Reduced Row Echelon Form Calculator

Select the matrix size and enter its values to calculate the reduced row echelon form (RREF) step by step.

Quick Examples

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How to Use the Calculator

This online reduced row echelon form solver is a linear algebra tool designed to analyze and transform general matrices. By processing the information, the mathematical engine not only gives you the reduced row echelon form (RREF) but also extracts the matrix's fundamental properties and generates the step-by-step row operations.

How to set up and enter your data:

  1. Select the dimensions: Use the drop-down menus to define the size of your matrix. The system supports configurations from 1 to 8 rows or columns, allowing you to work with both square and rectangular matrices.
  2. Enter the elements: Fill in each cell with the corresponding values. The tool processes integers, decimals, and exact fractions. To ensure maximum precision and avoid rounding errors, the algorithm will keep the values in fractional format throughout the entire calculation.

Once the calculation is executed, the tool will organize the information into two analytical blocks:

1. Quick answer and matrix properties

In the first highlighted block, the system will show you the fully reduced final matrix. From this result, the calculator will deduce and list the following key data:

  • Rank: The total number of non-zero rows present in the echelon matrix.
  • Pivot columns: The identification of the columns that contain the leading ones (pivots).
  • Nullity: The dimension of the null space or kernel of the matrix (the total number of columns minus the rank).
  • Invertibility (square matrices only): If your matrix has the same number of rows and columns, the system will indicate whether it is a regular (invertible) or singular (non-invertible) matrix based on its rank.

2. Step-by-step solution

Below the properties, you will find the detailed algebraic breakdown. The system will show you how the original matrix is transformed using elementary row operations:

  • Creating pivots: You will observe how the algorithm divides an entire row by a specific scalar to make the leading entry of each column a 1.
  • Column reduction: The procedure details the row additions and subtractions (e.g., R3 = R3 + 2R1) necessary to systematically eliminate all elements above and below each pivot, until the reduced row echelon form is achieved.

Note: This tool is focused on the analysis of general matrices. If your goal is to find the solution to a system of linear equations, we recommend using our Gauss-Jordan elimination calculator, which has specific options (such as augmented matrices) and a solution report designed for that purpose.

Solved Exercises

The following are examples of problems solved by the calculator.

Exercise 1

Find the reduced row echelon form of the following 3x3 matrix:

$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 0 & 1 & 4 \\[0.5em] 5 & 6 & 0 \end{bmatrix} $$

Result

The reduced row echelon form (RREF) matrix is:

$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 \\[0.5em] 0 & 1 & 0 \\[0.5em] 0 & 0 & 1 \end{bmatrix} $$

Number of non-zero rows (rank): 3

Pivot columns: 1, 2, 3

Nullity: 0

Invertibility: The matrix is invertible (nonsingular).

Step-by-step solution

1. Initial matrix.

$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 0 & 1 & 4 \\[0.5em] 5 & 6 & 0 \end{bmatrix} $$

2. Reduction of column 1.

$$ \displaystyle R_{3} \rightarrow R_{3} - 5R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 2 & 3 \\[0.5em] 0 & 1 & 4 \\[0.5em] 0 & -4 & -15 \end{bmatrix} $$

3. Reduction of column 2.

$$ \displaystyle R_{1} \rightarrow R_{1} - 2R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -5 \\[0.5em] 0 & \boldsymbol{1} & 4 \\[0.5em] 0 & -4 & -15 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} + 4R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -5 \\[0.5em] 0 & \boldsymbol{1} & 4 \\[0.5em] 0 & 0 & 1 \end{bmatrix} $$

4. Reduction of column 3.

$$ \displaystyle R_{1} \rightarrow R_{1} + 5R_{3} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 \\[0.5em] 0 & 1 & 4 \\[0.5em] 0 & 0 & \boldsymbol{1} \end{bmatrix} $$
$$ \displaystyle R_{2} \rightarrow R_{2} - 4R_{3} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 \\[0.5em] 0 & 1 & 0 \\[0.5em] 0 & 0 & \boldsymbol{1} \end{bmatrix} $$

No further operations are possible. The matrix is in its reduced row echelon form.

