Row Echelon Form (REF) & RREF

Row Echelon Form

Fact: A ladder hides in the name

Echelon is the French échelon, a rung of a ladder (échelle). Armies use it for troops drawn up in staggered steps, each unit set off to one side of the one ahead: the same offset staircase the pivots form.

A matrix is in row echelon form (REF) if it satisfies three conditions. Every zero row (a row of all zeros) sits at the bottom of the matrix. The leading entry of each nonzero row — called the pivot — is strictly to the right of the pivot in the row above. Every entry below a pivot is zero.

Pivot Definition

\text{pivot} = \text{leading nonzero entry of a row in echelon form}

Pivot definition →

These conditions create a staircase pattern that descends from upper-left to lower-right:

Row Echelon Form

\begin{pmatrix} \boxed{2} & 3 & -1 & 4 & 7 \\ 0 & \boxed{1} & 5 & 0 & 2 \\ 0 & 0 & 0 & \boxed{4} & -3 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

The boxed entries are pivots. Each sits to the right of and below the previous one, with zeros filling in below and to the left. The bottom row is all zeros. In general schematic form, with

\boxed{\ast}

marking pivots and

\bullet

any entry:

Row Echelon Form Definition

\text{REF: } \begin{pmatrix} \boxed{\ast} & \bullet & \bullet & \bullet & \bullet \\ 0 & \boxed{\ast} & \bullet & \bullet & \bullet \\ 0 & 0 & 0 & \boxed{\ast} & \bullet \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

Row echelon form definition →

REF is not unique. Different sequences of row operations applied to the same matrix can produce different echelon forms — the pivots may have different values, and the non-pivot entries above the pivots may differ.

THEOREM

What remains the same across all echelon forms of a given matrix is the set of pivot positions (which columns contain pivots).

Echelon form: a staircase of pivots

Each pivot sits strictly to the right of the one above it and every entry below a pivot is zero, which is what makes the shape a staircase. Rows of all zeros, if any, have sunk to the bottom. The form reached depends on the route taken through the elimination. Try a different pivot order and compare on the Gaussian elimination visualizer.

This is as far as plain Gaussian elimination goes; everything below this section is about what the extra reduction buys.

Echelon Form Notation

Notation

Echelon Form Notation

Two acronyms that differ by one word and by whether the article in front of them may be "the", a marked entry that is a position rather than a value, and a solution written with letters that stand for nothing in particular. All of them are catalogued among the linear algebra symbols.

The augmented bar

[A \mid \mathbf{b}]

, the row-operation arrows and

A \sim B

row equivalence come from Gaussian elimination notation; the

m \times n

shape and entry brackets from matrix operation notation;

\operatorname{rank}

from rank notation.

REF, RREF

row echelon form; reduced row echelon form

Two acronyms one word apart, and the extra word changes the grammar around them. REF is not unique — as this section notes, different row operations give different echelon forms — so a matrix has an REF. RREF is unique, which is what licenses the definite article: the RREF of a matrix, the subject of Uniqueness of RREF below.

Also writtenSpelled out in full in most European texts, which avoids the acronym entirely; some sources write

\operatorname{rref}(A)

as an operator, treating the unique reduced form as a function of the matrix — a spelling only the reduced version can support, precisely because it is unique.

Do not confuseUsing "the" with REF. Saying the echelon form of a matrix quietly asserts a uniqueness that only the reduced form has; what every echelon form does share is the set of pivot positions, not the entries.

\boxed{2}

,

\boxed{\ast}

a boxed entry marks a pivot — the leading nonzero entry of its row

The box is presentational, not algebraic: it highlights which entries are pivots without changing them, and the schematic form replaces the values with

\ast

to show only the staircase. Circling is the common alternative on a blackboard. A pivot position is a location in the array; a pivot column is a column containing one.

CasesIn RREF every pivot is forced to

1

— the "leading 1" — so the box becomes redundant and texts drop it; in plain REF the pivot can be any nonzero value, which is exactly why it needs marking there and not here.

