Methods Kramer and Gaussian one of the most popular solutions SLAU. In addition, in some cases it is advisable to use specific methods. The session is close, and now is the time to repeat or master them from scratch. Today we deal with the solution by the Cramer method. After all, the solution of the system linear equations Cramer's method is a very useful skill.

Systems of linear algebraic equations

The system of linear algebraic equations is a system of equations of the form:

Value set x , at which the equations of the system turn into identities, is called the solution of the system, a and b are real coefficients. A simple system consisting of two equations with two unknowns can be solved mentally or by expressing one variable in terms of the other. But there can be much more than two variables (x) in SLAE, and simple school manipulations are indispensable here. What to do? For example, solve SLAE by Cramer's method!

So let the system be n equations with n unknown.

Such a system can be rewritten in matrix form

Here A is the main matrix of the system, X and B , respectively, column matrices of unknown variables and free members.

SLAE solution by Cramer's method

If the determinant of the main matrix is ​​not equal to zero (the matrix is ​​nonsingular), the system can be solved using the Cramer method.

According to the Cramer method, the solution is found by the formulas:

Here delta is the determinant of the main matrix, and delta x n-th - the determinant obtained from the determinant of the main matrix by replacing the n-th column with a column of free members.

This is the whole point of Cramer's method. Substituting the values ​​found by the above formulas x into the desired system, we are convinced of the correctness (or vice versa) of our solution. To make it easier for you to grasp the point, here is an example. detailed solution SLAE by Cramer's method:

Even if you don't succeed the first time, don't be discouraged! With a little practice, you'll start popping SLOWs like nuts. Moreover, now it is absolutely not necessary to pore over a notebook, solving cumbersome calculations and writing on the rod. It is easy to solve SLAE by the Cramer method online, just by substituting the coefficients into the finished form. try out online calculator solutions by the Cramer method can be, for example, on this site.

And if the system turned out to be stubborn and does not give up, you can always ask our authors for help, for example, to buy a synopsis. If there are at least 100 unknowns in the system, we will definitely solve it correctly and just in time!

Let a system of three linear equations be given:

To solve a system of linear equations by the Cramer method, the main determinant of the system  is compiled from the coefficients of the unknowns. For system (1), the main determinant has the form
.

Next, the determinants are compiled with respect to the variables
,,. To do this, in the main determinant, instead of a column of coefficients for the corresponding variable, a column of free members is written, that is

,
,
.

Then the solution of the system is found by the Cramer formulas

,
,

It should be noted that the system has a unique solution
if the main determinant
. If
and
= 0,= 0,= 0, then the system has an infinite number of solutions, which cannot be found by Cramer's formulas. If
and
0, or 0, or 0, then the system of equations is inconsistent, that is, it has no solutions.

Example

Solution:

1) Compose and calculate the main determinant of the system, consisting of coefficients for unknowns.

.

Therefore, the system has a unique solution.

2) Compose and calculate auxiliary determinants, replacing the corresponding column in  with a column of free members.

Using Cramer's formulas, we find the unknowns:

,
,
.

We will check to make sure the solution is correct

Those.
.

, i.e.

, i.e.

Answer: .

Example

Solve the system of equations by Cramer's method:

Solution:

1) Compose and calculate the main determinant of the system from the coefficients of the unknowns:

.

Therefore, the system does not have a unique solution.

2) Compose and calculate auxiliary determinants, replacing the corresponding column in  with a column of free members:

,
, hence the system is inconsistent.

Answer: the system is inconsistent.

Gauss method

The Gauss method consists of two stages. The first stage consists in the successive elimination of variables from the equations of the system using actions that do not violate the equivalence of the system. For example, consider the first two equations of system (1).

(1)

It is necessary by adding these two equations to obtain an equation in which there is no variable . Multiply the first equation by , and the second on (
) and add the resulting equations

We replace the coefficient before y, z and a free member on ,and accordingly, we obtain a new pair of equations

Note that there is no variable in the second equation x.

Having carried out similar actions on the first and third equations of system (1), and then on the second and third equations obtained as a result of addition, we transform system (1) to the form

(2)

This result is possible if the system has a unique solution. In this case, the solution is found using the reverse Gauss method (second stage). From the last equation of system (2) we find the unknown variable z, then from the second equation we find y, a x respectively from the first, substituting in them already found unknowns.

