Gauss-Jordan Elimination Calculator

Select the size of the augmented matrix and enter the coefficients of the linear system to solve it using the Gauss-Jordan elimination method step by step.

Quick Examples

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How to Use the Calculator

This online Gauss-Jordan elimination solver is a linear algebra tool specially designed to find the solution to systems of linear equations. By processing the information through an augmented matrix, the tool classifies the system and generates a detailed step-by-step resolution with all row operations.

How to set up and enter your data:

  1. Select the dimensions: Use the dropdown menus to adjust the size of your system. The solver supports from 1 to 8 rows (equations) and from 1 to 8 columns (variables + constant terms).
  2. Enter the coefficients: Fill each cell in the grid with the values of your problem. The calculator processes integers, decimals, and exact fractions. To avoid rounding errors, the solver will keep numbers in their fractional form throughout the entire mathematical resolution.

Once the data is processed, the tool will structure the results into two main sections:

1. Quick answer and system analysis
In the first highlighted block, you will find the resulting matrix after applying the method to the augmented matrix. The tool will mathematically analyze this final matrix and provide a diagnosis:

  • Consistent and independent system: If there is a unique solution, it will give you the exact values for each unknown (for example, x1, x2, x3).
  • Consistent and dependent system: If there are infinitely many solutions, it will show you the parametric equations that define the solution set.
  • Inconsistent system: It will notify you if the intersection of equations generates a mathematical contradiction (such as 0 = 1), indicating that the problem has no solution.

2. Step-by-step resolution
Next, the complete analytical development will be displayed. You will be able to observe how the algorithm advances column by column by applying elementary row operations:

  • Choosing the pivot: At each step, the system will identify the element on the main diagonal. It will show you the number by which that entire row is divided to convert the pivot to 1.
  • Eliminating elements: You will see the row additions and subtractions (for example, R2 = R2 - 3R1) used to convert all other elements above and below the pivot to 0.
  • Row echelon form conversion: The process will continue iteratively until the section corresponding to the coefficient matrix (before the dividing line) is completely in reduced row echelon form.

Note on the working mode: By default, the calculator has the "Augmented matrix" toggle enabled, as this is the usual way to solve systems of linear equations (where the last column acts as constants and is not pivoted on). If your purely matrix-based goal is to find the reduced row echelon form of a general matrix, simply uncheck this box. By doing so, the dividing line will disappear, and the algorithm will continue the Gauss-Jordan method until the end of the last available column.

Solved Exercises

The following are examples of problems solved by the calculator.

Exercise 1

Solve the 3x3 linear system with the following augmented matrix using the Gauss-Jordan method:

$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\[0.5em] 2 & -1 & 1 & 3 \\[0.5em] 1 & 2 & -1 & 2 \end{array} \right] $$

Result

After applying the Gauss-Jordan elimination method, the resulting matrix is:

$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\[0.5em] 0 & 1 & 0 & 2 \\[0.5em] 0 & 0 & 1 & 3 \end{array} \right] $$

The system is consistent and independent and has a unique solution:

$$ \displaystyle x_{1} = 1, \quad x_{2} = 2, \quad x_{3} = 3 $$

Step-by-step resolution

1. Initial matrix

We identify the matrix we are going to operate on.

$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\[0.5em] 2 & -1 & 1 & 3 \\[0.5em] 1 & 2 & -1 & 2 \end{array} \right] $$

2. Clear column 1

The pivot at position (1, 1) is 1. We will use row 1 to create zeros in the rest of the column.

$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & 1 & 1 & 6 \\[0.5em] 2 & -1 & 1 & 3 \\[0.5em] 1 & 2 & -1 & 2 \end{array} \right] $$

We subtract 2 times row 1 from row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} - 2R_{1} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & 1 & 1 & 6 \\[0.5em] 0 & -3 & -1 & -9 \\[0.5em] 1 & 2 & -1 & 2 \end{array} \right] $$

We subtract row 1 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - R_{1} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & 1 & 1 & 6 \\[0.5em] 0 & -3 & -1 & -9 \\[0.5em] 0 & 1 & -2 & -4 \end{array} \right] $$

3. Clear column 2

We convert the pivot at position (2, 2) to 1. To do this, we multiply row 2 by -1/3. We will use row 2 to create zeros in the rest of the column.

$$ \displaystyle R_{2} \rightarrow -\frac{1}{3}R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 1 & 1 & 6 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 3 \\[0.5em] 0 & 1 & -2 & -4 \end{array} \right] $$

