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

Gaussian Elimination - dCode

Tag(s) : Matrix, Symbolic Computation

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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.

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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