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Row echelon form

In linear algebra, the row echelon form (REF) of a matrix is a structured arrangement achieved through elementary row operations, where all zero rows, if any, are positioned at the bottom, the leading nonzero entry (pivot) in each nonzero row is located strictly to the right of the pivot in the row above it, and all entries below each pivot are zeros.^[1] This form simplifies the analysis of matrices by revealing their rank and facilitating the solution of associated systems of linear equations.^[2] The process to obtain REF, known as Gaussian elimination, involves three types of row operations: swapping rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another, which do not alter the solution set of the linear system represented by the matrix.^[3] Unlike the stricter reduced row echelon form (RREF), which additionally requires each pivot to be 1 and the only nonzero entry in its column, REF allows pivots to be any nonzero value and permits nonzero entries above pivots.^[1] The reduced row echelon form of a matrix is unique, making these forms essential tools for computational efficiency in matrix theory.^[4] Row echelon form plays a central role in determining key properties of matrices, such as the dimension of the solution space (nullity) and the number of free variables in linear systems, as the number of nonzero rows equals the rank of the matrix.^[5] It is foundational in applications ranging from solving overdetermined systems in engineering to computing inverses and bases for subspaces in abstract algebra.^[6]

Definitions

Row Echelon Form

In linear algebra, a matrix is in row echelon form if it satisfies specific structural conditions that facilitate the analysis of linear systems.^[3] These conditions ensure that the matrix resembles a staircase pattern, with nonzero rows positioned above any zero rows and leading entries aligned in a way that simplifies row operations.^[4] The precise criteria for a matrix to be in row echelon form are as follows:^[1]

All zero rows, if any, are at the bottom of the matrix.
The leading entry (also called the pivot) in each nonzero row is nonzero, and this pivot is the leftmost nonzero entry in that row.
Each pivot is in a column where all entries below it are zero, and the pivot positions strictly increase from left to right across successive nonzero rows.

This form emphasizes the hierarchical structure of the rows, where each pivot marks the start of a nonzero segment that shifts rightward.^[7] A simple example is the 2×2 identity matrix, which is already in row echelon form:

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Here, the pivots are the 1's in positions (1,1) and (2,2), with zeros below each (trivially satisfied in the second row) and increasing column indices.^[4] Another example is the following 3×3 upper triangular matrix:

\begin{pmatrix} 2 & 2 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & 0 \end{pmatrix}

The pivots are at (1,1)=2 and (2,2)=3, with zeros below them in their columns, and the zero row at the bottom.^[3] In this context, the rows containing pivots are called pivot rows, and the columns containing pivots are pivot columns; these identify the structurally significant parts of the matrix without further computation.^[1] Reduced row echelon form extends this by requiring zeros above each pivot as well, ensuring a unique representation for each matrix rank.^[4]

Reduced Row Echelon Form

The reduced row echelon form (RREF), also known as row-reduced echelon form, refines the row echelon form by imposing stricter conditions to achieve a canonical representation of the matrix.^[8] Specifically, a matrix is in RREF if it is in row echelon form and satisfies two additional criteria: the leading entry (pivot) in each nonzero row is 1, and each such pivot is the only nonzero entry in its entire column, meaning all other entries in the pivot columns are zero.^[3] These requirements ensure that the pivots stand out distinctly, with zeros both above and below them, facilitating unique identification of the matrix's structure.^[9] This form builds directly on row echelon form by eliminating nonzero entries above the pivots, which are permitted in the basic version. Every matrix in reduced row echelon form is necessarily in row echelon form, but the reverse does not hold, as the basic form allows leading entries other than 1 and nonzero values above pivots.^[10] For illustration, consider the following 3×4 augmented matrix representing a system of linear equations, which is already in row echelon form but not reduced:

\begin{bmatrix} 1 & 2 & 0 & | & 5 \\ 0 & 1 & 3 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}

To obtain RREF, scale the second row if needed (here, the leading entry is already 1) and subtract 2 times the second row from the first row, yielding:

\begin{bmatrix} 1 & 0 & -4 & | & -3 \\ 0 & 1 & 3 & | & 4 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}

This final matrix is in reduced row echelon form, with pivots at (1,1) and (2,2) as the sole nonzeros in their columns.^[11]