Sometimes, as a result of adding two equations, the total equation can take one of the following forms:

BUT)
, where
. This means that the system being solved is inconsistent.

B), that is
. Such an equation is excluded from the system, as a result, the number of equations in the system becomes less than the number of variables, and the system has an infinite number of solutions, the finding of which will be shown by an example.

Example

Solution:

Consider the following method for implementing the first stage of the solution by the Gauss method. Let us write down three rows of coefficients for the unknown and free terms corresponding to the three equations of the system. We separate the free terms from the coefficients with a vertical line, and draw a horizontal line under the third line.

We circle the first line, which corresponds to the first equation of the system - the coefficients in this equation will remain unchanged. Instead of the second line (equation), you need to get a line (equation), where the coefficient at equals zero. To do this, we multiply all the numbers in the first row by (-2) and add them to the corresponding numbers in the second row. We write the resulting amounts under the horizontal line (fourth line). In order to instead of the third line (equation) also get a line (equation) in which the coefficient at equals zero, we multiply all the numbers in the first row by (-5) and add them to the corresponding numbers in the third row. We write the resulting amounts in the fifth line and draw a new horizontal line under it. The fourth line (or the fifth - optionally) will be circled. The row with smaller coefficients is selected. In this line, the coefficients will remain unchanged. Instead of the fifth line, you need to get a line where two coefficients are already equal to zero. Multiply the fourth row by 3 and add it to the fifth. We write the amount under the horizontal line (sixth line) and circle it.

All described actions are shown in Table 1 using arithmetic signs and arrows. We write the rows circled in the table again in the form of equations (3) and, using the reverse move of the Gauss method, we find the values ​​of the variables x, y and z.

Table 1

We restore the system of equations obtained as a result of our transformations:

(3)

Reverse Gauss method

From the third equation
find
.

Into the second equation of the system
substitute the found value
, we get
or
.

From the first equation
, substituting the already found values ​​of the variables, we obtain
, that is
.

To make sure that the solution is correct, a check must be made in all three equations of the system.

Examination:

, we get

Get

Get

This means that the system is correct.

Answer:
,
,
.

Example

Solve the system using the Gauss method:

Solution:

The order of actions in this example is similar to the order in the previous example, and the specific actions are indicated in Table 2.

As a result of the transformations, we obtain an equation of the form , therefore, the given system is inconsistent.

Answer: the system is inconsistent.

Example

Solve the system using the Gauss method:

Solution:

Table 3

As a result of the transformations, we obtain an equation of the form , which is excluded from consideration. Thus, we have a system of equations in which the number of unknowns is 3, and the number of equations is 2.

The system has an infinite number of solutions. To find these solutions, we introduce one free variable. (The number of free variables is always equal to the difference between the number of unknowns and the number of equations remaining after the transformation of the system. In our case, 3 - 2 = 1).

Let
is a free variable.

Then from the second equation we find
, where
and then find x from the first equation
or
.

In this way,
;
;
.

Let's make a check in the equations that were not involved in finding and , that is, in the second and third equations of the original system.

Examination:

or , we get
.

or , we get
.

The system is correct. Giving an arbitrary constant various meanings, we will get different values x, y and z.

Answer:
;
;
.

2. Solving systems of equations by the matrix method (using the inverse matrix).
3. Gauss method for solving systems of equations.

Cramer's method.

Cramer's method is used to solve systems of linear algebraic equations ( SLAU).

Formulas on the example of a system of two equations with two variables.
Given: Solve the system by Cramer's method

Concerning Variables X and at.
Solution:
Find the determinant of the matrix, composed of the coefficients of the system Calculation of determinants. :

Let's apply Cramer's formulas and find the values ​​of the variables:
and .
Example 1:
Solve the system of equations:

regarding variables X and at.
Solution:


Let's replace the first column in this determinant with a column of coefficients from the right side of the system and find its value:

Let's do a similar action, replacing the second column in the first determinant:

Applicable Cramer's formulas and find the values ​​of the variables:
and .
Answer:
Comment: This method can be used to solve systems of higher dimensions.

Comment: If it turns out that , and it is impossible to divide by zero, then they say that the system does not have a unique solution. In this case, the system has either infinitely many solutions or no solutions at all.

