=4x_2 = 4.

Frequently Asked Questions

What is the difference between row-echelon form and reduced row-echelon form?
Row-echelon form (REF) requires leading entries in a staircase pattern with zeros below each pivot, but the leading entries do not need to be 1, and entries above the pivots can be nonzero. Reduced row-echelon form (RREF) adds two extra conditions: every leading entry must be exactly 1, and it must be the only nonzero entry in its column. REF is reached by Gaussian elimination; RREF is reached by Gauss-Jordan elimination.
Is the reduced row-echelon form of a matrix unique?
Yes. While the sequence of row operations you use may differ, every matrix has exactly one RREF. This is a proven theorem in linear algebra. Row-echelon form, by contrast, is not unique — the same matrix can be reduced to different REF results depending on the operations chosen.
When should you use RREF instead of REF?
Use RREF when you want to read off the solution directly from the matrix without needing back-substitution. REF combined with back-substitution is often faster by hand for small systems, but RREF is more systematic and is what calculators and computer algebra systems typically produce.

Reduced Row-Echelon Form (RREF) vs. Row-Echelon Form (REF)

Reduced Row-Echelon Form (RREF)Row-Echelon Form (REF)
Leading entriesMust be 1 (called pivots)Can be any nonzero number
Entries above pivotsMust all be 0Can be nonzero
Entries below pivotsMust all be 0Must all be 0
UniquenessUnique for every matrixNot unique
Method to obtainGauss-Jordan eliminationGaussian elimination
Reading the solutionRead directly from the matrixRequires back-substitution

Why It Matters

RREF appears throughout linear algebra courses, from solving systems of linear equations to finding the rank of a matrix, determining whether a set of vectors is linearly independent, and computing inverse matrices. It is also the standard output of matrix-solving functions on graphing calculators and software like MATLAB or Python's NumPy. Mastering RREF is essential because it provides a systematic, algorithmic approach to problems that would otherwise require guesswork or case-by-case analysis.

Common Mistakes

Mistake: Stopping at row-echelon form and calling it RREF.
Correction: After getting zeros below each pivot (REF), you must continue: scale each pivot to 1 and then eliminate all entries above each pivot so that every pivot column contains a single 1 and zeros elsewhere.
Mistake: Forgetting to identify free variables when a column has no pivot.
Correction: Any column without a pivot corresponds to a free variable. Assign it a parameter (like t) and express the pivot variables in terms of it. Missing this step leads to incorrectly claiming a unique solution when infinitely many exist.

Related Terms