Properties

Properties of Row Echelon Form

A matrix in row echelon form has the property that the number of its nonzero rows equals its rank, since each nonzero row features exactly one pivot (the first nonzero entry), and the count of such pivots defines the dimension of the row space, which is the rank.^[12] The positions of these pivots further reveal a basis for the column space of the original matrix: the columns of the original matrix that correspond to the pivot columns in the echelon form form a basis for that space.^[13] Algebraically, a square matrix in row echelon form lacks full rank—and thus has determinant zero—if and only if it contains at least one all-zero row, as this introduces a zero on the "diagonal" in the staircase structure, making the matrix singular.^[14] For a nonsingular square matrix reduced to row echelon form via elementary row operations, the determinant equals (-1)^k times the product of the pivot entries, where k is the number of row interchanges. This product of pivots thereby relates to the product of the matrix's eigenvalues (with multiplicity), as the determinant is precisely that product; in singular cases, the zero determinant reflects at least one zero eigenvalue, while the product of nonzero pivots connects to the product of nonzero eigenvalues when elimination avoids row scaling. The structural features of row echelon form, such as rank and pivot locations, are invariant under elementary row operations, as these operations preserve the row space and linear dependence relations among columns.^[15] The form itself remains a row echelon form under type III operations (adding a multiple of one row to another below it), which maintain zeros below pivots without altering leading positions; row scaling by a nonzero scalar preserves the form by keeping pivots nonzero, though scaling a pivot row to zero would eliminate a leading entry and violate the structure.^[16] Consider the $4 \times 4 matrix

\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 1 & 1 & 1 & 1 \end{pmatrix},

which after eliminating below the first pivot becomes

\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & -2 & -3 \end{pmatrix}.

Swapping the second and fourth rows, then scaling the new second row by -1 yields the row echelon form

\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.

Here, there are two nonzero rows and two pivots (in columns 1 and 2), confirming the rank is 2.

Properties of Reduced Row Echelon Form

The reduced row echelon form (RREF) of a matrix is unique, meaning that every matrix is row equivalent to exactly one matrix in RREF through elementary row operations.^[17] This canonical form distinguishes it from the row echelon form, which is not unique for a given matrix.^[4] In the RREF of a matrix, the pivot positions—locations of the leading 1s in each nonzero row—clearly identify the dependent and free variables in the corresponding system of equations. Pivot columns correspond to dependent variables, while non-pivot columns indicate free variables that can take arbitrary values, allowing parametrization of the solution set.^[18]^[19] The columns of the original matrix that correspond to the pivot columns in its RREF form a basis for the column space of the matrix, as these columns are linearly independent and span the column space.^[20]^[21] For example, consider the matrix

A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{pmatrix}.

Its RREF is

\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix},

where the pivot columns are the first and second, so the first and second columns of A form a basis for the column space. For the equation A\mathbf{x} = \mathbf{0}, the free variable is x_3 = t, yielding the parametrized solution \mathbf{x} = \begin{pmatrix} t \\ -2t \\ t \end{pmatrix} = t \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.^[20]^[18]

Applications

Systems of Linear Equations

In the context of systems of linear equations, the row echelon form of an augmented matrix simplifies the process of determining solutions by organizing the equations into a triangular structure, where each leading nonzero entry (pivot) is to the right of the pivot in the row above, zeros appear below each pivot, and any zero rows are at the bottom. This form enables the use of back-substitution, a method that solves for variables starting from the bottom row and proceeding upward, substituting each value into the equations above.^[22]^[3] The type of solution depends on the structure of the row echelon form. A system has a unique solution if the number of pivots equals the number of variables, indicating full rank and no free variables. Infinite solutions arise when there are fewer pivots than variables, leaving free variables that can take any value, parameterized by those freedoms. The system is inconsistent, with no solution, if a row consists of zeros in the coefficient columns but a nonzero entry in the augmented column, representing an equation like $0 = b where b \neq 0.^[23]^[24] For illustration, consider the augmented matrix of a 3x3 system already reduced to row echelon form:

\begin{pmatrix} 1 & 2 & 3 & | & 6 \\ 0 & 1 & 1 & | & 4 \\ 0 & 0 & 1 & | & 2 \end{pmatrix}

This corresponds to the equations x + 2y + 3z = 6, y + z = 4, and z = 2. Back-substitution begins with the last row: z = 2. Substituting into the second row gives y + 2 = 4, so y = 2. Finally, the first row yields x + 2(2) + 3(2) = 6, so x + 4 + 6 = 6, hence x = -4. This yields the unique solution x = -4, y = 2, z = 2.^[1]^[22] To detect inconsistency, examine rows like \begin{pmatrix} 0 & 0 & 0 & | & 5 \end{pmatrix}, which implies $0 = 5, indicating no solution exists for the system. Reduced row echelon form can further simplify expressing infinite solutions in parametric form.^[23]^[3]

