Since the beginning of the 16th-18th centuries, mathematicians have intensively begun to study functions, thanks to which so much in our lives has changed. Computer technology without this knowledge it simply would not exist. Various concepts, theorems, and solution techniques have been created to solve complex problems, linear equations, and functions. One of such universal and rational methods and techniques for solving linear equations and their systems became the Gaussian method. Matrices, their rank, determinant - everything can be calculated without using complex operations.

What is SLAU

In mathematics there is the concept of SLAE - a system of linear algebraic equations. What is she like? This is a set of m equations with the required n unknown quantities, usually denoted as x, y, z, or x 1, x 2 ... x n, or other symbols. Solving a given system using the Gaussian method means finding all the unknown unknowns. If a system has the same number of unknowns and equations, then it is called an nth order system.

The most popular methods for solving SLAEs

IN educational institutions Secondary education students study various methods for solving such systems. Most often this simple equations, consisting of two unknowns, so any existing method It won’t take much time to find the answer to them. This can be like a substitution method, when another is derived from one equation and substituted into the original one. Or the method of term-by-term subtraction and addition. But the Gauss method is considered the easiest and most universal. It makes it possible to solve equations with any number of unknowns. Why is this particular technique considered rational? It's simple. The good thing about the matrix method is that it does not require rewriting unnecessary symbols several times as unknowns; it is enough to perform arithmetic operations on the coefficients - and you will get a reliable result.

Where are SLAEs used in practice?

The solution to SLAEs are the points of intersection of lines on the graphs of functions. In our high-tech computer age, people who are closely associated with the development of games and other programs need to know how to solve such systems, what they represent and how to check the correctness of the resulting result. Most often, programmers develop special linear algebra calculator programs, which also includes a system of linear equations. The Gauss method allows you to calculate all existing solutions. Other simplified formulas and techniques are also used.

SLAU compatibility criterion

Such a system can only be solved if it is compatible. For clarity, let us represent the SLAE in the form Ax=b. It has a solution if rang(A) equals rang(A,b). In this case, (A,b) is an extended form matrix that can be obtained from matrix A by rewriting it with free terms. It turns out that solving linear equations using the Gaussian method is quite easy.

Perhaps some of the symbols are not entirely clear, so it is necessary to consider everything with an example. Let's say there is a system: x+y=1; 2x-3y=6. It consists of only two equations, in which there are 2 unknowns. The system will have a solution only if the rank of its matrix is ​​equal to the rank of the extended matrix. What is rank? This is the number of independent lines of the system. In our case, the rank of the matrix is ​​2. Matrix A will consist of coefficients located near the unknowns, and the coefficients located behind the “=” sign also fit into the extended matrix.

Why can SLAEs be represented in matrix form?

Based on the compatibility criterion according to the proven Kronecker-Capelli theorem, a system of linear algebraic equations can be represented in matrix form. Using the Gaussian cascade method, you can solve the matrix and get a single reliable answer for the entire system. If the rank of an ordinary matrix is ​​equal to the rank of its extended matrix, but is less than the number of unknowns, then the system has an infinite number of answers.

Matrix transformations

Before moving on to solving matrices, you need to know what actions can be performed on their elements. There are several elementary transformations:

• By rewriting the system in matrix form and solving it, you can multiply all elements of the series by the same coefficient.
• In order to transform the matrix into canonical form, you can swap two parallel rows. The canonical form implies that all matrix elements that are located along the main diagonal become ones, and the remaining ones become zeros.
• The corresponding elements of parallel rows of the matrix can be added to one another.

Jordan-Gauss method

The essence of solving systems of linear homogeneous and inhomogeneous equations The Gaussian method is to gradually eliminate the unknowns. Let's say we have a system of two equations in which there are two unknowns. To find them, you need to check the system for compatibility. The equation is solved very simply by the Gauss method. It is necessary to write down the coefficients located near each unknown in matrix form. To solve the system, you will need to write out the extended matrix. If one of the equations contains a smaller number of unknowns, then “0” must be put in place of the missing element. All apply to the matrix known methods transformations: multiplication, division by a number, adding corresponding elements of series to each other and others. It turns out that in each row it is necessary to leave one variable with the value “1”, the rest should be reduced to zero. For a more precise understanding, it is necessary to consider the Gauss method with examples.

A simple example of solving a 2x2 system

To begin with, let's take a simple system of algebraic equations, in which there will be 2 unknowns.

Let's rewrite it into an extended matrix.