Example 2(an infinite number of solutions):

Solve the system of equations:

regarding variables X and at.
Solution:
Find the determinant of the matrix, composed of the coefficients of the system:

Solving systems by the substitution method.

The first of the equations of the system is an equality that is true for any values ​​of the variables (because 4 is always equal to 4). So there is only one equation left. This is a relationship equation between variables.
We got that the solution of the system is any pair of values ​​of variables related by equality .
The general solution is written like this:
Particular solutions can be determined by choosing an arbitrary value of y and calculating x from this relationship equation.

etc.
There are infinitely many such solutions.
Answer: common decision
Private Solutions:

Example 3(no solutions, the system is inconsistent):

Solve the system of equations:

Solution:
Find the determinant of the matrix, composed of the coefficients of the system:

You can't use Cramer's formulas. Let's solve this system by the substitution method

The second equation of the system is an equality that is not valid for any values ​​of the variables (of course, since -15 is not equal to 2). If one of the equations of the system is not true for any values ​​of the variables, then the whole system has no solutions.
Answer: no solutions

In the first part, we looked at some theoretical material, the substitution method, and the method of term-by-term addition of system equations. To everyone who came to the site through this page, I recommend that you read the first part. Perhaps, some visitors will find the material too simple, but in the course of solving systems of linear equations, I made a number of very important remarks and conclusions regarding the solution of mathematical problems in general.

And now we will analyze Cramer's rule, as well as the solution of a system of linear equations using the inverse matrix (matrix method). All materials are presented simply, in detail and clearly, almost all readers will be able to learn how to solve systems using the above methods.

We first consider Cramer's rule in detail for a system of two linear equations in two unknowns. What for? “After all, the simplest system can be solved by the school method, by term-by-term addition!

The fact is that even if sometimes, but there is such a task - to solve a system of two linear equations with two unknowns using Cramer's formulas. Secondly, a simpler example will help you understand how to use Cramer's rule for a more complex case - a system of three equations with three unknowns.

In addition, there are systems of linear equations with two variables, which it is advisable to solve exactly according to Cramer's rule!

Consider the system of equations

At the first step, we calculate the determinant , it is called the main determinant of the system.

Gauss method.

If , then the system has a unique solution, and to find the roots, we must calculate two more determinants:
and

In practice, the above qualifiers can also be denoted by the Latin letter.

The roots of the equation are found by the formulas:
,

Example 7

Solve a system of linear equations

Solution: We see that the coefficients of the equation are quite large, on the right side there are decimal fractions with a comma. The comma is a rather rare guest in practical tasks in mathematics, I took this system from an econometric problem.

How to solve such a system? You can try to express one variable in terms of another, but in this case, you will surely get terrible fancy fractions that are extremely inconvenient to work with, and the design of the solution will look just awful. You can multiply the second equation by 6 and subtract term by term, but the same fractions will appear here.

What to do? In such cases, Cramer's formulas come to the rescue.

;

;

Answer: ,

Both roots have infinite tails and are found approximately, which is quite acceptable (and even commonplace) for econometrics problems.

Comments are not needed here, since the task is solved according to ready-made formulas, however, there is one caveat. When using this method, compulsory The fragment of the assignment is the following fragment: "so the system has a unique solution". Otherwise, the reviewer may punish you for disrespecting Cramer's theorem.

It will not be superfluous to check, which is convenient to carry out on a calculator: we substitute the approximate values ​​\u200b\u200bin the left side of each equation of the system. As a result, with a small error, numbers that are on the right side should be obtained.

Example 8

Express your answer in ordinary improper fractions. Make a check.

This is an example for an independent solution (example of fine design and answer at the end of the lesson).

We turn to the consideration of Cramer's rule for a system of three equations with three unknowns:

We find the main determinant of the system:

If , then the system has infinitely many solutions or is inconsistent (has no solutions). In this case, Cramer's rule will not help, you need to use the Gauss method.

If , then the system has a unique solution, and to find the roots, we must calculate three more determinants:
, ,

And finally, the answer is calculated by the formulas:

As you can see, the “three by three” case is fundamentally no different from the “two by two” case, the column of free terms sequentially “walks” from left to right along the columns of the main determinant.

Example 9

Solve the system using Cramer's formulas.

Solution: Let's solve the system using Cramer's formulas.

, so the system has a unique solution.

Answer: .