Rank and Solutions

In the context of a system of linear equations Ax = b, where A is an m \times n matrix, the row echelon form of the augmented matrix [A | b] provides a direct way to determine the rank of A. The rank r of A is equal to the number of pivot positions (or nonzero rows) in the row echelon form of A.^[12] This rank r represents the dimension of the column space (or row space) of A, indicating the maximum number of linearly independent rows or columns.^[25] For the homogeneous system Ax = 0, the solution set is the null space of A, a subspace of \mathbb{R}^n. The dimension of this null space, known as the nullity of A, is given by n - r, which corresponds to the number of free variables in the row echelon form.^[26] If r = n (full column rank), the nullity is zero, and the only solution is the trivial solution x = 0.^[27] Otherwise, when r < n, the solution space has dimension greater than zero, yielding infinitely many solutions parametrized by the free variables. For the non-homogeneous system Ax = b with b \neq 0, the system is consistent if and only if the rank of A equals the rank of the augmented matrix [A | b].^[12] In this case, the solution set forms an affine subspace: it consists of a particular solution x_p plus the null space of A, translated by x_p.^[28] The dimension of this affine space remains the nullity n - r, reflecting the same degrees of freedom as in the homogeneous case. If the ranks differ, no solutions exist. Consider the rank-deficient $2 \times 3 matrix A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{pmatrix}, which row reduces to the echelon form \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{pmatrix}, yielding rank r = 1 and nullity n - r = 2. For the homogeneous system Ax = 0, the solutions are parametrized by two free variables, say x_2 = s and x_3 = t, with x_1 = -2s - 3t, giving infinitely many solutions in a 2-dimensional subspace.^[26] For a consistent non-homogeneous system, such as with b = \begin{pmatrix} 3 \\ 6 \end{pmatrix}, a particular solution is x_p = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, and the general solution is x = x_p + (homogeneous solutions), forming a 2-dimensional affine plane.^[28]

Computation

Gaussian Elimination to Row Echelon Form

Gaussian elimination is an algorithmic process that applies elementary row operations to an augmented matrix representing a system of linear equations, transforming it into row echelon form without altering the solution set.^[29] The three elementary row operations are: (1) interchanging any two rows, (2) multiplying all entries in a row by a nonzero scalar, and (3) adding a scalar multiple of one row to another row. To illustrate, consider the matrix

\begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 1 \\ 4 & 0 & 2 \end{pmatrix}.

Interchanging the first and third rows yields

\begin{pmatrix} 4 & 0 & 2 \\ 0 & 3 & 1 \\ 2 & 1 & 0 \end{pmatrix}.

^[30] To eliminate the entry below the pivot in column 1, subtract \frac{1}{2} times the first row from the third row, resulting in

\begin{pmatrix} 4 & 0 & 2 \\ 0 & 3 & 1 \\ 0 & 1 & -1 \end{pmatrix}. $$ For column 2, the entry in row 2 is 3 (nonzero), so no swap is needed. To eliminate below in column 2, subtract $ \frac{1}{3} $ times the second row from the third row, yielding

\begin{pmatrix} 4 & 0 & 2 \ 0 & 3 & 1 \ 0 & 0 & -\frac{4}{3} \end{pmatrix}.

The Gaussian elimination algorithm proceeds column by column on an $ m \times n $ augmented matrix $ A $, aiming to create zeros below each pivot position. The steps are: 1. For each column $ k = 1, 2, \dots, \min(m, n) $: - Identify the pivot row $ i \geq k $ such that $ a_{ik} \neq 0 $; if no such row exists, skip to the next column. - Swap row $ k $ with row $ i $ to place the pivot in position $ (k, k) $. - Scale row $ k $ by dividing by the pivot entry $ a_{kk} $ to make the pivot equal to 1. - For each row $ j = k+1, k+2, \dots, m $, add $ -a_{jk} $ times row $ k $ to row $ j $ to eliminate entries below the pivot.[](https://math.mit.edu/~dav/gauss19.pdf) This forward elimination phase continues until the matrix is in row echelon form, where all entries below pivots are zero.[](https://tutorial.math.lamar.edu/classes/de/la_systems.aspx) Pseudocode for the algorithm on an $ m \times n $ matrix is as follows: ``` for k = 1 to min(m, n) # Find pivot i = k while i <= m and A[i][k] == 0 i = i + 1 if i > m continue # No pivot in this column # Swap rows swap rows k and i # Scale pivot row for j = k to n A[k][j] = A[k][j] / A[k][k] # Eliminate below for i = k+1 to m factor = A[i][k] for j = k to n A[i][j] = A[i][j] - factor * A[k][j] ``` This implementation assumes real numbers and handles the augmented part uniformly up to column $ n $.[](https://people.tamu.edu/~yvorobets/MATH304-2010A/Lect1-02web.pdf) As an example, apply the algorithm to the augmented matrix for the system $ x + 2y + 3z = 1 $, $ 2x + 5y + 3z = 2 $, $ 3x + 6y + 6z = 4 $:

