Gauss Jordan Elimination Calculator

Gauss-Jordan Elimination Calculator

Use this calculator to find the reduced row echelon form of a matrix or solve a system of linear equations using the Gauss-Jordan elimination method. Enter the coefficients of your augmented matrix below.

Enter Augmented Matrix (e.g., for 3 equations, 3 variables):

Each row represents an equation. The last column represents the constants on the right side of the equations.

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Understanding Gauss-Jordan Elimination

Gauss-Jordan elimination is a powerful algorithm in linear algebra used to solve systems of linear equations, find the inverse of a matrix, and determine the rank of a matrix. It's an extension of Gaussian elimination, which brings a matrix to row echelon form. Gauss-Jordan takes it a step further to reduced row echelon form (RREF).

What is Reduced Row Echelon Form (RREF)?

A matrix is in RREF if it satisfies the following conditions:

1. Any row consisting entirely of zeros is at the bottom of the matrix.
2. For each non-zero row, the first non-zero entry (called the leading entry or pivot) is 1.
3. Each leading 1 is in a column to the right of the leading 1 of the row above it.
4. Each column that contains a leading 1 has zeros everywhere else in that column.

How Gauss-Jordan Elimination Works

The method involves applying a series of elementary row operations to an augmented matrix until it is in RREF. The elementary row operations are:

1. Swapping two rows: Interchanging the positions of any two rows.
2. Multiplying a row by a non-zero scalar: Multiplying all entries in a row by a non-zero number.
3. Adding a multiple of one row to another row: Replacing a row with the sum of that row and a multiple of another row.

The process typically proceeds column by column, from left to right. For each column, a "pivot" (a leading 1) is created, and then all other entries in that column are made zero.

Steps for Solving a System of Linear Equations:

1. Form the Augmented Matrix: Write the system of equations as an augmented matrix, where the coefficients of the variables form the left part, and the constants form the right part, separated by a vertical line. For example, for the system:

   x + y + z = 6
       2y + 5z = -4
   2x + 5y - z = 27
                   

   The augmented matrix is:

   [ 1  1  1 |  6 ]
   [ 0  2  5 | -4 ]
   [ 2  5 -1 | 27 ]
                   

2. Forward Elimination (Gaussian Elimination part):
   • Work from left to right, column by column.
   • For each column, find a non-zero entry (pivot) in the current row or below it. If necessary, swap rows to bring this pivot to the current row.
   • Divide the current row by the pivot to make the pivot element 1.
   • Use this leading 1 to eliminate (make zero) all entries below it in the same column by adding multiples of the current row to the rows below.
   At the end of this phase, the matrix will be in row echelon form.
3. Backward Elimination (Jordan part):
   • Work from right to left, column by column, starting from the last pivot.
   • Use each leading 1 to eliminate (make zero) all entries above it in the same column.
   After this phase, the matrix will be in reduced row echelon form.
4. Interpret the Result:
   • If the left part of the augmented matrix becomes an identity matrix, the solutions for the variables are directly given by the rightmost column.
   • If a row like [0 0 ... 0 | k] appears where k is a non-zero number, there is no solution.
   • If a row like [0 0 ... 0 | 0] appears, and there are fewer leading 1s than variables, there are infinitely many solutions (some variables are free variables).

Example Calculation:

Let's use the example system provided in the calculator's default values:

x + y + z = 6
    2y + 5z = -4
2x + 5y - z = 27
        

The augmented matrix is:

[ 1  1  1 |  6 ]
[ 0  2  5 | -4 ]
[ 2  5 -1 | 27 ]
        

Applying Gauss-Jordan elimination steps:

1. R3 = R3 – 2*R1

   [ 1  1  1 |  6 ]
   [ 0  2  5 | -4 ]
   [ 0  3 -3 | 15 ]
                   

2. R2 = R2 / 2

   [ 1  1  1 |  6 ]
   [ 0  1  2.5 | -2 ]
   [ 0  3 -3 | 15 ]
                   

3. R1 = R1 – R2

   [ 1  0 -1.5 |  8 ]
   [ 0  1  2.5 | -2 ]
   [ 0  3 -3 | 15 ]
                   

4. R3 = R3 – 3*R2

   [ 1  0 -1.5 |  8 ]
   [ 0  1  2.5 | -2 ]
   [ 0  0 -10.5 | 21 ]
                   

5. R3 = R3 / -10.5

   [ 1  0 -1.5 |  8 ]
   [ 0  1  2.5 | -2 ]
   [ 0  0  1 | -2 ]
                   

6. R1 = R1 + 1.5*R3

   [ 1  0  0 |  5 ]
   [ 0  1  2.5 | -2 ]
   [ 0  0  1 | -2 ]
                   

7. R2 = R2 – 2.5*R3

   [ 1  0  0 |  5 ]
   [ 0  1  0 |  3 ]
   [ 0  0  1 | -2 ]
                   

The reduced row echelon form is:

[ 1  0  0 |  5 ]
[ 0  1  0 |  3 ]
[ 0  0  1 | -2 ]
        

This directly gives the solution: x = 5, y = 3, z = -2.

" + value + "

No Solution:

Infinite Solutions:

"; for (var i = 0; i < rows; i++) { var leadingOneCol = -1; for (var j = 0; j < numVariables; j++) { if (pivotCols[j] && Math.abs(rrefMatrix[i][j] – 1) < 1e-9) { leadingOneCol = j; break; } } if (leadingOneCol !== -1) { var equation = String.fromCharCode(120 + leadingOneCol) + " = " + rrefMatrix[i][numVariables].toFixed(4); for (var j = 0; j 1e-9) { var sign = (rrefMatrix[i][j] < 0) ? "+" : "-"; equation += " " + sign + " " + Math.abs(rrefMatrix[i][j]).toFixed(4) + String.fromCharCode(120 + j); } } solutionText += "
• " + equation + "
"; } } solutionText += "

Unique Solution:

"; for (var k = 0; k < numVariables; k++) { var variableName = String.fromCharCode(120 + k); solutionText += "
• " + variableName + " = " + solutions[k].toFixed(4) + "
"; } solutionText += "

Unique Solution:

"; for (var i = 0; i < numVariables; i++) { var variableName = String.fromCharCode(120 + i); // x, y, z… solutionText += "
• " + variableName + " = " + solutions[i].toFixed(4) + "
"; } solutionText += "