Exercise 2

Determine the reduced row echelon matrix equivalent to the given matrix:

$$ \displaystyle \begin{bmatrix} 1 & -2 & 1 & -1 \\[0.5em] 2 & -3 & 3 & 1 \\[0.5em] 1 & -1 & 2 & 2 \end{bmatrix} $$

Result

The reduced row echelon form matrix is:

$$ \displaystyle \begin{bmatrix} 1 & 0 & 3 & 5 \\[0.5em] 0 & 1 & 1 & 3 \\[0.5em] 0 & 0 & 0 & 0 \end{bmatrix} $$

Number of non-zero rows (rank): 2

Pivot columns: 1, 2

Nullity: 2

Step-by-step solution

1. Initial matrix.

$$ \displaystyle \begin{bmatrix} 1 & -2 & 1 & -1 \\[0.5em] 2 & -3 & 3 & 1 \\[0.5em] 1 & -1 & 2 & 2 \end{bmatrix} $$

2. Reduction of column 1.

$$ \displaystyle R_{2} \rightarrow R_{2} - 2R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & -2 & 1 & -1 \\[0.5em] 0 & 1 & 1 & 3 \\[0.5em] 1 & -1 & 2 & 2 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} - R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & -2 & 1 & -1 \\[0.5em] 0 & 1 & 1 & 3 \\[0.5em] 0 & 1 & 1 & 3 \end{bmatrix} $$

3. Reduction of column 2.

$$ \displaystyle R_{1} \rightarrow R_{1} + 2R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 3 & 5 \\[0.5em] 0 & \boldsymbol{1} & 1 & 3 \\[0.5em] 0 & 1 & 1 & 3 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} - R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 3 & 5 \\[0.5em] 0 & \boldsymbol{1} & 1 & 3 \\[0.5em] 0 & 0 & 0 & 0 \end{bmatrix} $$

The process has finished. The matrix is now expressed in its reduced row echelon form.

Exercise 3

Obtain the corresponding reduced row echelon form for the following 4x4 matrix:

$$ \displaystyle \begin{bmatrix} 1 & 0 & 2 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 2 & 1 & 3 & 1 \\[0.5em] 1 & 0 & 1 & 1 \end{bmatrix} $$

Result

The reduced row echelon form (RREF) matrix is:

$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 & 0 \\[0.5em] 0 & 1 & 0 & 0 \\[0.5em] 0 & 0 & 1 & 0 \\[0.5em] 0 & 0 & 0 & 1 \end{bmatrix} $$

Number of non-zero rows (rank): 4

Pivot columns: 1, 2, 3, 4

Nullity: 0

Invertibility: The matrix is invertible (nonsingular).

Step-by-step solution

1. Initial matrix.

$$ \displaystyle \begin{bmatrix} 1 & 0 & 2 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 2 & 1 & 3 & 1 \\[0.5em] 1 & 0 & 1 & 1 \end{bmatrix} $$

2. Reduction of column 1.

$$ \displaystyle R_{3} \rightarrow R_{3} - 2R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 0 & 2 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 0 & 1 & -1 & -1 \\[0.5em] 1 & 0 & 1 & 1 \end{bmatrix} $$
$$ \displaystyle R_{4} \rightarrow R_{4} - R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 0 & 2 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 0 & 1 & -1 & -1 \\[0.5em] 0 & 0 & -1 & 0 \end{bmatrix} $$

3. Reduction of column 2.

$$ \displaystyle R_{3} \rightarrow R_{3} - R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 2 & 1 \\[0.5em] 0 & \boldsymbol{1} & -1 & 0 \\[0.5em] 0 & 0 & 0 & -1 \\[0.5em] 0 & 0 & -1 & 0 \end{bmatrix} $$

4. Reduction of column 3.

$$ \displaystyle R_{3} \leftrightarrow R_{4} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 2 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 0 & 0 & \boldsymbol{-1} & 0 \\[0.5em] 0 & 0 & 0 & -1 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow -R_{3} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 2 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 0 & 0 & \boldsymbol{1} & 0 \\[0.5em] 0 & 0 & 0 & -1 \end{bmatrix} $$
$$ \displaystyle R_{1} \rightarrow R_{1} - 2R_{3} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 & 1 \\[0.5em] 0 & 1 & -1 & 0 \\[0.5em] 0 & 0 & \boldsymbol{1} & 0 \\[0.5em] 0 & 0 & 0 & -1 \end{bmatrix} $$
$$ \displaystyle R_{2} \rightarrow R_{2} + R_{3} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 & 1 \\[0.5em] 0 & 1 & 0 & 0 \\[0.5em] 0 & 0 & \boldsymbol{1} & 0 \\[0.5em] 0 & 0 & 0 & -1 \end{bmatrix} $$

5. Reduction of column 4.

$$ \displaystyle R_{4} \rightarrow -R_{4} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 & 1 \\[0.5em] 0 & 1 & 0 & 0 \\[0.5em] 0 & 0 & 1 & 0 \\[0.5em] 0 & 0 & 0 & \boldsymbol{1} \end{bmatrix} $$
$$ \displaystyle R_{1} \rightarrow R_{1} - R_{4} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & 0 & 0 \\[0.5em] 0 & 1 & 0 & 0 \\[0.5em] 0 & 0 & 1 & 0 \\[0.5em] 0 & 0 & 0 & \boldsymbol{1} \end{bmatrix} $$

This concludes the procedure. The matrix has reached its reduced row echelon form.

