RREF Calculator

Reduce any matrix to reduced row echelon form, step by step, with the rank and the solution of the linear system spelled out.

Size of the matrix:

Matrix A

Your input

Find the reduced row echelon form of $[\begin{matrix} 1 & 5 & 1 \\ 2 & 11 & 5 \end{matrix}]$ .

The reduced row echelon form is $[\begin{matrix} 1 & 0 & −14 \\ 0 & 1 & 3 \end{matrix}]$ .

Solution

Swap row 1 and row 2: $R_{1} \leftrightarrow R_{2}$

[\begin{matrix} 2 & 11 & 5 \\ 1 & 5 & 1 \end{matrix}]

Multiply row 1 by 1/2: $R_{1} = R_{1} \times \frac{1}{2}$

[\begin{matrix} 1 & \frac{11}{2} & \frac{5}{2} \\ 1 & 5 & 1 \end{matrix}]

Subtract row 1 multiplied by 1 from row 2: $R_{2} = R_{2} - 1 R_{1}$

[\begin{matrix} 1 & \frac{11}{2} & \frac{5}{2} \\ 0 & - \frac{1}{2} & - \frac{3}{2} \end{matrix}]

Multiply row 2 by -2: $R_{2} = R_{2} \times −2$

[\begin{matrix} 1 & \frac{11}{2} & \frac{5}{2} \\ 0 & 1 & 3 \end{matrix}]

Subtract row 2 multiplied by 11/2 from row 1: $R_{1} = R_{1} - \frac{11}{2} R_{2}$

[\begin{matrix} 1 & 0 & −14 \\ 0 & 1 & 3 \end{matrix}]

Answer

The reduced row echelon form is $[\begin{matrix} 1 & 0 & −14 \\ 0 & 1 & 3 \end{matrix}]$ .

Rank 2. If the last column is a right-hand side: unique solution x = (-14, 3).

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How to use the RREF calculator

Set the size of the matrix with the row and column fields.
Fill every cell - integers, decimals, and fractions like 1/2 all work; the write-up updates as you type.
Follow each row operation, then read the final form under Answer. The worked example uses the matrix the page opens with.

What it is for

Reduced row echelon form is the finished shape of Gaussian elimination. It answers whether a linear system has one solution, no solution, or infinitely many, gives the rank directly, and exposes the free variables that parametric solutions come from.

How it works

Gauss-Jordan elimination walks the columns left to right. In each pivot column it picks the largest remaining entry, swaps it into place, scales the pivot row to a leading 1, and clears the column everywhere else. Every swap, scale, and subtraction is recorded and replayed in the Solution section so you can follow the arithmetic cell by cell. The matrix shown after each sentence is the matrix after that operation.

Reduced row echelon form

A matrix is in reduced row echelon form when each pivot is 1, each pivot is the only nonzero entry in its column, and each pivot sits to the right of the pivot above it. Row echelon form only requires the pivots to be 1 with zeros below them. Reduced form also clears the entries above each pivot.

The rank is the number of pivot columns. If you treat the last column as the right-hand side of a linear system, a pivot in that column means no solution. Columns without a pivot, other than the right-hand side, are free variables, and each free variable gives a family of solutions instead of a single one.

Worked example

The page opens on this 2×3 matrix. The same steps are written out above, and they stay in the page source.

Start with rows [1, 5, 1] and [2, 11, 5]. Column 1’s largest entry is 2, so the rows swap. Divide the new first row by 2 and the pivot becomes 1, with 11/2 and 5/2 beside it. Subtract that row from row 2:

[1, 11/2, 5/2]
[0, −1/2, −3/2]

Multiply row 2 by −2. The matrix is now in row echelon form: each pivot is 1, and every entry below a pivot is 0.

[1, 11/2, 5/2]
[0, 1, 3]

11/2 is still above the second pivot, so this is not reduced yet. Subtract 11/2 times row 2 from row 1. That clears the column and gives the reduced row echelon form:

[1, 0, −14]
[0, 1, 3]

The pivot columns are the first and the second, so the rank is 2. No column is free. Read the last column as the right-hand side and the system has one solution: x = −14, y = 3.

A free variable

Rows [1, 2, 5] and [2, 4, 10]. Column 1’s largest entry is 2, so the rows swap and the new first row is divided by 2. The second row then cancels, and the reduced form is [1, 2, 5] over [0, 0, 0]. There is one pivot, in column 1, so the rank is 1. Column 2 has no pivot. If the last column is the right-hand side, y is free and x = 5 − 2y.

No solution

Rows [1, 2, 3] and [2, 4, 7]. Column 1’s largest entry is 2, so the rows swap. After scaling and clearing, the reduced form is [1, 2, 0] over [0, 0, 1]. The last column contains a pivot, which means the equations require 0 = 1. The rank is 2, and the system has no solution.

Step-by-step guide

For the full walkthrough with all three cases side by side, see the guide How to solve a system of equations with RREF - each example links back here so you can verify it live.