Actually, there is nothing special to comment here again, in view of the fact that the decision is made according to ready-made formulas. But there are a couple of notes.

It happens that as a result of calculations, “bad” irreducible fractions are obtained, for example: .
I recommend the following "treatment" algorithm. If there is no computer at hand, we do this:

1) There may be a mistake in the calculations. As soon as you encounter a “bad” shot, you must immediately check whether is the condition rewritten correctly. If the condition is rewritten without errors, then you need to recalculate the determinants using the expansion in another row (column).

2) If no errors were found as a result of the check, then most likely a typo was made in the condition of the assignment. In this case, calmly and CAREFULLY solve the task to the end, and then make sure to check and draw it up on a clean copy after the decision. Of course, checking a fractional answer is an unpleasant task, but it will be a disarming argument for the teacher, who, well, really likes to put a minus for any bad thing like. How to deal with fractions is detailed in the answer for Example 8.

If you have a computer at hand, then use an automated program to check it, which can be downloaded for free at the very beginning of the lesson. By the way, it is most advantageous to use the program right away (even before starting the solution), you will immediately see the intermediate step at which you made a mistake! The same calculator automatically calculates the solution of the system using the matrix method.

Second remark. From time to time there are systems in the equations of which some variables are missing, for example:

Here in the first equation there is no variable , in the second there is no variable . In such cases, it is very important to correctly and CAREFULLY write down the main determinant:
– zeros are put in place of missing variables.
By the way, it is rational to open determinants with zeros in the row (column) in which zero is located, since there are noticeably fewer calculations.

Example 10

Solve the system using Cramer's formulas.

This is an example for self-solving (finishing sample and answer at the end of the lesson).

For the case of a system of 4 equations with 4 unknowns, Cramer's formulas are written according to similar principles. You can see a live example in the Determinant Properties lesson. Reducing the order of the determinant - five 4th order determinants are quite solvable. Although the task is already very reminiscent of a professor's shoe on the chest of a lucky student.

Solution of the system using the inverse matrix

The inverse matrix method is essentially a special case matrix equation(See Example No. 3 of the specified lesson).

To study this section, you need to be able to expand the determinants, find the inverse matrix and perform matrix multiplication. Relevant links will be given as the explanation progresses.

Example 11

Solve the system with the matrix method

Solution: We write the system in matrix form:
, where

Please look at the system of equations and the matrices. By what principle we write elements into matrices, I think everyone understands. The only comment: if some variables were missing in the equations, then zeros would have to be put in the corresponding places in the matrix.

We find the inverse matrix by the formula:
, where is the transposed matrix of algebraic complements of the corresponding elements of the matrix .

First, let's deal with the determinant:

Here the determinant is expanded by the first line.

Attention! If , then the inverse matrix does not exist, and it is impossible to solve the system by the matrix method. In this case, the system is solved by the elimination of unknowns (Gauss method).

Now you need to calculate 9 minors and write them into the matrix of minors

Reference: It is useful to know the meaning of double subscripts in linear algebra. The first digit is the line number in which the element is located. The second digit is the number of the column in which the element is located:

That is, a double subscript indicates that the element is in the first row, third column, while, for example, the element is in the 3rd row, 2nd column

Let the system of linear equations contain as many equations as the number of independent variables, i.e. has the form

Such systems of linear equations are called quadratic. The determinant composed of the coefficients of the independent variables of the system (1.5) is called the main determinant of the system. We will denote it with the Greek letter D. Thus,

If in the main determinant an arbitrary ( j th) column, replace it with the column of free members of the system (1.5), then we can get more n auxiliary determinants:

(j = 1, 2, …, n). (1.7)

Cramer's rule solving quadratic systems of linear equations is as follows. If the principal determinant D of system (1.5) is nonzero, then the system has a unique solution, which can be found by the formulas:

Example 1.5. Solve the system of equations using Cramer's method

Let us calculate the main determinant of the system:

Since D¹0, the system has a unique solution that can be found using formulas (1.8):

In this way,

Matrix Actions

1. Multiplication of a matrix by a number. The operation of multiplying a matrix by a number is defined as follows.

2. In order to multiply a matrix by a number, you need to multiply all its elements by this number. That is

Example 1.6. .

Matrix addition.

This operation is introduced only for matrices of the same order.

