In linear algebra, Gaussian elimination is an efficient algorithm for solving systems of linear equations, finding the rank of a matrix, and calculating the inverse of an invertible square matrix. Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss.
Elementary row operations are used to reduce a matrix to row echelon form. An extension of this algorithm, Gauss–Jordan elimination, reduces the matrix further to reduced row echelon form. Gaussian elimination alone is sufficient for many applications.
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The method of Gaussian elimination appears in Chapter Eight, Rectangular Arrays, of the important Chinese mathematical text Jiuzhang suanshu or The Nine Chapters on the Mathematical Art. Its use is illustrated in eighteen problems, with two to five equations. The first reference to the book by this title is dated to 179 CE, but parts of it were written as early as approximately 150 BCE.[1] It was commented on by Liu Hui in the 3rd century.
However, the method was invented in Europe independently by Carl Friedrich Gauss when developing the method of least squares in his 1809 publication Theory of Motion of Heavenly Bodies.[2]
The process of Gaussian elimination has two parts. The first part (Forward Elimination) reduces a given system to either triangular or echelon form, or results in a degenerate equation with no solution, indicating the system has no solution. This is accomplished through the use of elementary row operations. The second step uses back substitution to find the solution of the system above.
Stated equivalently for matrices, the first part reduces a matrix to row echelon form using elementary row operations while the second reduces it to reduced row echelon form, or row canonical form.
Another point of view, which turns out to be very useful to analyze the algorithm, is that Gaussian elimination computes a matrix decomposition. The three elementary row operations used in the Gaussian elimination (multiplying rows, switching rows, and adding multiples of rows to other rows) amount to multiplying the original matrix with invertible matrices from the left. The first part of the algorithm computes an LU decomposition, while the second part writes the original matrix as the product of a uniquely determined invertible matrix and a uniquely determined reduced row-echelon matrix.
Suppose the goal is to find and describe the solution(s), if any, of the following system of linear equations:
The algorithm is as follows: eliminate from all equations below
, and then eliminate
from all equations below
. This will put the system into triangular form. Then, using back-substitution, each unknown can be solved for.
In our example, we eliminate from
by adding
to
, and then we eliminate
from
by adding
to
. Formally:
The result is:
Now we eliminate from
by adding
to
:
The result is:
This result is a system of linear equations in triangular form, and so the first part of the algorithm is complete.
The second part, back-substitution, consists of solving for the unknowns in reverse order. Thus, we can easily see that
Then, can be substituted into
, which can then be solved easily to obtain
Next, and
can be substituted into
, which can be solved to obtain
Thus, the system is solved.
This algorithm works for any system of linear equations. It is possible that the system cannot be reduced to triangular form, yet still have at least one valid solution: for example, if had not occurred in
and
after our first step above, the algorithm would have been unable to reduce the system to triangular form. However, it would still have reduced the system to echelon form. In this case, the system does not have a unique solution, as it contains at least one free variable. The solution set can then be expressed parametrically (that is, in terms of the free variables, so that if values for the free variables are chosen, a solution will be generated).
In practice, one does not usually deal with the actual systems in terms of equations but instead makes use of the augmented matrix (which is also suitable for computer manipulations). This, then, is the Gaussian Elimination algorithm applied to the augmented matrix of the system above, beginning with:
which, at the end of the first part of the algorithm looks like this:
That is, it is in row echelon form.
At the end of the algorithm, if the Gauss–Jordan elimination is applied we are left with
That is, it is in reduced row echelon form, or row canonical form.
Suppose is a
matrix and you need to calculate its inverse. The
identity matrix is augmented to the right of
, forming a
matrix (the block matrix
). Through application of elementary row operations and the Gaussian elimination algorithm, the left block of
can be reduced to the identity matrix
, which leaves
in the right block of
.
If the algorithm is unable to reduce to triangular form, then
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.[3]
The Gaussian elimination algorithm can be applied to any matrix
. If we get "stuck" in a given column, we move to the next column. In this way, for example, some
matrices can be transformed to a matrix that has a reduced row echelon form like
(the *'s are arbitrary entries). This echelon matrix contains a wealth of information about
: the rank of
is 5 since there are 5 non-zero rows in
; the vector space spanned by the columns of
has a basis consisting of the first, third, fourth, seventh and ninth column of
(the columns of the ones in
), and the *'s tell you how the other columns of
can be written as linear combinations of the basis columns.
Gaussian elimination to solve a system of n equations for n unknowns requires n(n+1) / 2 divisions, (2n3 + 3n2 − 5n)/6 multiplications, and (2n3 + 3n2 − 5n)/6 subtractions,[4] for a total of approximately 2n3 / 3 operations. So it has a complexity of .
This algorithm can be used on a computer for systems with thousands of equations and unknowns. However, the cost becomes prohibitive for systems with millions of equations. These large systems are generally solved using iterative methods. Specific methods exist for systems whose coefficients follow a regular pattern (see system of linear equations).
The Gaussian elimination can be performed over any field.
Gaussian elimination is numerically stable for diagonally dominant or positive-definite matrices. For general matrices, Gaussian elimination is usually considered to be stable in practice if you use partial pivoting as described below, even though there are examples for which it is unstable.[5]
As explained above, Gaussian elimination writes a given m × n matrix A uniquely as a product of an invertible m × m matrix S and a row-echelon matrix T. Here, S is the product of the matrices corresponding to the row operations performed.
The formal algorithm to compute from
follows. We write
for the entry in row
, column
in matrix
. The transformation is performed "in place", meaning that the original matrix
is lost and successively replaced by
.
i := 1
j := 1
while (i ≤ m and j ≤ n) do
Find pivot in column j, starting in row i:
maxi := i
for k := i+1 to m do
if abs(A[k,j]) > abs(A[maxi,j]) then
maxi := k
end if
end for
if A[maxi,j] ≠ 0 then
swap rows i and maxi, but do not change the value of i
Now A[i,j] will contain the old value of A[maxi,j].
divide each entry in row i by A[i,j]
Now A[i,j] will have the value 1.
for u := i+1 to m do
subtract A[u,j] * row i from row u
Now A[u,j] will be 0, since A[u,j] - A[i,j] * A[u,j] = A[u,j] - 1 * A[u,j] = 0.
end for
i := i + 1
end if
j := j + 1
end 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 "partial pivoting" improves the numerical stability of the algorithm; some variants are also in use.
The column currently being transformed is called the pivot column. Proceed from left to right, letting the pivot column be the first column, then the second column, etc. and finally the last column before the vertical line. For each pivot column, do the following two steps before moving on to the next pivot column:
Upon completion of this procedure the augmented matrix will be in row-echelon form and may be solved by back-substitution.