RREF Calculator (Row Reduction)
Every row operation in exact fractions, from the first pivot to the reduced form.
Try before you buy.
- Free preview: the matrix you entered and what the full result holds, with the reduced row echelon form, the row operations, rank, bases and solutions hidden.
- Locked until you unlock it: download and copy.
- Unlock: Pro pass, ₹179 for 30 days, a one-time payment that never renews.
Ways to unlock shows how to get the full result.
Printing this result is locked in the free preview.
Result
This is the last result worked out. Fix the input above to update it.
Locked in the free preview. Opens the ways to unlock this result.
Locked in the free preview. Batch runs unlock with a pass.
Locked in the free preview. Query results unlock with a pass.
About the RREF Calculator (Row Reduction)
Type a matrix, an augmented system Ax = b (as a matrix or as the equations themselves) or a set of vectors, and the calculator row-reduces it in exact fractions — no rounding, ever — listing every row operation (swap, scale, add a multiple of a row) with the matrix it produces and the pivot it works on. It gives the row echelon form after elimination and the reduced row echelon form (RREF), the pivot positions, rank and nullity, and bases for the column space, row space, null space and left null space.
For a system it says whether there is no solution, exactly one or infinitely many, and writes the general solution with its parameters. For vectors it tests linear independence, finds the dependency relations, describes what they span (with the equations of a plane or line through the origin) and checks whether another vector is in that span.
How to use it
- Choose Matrix, System Ax = b or Vectors. For a system, enter the augmented matrix (the last column is b) or switch to Equations and type one equation per line, such as 2x + y − z = 8.
- Set the number of rows and columns and type the entries — whole numbers, decimals or fractions such as −3/4. You can paste a block from a spreadsheet into a cell, or switch to Text and paste rows, including [[1, 2], [3, 4]] or a LaTeX bmatrix.
- Pick the method: elimination, then back substitution (zeros below each pivot first, then each pivot made 1 and cleared above) or Gauss–Jordan (each column cleared above and below at once).
- Read the reduced form, then the steps: each one names the operation, why it is done and shows the matrix after it, with the changed rows and the pivot marked.
- Copy the RREF as text or the whole working as LaTeX, or download the working as a text file — with a Pro pass or after unlocking this result; without one you see the free preview.
Examples
x + 2y − z = −4 2x + 3y − z = −11 −2x − 3z = 22
RREF [[1, 0, 0, −8], [0, 1, 0, 1], [0, 0, 1, −2]]: x = −8, y = 1, z = −2
A pivot in every coefficient column and none in the b column, so the solution is unique; the last column of the RREF is the solution.
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
RREF [[1, 0, −1], [0, 1, 2], [0, 0, 0]]; rank 2, null space spanned by (1, −2, 1)
Operations: R₂ → R₂ − 4R₁, R₃ → R₃ − 7R₁, R₃ → R₃ − 2R₂, R₂ → (−1/3)R₂, R₁ → R₁ − 2R₂.
x + 2y + 3z = 4 2x + 4y + 6z = 8
x = 4 − 2y − 3z, with y and z any values
Rank 1 with 3 unknowns leaves two parameters: x = (4, 0, 0) + y(−2, 1, 0) + z(−3, 0, 1).
(1, 2, 3), (4, 5, 6), (7, 8, 9)
Dependent: v₁ − 2v₂ + v₃ = 0; they span the plane x − 2y + z = 0
The matrix with these vectors as columns has rank 2, so only two of them are needed for a basis of the span.
Common uses
- Checking row-reduction homework step by step, with the operations written the way courses write them (R₂ → R₂ − 2R₁).
- Finding the rank, the pivot columns and the special solutions of the null space for a linear algebra assignment.
- Solving a system of linear equations and reading off whether it is consistent and how many parameters its solution has.
- Deciding whether a set of vectors is a basis, and finding the equation of the plane two vectors span.