Do not confuseA pivot's value with its position. What is invariant across all echelon forms of a matrix is where the pivots sit, not what they are — so "the pivots of

A

" means the positions, and it is the positions that determine rank.

\mathbf{x} = \mathbf{p} + t\mathbf{v}_1 + s\mathbf{v}_2

the general solution: a particular vector plus free directions

Each free column contributes one letter, and the letters are arbitrary:

t

,

s

, or

t_1, t_2, \ldots

when there are many — Parametric Vector Form below assembles them. A free variable is one whose column holds no pivot, so the pivot pattern decides how many letters the answer needs.

CasesThe count is forced, not chosen: free variables

= n - \operatorname{rank}(A)

, so the number of parameters is fixed even though their names are not. A unique solution is the case where that count is zero and the parameter part disappears entirely.

Do not confuseReading the parameters as unknowns still to be solved for. They are not — they range over all scalars, and the expression is the whole solution set at once rather than an equation waiting to be finished.

All linear algebra symbols →↑ Linear Systems

Reduced Row Echelon Form

A matrix is in reduced row echelon form (RREF) if it satisfies two additional conditions beyond REF. Each pivot is equal to

1

. Each pivot is the only nonzero entry in its column — all entries above the pivot are also zero, not just below.

Applying these conditions to the previous example:

Reduced Row Echelon Form

\begin{pmatrix} 1 & 0 & -8 & 0 & 2 \\ 0 & 1 & 5 & 0 & 2 \\ 0 & 0 & 0 & 1 & -\frac{3}{4} \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

Every pivot is

1

, and every other entry in a pivot column is

0

. The pivot columns are clean unit vectors within the matrix. The non-pivot columns (columns

3

and

5

in this example) can contain anything. In general schematic form:

Reduced Row Echelon Form Definition

\text{RREF: } \begin{pmatrix} \boxed{1} & 0 & \bullet & 0 & \bullet \\ 0 & \boxed{1} & \bullet & 0 & \bullet \\ 0 & 0 & 0 & \boxed{1} & \bullet \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}

Reduced row echelon form definition →

RREF is achieved from REF by Gauss-Jordan elimination: scale each pivot row so the pivot becomes

1

, then use row addition to eliminate all entries above each pivot.

Reduced form: pivots alone in their columns

Every pivot is one and is the only non-zero entry in its column. Unlike the echelon form above, this one does not depend on how the elimination was carried out — every route ends at the same matrix, which is what makes uniqueness claims about it possible. Verify that by taking a different route on the Gaussian elimination visualizer.

Uniqueness is what lets the reduced form be used as a canonical description of the matrix rather than merely a convenient one.

Uniqueness of RREF

Every matrix has exactly one RREF. No matter which sequence of row operations is used to reach it, the result is the same:

RREF Uniqueness

\text{RREF}(A) \text{ is unique}

Rref uniqueness →

This is not true of REF — different reduction paths can produce different row echelon forms of the same matrix, with different values in the non-pivot entries. But RREF is canonical: it is the unique representative of the matrix's row-equivalence class in reduced form.

The uniqueness of RREF implies that the pivot positions are intrinsic to the matrix. They do not depend on how the reduction is carried out. This means the rank (the number of pivots), the free variables (the non-pivot columns), and the entire solution structure are determined by the matrix itself, not by any particular algorithm applied to it.

Aspect	Row Echelon Form (REF)	Reduced Row Echelon Form (RREF)
Defining conditions	zero rows at the bottom; pivots staircase strictly right; entries below each pivot are 0	all REF conditions + each pivot equals 1 + each pivot is the only nonzero entry in its column
Pivot values	any nonzero number	always 1
Entries above pivots	anything (not constrained)	always 0
Uniqueness	not unique — different row-operation paths give different REFs (pivot positions stay the same)	unique — every matrix has exactly one RREF, regardless of reduction path
Reached by	forward elimination only	Gauss-Jordan (forward + backward, plus pivot scaling)
Reading the solution	by back substitution from the bottom row upward	by direct inspection — each pivot variable is already isolated

However a matrix is reduced, its RREF comes out the same, so whatever it shows belongs to the matrix itself.