We subtract row 2 from row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} - R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & 3 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 3 \\[0.5em] 0 & 1 & -2 & -4 \end{array} \right] $$

We subtract row 2 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & 3 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 3 \\[0.5em] 0 & 0 & -\frac{7}{3} & -7 \end{array} \right] $$

4. Clear column 3

We convert the pivot at position (3, 3) to 1. To do this, we multiply row 3 by -3/7. We will use row 3 to create zeros in the rest of the column.

$$ \displaystyle R_{3} \rightarrow -\frac{3}{7}R_{3} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & \frac{2}{3} & 3 \\[0.5em] 0 & 1 & \frac{1}{3} & 3 \\[0.5em] 0 & 0 & \boldsymbol{1} & 3 \end{array} \right] $$

We subtract 2/3 times row 3 from row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} - \frac{2}{3}R_{3} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\[0.5em] 0 & 1 & \frac{1}{3} & 3 \\[0.5em] 0 & 0 & \boldsymbol{1} & 3 \end{array} \right] $$

We subtract 1/3 times row 3 from row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} - \frac{1}{3}R_{3} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\[0.5em] 0 & 1 & 0 & 2 \\[0.5em] 0 & 0 & \boldsymbol{1} & 3 \end{array} \right] $$
Exercise 2

Calculate the solution to the linear system by applying Gauss-Jordan elimination to the augmented matrix:

$$ \displaystyle \left[ \begin{array}{ccc|c} 0 & 2 & 3 & 7 \\[0.5em] 1 & -1 & 1 & 3 \\[0.5em] 2 & 1 & -1 & 4 \end{array} \right] $$

Result

After applying the Gauss Jordan elimination method, the resulting matrix is:

$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{7}{3} \\[0.5em] 0 & 1 & 0 & 1 \\[0.5em] 0 & 0 & 1 & \frac{5}{3} \end{array} \right] $$

The system is consistent and independent and has a unique solution:

$$ \displaystyle x_{1} = \frac{7}{3}, \quad x_{2} = 1, \quad x_{3} = \frac{5}{3} $$

Step-by-step resolution

1. Initial matrix

We identify the matrix we are going to operate on.

$$ \displaystyle \left[ \begin{array}{ccc|c} 0 & 2 & 3 & 7 \\[0.5em] 1 & -1 & 1 & 3 \\[0.5em] 2 & 1 & -1 & 4 \end{array} \right] $$

2. Clear column 1

The element at position (1, 1) is zero. We look for a lower row with a non-zero value and swap them. We swap row 1 with row 2.

$$ \displaystyle R_{1} \leftrightarrow R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & -1 & 1 & 3 \\[0.5em] 0 & 2 & 3 & 7 \\[0.5em] 2 & 1 & -1 & 4 \end{array} \right] $$

The pivot at position (1, 1) is 1. We will use row 1 to create zeros in the rest of the column.

$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & -1 & 1 & 3 \\[0.5em] 0 & 2 & 3 & 7 \\[0.5em] 2 & 1 & -1 & 4 \end{array} \right] $$

We subtract 2 times row 1 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - 2R_{1} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & -1 & 1 & 3 \\[0.5em] 0 & 2 & 3 & 7 \\[0.5em] 0 & 3 & -3 & -2 \end{array} \right] $$

3. Clear column 2

We convert the pivot at position (2, 2) to 1. To do this, we multiply row 2 by 1/2. We will use row 2 to create zeros in the rest of the column.

$$ \displaystyle R_{2} \rightarrow \frac{1}{2}R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & -1 & 1 & 3 \\[0.5em] 0 & \boldsymbol{1} & \frac{3}{2} & \frac{7}{2} \\[0.5em] 0 & 3 & -3 & -2 \end{array} \right] $$

We add row 2 to row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} + R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & \frac{5}{2} & \frac{13}{2} \\[0.5em] 0 & \boldsymbol{1} & \frac{3}{2} & \frac{7}{2} \\[0.5em] 0 & 3 & -3 & -2 \end{array} \right] $$

We subtract 3 times row 2 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - 3R_{2} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & \frac{5}{2} & \frac{13}{2} \\[0.5em] 0 & \boldsymbol{1} & \frac{3}{2} & \frac{7}{2} \\[0.5em] 0 & 0 & -\frac{15}{2} & -\frac{25}{2} \end{array} \right] $$

4. Clear column 3

We convert the pivot at position (3, 3) to 1. To do this, we multiply row 3 by -2/15. We will use row 3 to create zeros in the rest of the column.