Exercise 4

Find the row reduction for the following matrix:

$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 4 & 5 & 6 \\[0.5em] 7 & 8 & 9 \end{bmatrix} $$

Result

The reduced row echelon form matrix is:

$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & 1 & 2 \\[0.5em] 0 & 0 & 0 \end{bmatrix} $$

Number of non-zero rows (rank): 2

Pivot columns: 1, 2

Nullity: 1

Invertibility: The matrix is not invertible (singular).

Step-by-step solution

1. Initial matrix.

$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 4 & 5 & 6 \\[0.5em] 7 & 8 & 9 \end{bmatrix} $$

2. Reduction of column 1.

$$ \displaystyle R_{2} \rightarrow R_{2} - 4R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 2 & 3 \\[0.5em] 0 & -3 & -6 \\[0.5em] 7 & 8 & 9 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} - 7R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 2 & 3 \\[0.5em] 0 & -3 & -6 \\[0.5em] 0 & -6 & -12 \end{bmatrix} $$

3. Reduction of column 2.

$$ \displaystyle R_{2} \rightarrow -\frac{1}{3}R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & -6 & -12 \end{bmatrix} $$
$$ \displaystyle R_{1} \rightarrow R_{1} - 2R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & -6 & -12 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} + 6R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & 0 & 0 \end{bmatrix} $$

There are no more pivot columns to reduce. We have obtained the reduced row echelon matrix.

Exercise 5

Calculate the reduced row echelon form (RREF) of this matrix:

$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 2 & 3 & 4 \\[0.5em] 3 & 4 & 5 \\[0.5em] 4 & 5 & 6 \end{bmatrix} $$

Result

The reduced row echelon form matrix is:

$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & 1 & 2 \\[0.5em] 0 & 0 & 0 \\[0.5em] 0 & 0 & 0 \end{bmatrix} $$

Number of non-zero rows (rank): 2

Pivot columns: 1, 2

Nullity: 1

Step-by-step solution

1. Initial matrix.

$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 2 & 3 & 4 \\[0.5em] 3 & 4 & 5 \\[0.5em] 4 & 5 & 6 \end{bmatrix} $$

2. Reduction of column 1.

$$ \displaystyle R_{2} \rightarrow R_{2} - 2R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 2 & 3 \\[0.5em] 0 & -1 & -2 \\[0.5em] 3 & 4 & 5 \\[0.5em] 4 & 5 & 6 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} - 3R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 2 & 3 \\[0.5em] 0 & -1 & -2 \\[0.5em] 0 & -2 & -4 \\[0.5em] 4 & 5 & 6 \end{bmatrix} $$
$$ \displaystyle R_{4} \rightarrow R_{4} - 4R_{1} $$
$$ \displaystyle \begin{bmatrix} \boldsymbol{1} & 2 & 3 \\[0.5em] 0 & -1 & -2 \\[0.5em] 0 & -2 & -4 \\[0.5em] 0 & -3 & -6 \end{bmatrix} $$

3. Reduction of column 2.

$$ \displaystyle R_{2} \rightarrow -R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 2 & 3 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & -2 & -4 \\[0.5em] 0 & -3 & -6 \end{bmatrix} $$
$$ \displaystyle R_{1} \rightarrow R_{1} - 2R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & -2 & -4 \\[0.5em] 0 & -3 & -6 \end{bmatrix} $$
$$ \displaystyle R_{3} \rightarrow R_{3} + 2R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & 0 & 0 \\[0.5em] 0 & -3 & -6 \end{bmatrix} $$
$$ \displaystyle R_{4} \rightarrow R_{4} + 3R_{2} $$
$$ \displaystyle \begin{bmatrix} 1 & 0 & -1 \\[0.5em] 0 & \boldsymbol{1} & 2 \\[0.5em] 0 & 0 & 0 \\[0.5em] 0 & 0 & 0 \end{bmatrix} $$

The calculation ends here as the matrix has reached its reduced row echelon form.

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Daniel Machado

Professor of Mathematics, graduated from the Faculty of Exact, Chemical and Natural Sciences of the National University of Misiones (UNAM). Developer and creator of RigelUp, dedicated to building tools for mathematical learning.

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