Kramer's method for a system of equations of the 4th order. Cramer's rule

In order to master this paragraph, you must be able to open the qualifiers "two by two" and "three by three". If the qualifiers are bad, please study the lesson How to calculate the determinant?

First, we take a closer look at Cramer's rule for a system of two linear equations with two unknowns. What for? - After all the simplest system can be solved by the school method, the method of term addition!

The fact is that, even if sometimes, such a task is encountered - to solve a system of two linear equations with two unknowns according to Cramer's formulas. Second, 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 main determinant of the system.

Gauss method.

If, then the system has only decision, and to find the roots, we have to calculate two more determinants:
and

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

We find the roots of the equation by the formulas:
,

Example 7

Solve a system of linear equations

Solution: We see that the coefficients of the equation are large enough, on the right side there are decimals with a comma. The comma is a rather rare guest in practical assignments 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, which 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 perform term-by-term subtraction, 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 common) for econometric 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 a fragment of the assignment is the following fragment: "Which means that the system has the only 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 approximate values ​​into the left side of each equation in the system. As a result, with a small error, you should get numbers that are in the right parts.

Example 8

The answer is presented in ordinary irregular fractions. Make a check.

This is an example for independent decision(example of finishing and the answer at the end of the lesson).

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

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 Gaussian 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 using the formulas:

As you can see, the case "three by three" is fundamentally no different from the case "two by two", the column of free members 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.

, which means that the system has a unique solution.

Answer: .

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

It so happens that as a result of calculations "bad" irreducible fractions are obtained, for example:.
I recommend the following "cure" algorithm. If you don't have a computer at hand, we do this:

1) There may be a calculation error. As soon as you are faced with a "bad" fraction, you should immediately check is the condition rewritten correctly... If the condition is rewritten without errors, then it is necessary to recalculate the determinants using the expansion by another row (column).

2) If no errors were found as a result of checking, then most likely there was a typo in the task condition. In this case, calmly and CAREFULLY we solve the task to the end, and then be sure to check and make it out on a clean copy after the decision. Of course, checking a fractional answer is an unpleasant lesson, but it will be a disarming argument for a teacher who, well, very much loves to put a minus for any byaka like. How to handle 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 profitable 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 matrix method.

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

Here, the first equation is missing a variable, the second is missing a 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 according to the row (column) in which there is zero, since the calculations are much less.

Example 10

Solve the system using Cramer's formulas.

This is an example for an independent solution (a sample of finishing and the 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. A live example can be found in the Determinant Properties lesson. Lowering the order of the determinant - five determinants of the 4th order are quite solvable. Although the task is already quite reminiscent of the professor's boot on the chest of a lucky student.

Solving the system using the inverse matrix

Method inverse matrix Is essentially a special case matrix equation(see Example # 3 of the specified lesson).

To study this section, you must be able to expand determinants, find the inverse matrix, and perform matrix multiplication. Relevant links will be provided along the way.

Example 11

Solve system with matrix method

Solution: Let's write the system in matrix form:
, where

Please take a 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, we deal with the determinant:

Here the qualifier is expanded on 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 method of 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 this element is located. The second digit is the number of the column in which this element is located:

That is, a double subscript indicates that the item is on the first row, third column, and, for example, the item is on row 3, column 2

In the course of the solution, it is better to describe the calculation of the minors in detail, although, with some experience, they can be accustomed to counting with errors orally.

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 for the example of a system of two equations in two variables.
Given: Solve the system by Cramer's method

Variables X and at.
Solution:
Let us find the determinant of the matrix, composed of the coefficients of the system Calculation of determinants. :

We 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 the column of coefficients from the right-hand 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 single solution. In this case, the system has either infinitely many solutions or no solutions at all.

Example 2(infinite number of solutions):

Solve the system of equations:

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

Solution of systems by substitution method.