$$ \displaystyle R_{3} \rightarrow -\frac{2}{15}R_{3} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & \frac{5}{2} & \frac{13}{2} \\[0.5em] 0 & 1 & \frac{3}{2} & \frac{7}{2} \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{5}{3} \end{array} \right] $$

We subtract 5/2 times row 3 from row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} - \frac{5}{2}R_{3} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{7}{3} \\[0.5em] 0 & 1 & \frac{3}{2} & \frac{7}{2} \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{5}{3} \end{array} \right] $$

We subtract 3/2 times row 3 from row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} - \frac{3}{2}R_{3} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{7}{3} \\[0.5em] 0 & 1 & 0 & 1 \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{5}{3} \end{array} \right] $$
Exercise 3

Find the result of the linear system using the Gauss Jordan method with the augmented matrix:

$$ \displaystyle \left[ \begin{array}{cc|c} 2 & 3 & 8 \\[0.5em] 1 & -1 & -1 \end{array} \right] $$

Result

After applying the Gauss-Jordan elimination method, the resulting matrix is:

$$ \displaystyle \left[ \begin{array}{cc|c} 1 & 0 & 1 \\[0.5em] 0 & 1 & 2 \end{array} \right] $$

The system is consistent and independent and has a unique solution:

$$ \displaystyle x_{1} = 1, \quad x_{2} = 2 $$

Step-by-step resolution

1. Initial matrix

We identify the matrix we are going to operate on.

$$ \displaystyle \left[ \begin{array}{cc|c} 2 & 3 & 8 \\[0.5em] 1 & -1 & -1 \end{array} \right] $$

2. Clear column 1

We convert the pivot at position (1, 1) to 1. To do this, we multiply row 1 by 1/2. We will use row 1 to create zeros in the rest of the column.

$$ \displaystyle R_{1} \rightarrow \frac{1}{2}R_{1} $$
$$ \displaystyle \left[ \begin{array}{cc|c} \boldsymbol{1} & \frac{3}{2} & 4 \\[0.5em] 1 & -1 & -1 \end{array} \right] $$

We subtract row 1 from row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} - R_{1} $$
$$ \displaystyle \left[ \begin{array}{cc|c} \boldsymbol{1} & \frac{3}{2} & 4 \\[0.5em] 0 & -\frac{5}{2} & -5 \end{array} \right] $$

3. Clear column 2

We convert the pivot at position (2, 2) to 1. To do this, we multiply row 2 by -2/5. We will use row 2 to create zeros in the rest of the column.

$$ \displaystyle R_{2} \rightarrow -\frac{2}{5}R_{2} $$
$$ \displaystyle \left[ \begin{array}{cc|c} 1 & \frac{3}{2} & 4 \\[0.5em] 0 & \boldsymbol{1} & 2 \end{array} \right] $$

We subtract 3/2 times row 2 from row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} - \frac{3}{2}R_{2} $$
$$ \displaystyle \left[ \begin{array}{cc|c} 1 & 0 & 1 \\[0.5em] 0 & \boldsymbol{1} & 2 \end{array} \right] $$
Exercise 4

Determine the solution (if it exists) to the linear system using the Gauss-Jordan method for the 3x4 augmented matrix:

$$ \displaystyle \left[ \begin{array}{ccc|c} 2 & -1 & 3 & 4 \\[0.5em] 4 & -2 & 6 & 8 \\[0.5em] 6 & -3 & 9 & 12 \end{array} \right] $$

Result

After applying the Gauss Jordan elimination method, the resulting matrix is:

$$ \displaystyle \left[ \begin{array}{ccc|c} 1 & -\frac{1}{2} & \frac{3}{2} & 2 \\[0.5em] 0 & 0 & 0 & 0 \\[0.5em] 0 & 0 & 0 & 0 \end{array} \right] $$

The system is consistent and dependent. There are fewer valid equations (non-zero rows) than variables, meaning there are infinitely many solutions.