In order to add two matrices, it is necessary to add the corresponding elements of the other matrix to the elements of one matrix:

(1.10)
The operation of matrix addition has the properties of associativity and commutativity.

Example 1.7. .

Matrix multiplication.

If the number of matrix columns BUT matches the number of matrix rows AT, then for such matrices the operation of multiplication is introduced:

Thus, when multiplying the matrix BUT dimensions m´ n to matrix AT dimensions n´ k we get a matrix FROM dimensions m´ k. In this case, the elements of the matrix FROM are calculated according to the following formulas:

Problem 1.8. Find, if possible, the product of matrices AB and BA:

Solution. 1) To find a work AB, you need matrix rows A multiply by matrix columns B:

2) Artwork BA does not exist, because the number of columns of the matrix B does not match the number of matrix rows A.

Inverse matrix. Solving systems of linear equations in a matrix way

Matrix A- 1 is called the inverse of a square matrix BUT if the equality holds:

where through I denotes the identity matrix of the same order as the matrix BUT:

For a square matrix to have an inverse, it is necessary and sufficient that its determinant be nonzero. The inverse matrix is ​​found by the formula:

where A ij- algebraic additions to elements aij matrices BUT(note that algebraic additions to the rows of the matrix BUT are arranged in the inverse matrix in the form of corresponding columns).

Example 1.9. Find inverse matrix A- 1 to matrix

We find the inverse matrix by formula (1.13), which for the case n= 3 looks like:

Let's find det A = | A| = 1 x 3 x 8 + 2 x 5 x 3 + 2 x 4 x 3 - 3 x 3 x 3 - 1 x 5 x 4 - 2 x 2 x 8 = 24 + 30 + 24 - 27 - 20 - 32 = - 1. Since the determinant of the original matrix is ​​different from zero, then the inverse matrix exists.

1) Find algebraic additions A ij:

For the convenience of finding the inverse matrix, we placed the algebraic additions to the rows of the original matrix in the corresponding columns.

From the obtained algebraic additions, we compose a new matrix and divide it by the determinant det A. Thus, we will get the inverse matrix:

Quadratic systems of linear equations with a non-zero principal determinant can be solved using an inverse matrix. For this, system (1.5) is written in matrix form:

Multiplying both sides of equality (1.14) on the left by A- 1 , we get the solution of the system:

Thus, in order to find a solution to a square system, you need to find the inverse matrix to the main matrix of the system and multiply it on the right by the column matrix of free terms.

Problem 1.10. Solve a system of linear equations

using an inverse matrix.

Solution. We write the system in matrix form: ,

where is the main matrix of the system, is the column of unknowns, and is the column of free members. Since the main determinant of the system is , then the main matrix of the system BUT has an inverse matrix BUT-one . To find the inverse matrix BUT-1 , calculate the algebraic complements to all elements of the matrix BUT:

From the obtained numbers we compose a matrix (moreover, algebraic additions to the rows of the matrix BUT write in the appropriate columns) and divide it by the determinant D. Thus, we have found the inverse matrix:

The solution of the system is found by the formula (1.15):

In this way,

Solving Systems of Linear Equations by Ordinary Jordan Exceptions

Let an arbitrary (not necessarily square) system of linear equations be given:

It is required to find a solution to the system, i.e. such a set of variables that satisfies all the equalities of the system (1.16). AT general case system (1.16) can have not only one solution, but also an infinite number of solutions. It may also have no solutions at all.

When solving such problems, the well-known method of eliminating unknowns from the school course, which is also called the method of ordinary Jordanian eliminations, is used. essence this method lies in the fact that in one of the equations of system (1.16) one of the variables is expressed in terms of other variables. Then this variable is substituted into other equations of the system. The result is a system that contains one equation and one less variable than the original system. The equation from which the variable was expressed is remembered.

This process is repeated until one last equation remains in the system. In the process of eliminating unknowns, some equations can turn into true identities, for example. Such equations are excluded from the system, since they are valid for any values ​​of the variables and, therefore, do not affect the solution of the system. If, in the process of eliminating unknowns, at least one equation becomes an equality that cannot be satisfied for any values ​​of the variables (for example, ), then we conclude that the system has no solution.