Pivot Columns and Free Columns

Pivot columns are the columns that contain a pivot position in the echelon form. Free columns are everything else.

The number of pivot columns equals the rank of the matrix. The number of free columns in the coefficient matrix

A

(not counting the augmented column) is the nullity — the dimension of the null space:

Free Variables Count

\text{(number of free variables)} = n - \text{rank}(A)

Free variables count →

In the context of a linear system, each pivot column corresponds to a pivot variable: a variable whose value is determined once the free variables are assigned. Each free column corresponds to a free variable: a parameter that can take any real value, generating a family of solutions.

For the RREF example above, columns

1

,

2

, and

4

are pivot columns, while columns

3

and

5

are free. The variables

x_1

,

x_2

,

x_4

are pivot variables determined by

x_3

and

x_5

, which are free.

A free column becomes a direction

A reduced form with fewer pivots than unknowns. The pivot columns fix their variables; the column without a pivot is free, and its variable can take any value. That one free variable turns the solution into a particular point p plus every multiple of a direction v₁ — one free column, one dimension of solutions. Compare the unique and inconsistent cases on the solution sets visualizer.

Count the free columns and you have counted the dimensions of the solution set.

Reading Solutions from REF

In row echelon form, the solution is extracted by back substitution: solve from the bottom nonzero row upward, one pivot variable at a time.

Worked Example

Reading Solutions from REF

\left(\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 1 & 4 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right)

Row

3

:

x_3 = 2

.

Row

2

:

x_2 + 4(2) = -1

, so

x_2 = -9

.

Row

1

:

x_1 + 3(-9) - 2(2) = 5

, so

x_1 = 5 + 27 + 4 = 36

.

Pitfall: Reading the answer straight off the REF

Taking the last column as the solution gives

x_1 = 5

,

x_2 = -1

,

x_3 = 2

, but row

1

then reads

5 + 3(-1) - 2(2) = -2 \neq 5

. The entries above the pivots still have to be cleared by back substitution, which gives

x_1 = 36

,

x_2 = -9

,

x_3 = 2

.

Each step uses values computed in previous steps. The process requires

n

substitutions for an

n \times n

system with a unique solution. When free variables are present, they appear as unresolved symbols carried through the substitution, producing a parametric expression.

Reading Solutions from RREF

In RREF, each pivot variable is already isolated. The solution is visible by inspection — no back substitution is needed.

Worked Example

The same system in RREF:

Reading Solutions from RREF

\left(\begin{array}{ccc|c} 1 & 0 & 0 & 36 \\ 0 & 1 & 0 & -9 \\ 0 & 0 & 1 & 2 \end{array}\right)

Reading directly:

x_1 = 36

,

x_2 = -9

,

x_3 = 2

.

When free variables are present, RREF isolates each pivot variable in terms of the free variables:

Reading Solutions from RREF

\left(\begin{array}{cccc|c} 1 & 0 & 3 & 0 & 7 \\ 0 & 1 & -2 & 0 & 1 \\ 0 & 0 & 0 & 1 & 4 \end{array}\right)

Column

3

is free. Reading off:

x_1 = 7 - 3x_3

,

x_2 = 1 + 2x_3

,

x_4 = 4

, and

x_3

is free. Setting

x_3 = t

gives the parametric solution without any backward solving.

Parametric Vector Form

When free variables exist, the general solution is best expressed in parametric vector form. Each free variable becomes a parameter, and the solution is written as a particular solution plus a linear combination of direction vectors — one per free variable.

For the system with RREF

Parametric Vector Form

\left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 5 \\ 0 & 0 & 1 & 3 & -2 \end{array}\right)

Pivots in columns

1

and

3

. Free variables:

x_2 = s

,

x_4 = t

. Reading off:

x_1 = 5 - 2s + t

,

x_3 = -2 - 3t

.