The general solution parameterized in terms of the free variables is:

$$ \displaystyle x_{1} = -\frac{3 x_{3}}{2}+\frac{x_{2}}{2}+2, \quad x_{2} \in \mathbb{R}, \quad x_{3} \in \mathbb{R} $$

Step-by-step resolution

1. Initial matrix

We identify the matrix we are going to operate on.

$$ \displaystyle \left[ \begin{array}{ccc|c} 2 & -1 & 3 & 4 \\[0.5em] 4 & -2 & 6 & 8 \\[0.5em] 6 & -3 & 9 & 12 \end{array} \right] $$

2. Clear column 1

We convert the pivot at position (1, 1) to 1. To do this, we multiply row 1 by 1/2. We will use row 1 to create zeros in the rest of the column.

$$ \displaystyle R_{1} \rightarrow \frac{1}{2}R_{1} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & -\frac{1}{2} & \frac{3}{2} & 2 \\[0.5em] 4 & -2 & 6 & 8 \\[0.5em] 6 & -3 & 9 & 12 \end{array} \right] $$

We subtract 4 times row 1 from row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} - 4R_{1} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & -\frac{1}{2} & \frac{3}{2} & 2 \\[0.5em] 0 & 0 & 0 & 0 \\[0.5em] 6 & -3 & 9 & 12 \end{array} \right] $$

We subtract 6 times row 1 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - 6R_{1} $$
$$ \displaystyle \left[ \begin{array}{ccc|c} \boldsymbol{1} & -\frac{1}{2} & \frac{3}{2} & 2 \\[0.5em] 0 & 0 & 0 & 0 \\[0.5em] 0 & 0 & 0 & 0 \end{array} \right] $$
Exercise 5

Find the solution to the 4x4 linear system using Gauss Jordan elimination from the augmented matrix:

$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & -1 & 1 & 6 \\[0.5em] -1 & 1 & 2 & -1 & -3 \\[0.5em] 2 & -1 & 1 & 3 & 9 \\[0.5em] 1 & 1 & 1 & 1 & 10 \end{array} \right] $$

Result

After applying the Gauss-Jordan elimination method, the resulting matrix is:

$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & 0 & \frac{109}{7} \\[0.5em] 0 & 1 & 0 & 0 & \frac{2}{7} \\[0.5em] 0 & 0 & 1 & 0 & \frac{15}{7} \\[0.5em] 0 & 0 & 0 & 1 & -8 \end{array} \right] $$

The system is consistent and independent and has a unique solution:

$$ \displaystyle x_{1} = \frac{109}{7}, \quad x_{2} = \frac{2}{7}, \quad x_{3} = \frac{15}{7}, \quad x_{4} = -8 $$

Step-by-step resolution

1. Initial matrix

We identify the matrix we are going to operate on.

$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & -1 & 1 & 6 \\[0.5em] -1 & 1 & 2 & -1 & -3 \\[0.5em] 2 & -1 & 1 & 3 & 9 \\[0.5em] 1 & 1 & 1 & 1 & 10 \end{array} \right] $$

2. Clear column 1

The pivot at position (1, 1) is 1. We will use row 1 to create zeros in the rest of the column.

$$ \displaystyle \left[ \begin{array}{cccc|c} \boldsymbol{1} & 2 & -1 & 1 & 6 \\[0.5em] -1 & 1 & 2 & -1 & -3 \\[0.5em] 2 & -1 & 1 & 3 & 9 \\[0.5em] 1 & 1 & 1 & 1 & 10 \end{array} \right] $$

We add row 1 to row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} + R_{1} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} \boldsymbol{1} & 2 & -1 & 1 & 6 \\[0.5em] 0 & 3 & 1 & 0 & 3 \\[0.5em] 2 & -1 & 1 & 3 & 9 \\[0.5em] 1 & 1 & 1 & 1 & 10 \end{array} \right] $$

We subtract 2 times row 1 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - 2R_{1} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} \boldsymbol{1} & 2 & -1 & 1 & 6 \\[0.5em] 0 & 3 & 1 & 0 & 3 \\[0.5em] 0 & -5 & 3 & 1 & -3 \\[0.5em] 1 & 1 & 1 & 1 & 10 \end{array} \right] $$

We subtract row 1 from row 4.

$$ \displaystyle R_{4} \rightarrow R_{4} - R_{1} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} \boldsymbol{1} & 2 & -1 & 1 & 6 \\[0.5em] 0 & 3 & 1 & 0 & 3 \\[0.5em] 0 & -5 & 3 & 1 & -3 \\[0.5em] 0 & -1 & 2 & 0 & 4 \end{array} \right] $$

3. Clear column 2

We convert the pivot at position (2, 2) to 1. To do this, we multiply row 2 by 1/3. We will use row 2 to create zeros in the rest of the column.

$$ \displaystyle R_{2} \rightarrow \frac{1}{3}R_{2} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 2 & -1 & 1 & 6 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 0 & 1 \\[0.5em] 0 & -5 & 3 & 1 & -3 \\[0.5em] 0 & -1 & 2 & 0 & 4 \end{array} \right] $$