\begin{pmatrix} 1 & 2 & 3 & | & 1 \ 2 & 5 & 3 & | & 2 \ 3 & 6 & 6 & | & 4 \end{pmatrix}. $$ The first pivot is already 1 at position (1,1). Eliminate below: subtract 2 times row 1 from row 2, yielding row 2 as [0, 1, -3 | 0]; subtract 3 times row 1 from row 3, yielding row 3 as [0, 0, -3 | 1]. The matrix is now

\begin{pmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 1 & -3 & | & 0 \\ 0 & 0 & -3 & | & 1 \end{pmatrix}.

^[31] For column 2, the pivot is 1 at (2,2). No elimination needed below since row 3 has zero there. For column 3, scale row 3 by -\frac{1}{3} to make the pivot 1, resulting in row 3 as [0, 0, 1 | -\frac{1}{3}]. The final row echelon form is

\begin{pmatrix} 1 & 2 & 3 & | & 1 \\ 0 & 1 & -3 & | & 0 \\ 0 & 0 & 1 & | & -\frac{1}{3} \end{pmatrix}. $$ This form reveals the structure for further analysis, such as achieving reduced row echelon form by eliminating above pivots.[](https://math.mit.edu/~dav/gauss19.pdf) ### Reduction to Reduced Row Echelon Form Once the matrix has been transformed into row echelon form via the forward elimination phase of Gaussian elimination, the process continues with backward elimination to obtain the reduced row echelon form (RREF). This phase begins at the bottommost pivot and proceeds upward, systematically eliminating all non-zero entries above each pivot position. For each pivot row, first scale the row (if necessary) so that the pivot entry is 1 by dividing the entire row by the pivot value; then, for every row above the pivot row, subtract an appropriate multiple of the pivot row to zero out the entry in the pivot column. These operations ensure that each leading entry (pivot) is 1 and the only non-zero element in its column, while preserving the row space of the original matrix.[](https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/dddb31dfe72d2e2e2fd09e74713b7775_MIT18_06SCF11_Ses1.7sum.pdf) The complete algorithm for reduction to RREF integrates the forward and backward phases into a single procedure. It starts with the standard [Gaussian elimination](/page/Gaussian_elimination) steps to achieve row echelon form by introducing zeros below pivots from top to bottom. This is immediately followed by the backward phase described above, processing pivots from bottom to top to introduce zeros above pivots and normalize them to 1. Row swaps may occur during the forward phase if a pivot is zero, but no additional swaps are typically needed in the backward phase. The resulting RREF uniquely represents the row equivalence class of the matrix.[](https://textbooks.math.gatech.edu/ila/row-reduction.html) For an $m \times n$ matrix, the time complexity of this full row reduction algorithm is $O(m n \min(m, n))$, accounting for the arithmetic operations across both phases.[](https://www-users.cse.umn.edu/~olver/num_/lng.pdf) Consider the following $3 \times 3$ matrix already in row echelon form after forward elimination:

\begin{pmatrix} 1 & 2 & 0 \ 0 & 1 & 3 \ 0 & 0 & 0 \end{pmatrix}

The second pivot is already 1, so proceed to eliminate above it: subtract 2 times row 2 from row 1, yielding

\begin{pmatrix} 1 & 0 & -6 \ 0 & 1 & 3 \ 0 & 0 & 0 \end{pmatrix}.