To solve this system of linear equations, only two operations are required. We need to bring the matrix to canonical form so that there are ones along the main diagonal. So, transferring from the matrix form back to the system, we get the equations: 1x+0y=b1 and 0x+1y=b2, where b1 and b2 are the resulting answers in the solution process.

1. The first action when solving an extended matrix will be this: the first row must be multiplied by -7 and added corresponding elements to the second row in order to get rid of one unknown in the second equation.
2. Since solving equations using the Gauss method involves reducing the matrix to canonical form, then it is necessary to perform the same operations with the first equation and remove the second variable. To do this, we subtract the second line from the first and get the required answer - the solution of the SLAE. Or, as shown in the figure, we multiply the second row by a factor of -1 and add the elements of the second row to the first row. It is the same.

As we can see, our system was solved by the Jordan-Gauss method. We rewrite it in the required form: x=-5, y=7.

An example of a 3x3 SLAE solution

Suppose we have a more complex system of linear equations. The Gauss method makes it possible to calculate the answer even for the most seemingly confusing system. Therefore, in order to delve deeper into the calculation methodology, you can move on to a more complex example with three unknowns.

As in the previous example, we rewrite the system in the form of an extended matrix and begin to bring it to its canonical form.

To solve this system, you will need to perform much more actions than in the previous example.

1. First you need to make the first column one unit element and the rest zeros. To do this, multiply the first equation by -1 and add the second equation to it. It is important to remember that we rewrite the first line in its original form, and the second in a modified form.
2. Next, we remove this same first unknown from the third equation. To do this, multiply the elements of the first row by -2 and add them to the third row. Now the first and second lines are rewritten in their original form, and the third - with changes. As you can see from the result, we got the first one at the beginning of the main diagonal of the matrix and the remaining zeros. A few more steps, and the system of equations by the Gaussian method will be reliably solved.
3. Now you need to perform operations on other elements of the rows. The third and fourth actions can be combined into one. We need to divide the second and third lines by -1 to get rid of the minus ones on the diagonal. We have already brought the third line to the required form.
4. Next we bring the second line to canonical form. To do this, we multiply the elements of the third row by -3 and add them to the second row of the matrix. From the result it is clear that the second line is also reduced to the form we need. It remains to perform a few more operations and remove the coefficients of the unknowns from the first line.
5. To make 0 from the second element of a row, you need to multiply the third row by -3 and add it to the first row.
6. The next decisive step will be to add the necessary elements of the second row to the first row. This way we get the canonical form of the matrix, and, accordingly, the answer.

As you can see, solving equations using the Gauss method is quite simple.

An example of solving a 4x4 system of equations

Some more complex systems equations can be solved by the Gaussian method using computer programs. It is necessary to enter the coefficients for the unknowns into the existing empty cells, and the program itself will step by step calculate the required result, describing in detail each action.

Described below step-by-step instruction solutions to this example.

In the first step, free coefficients and numbers for unknowns are entered into empty cells. Thus, we get the same extended matrix that we write manually.

And all the necessary arithmetic operations are performed to bring the extended matrix to its canonical form. It is necessary to understand that the answer to a system of equations is not always integers. Sometimes the solution may be from fractional numbers.

Checking the correctness of the solution

The Jordan-Gauss method provides for checking the correctness of the result. In order to find out whether the coefficients are calculated correctly, you just need to substitute the result into the original system of equations. Left-hand side equations must match right side, located behind the equal sign. If the answers do not match, then you need to recalculate the system or try to apply to it another method of solving SLAEs known to you, such as substitution or term-by-term subtraction and addition. After all, mathematics is a science that has a huge number of different solution methods. But remember: the result should always be the same, no matter what solution method you used.

Gauss method: the most common errors when solving SLAEs

When solving linear systems of equations, errors most often occur such as incorrect transfer of coefficients into matrix form. There are systems in which some unknowns are missing from one of the equations; then, when transferring data to an extended matrix, they can be lost. As a result, when solving this system, the result may not correspond to the actual one.

Another major mistake may be incorrectly writing out the final result. It is necessary to clearly understand that the first coefficient will correspond to the first unknown from the system, the second - to the second, and so on.

The Gauss method describes in detail the solution of linear equations. Thanks to it, it is easy to carry out the necessary operations and find the right result. In addition, this is a universal tool for finding a reliable answer to equations of any complexity. Maybe that's why it is so often used when solving SLAEs.

Two systems of linear equations are called equivalent if the set of all their solutions coincides.