We subtract 2 times row 2 from row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} - 2R_{2} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & -\frac{5}{3} & 1 & 4 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 0 & 1 \\[0.5em] 0 & -5 & 3 & 1 & -3 \\[0.5em] 0 & -1 & 2 & 0 & 4 \end{array} \right] $$

We add 5 times row 2 to row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} + 5R_{2} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & -\frac{5}{3} & 1 & 4 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 0 & 1 \\[0.5em] 0 & 0 & \frac{14}{3} & 1 & 2 \\[0.5em] 0 & -1 & 2 & 0 & 4 \end{array} \right] $$

We add row 2 to row 4.

$$ \displaystyle R_{4} \rightarrow R_{4} + R_{2} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & -\frac{5}{3} & 1 & 4 \\[0.5em] 0 & \boldsymbol{1} & \frac{1}{3} & 0 & 1 \\[0.5em] 0 & 0 & \frac{14}{3} & 1 & 2 \\[0.5em] 0 & 0 & \frac{7}{3} & 0 & 5 \end{array} \right] $$

4. Clear column 3

We convert the pivot at position (3, 3) to 1. To do this, we multiply row 3 by 3/14. We will use row 3 to create zeros in the rest of the column.

$$ \displaystyle R_{3} \rightarrow \frac{3}{14}R_{3} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & -\frac{5}{3} & 1 & 4 \\[0.5em] 0 & 1 & \frac{1}{3} & 0 & 1 \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & \frac{7}{3} & 0 & 5 \end{array} \right] $$

We add 5/3 times row 3 to row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} + \frac{5}{3}R_{3} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & \frac{19}{14} & \frac{33}{7} \\[0.5em] 0 & 1 & \frac{1}{3} & 0 & 1 \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & \frac{7}{3} & 0 & 5 \end{array} \right] $$

We subtract 1/3 times row 3 from row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} - \frac{1}{3}R_{3} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & \frac{19}{14} & \frac{33}{7} \\[0.5em] 0 & 1 & 0 & -\frac{1}{14} & \frac{6}{7} \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & \frac{7}{3} & 0 & 5 \end{array} \right] $$

We subtract 7/3 times row 3 from row 4.

$$ \displaystyle R_{4} \rightarrow R_{4} - \frac{7}{3}R_{3} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & \frac{19}{14} & \frac{33}{7} \\[0.5em] 0 & 1 & 0 & -\frac{1}{14} & \frac{6}{7} \\[0.5em] 0 & 0 & \boldsymbol{1} & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & 0 & -\frac{1}{2} & 4 \end{array} \right] $$

5. Clear column 4

We convert the pivot at position (4, 4) to 1. To do this, we multiply row 4 by -2. We will use row 4 to create zeros in the rest of the column.

$$ \displaystyle R_{4} \rightarrow -2R_{4} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & \frac{19}{14} & \frac{33}{7} \\[0.5em] 0 & 1 & 0 & -\frac{1}{14} & \frac{6}{7} \\[0.5em] 0 & 0 & 1 & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & 0 & \boldsymbol{1} & -8 \end{array} \right] $$

We subtract 19/14 times row 4 from row 1.

$$ \displaystyle R_{1} \rightarrow R_{1} - \frac{19}{14}R_{4} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & 0 & \frac{109}{7} \\[0.5em] 0 & 1 & 0 & -\frac{1}{14} & \frac{6}{7} \\[0.5em] 0 & 0 & 1 & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & 0 & \boldsymbol{1} & -8 \end{array} \right] $$

We add 1/14 times row 4 to row 2.

$$ \displaystyle R_{2} \rightarrow R_{2} + \frac{1}{14}R_{4} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & 0 & \frac{109}{7} \\[0.5em] 0 & 1 & 0 & 0 & \frac{2}{7} \\[0.5em] 0 & 0 & 1 & \frac{3}{14} & \frac{3}{7} \\[0.5em] 0 & 0 & 0 & \boldsymbol{1} & -8 \end{array} \right] $$

We subtract 3/14 times row 4 from row 3.

$$ \displaystyle R_{3} \rightarrow R_{3} - \frac{3}{14}R_{4} $$
$$ \displaystyle \left[ \begin{array}{cccc|c} 1 & 0 & 0 & 0 & \frac{109}{7} \\[0.5em] 0 & 1 & 0 & 0 & \frac{2}{7} \\[0.5em] 0 & 0 & 1 & 0 & \frac{15}{7} \\[0.5em] 0 & 0 & 0 & \boldsymbol{1} & -8 \end{array} \right] $$

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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.

Last updated: (2 weeks ago)