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Gaussian Elimination

Tool to apply the gaussian elimination method and get the row reduced echelon form, with steps, details, inverse matrix and vector solution.

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Gaussian Elimination -

Tag(s) : Matrix, Symbolic Computation

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Gaussian Elimination

Gaussian Elimination Calculator


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Equation System to Matrix Converter




See also: Equation Solver

Tool to apply the gaussian elimination method and get the row reduced echelon form, with steps, details, inverse matrix and vector solution.

Answers to Questions

What is the Gaussian Elimination method?

The Gaussian elimination algorithm (also called Gauss-Jordan, or pivot method) makes it possible to find the solutions of a system of linear equations, and to determine the inverse of a matrix.

The algorithm works on the rows of the matrix, by exchanging or multiplying the rows between them (up to a factor).

At each step, the algorithm aims to introduce into the matrix, on the elements outside the diagonal, zero values.

How to calculate the solutions of a linear equation system with Gauss?

From a system of linear equations, the first step is to convert the equations into a matrix.

Example: $$ \left\{ \begin{array}{} x&-&y&+&2z&=&5\\3x&+&2y&+&z&=&10\\2x&-&3y&-&2z&=&-10\\\end{array} \right. $$ can be written under multiplication">matrix multiplication form: $$ \left( \begin{array}{ccc} 1 & -1 & 2 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right) . \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 5 \\ 10 \\ -10 \end{array} \right) $$ that corresponds to the (augmented) matrix $$ \left( \begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 3 & 2 & 1 & 10 \\ 2 & -3 & 2 & -10 \end{array} \right) $$

Then, for each element outside the non-zero diagonal, perform an adequate calculation by adding or subtracting the other lines so that the element becomes 0.

Example: Subtract 3 times (Row 1) to (Row 2) such as the element in row 2, column 1 becomes 0: $$ \left( \begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 0 & 5 & -5 & -5 \\ 2 & -3 & -2 & -10 \end{array} \right) $$
Subtract 2 times (Row 1) to (Row 3) such as the element in row 3, column 1 becomes 0: $$ \left( \begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 0 & 5 & -5 & -5 \\ 0 & -1 & -6 & -20 \end{array} \right) $$
Subtract 1/5 times (Row 2) to (Row 3) such as the element in row 3, column 2 becomes 0: $$ \left( \begin{array}{ccc|c} 1 & -1 & 2 & 5 \\ 0 & 5 & -5 & -5 \\ 0 & 0 & -7 & -21 \end{array} \right) $$
Subtract 1/5 times (Row 2) to (Row 1) such as the element in row 1, column 2 becomes 0: $$ \left( \begin{array}{ccc|c} 1 & 0 & 1 & 4 \\ 0 & 5 & -5 & -5 \\ 0 & 0 & -7 & -21 \end{array} \right) $$
Subtract 1/7 times (Row 3) to (Row 1) such as the element in row 1, column 3 becomes 0: $$ \left( \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 5 & -5 & -5 \\ 0 & 0 & -7 & -21 \end{array} \right) $$
Subtract 5/7 times (Row 3) to (Row 2) such as the element in row 2, column 3 becomes 0: $$ \left( \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 5 & 0 & 10 \\ 0 & 0 & -7 & -21 \end{array} \right) $$

Simplify each line by dividing the value on the diagonal

Example: $$ \left( \begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 3 \end{array} \right) $$

The result vector is the last column.

Example: $ {1,2,3} $ that corresponds to $ {x,y,z} $ so $ x=1, y=2, z=3 $

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