Elementary transformations of a system of equations are:

1. Deleting trivial equations from the system, i.e. those for which all coefficients are equal to zero;
2. Multiplying any equation by a number other than zero;
3. Adding to any i-th equation any j-th equation multiplied by any number.

A variable x i is called free if this variable is not allowed, but the entire system of equations is allowed.

Theorem. Elementary transformations transform a system of equations into an equivalent one.

The meaning of the Gaussian method is to transform the original system of equations and obtain an equivalent resolved or equivalent inconsistent system.

So, the Gaussian method consists of the following steps:

1. Let's look at the first equation. Let's choose the first non-zero coefficient and divide the entire equation by it. We obtain an equation in which some variable x i enters with a coefficient of 1;
2. Let's subtract this equation from all the others, multiplying it by such numbers that the coefficients of the variable x i in the remaining equations are zeroed. We obtain a system resolved with respect to the variable x i and equivalent to the original one;
3. If trivial equations arise (rarely, but it happens; for example, 0 = 0), we cross them out of the system. As a result, there are one fewer equations;
4. We repeat the previous steps no more than n times, where n is the number of equations in the system. Each time we select a new variable for “processing”. If inconsistent equations arise (for example, 0 = 8), the system is inconsistent.

As a result, after a few steps we will obtain either a resolved system (possibly with free variables) or an inconsistent one. Allowed systems fall into two cases:

1. The number of variables is equal to the number of equations. This means that the system is defined;
2. Number of variables more number equations. We collect all the free variables on the right - we get formulas for the allowed variables. These formulas are written in the answer.

That's all! System of linear equations solved! This is a fairly simple algorithm, and to master it you do not have to contact a higher mathematics tutor. Let's look at an example:

Task. Solve the system of equations:

Description of steps:

1. Subtract the first equation from the second and third - we get the allowed variable x 1;
2. We multiply the second equation by (−1), and divide the third equation by (−3) - we get two equations in which the variable x 2 enters with a coefficient of 1;
3. We add the second equation to the first, and subtract from the third. We get the allowed variable x 2 ;
4. Finally, we subtract the third equation from the first - we get the allowed variable x 3;
5. We have received an approved system, write down the response.

The general solution to a simultaneous system of linear equations is new system, equivalent to the original one, in which all allowed variables are expressed in terms of free ones.

When might a general solution be needed? If you have to do fewer steps than k (k is how many equations there are). However, the reasons why the process ends at some step l< k , может быть две:

1. After the lth step, we obtained a system that does not contain an equation with number (l + 1). In fact, this is good, because... the authorized system is still obtained - even a few steps earlier.
2. After the lth step, we obtained an equation in which all coefficients of the variables are equal to zero, and the free coefficient is different from zero. This is a contradictory equation, and, therefore, the system is inconsistent.

It is important to understand that the emergence of an inconsistent equation using the Gaussian method is a sufficient basis for inconsistency. At the same time, we note that as a result of the lth step, no trivial equations can remain - all of them are crossed out right in the process.

Description of steps:

1. Subtract the first equation, multiplied by 4, from the second. We also add the first equation to the third - we get the allowed variable x 1;
2. Subtract the third equation, multiplied by 2, from the second - we get the contradictory equation 0 = −5.

So, the system is inconsistent because an inconsistent equation has been discovered.

Task. Explore compatibility and find a general solution to the system:

Description of steps:

1. We subtract the first equation from the second (after multiplying by two) and the third - we get the allowed variable x 1;
2. Subtract the second equation from the third. Since all the coefficients in these equations are the same, the third equation will become trivial. At the same time, multiply the second equation by (−1);
3. Subtract the second from the first equation - we get the allowed variable x 2. The entire system of equations is now also resolved;
4. Since the variables x 3 and x 4 are free, we move them to the right to express the allowed variables. This is the answer.

So, the system is consistent and indeterminate, since there are two allowed variables (x 1 and x 2) and two free ones (x 3 and x 4).

One of the simplest ways to solve a system of linear equations is a technique based on the calculation of determinants ( Cramer's rule). Its advantage is that it allows you to immediately record the solution; it is especially convenient in cases where the coefficients of the system are not numbers, but some parameters. Its disadvantage is the cumbersomeness of calculations in the case large number equations; moreover, Cramer's rule is not directly applicable to systems in which the number of equations does not coincide with the number of unknowns. In such cases, it is usually used Gaussian method.