If in the course of solving inconsistent equations did not arise, then one of the remaining variables in it is found from the last equation. If only one variable remains in the last equation, then it is expressed as a number. If other variables remain in the last equation, then they are considered parameters, and the variable expressed through them will be a function of these parameters. Then the so-called "reverse move" is made. The found variable is substituted into the last memorized equation and the second variable is found. Then the two found variables are substituted into the penultimate memorized equation and the third variable is found, and so on, up to the first memorized equation.

As a result, we get the solution of the system. This solution will be the only one if the found variables are numbers. If the first found variable, and then all the others depend on the parameters, then the system will have an infinite number of solutions (each set of parameters corresponds to a new solution). Formulas that allow finding a solution to the system depending on a particular set of parameters are called the general solution of the system.

Example 1.11.

x

After memorizing the first equation and bringing similar terms in the second and third equations, we arrive at the system:

Express y from the second equation and substitute it into the first equation:

Remember the second equation, and from the first we find z:

Making the reverse move, we successively find y and z. To do this, we first substitute into the last memorized equation , from which we find y:

Then we substitute and into the first memorized equation , from which we find x:

Problem 1.12. Solve a system of linear equations by eliminating unknowns:

Solution. Let us express the variable from the first equation x and substitute it into the second and third equations:

In this system, the first and second equations contradict each other. Indeed, expressing y from the first equation and substituting it into the second equation, we get that 14 = 17. This equality is not satisfied, for any values ​​of the variables x, y, and z. Consequently, system (1.17) is inconsistent, i.e., has no solution.

Readers are invited to independently verify that the main determinant of the original system (1.17) is equal to zero.

Consider a system that differs from system (1.17) by only one free term.

Problem 1.13. Solve a system of linear equations by eliminating unknowns:

Solution. As before, we express the variable from the first equation x and substitute it into the second and third equations:

Remember the first equation and give similar terms in the second and third equations. We arrive at the system:

expressing y from the first equation and substituting it into the second equation , we get the identity 14 = 14, which does not affect the solution of the system, and, therefore, it can be excluded from the system.

In the last memorized equality, the variable z will be considered as a parameter. We believe . Then

Substitute y and z into the first memorized equality and find x:

Thus, system (1.18) has an infinite set of solutions, and any solution can be found by formulas (1.19) by choosing an arbitrary value of the parameter t:

(1.19)
Thus, the solutions of the system, for example, are the following sets of variables (1; 2; 0), (2; 26; 14), etc. Formulas (1.19) express the general (any) solution of system (1.18).

In the case when the original system (1.16) has a sufficiently large number of equations and unknowns, the indicated method of ordinary Jordan eliminations seems cumbersome. However, it is not. It suffices to derive the algorithm for recalculating the coefficients of the system at one step in a general form and formalize the solution of the problem in the form of special Jordan tables.

Let a system of linear forms (equations) be given:

, (1.20)
where x j- independent (desired) variables, aij- constant coefficients
(i = 1, 2,…, m; j = 1, 2,…, n). Right parts of the system y i (i = 1, 2,…, m) can be both variables (dependent) and constants. It is required to find solutions to this system by eliminating unknowns.

Let us consider the following operation, hereinafter referred to as "one step of ordinary Jordan exceptions". From an arbitrary ( r th) equality, we express an arbitrary variable ( x s) and substitute into all other equalities. Of course, this is only possible if a rs¹ 0. Coefficient a rs is called the resolving (sometimes guiding or main) element.

We will get the following system:

From s th equality of system (1.21), we will subsequently find the variable x s(after other variables are found). S The th line is remembered and subsequently excluded from the system. The remaining system will contain one equation and one less independent variable than the original system.

Let us calculate the coefficients of the resulting system (1.21) in terms of the coefficients of the original system (1.20). Let's start with r th equation, which, after expressing the variable x s through the rest of the variables will look like this:

Thus, the new coefficients r th equation are calculated by the following formulas:

(1.23)
Let us now calculate the new coefficients b ij(i¹ r) of an arbitrary equation. To do this, we substitute the variable expressed in (1.22) x s in i th equation of system (1.20):

After bringing like terms, we get:

(1.24)
From equality (1.24) we obtain formulas by which the remaining coefficients of system (1.21) are calculated (with the exception of r th equation):

(1.25)
The transformation of systems of linear equations by the method of ordinary Jordanian eliminations is presented in the form of tables (matrices). These tables are called "Jordan tables".