Parametric Vector Form

\mathbf{x} = \begin{pmatrix} 5 \\ 0 \\ -2 \\ 0 \end{pmatrix} + s\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ 0 \\ -3 \\ 1 \end{pmatrix}

Pitfall: Copying the free column without changing sign

Taking the entries of column

2

as they stand gives the direction

(2, 1, 0, 0)

, and at

s = 1

the point

(7, 1, -2, 0)

, where row

1

gives

7 + 2(1) - 0 = 9 \neq 5

. Moving

2x_2

to the right side flips the sign, so the direction is

(-2, 1, 0, 0)

.

The first vector is a particular solution

\mathbf{x}_p

. The two direction vectors form a basis for the null space of the coefficient matrix. The solution set is the null space translated by

\mathbf{x}_p

— an affine subspace of dimension

2

in

\mathbb{R}^4

.

For a homogeneous system (

\mathbf{b} = \mathbf{0}

), the particular solution is

\mathbf{0}

and the general solution is purely a null-space combination.

Every solution set of a consistent linear system is one particular solution shifted by the whole null space.

Detecting Inconsistency

A system is inconsistent — has no solution — if and only if the echelon form of the augmented matrix contains a row of the form

Detecting Inconsistency

[0 \quad 0 \quad \cdots \quad 0 \mid d] \qquad \text{with } d \neq 0

This row represents the equation

0 = d

, which no values of the unknowns can satisfy.

THEOREM

In terms of rank, inconsistency occurs when

\text{rank}([A \mid \mathbf{b}]) > \text{rank}(A)

.

The extra column

\mathbf{b}

introduces a new pivot that the coefficient matrix alone does not have — meaning

\mathbf{b}

is not in the column space of

A

.

Pitfall: Reading a zero row as a contradiction

Declaring

\left(\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}\right)

inconsistent because of its zero row gives no solution. That row says

0 = 0

, which every

\mathbf{x}

satisfies, and the system has infinitely many solutions,

x_1 = 3 - 2t

,

x_2 = t

.

If no such contradictory row appears, the system is consistent. The distinction between unique and infinite solutions then depends on whether free variables exist: rank

= n

gives a unique solution, rank

< n

gives infinitely many.

The contradiction row

All coefficients gone, constant still present: the row claims zero equals a non-zero number. Detecting inconsistency needs no separate test, because reduction surfaces this row on its own if the system has no solution. Force one to appear on the Gaussian elimination visualizer.

Spotting this row early saves the rest of the reduction, since no later step can repair it.

Echelon Form of Special Matrices

Several matrix types are already in echelon form or reach it with minimal work.

The identity matrix

I_n

is in RREF:

n

pivots, each equal to

1

, each alone in its column. Rank

n

, no free variables.

Any diagonal matrix is in REF once its zero rows are moved to the bottom. If all diagonal entries are nonzero, scaling each to

1

produces RREF. If some diagonal entries are zero, the corresponding rows are zero rows and the rank drops.

An upper triangular matrix with no zeros on the diagonal is already in REF. Its pivots are the diagonal entries, and its rank is

n

; with zeros on the diagonal, further elimination may be needed and the rank is at least the number of nonzero diagonal entries.

A lower triangular matrix is not in REF (the staircase goes the wrong way), but forward elimination converts it quickly — when the diagonal entries are nonzero, each elimination step changes only the entry it clears.

The zero matrix is in both REF and RREF. It has rank

0

, and the corresponding homogeneous system

O\mathbf{x} = \mathbf{0}

has every variable free.

Matrix type	Echelon status	Pivot structure	Rank
Identity I_n	already in RREF	n pivots, each equal to 1, each alone in its column	n
Diagonal — all entries nonzero	in REF; reaches RREF after scaling each row	one pivot per row, equal to the diagonal entry	n
Diagonal — some entries zero	in REF once zero rows are moved to the bottom	a pivot only where the diagonal entry is nonzero	number of nonzero diagonal entries
Upper triangular, no zero on the diagonal	already in REF	nonzero diagonal entries serve as pivots	number of nonzero diagonal entries
Lower triangular	NOT in REF — the staircase goes the wrong direction	forward elimination converts it quickly; with a nonzero diagonal, each elimination changes only the entry it clears	depends on the matrix
Zero matrix O	already in both REF and RREF	no pivots; every column is free	0

Echelon Form and Rank

THEOREM

The rank of any matrix equals the number of pivots in any echelon form.

