# Linear system/Elimination method/Introduction/Section

Systems of linear equations are best solved by the *elimination method*, where successively a variable gets eliminated, and in the end we get an equivalent simple system which can be solved directly
(or read of that there is no solution).
For small systems, also the
substitution method
or the
equating method
are useful.

Let denote a field, and let two (inhomogeneous) systems of linear equations,

with respect to the same set of variables, be given. The systems are called*equivalent*, if their solution sets are identical.

Let be a field, and let

be an inhomogeneous system of linear equations over . Then the following manipulations on this system yield an equivalent system.

- Exchanging two equations.
- The multiplication of an equation by a scalar .
- The omitting of an equation, if it occurs twice.
- The duplication of an equation (in the sense to write down the equation again).
- The omitting or adding of a zero row (zero equation).
- The replacement of an equation by the equation which arises if we add to another equation of the system.

Most of the statements are immediately clear. (2) follows from the fact that if

holds, then also

holds for every . If , then this implication can be reversed by multiplication with .

(6). Let be the equation

and be the equation

If a tuple satisfies both equations, then it also satisfies the equation . And if the tuple satisfies the equations and , then it also satisfies the equation and .

For finding the solution of a linear system, the manipulations (2) and (6) are most important, where in general these two steps are combined, and the equation is replaced by an equation of the form
(with
).
Here,
has to be chosen is such a way that the new equation contains one variable less than the old equation. This process is called *elimination of a Variable*. This elimination is not only applied to one equation, but for all equations except one
(suitable chosen)
"working row“ , and with a fixed "working variable“. The following *elimination lemma* describes this step.

Let denote a
field,
and let denote an
(inhomogeneous)
system of linear equations in the variables . Suppose that is a variable which occurs in at least one equation with a coefficient
. Then every equation , different from ,^{[1]}
can be replaced by an equation in which does not occur any more, and such that the new system of equations , which consists of and the equations , is
equivalent
with the system .

Changing the numbering, we may assume . Let be the equation

(with ), and be the equation

Then the equation

has the form

and does not occur in it. Because of , the systems are equivalent.

Every (inhomogeneous) system of linear equations over a field can be transformed, by the manipulations described in fact, to an equivalent linear system of the form

where in each row, the first coefficient is different from . Here, either , and the last row can be omitted, or , and then the system has no solution at all.

This follows directly from the elimination lemma, by eliminating successively variables. Elimination is applied firstly to the first variable (in the given ordering), say , which occurs at all in at least one equation with a coefficient (if it only occurs in one equation, that the elimination step is done). This elimination process is applied as long as the new subsystem (without the working equation used in the elimination step before) has at least one equation with a coefficient for one variable different from . After this, we have in the end only equations without variables, and they are either only zero equations, or there is no solution.

Let an inhomogeneous system of linear equations in triangular form

with over a field be given, where the diagonal elements are all not . Then the solutions are in bijection with the tuples . The entries can be chosen arbitrarily, and they determine a unique solution, and every solution is of this form.

This is clear, as when the tuple is given, the rows are determined successively from bottom to top.

For
,
there are no free variables, and the linear system has exactly one solution.

We want to solve the inhomogeneous linear system

over (or over ). Firstly, we eliminate by keeping the first row , replacing the second row by , and replacing the third row by . This yields

Now, we can eliminate from the
(new)
third row, with the help of the second row. Because of the fractions, we rather eliminate
(which eliminates also ).
We leave the first and the second row as they are, and we replace the third row by . This yields the system, in a new ordering of the variables,^{[2]}

Now we can choose an arbitrary (free) value for . The third row determines then uniquely, we must have

In the second equation, we can choose arbitrarily, this determines via

The first row determines , namely

Hence, the solution set is

A particularly simple solution is obtained by equating the free variables and with . This yields the special solution

The general solution set can also be written as

Here,

is a description of the general solution of the corresponding homogeneous linear system.