The first pivot is 1 and now has zeros above and below (trivially, as it is the top row), completing the reduction to RREF.[](https://sites.millersville.edu/bikenaga/linear-algebra/row-reduction/row-reduction.html) ## Advanced Concepts ### Uniqueness of Reduced Forms The reduced row echelon form (RREF) of any matrix is unique, meaning that every matrix is row equivalent to exactly one matrix in RREF. This property ensures that the RREF serves as a canonical representative for all matrices sharing the same row space, regardless of the specific sequence of elementary row operations used to obtain it.[](http://linear.ups.edu/html/section-RREF.html)[](https://web.ma.utexas.edu/users/jmeth/DEFall18/LinAlg3.html) To establish this uniqueness, consider two matrices $ R $ and $ S $, both in RREF and row equivalent to the same original matrix $ A $. Since row equivalence preserves the row space, $ R $ and $ S $ generate the same subspace. The pivot positions—the columns containing the leading 1's in the nonzero rows of $ R $ and $ S $—must coincide. These positions are uniquely determined by the dimension of the row space (the rank) and the specific linear dependencies among the rows; if $ R $ had a pivot in a column where $ S $ did not, applying the row operations transforming $ S $ to $ R $ would introduce a leading entry in that column for $ S $, contradicting $ S $'s RREF properties, and vice versa. With matching pivot positions established, the requirements of RREF further dictate that all entries above and below each pivot must be zero, and each pivot entry is exactly 1, with no other nonzero entries in those columns. These constraints, combined with the fact that the rows of $ R $ and $ S $ are linear combinations of each other, force every entry in $ R $ to equal the corresponding entry in $ S $. Therefore, $ R = S $.[](https://people.eecs.berkeley.edu/~wkahan/MathH110/RREF1.pdf)[](https://ximera.osu.edu/laode/linearAlgebra/solvingLinearEquations/uniquenessOfReducedEchelonForm) This uniqueness implies that the RREF provides a standard representative for expressing linear dependence relations among the rows of the original matrix. In particular, the positions of the pivots identify the linearly independent rows, while the nonzero entries above the pivots encode the coefficients of the dependencies among all rows, offering a unique, normalized description of the row space's structure.[](http://linear.ups.edu/html/section-RREF.html)[](https://people.tamu.edu/~yvorobets//MATH423-2012A/Lect2-04web.pdf) The foundations of this form trace back to the elimination method employed by Carl Friedrich Gauss in the early 19th century, initially for solving linear systems in astronomical computations, with the uniqueness aspect formalized in subsequent developments of linear algebra theory.[](https://www.cis.upenn.edu/~cis6100/Notices-06-11-Gausselim.pdf) ### Affine Spaces of Echelon Forms The set of all matrices in reduced row echelon form with a fixed rank $ r $ and fixed pivot positions forms an affine subspace of the vector space of all $ m \times n $ matrices over the field. This affine structure arises because the pivot columns are fixed to contain the canonical 1's and 0's, while the entries in the non-pivot columns occupy specific positions determined by the echelon shape, parametrizing a translate of a vector subspace corresponding to those free positions. The affine dimension of this space equals the number of non-pivot entries above the "staircase" defined by the pivot positions, which counts the allowable arbitrary entries in the non-pivot columns for each pivot row without violating the leading 1 condition. For pivot positions $ j_1 < j_2 < \cdots < j_r $, the dimension is $ \sum_{i=1}^r (n - j_i - (r - i)) $, reflecting the free spots to the right of each pivot in non-pivot columns. In the standard case of consecutive pivots in the first $ r $ columns, this simplifies to $ r(n - r) $. Consider 3×3 matrices of rank 2. If the pivot positions are columns 1 and 2, the reduced row echelon forms are

\begin{pmatrix} 1 & 0 & a \ 0 & 1 & b \ 0 & 0 & 0 \end{pmatrix},

where $ a, b $ are arbitrary scalars, yielding an affine space of dimension 2 parametrized by these non-pivot entries above the staircase in column 3, rows 1 and 2. If instead the pivots are in columns 1 and 3, the forms are

\begin{pmatrix} 1 & a & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{pmatrix},

with only $ a $ free (in row 1, column 2, above the staircase), giving dimension 1. These examples illustrate how the affine structure and dimension depend on the specific pivot configuration.

References

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May 18, 2008 · Gaussian Elimination is a simple, systematic algorithm to solve systems of linear equations. It is the workhorse of linear algebra, and, as such ...
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Uniqueness of Reduced Echelon Form - Ximera
In this section we prove Theorem ??, which states that every matrix is row equivalent to precisely one reduced echelon form matrix. Proof of Theorem ??:
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[PDF] MATH 423 Linear Algebra II Lecture 17: Reduced row echelon form ...
Reduced row echelon form. A matrix is said to be in the reduced row echelon form if. (i) it is in the row echelon form (i.e., leading entries shift to.
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[PDF] Mathematicians of Gaussian Elimination - CIS UPenn
Gaussian elimination is universally known as “the” method for solving simultaneous linear equations. As. Leonhard Euler remarked, it is the.
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Low-Rank Matrix Recovery using Gabidulin Codes in Characteristic ...
... matrix completion [27] . ... Systematic encoders ... reduced row echelon form of a k×n matrix. ... machine learning. In many applications where low-rank ...