This is the standard computational method for determining rank: row reduce and count pivots.

Fact: Rank 1 on paper, rank 2 on a computer

The rows

(7, 21)

and

(9, 27)

are proportional, yet

R_2 - \frac{9}{7}R_1

in double precision leaves about

3.6 \times 10^{-15}

where

0

belongs, because

\frac{9}{7}

has no exact binary form. Numerical software therefore counts a pivot only above a small tolerance.

From the rank, everything else follows. The dimension of the column space is

r

. The dimension of the row space is

r

. The number of free variables is

n - r

. The dimension of the null space is

n - r

. The system

A\mathbf{x} = \mathbf{b}

is consistent if and only if the augmented matrix has the same rank

r

as

A

. The solution, when it exists, is unique if

r = n

and infinite if

r < n

.

The pivot positions also identify bases for the fundamental subspaces. The original columns at pivot positions form a basis for the column space. The nonzero rows of the echelon form are a basis for the row space. The parametric solution of

A\mathbf{x} = \mathbf{0}

gives a basis for the null space. Everything flows from counting and locating the pivots.

Pitfall: Taking basis columns from the reduced matrix

For

A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}

, with reduced form

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

, taking pivot column

1

of the reduced form gives

(1, 0)

, which is not in

\text{Col}(A)

at all. The basis is column

1

of

A

itself,

(1, 2)

.

Each basis is read after the same single reduction. The reduction locates the pivots, and that one list of positions then serves the column space, the row space and the null space. What differs from one basis to the next is the matrix the vectors are copied from, and that choice follows from how row operations act on rows and on columns.

Summary: Everything Readable from RREF

The RREF of A is a single canonical form, but it carries a remarkable amount of structural information about the matrix — the rank, the nullity, bases for the column, row, and null spaces, and the consistency, uniqueness, and invertibility verdicts for any system A·x = b. The table below collects each property that can be extracted directly from RREF alongside the rule for reading it and the underlying reason.

The readings below are grouped by where in the reduced form each answer sits, because that is what makes them one computation rather than eight. Two come from counting pivots, three from noting which positions hold them, and three from the augmented column. Nothing needs a second reduction, and nothing needs the original matrix again except where the entry says so explicitly.

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Linear systems · summary

Everything readable from the RREF

One reduction answers all of these. What differs is where in the reduced form the answer sits — the pivot positions, the free columns, or the augmented column.

8readings

From the pivot count2

1

Rank§ 10

count them; nothing else needed

\operatorname{rank}(A) =

number of pivots

Pivots correspond to independent rows and to independent columns simultaneously, which is why row rank equals column rank — they are the same pivots counted two ways.

2

Nullity§ 10

the number of free columns

n - \operatorname{rank}(A)

Every column without a pivot is a free variable, and every free variable contributes one dimension to the null space. Rank-nullity is not a separate theorem here so much as a restatement of the column count.

From the pivot positions — bases3

3

Column space basis§ 4

not of the reduced form

pivot columns of the original

A

Row operations change the column space but preserve which columns are dependent on which. So the reduction tells you which columns to take, and you take them from

A

— the single most common error on this page.

4

Row space basis§ 4

of the reduced form, not the original

nonzero rows of the RREF

The exact opposite of the rule above, and the reason both are stated here together. Row operations preserve the row space itself, so the reduced rows still span it — and they are independent, which the original rows may not be.

5

Null space basis§ 7

set each free variable to

1

in turn

one vector per free variable

Read from the parametric vector form. The count matches the nullity above, and each basis vector is the solution obtained by turning one free variable on and the rest off.