Systems of linear equations having the same set of solutions are called equivalent. Obviously, many solutions linear system does not change if any equations are swapped, or if one of the equations is multiplied by some non-zero number, or if one equation is added to another.

Gauss method (method sequential elimination unknown) is that with the help of elementary transformations the system is reduced to an equivalent system of a step type. First, using the 1st equation, we eliminate x 1 of all subsequent equations of the system. Then, using the 2nd equation, we eliminate x 2 from the 3rd and all subsequent equations. This process, called direct Gaussian method, continues until there is only one unknown left on the left side of the last equation x n. After this it is done inverse of the Gaussian method– solving the last equation, we find x n; after that, using this value, from the penultimate equation we calculate x n–1, etc. We find the last one x 1 from the first equation.

It is convenient to carry out Gaussian transformations by performing transformations not with the equations themselves, but with the matrices of their coefficients. Consider the matrix:

called expanded matrix of the system, because, in addition to the main matrix of the system, it includes a column of free terms. The Gaussian method is based on reducing the main matrix of the system to triangular view(or trapezoidal form in the case of non-square systems) using elementary row transformations (!) of the extended matrix of the system.

Example 5.1. Solve the system using the Gaussian method:

Solution. Let's write out the extended matrix of the system and, using the first row, after that we will reset the remaining elements:

we get zeros in the 2nd, 3rd and 4th rows of the first column:

Now we need all elements in the second column below the 2nd row to be equal to zero. To do this, you can multiply the second line by –4/7 and add it to the 3rd line. However, in order not to deal with fractions, let's create a unit in the 2nd row of the second column and only

Now, to get a triangular matrix, you need to reset the element of the fourth row of the 3rd column; to do this, you can multiply the third row by 8/54 and add it to the fourth. However, in order not to deal with fractions, we will swap the 3rd and 4th rows and the 3rd and 4th columns and only after that we will reset the specified element. Note that when rearranging the columns, the corresponding variables change places and this must be remembered; other elementary transformations with columns (addition and multiplication by a number) cannot be performed!

The last simplified matrix corresponds to a system of equations equivalent to the original one:

From here, using the inverse of the Gaussian method, we find from the fourth equation x 3 = –1; from the third x 4 = –2, from the second x 2 = 2 and from the first equation x 1 = 1. In matrix form, the answer is written as

We considered the case when the system is definite, i.e. when there is only one solution. Let's see what happens if the system is inconsistent or uncertain.

Example 5.2. Explore the system using the Gaussian method:

Solution. We write out and transform the extended matrix of the system

We write a simplified system of equations:

Here, in the last equation it turned out that 0=4, i.e. contradiction. Consequently, the system has no solution, i.e. she incompatible. à

Example 5.3. Explore and solve the system using the Gaussian method:

Solution. We write out and transform the extended matrix of the system:

As a result of the transformations, the last line contains only zeros. This means that the number of equations has decreased by one:

Thus, after simplifications, there are two equations left, and four unknowns, i.e. two unknown "extra". Let them be "superfluous", or, as they say, free variables, will x 3 and x 4 . Then

Believing x 3 = 2a And x 4 = b, we get x 2 = 1–a And x 1 = 2b–a; or in matrix form

A solution written in this way is called general, because, giving parameters a And b different meanings, it is possible to describe all possible solutions of the system. a

Solving systems of linear equations using the Gauss method. Suppose we need to find a solution to the system from n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.

The essence of the Gauss method consists of sequentially eliminating unknown variables: first eliminating x 1 from all equations of the system, starting from the second, is further excluded x 2 from all equations, starting with the third, and so on, until only the unknown variable remains in the last equation x n. This process of transforming system equations to sequentially eliminate unknown variables is called direct Gaussian method. After completing the forward progression of the Gaussian method, from the last equation we find x n, using this value from the penultimate equation we calculate xn-1, and so on, from the first equation we find x 1. The process of calculating unknown variables when moving from the last equation of the system to the first is called inverse of the Gaussian method.

Let us briefly describe the algorithm for eliminating unknown variables.

We will assume that , since we can always achieve this by rearranging the equations of the system. Eliminate the unknown variable x 1 from all equations of the system, starting from the second. To do this, to the second equation of the system we add the first, multiplied by , to the third equation we add the first, multiplied by , and so on, to nth to the equation we add the first one, multiplied by . The system of equations after such transformations will take the form

where and .

We would arrive at the same result if we expressed x 1 through other unknown variables in the first equation of the system and the resulting expression was substituted into all other equations. So the variable x 1 excluded from all equations, starting from the second.

