In mathematics, **Gaussian elimination** is an algorithm in linear algebra for determining the solutions of a system of linear equations, for determining the rank of a matrix, and for calculating the inverse of an invertible square matrix.

The method is named after the mathematician Carl Friedrich Gauss, but the method is described by Liu Hui's comments written in 263 A.D. to the Chinese book *Jiuzhang suanshu* or *The Nine Chapters on the Mathematical Art*.

The computation complexity of Gaussian elimination is O(*n*^{3}), that is, the number of operations required is proportional to *n*^{3} if the matrix size is *n*-by-*n*.

Table of contents |

2 Finding the inverse of a matrix |

2x | + | y | - | z | = | 8, |

-3x | - | y | + | 2z | = | -11, |

-2x | + | y | - | 2z | = | -3. |

This is called a *system of linear equations* for the unknowns `x`, `y`, and `z`. They are called *linear* because each term is either constant or is a constant times a single variable to the first power. The goal is to transform this system to an equivalent one so that we can easily read off the solution. The allowed operations to transform a system of equations to an equivalent one are as follows:

- multiply/divide an equation by a non-zero number
- switch two equations
- add a multiple of one equation to another one

In our example, we eliminate `x` from the second equation by adding 3/2 times the first equation to the second, and then we eliminate `x` from the third equation by adding the first equation to the third. The result is:

2x | + | y | - | z | = | 8, |

1/2 y | + | 1/2 z | = | 1, | ||

2y | + | z | = | 5. |

Now we eliminate *y* from the first equation by adding -2 times the second equation to the first, and then we eliminate *y* from the third equation by adding -4 times the second equation to the third:

2x | - | 2z | = | 6, | ||

1/2 y | + | 1/2 z | = | 1, | ||

- | z | = | 1. |

Finally, we eliminate *z* from the first equation by adding -2 times the third equation to the first, and then we eliminate *z* from the second equation by adding .5 times the third equation to the second:

2x | = | 4, | ||||

1/2 y | = | 3/2, | ||||

- | z | = | 1. |

We can now read off the solution by dividing: `x` = 2, `y` = 3 and `z` = -1.

This algorithm works generally, also for bigger systems. Sometimes it is necessary to switch two equations: for instance if *y* hadn't occurred in the second equation after our first step above, we would have switched the second and third equation and then eliminated *y* from the first equation. It is possible that the algorithm gets "stuck": for instance if *y* hadn't occurred in the second and the third equation after our first step above. In this case, the system doesn't have a unique solution.

When implemented on a computer, one would typically store the system as a coefficient matrix; our original system would then look like

and in the end we're left with

or, after dividing the rows by 2, .5 and -1, respectively:

This algorithm can be used on a computer for systems with thousands of equations and unknowns. For even larger systems whose coefficients follow a regular pattern, faster iterative methods have been developed. See system of linear equations.

Suppose *A* is a square *n*-by-*n* matrix and you need to calculate its inverse. You attach the *n*-by-*n* identity matrix to the right of *A*, which produces an *n*-by-2*n* matrix. Then you start the Gauss-Jordan algorithm on that matrix. When the algorithm finishes, the identity matrix will appear on the left; the inverse of *A* can then be found to the right of the identity matrix.

If the algorithm gets "stuck" as explained above, then *A* is not invertible.

In practice, inverting a matrix is rarely required. Most of the time, one is really after the solution of a particular system of linear equations.

(the *s are arbitrary entries). Note that the entries above and below and in front of the leading ones are all zero. This echelon matrix *T* contains a wealth of information about *A*: the rank of *A* is 5 since there are 5 non-zero rows in *T*; the vectorspace spanned by the rows of *A* has as basis the first, third, forth, seventh and ninth column of *A* (the rows of the ones in *T*), and the *'s tell you how the other rows of *A* can be written as linear combinations of the basis rows.

The Gaussian elimination can be performed over any field. The three basic operations used in the Gaussian elimination (multiplying rows, switching rows, and adding multiples of rows to other rows) amount to multiplying the original matrix *A* with invertible *m*-by-*m* matrices from the left. In general, we can say:

- To every
*m*-by-*n*matrix*A*over the field*K*there exists a uniquely determined invertible*m*-by-*m*matrix*S*and a uniquely determined reduced row echelon matrix*T*such that*A*=*ST*.

i=1 j=1 while (i <= m and j<= n) do# Find pivot in column j, starting in row i: max_val = abs(A[i,j]) max_ind = i for k=i+1 to m do if abs(A[k,j]) > max_val then max_val = abs(A[k,j]) max_ind = k end_if end_for if max_val <> 0 then switch rows i and max_ind divide row i by max_val for u = i+1 to m do add - A[u,j] * row i to row u end_for i = i + 1 end_if j = j + 1end_while

This algorithm differs slightly from the one discussed earlier, because before eliminating a variable, it first exchanges rows to move the entry with the largest absolute value to the "pivot position". Such a pivoting procedure improves the numerical stability of the algorithm; some variants are also in use.

Note that if the field is the real or complex numbers and floating point arithmetic is in use, the comparison "max_val <> 0" should be exchanged by "max_val > eps" for some small, machine-dependent constant eps, since it is never correct to compare floating point numbers to zero.