From the augmented column3

6

Consistency§ 8

\operatorname{rank}(A) = \operatorname{rank}([A \mid \mathbf{b}])

no row

[\,0 \cdots 0 \mid d\,]

,

d \neq 0

Such a row asserts

0 = d

, so the system has no solution. Equivalently

\mathbf{b}

lies outside the column space — the augmented column raises the rank, and that is what the contradiction row records.

7

Uniqueness§ 6

no free variables

a pivot in every column of

A

Consistency gives at least one solution, no free variables gives at most one. Both conditions are needed — an inconsistent system has none however many pivots it has.

8

Invertibility§ 9

square and full rank

\operatorname{rref}(A) = I_n

The two conditions above at once, for a square matrix: a pivot in every row and every column leaves the identity. This is one line of the invertibility equivalence, and reduction is how it is checked in practice.

Nothing below requires a second computation. That is the return on reducing: rank, nullity, three bases, consistency, uniqueness and invertibility all come out of a single echelon form, which is why Gaussian elimination is the one algorithm worth doing by hand.

Everything readable from the RREF·/linear-algebra/linear-systems/echelon-formLearn Math Class

The two basis rules are worth reading against each other. For the column space the reduction tells you which columns to take and you take them from the original matrix; for the row space you take the reduced rows themselves. They look symmetric and are not, because row operations preserve the row space while changing the column space — the pivots survive both, which is why one reduction serves both purposes.

The last three entries are the same question asked with increasing strictness: does a solution exist, is it unique, and does the matrix invert. Each adds a condition to the one before, and a square matrix satisfying all three reduces to the identity — which is the point at which reduction stops being a technique for solving systems and becomes a test of the matrix itself.

Common Pitfalls

Four mistakes cause many wrong answers when solutions and bases are read from an echelon form. Each is marked with a box where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.

Reading the answer straight off the REF

The last column of a row echelon form is taken as the solution, as if the matrix were already reduced. In REF a pivot row can still hold entries in the columns of later pivots, so the last column is not yet the answer.

Wrong working: for the system of the worked example,

\left(\begin{array}{ccc|c} 1 & 3 & -2 & 5 \\ 0 & 1 & 4 & -1 \\ 0 & 0 & 1 & 2 \end{array}\right) \quad \Longrightarrow \quad x_1 = 5, \; x_2 = -1, \; x_3 = 2

Right working: solve from the bottom row upward.

x_3 = 2, \qquad x_2 = -1 - 4(2) = -9, \qquad x_1 = 5 - 3(-9) + 2(2) = 36

Why it happens: in RREF the last column really is the solution, and the two forms look alike at a glance. The difference sits in the entries above the pivots, which REF does not clear.

How to check: substitute into the first row. The wrong values give

5 + 3(-1) - 2(2) = -2

, not

5

; the right ones give

36 + 3(-9) - 2(2) = 5

.

Copying the free column without changing sign

When a direction vector is built for a free variable, the entries of its column in the RREF are copied into the vector as they stand. Those entries sit on the left side of the equations, and moving them to the right side changes their sign.

Wrong working: for the RREF

\left(\begin{array}{cccc|c} 1 & 2 & 0 & -1 & 5 \\ 0 & 0 & 1 & 3 & -2 \end{array}\right)

copying columns

2

and

4

gives

\mathbf{x} = \begin{pmatrix} 5 \\ 0 \\ -2 \\ 0 \end{pmatrix} + s\begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} -1 \\ 0 \\ 3 \\ 1 \end{pmatrix}

Right working: write each row as an equation and solve for its pivot variable,

x_1 = 5 - 2s + t

and

x_3 = -2 - 3t

. The coefficients of

s

and

t

are the free-column entries with the sign changed:

\mathbf{x} = \begin{pmatrix} 5 \\ 0 \\ -2 \\ 0 \end{pmatrix} + s\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t\begin{pmatrix} 1 \\ 0 \\ -3 \\ 1 \end{pmatrix}

Why it happens: the particular solution is copied from the last column with no change of sign, because those entries already sit on the right side. The same copying is then applied to the free columns, whose entries sit on the left.