Next, we proceed in a similar way, but only with part of the resulting system, which is marked in the figure

To do this, to the third equation of the system we add the second, multiplied by , to the fourth equation we add the second, multiplied by , and so on, to nth to the equation we add the second one, multiplied by . The system of equations after such transformations will take the form

where and . So the variable x 2 excluded from all equations starting from the third.

Next we proceed to eliminating the unknown x 3, in this case we act similarly with the part of the system marked in the figure

So we continue the direct progression of the Gaussian method until the system takes the form

From this moment we begin the reverse of the Gaussian method: we calculate x n from the last equation as, using the obtained value x n we find xn-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Example.

Solve system of linear equations Gauss method.

One of the universal and effective methods for solving linear algebraic systems is Gaussian method , consisting in the sequential elimination of unknowns.

Recall that the two systems are called equivalent (equivalent) if the sets of their solutions coincide. In other words, systems are equivalent if every solution of one of them is a solution of the other and vice versa. Equivalent systems are obtained when elementary transformations equations of the system:

multiplying both sides of the equation by a number other than zero;

adding to some equation the corresponding parts of another equation, multiplied by a number other than zero;

rearranging two equations.

Let a system of equations be given

The process of solving this system using the Gaussian method consists of two stages. At the first stage (direct motion), the system, using elementary transformations, is reduced to stepwise , or triangular form, and at the second stage (reverse) there is a sequential, starting from the last variable number, determination of the unknowns from the resulting step system.

Let us assume that the coefficient of this system
, otherwise in the system the first row can be swapped with any other row so that the coefficient at was different from zero.

Let's transform the system by eliminating the unknown in all equations except the first. To do this, multiply both sides of the first equation by and add term by term with the second equation of the system. Then multiply both sides of the first equation by and add it to the third equation of the system. Continuing this process, we obtain the equivalent system

Here
– new values ​​of coefficients and free terms that are obtained after the first step.

Similarly, considering the main element
, exclude the unknown from all equations of the system, except the first and second. Let's continue this process as long as possible, and as a result we will get a stepwise system

,

Where ,
,…,– main elements of the system
.

If, in the process of reducing the system to a stepwise form, equations appear, i.e., equalities of the form
, they are discarded since they are satisfied by any set of numbers
. If at
will appear equation of the form, which has no solutions, then this indicates the incompatibility of the system.

At reverse stroke the first unknown is expressed from the last equation of the transformed step system through all the other unknowns
which are called free . Then the variable expression from the last equation of the system is substituted into the penultimate equation and the variable is expressed from it
. Variables are defined sequentially in a similar way
. Variables
, expressed through free variables, are called basic (dependent). The result is a general solution to the system of linear equations.

To find private solution systems, free unknown
V general decision arbitrary values ​​are assigned and variable values ​​are calculated
.

It is technically more convenient to subject to elementary transformations not the system equations themselves, but the extended matrix of the system

.

The Gauss method is a universal method that allows you to solve not only square, but also rectangular systems in which the number of unknowns
not equal to the number of equations
.

The advantage of this method is also that in the process of solving we simultaneously examine the system for compatibility, since, having given the extended matrix
to stepwise form, it is easy to determine the ranks of the matrix and extended matrix
and apply Kronecker-Capelli theorem .

Example 2.1 Solve the system using the Gauss method

Solution. Number of equations
and the number of unknowns
.

Let's create an extended matrix of the system by assigning coefficients to the right of the matrix free members column .

Let's present the matrix to a triangular view; To do this, we will obtain “0” below the elements located on the main diagonal using elementary transformations.

To get the "0" in the second position of the first column, multiply the first row by (-1) and add it to the second row.

We write this transformation as the number (-1) against the first line and denote it with an arrow going from the first line to the second line.

To get "0" in the third position of the first column, multiply the first row by (-3) and add to the third row; Let's show this action using an arrow going from the first line to the third.

.

In the resulting matrix, written second in the chain of matrices, we get “0” in the second column in the third position. To do this, we multiplied the second line by (-4) and added it to the third. In the resulting matrix, multiply the second row by (-1), and divide the third by (-8). All elements of this matrix lying below the diagonal elements are zeros.

Because , the system is collaborative and defined.

The system of equations corresponding to the last matrix has a triangular form:

From the last (third) equation
. Substitute into the second equation and get
.

Let's substitute
And
into the first equation, we find


.