Thus, problem (1.20) is associated with the following Jordan table:

Table 1.1

	x 1	x 2	…	x j	…	x s	…	x n
y 1 =	a 11	a 12		a 1j		a 1s		a 1n
…………………………………………………………………..
y i=	a i 1	a i 2		aij		a is		a in
…………………………………………………………………..
y r=	a r 1	a r 2		a rj		a rs		a rn
………………………………………………………………….
y n=	a m 1	a m 2		a mj		a ms		amn

Jordan table 1.1 contains the left head column, in which the right parts of the system (1.20) are written, and the top head line, in which the independent variables are written.

The remaining elements of the table form the main matrix of coefficients of system (1.20). If we multiply the matrix BUT to the matrix consisting of the elements of the upper header row, then we get the matrix consisting of the elements of the left header column. That is, in essence, the Jordan table is a matrix form of writing a system of linear equations: . In this case, the following Jordan table corresponds to system (1.21):

Table 1.2

	x 1	x 2	…	x j	…	y r	…	x n
y 1 =	b 11	b 12		b 1 j		b 1 s		b 1 n
…………………………………………………………………..
y i =	b i 1	b i 2		b ij		b is		b in
…………………………………………………………………..
x s =	br 1	br 2		b rj		brs		b rn
………………………………………………………………….
y n =	b m 1	b m 2		bmj		b ms		bmn

Permissive element a rs we will highlight in bold. Recall that in order to implement one step of Jordan exceptions, the resolving element must be nonzero. A table row containing a permissive element is called a permissive row. The column containing the enable element is called the enable column. When moving from a given table to the next table, one variable ( x s) from the top header row of the table is moved to the left header column and, conversely, one of the free members of the system ( y r) is moved from the left header column of the table to the top header row.

Let us describe the algorithm for recalculating the coefficients in passing from the Jordan table (1.1) to the table (1.2), which follows from formulas (1.23) and (1.25).

1. The enabling element is replaced by the inverse number:

2. The remaining elements of the permissive line are divided by the permissive element and change sign to the opposite:

3. The remaining elements of the enabling column are divided into the enabling element:

4. Elements that are not included in the resolving row and resolving column are recalculated according to the formulas:

The last formula is easy to remember if you notice that the elements that make up the fraction are at the intersection i-oh and r-th lines and j th and s-th columns (resolving row, resolving column and the row and column at the intersection of which the element to be recalculated is located). More precisely, when memorizing the formula, you can use the following diagram:

-21 -26 -13 -37

Performing the first step of the Jordanian exceptions, any element of Table 1.3 located in the columns x 1 ,…, x 5 (all specified elements are not equal to zero). You should not only select the enabling element in the last column, because need to find independent variables x 1 ,…, x 5 . We choose, for example, the coefficient 1 with a variable x 3 in the third row of table 1.3 (the enabling element is shown in bold). When moving to table 1.4, the variable x The 3 from the top header row is swapped with the constant 0 of the left header column (third row). At the same time, the variable x 3 is expressed in terms of the remaining variables.

string x 3 (Table 1.4) can, having previously remembered, be excluded from Table 1.4. Table 1.4 also excludes the third column with a zero in the upper header line. The point is that regardless of the coefficients of this column b i 3 all terms corresponding to it of each equation 0 b i 3 systems will be equal to zero. Therefore, these coefficients can not be calculated. Eliminating one variable x 3 and remembering one of the equations, we arrive at a system corresponding to Table 1.4 (with the line crossed out x 3). Choosing in table 1.4 as a resolving element b 14 = -5, go to table 1.5. In table 1.5, we remember the first row and exclude it from the table along with the fourth column (with zero at the top).

Table 1.5 Table 1.6

From the last table 1.7 we find: x 1 = - 3 + 2x 5 .

Sequentially substituting the already found variables into the memorized lines, we find the remaining variables:

Thus, the system has an infinite number of solutions. variable x 5 , you can assign arbitrary values. This variable acts as a parameter x 5 = t. We proved the compatibility of the system and found its general solution:

x 1 = - 3 + 2t

x 2 = - 1 - 3t

x 3 = - 2 + 4t . (1.27)
x 4 = 4 + 5t

x 5 = t

Giving parameter t different values, we get an infinite number of solutions to the original system. So, for example, the solution of the system is the following set of variables (- 3; - 1; - 2; 4; 0).