How to check: every direction vector must solve the homogeneous system. The wrong vector

(2, 1, 0, 0)

gives

2 + 2(1) - 0 = 4

in row

1

, not

0

; the right one gives

-2 + 2(1) - 0 = 0

.

Reading a zero row as a contradiction

A row of zeros in the reduced augmented matrix is taken as a sign that the system has no solution. Only a row with zeros on the left of the bar and a nonzero entry on the right is a contradiction.

Wrong working:

\left(\begin{array}{cc|c} 1 & 2 & 3 \\ 0 & 0 & 0 \end{array}\right) \quad \Longrightarrow \quad \text{a zero row, so no solution}

Right working: the second row is the equation

0x_1 + 0x_2 = 0

, true for every

\mathbf{x}

. Only the first row constrains the unknowns,

x_1 + 2x_2 = 3

. Column

2

is free, so

x_2 = t

and

x_1 = 3 - 2t

, infinitely many solutions. In rank terms,

\text{rank}(A) = \text{rank}([A \mid \mathbf{b}]) = 1 < 2

.

Why it happens: the contradiction row

[0 \; \cdots \; 0 \mid d]

and the zero row look almost the same, and both turn up at the bottom of the matrix. The difference is one entry, the one to the right of the bar.

How to check: look only at the last entry of each row whose coefficients are all zero. If it is

0

, the row can be dropped; if it is nonzero, the system is inconsistent.

Taking basis columns from the reduced matrix

The pivot columns of the reduced form are used as a basis for the column space of

A

. Row operations preserve which columns depend on which, but they change the columns themselves, so the reduced columns need not lie in

\text{Col}(A)

.

Wrong working:

A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \sim \begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix} \quad \Longrightarrow \quad \text{basis } \left\{ \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right\}

Right working: the only pivot is in column

1

, so the basis is column

1

of the original

A

:

\text{Col}(A) = \text{span}\left\{ \begin{pmatrix} 1 \\ 2 \end{pmatrix} \right\}

Why it happens: for the row space the reduced rows are the right answer, and the two rules look symmetric. They are not, because row operations preserve the row space but not the column space.

How to check: every column of

A

must be a combination of the basis vectors. Column

1

of

A

,

(1, 2)

, is not a multiple of

(1, 0)

, while both columns of

A

are multiples of

(1, 2)

.

Echelon Form FAQ

Can you speak of "the" echelon form of a matrix?

+

Only for the reduced form, which is unique. Plain row echelon form is not: different elimination paths produce different entries, so the definite article quietly asserts a uniqueness that only RREF has. What every echelon form of a matrix does share is the set of pivot positions, and those are what determine the rank.Read more →

Does a pivot have to equal $1$ ?

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In reduced row echelon form yes, where each pivot is forced to

1

and earns the name leading one. In plain echelon form a pivot can be any nonzero value, which is exactly why it needs marking there and not in RREF. Note that the phrase "the pivots of

A

" refers to positions rather than values.Read more →

Are the free parameters unknowns still waiting to be solved for?

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No, they range over all scalars, and the expression containing them is the entire solution set written at once rather than an equation left unfinished. Their number is forced even though their names are not: there are

n - \operatorname{rank}(A)

of them, and a unique solution is simply the case where that count reaches zero.Read more →

Reduced Row Echelon Form (RREF)

The Staircase That Reveals Everything

Row Echelon Form

Echelon Form Notation

Echelon Form Notation

Reduced Row Echelon Form

Uniqueness of RREF

Pivot Columns and Free Columns

Reading Solutions from REF

Worked Example

Reading Solutions from RREF

Worked Example

Parametric Vector Form

Detecting Inconsistency

Echelon Form of Special Matrices

Echelon Form and Rank

Summary: Everything Readable from RREF

Everything readable from the RREF

Common Pitfalls

Echelon Form FAQ

Can you speak of "the" echelon form of a matrix?

Does a pivot have to equal 111?

Are the free parameters unknowns still waiting to be solved for?

Does a pivot have to equal $1$ ?