Row echelon form and reduced row echelon form
A matrix is in row echelon form when every all-zero row is at the bottom and each row’s first non-zero entry (its pivot) sits to the right of the pivot above it. It is in reduced row echelon form when, in addition, every pivot is 1 and is the only non-zero entry in its column.
Elimination makes the zeros below each pivot (the forward phase); back substitution then scales each pivot to 1 and clears the entries above it, from the last pivot up (the backward phase). A row echelon form depends on the operations chosen, but every matrix has exactly one reduced row echelon form, which is why two people who reduce the same matrix differently still end with the same RREF.
What the pivots tell you
- Rank = the number of pivots = the number of independent columns = the number of independent rows.
- Nullity = the number of non-pivot columns; rank + nullity = the number of columns.
- A system Ax = b has no solution when the reduced augmented matrix has a row 0 = c with c ≠ 0 (a pivot in the b column); otherwise it has one solution when every column of A has a pivot, and infinitely many when some column has none: those variables are the parameters.
- A square matrix is invertible exactly when its RREF is the identity.
The four fundamental subspaces
For an m × n matrix A of rank r (Strang, Introduction to Linear Algebra, section 3.5):
- Column space C(A), dimension r: the pivot columns of the original matrix A (not of R) are a basis.
- Row space C(Aᵀ), dimension r: the non-zero rows of R are a basis.
- Null space N(A), dimension n − r: one special solution for each non-pivot column — set that variable to 1 and the other non-pivot variables to 0, then read the pivot variables from R.
- Left null space N(Aᵀ), dimension m − r: the solutions of Aᵀy = 0. For a set of vectors placed as columns, these give the linear equations that describe their span.
Limitations
- Matrices up to 10 rows and 12 columns, with rational entries: whole numbers, decimals and fractions. Irrational numbers such as √2 or π must be typed as decimals, and complex entries are not supported.
- Fractions are exact but can grow long in big matrices with awkward entries; switch to decimals to read them more easily (the work itself stays exact).
- The row echelon form shown is the one this sequence of operations produces; another valid sequence can give a different echelon form, but never a different reduced form.
- Pivots are chosen as in textbooks (the first non-zero entry, or a 1 when you allow the swap), not by size as numerical software does to limit rounding — there is no rounding here to limit.
Privacy
Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.
Frequently asked questions
What do I get without a pass?
Without a pass, RREF Calculator (Row Reduction) shows the matrix you entered and what the full result holds, with the reduced row echelon form, the row operations, rank, bases and solutions hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
What is the difference between REF and RREF?
Both have zeros below each pivot and all-zero rows at the bottom. The reduced form (RREF) also has every pivot equal to 1 with zeros above it as well, so each pivot is alone in its column. RREF is unique; REF is not.
How do I find the rank of a matrix?
Row-reduce it and count the pivots (the leading entries). [[1, 2, 3], [4, 5, 6], [7, 8, 9]] reduces to two non-zero rows, so its rank is 2. The rank is the same whether you count independent rows or columns.
How do I find a basis for the null space?
Reduce A to R. Each column without a pivot belongs to a non-pivot variable: set it to 1 and the other non-pivot variables to 0, and read the pivot variables from R with their signs changed. The vectors you get (the special solutions) are a basis of N(A).
Why must the column space basis come from the original matrix?
Row operations change the column space (they keep the row space and the null space). They do keep which columns are independent, so the pivot positions found in R pick out a basis — but the vectors themselves are the matching columns of the original A.
How can I tell if a system has no solution or infinitely many?
Reduce the augmented matrix [A | b]. A row that reads 0 0 … 0 | c with c ≠ 0 means no solution. Otherwise each column of A without a pivot gives a parameter, so a system with fewer pivots than unknowns has infinitely many solutions.
How do I check whether vectors are linearly independent?
Put them as the columns of a matrix and row-reduce. They are independent exactly when every column has a pivot. If one does not, the null space gives the dependency, such as v₁ − 2v₂ + v₃ = 0 for (1, 2, 3), (4, 5, 6) and (7, 8, 9).