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

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

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

Solve the system of equations:

Solution:
Let us find the determinant of the matrix, composed of the coefficients of the system:

Cramer's formulas cannot be applied. Let's solve this system by the substitution method

The second equation of the system is equality, which is not true 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 entire system has no solutions.
Answer: no solutions

With the number of equations the same as the number of unknowns with the main determinant of the matrix, which is not is zero, coefficients of the system (for such equations there is a solution and it is only one).

Cramer's theorem.

When the determinant of the matrix of a square system is nonzero, it means that the system is consistent and it has one solution and it can be found by Cramer's formulas:

where Δ - determinant of the system matrix,

Δ i is the determinant of the matrix of the system, in which instead of i The th column contains the column of the right-hand sides.

When the determinant of a system is zero, it means that the system can become joint or incompatible.

This method is usually used for small systems with large calculations and when it is necessary to determine one of the unknowns. The complexity of the method is that many determinants need to be calculated.

Description of Cramer's method.

There is a system of equations:

The system of 3 equations can be solved by the Cramer method, which was considered above for a system of 2 equations.

We compose the determinant from the coefficients of the unknowns:

It will be system identifier... When D ≠ 0, then the system is compatible. Now let's compose 3 additional determinants:

,,

We solve the system of Cramer's formulas:

Examples of solving systems of equations by Cramer's method.

Example 1.

Given the system:

Let's solve it using Cramer's method.

First, you need to calculate the determinant of the matrix of the system:

Because Δ ≠ 0, hence from Cramer's theorem the system is consistent and it has one solution. We calculate additional determinants. The determinant Δ 1 is obtained from the determinant Δ, replacing its first column with the column of free coefficients. We get:

In the same way, we obtain the determinant Δ 2 from the determinant of the matrix of the system by replacing the second column with the column of free coefficients:

Cramer's method is based on the use of determinants in solving systems of linear equations. This greatly speeds up the solution process.

Cramer's method can be used to solve a system of as many linear equations as there are unknowns in each equation. If the determinant of the system is not equal to zero, then Cramer's method can be used in the solution, if it is equal to zero, then it cannot. In addition, Cramer's method can be used to solve systems of linear equations that have a unique solution.

Definition... The determinant, composed of the coefficients of the unknowns, is called the system determinant and is denoted by (delta).

Determinants

are obtained by replacing the coefficients with the corresponding unknown free terms:

;

.

Cramer's theorem. If the determinant of the system is nonzero, then the system of linear equations has one unique solution, and the unknown is equal to the ratio of the determinants. The denominator contains the determinant of the system, and the numerator contains the determinant obtained from the determinant of the system by replacing the coefficients in this unknown with free terms. This theorem holds for a system of linear equations of any order.

Example 1. Solve a system of linear equations:

According to Cramer's theorem we have:

So, the solution to system (2):

online calculator, decisive method Cramer.

Three cases when solving systems of linear equations

As is clear from Cramer's theorems, when solving a system of linear equations, three cases may occur:

First case: a system of linear equations has a unique solution

(the system is consistent and definite)

Second case: a system of linear equations has an infinite number of solutions

(the system is consistent and undefined)

** ,

those. the coefficients of the unknowns and the free terms are proportional.

Third case: the system of linear equations has no solutions

(system inconsistent)

So the system m linear equations with n variables are called inconsistent if she has no solutions, and joint if it has at least one solution. A joint system of equations that has only one solution is called a certain, and more than one - undefined.

Examples of solving systems of linear equations by Cramer's method

Let the system be given

.

Based on Cramer's theorem

………….
,

where
-

system determinant. The rest of the determinants will be obtained by replacing the column with the coefficients of the corresponding variable (unknown) with free terms:

Example 2.

.

Therefore, the system is definite. To find its solution, we calculate the determinants

According to Cramer's formulas, we find:

So, (1; 0; -1) is the only solution to the system.

To check the solutions of the 3 X 3 and 4 X 4 systems of equations, you can use the online calculator that solves the Cramer method.

If in the system of linear equations in one or several equations there are no variables, then in the determinant the corresponding elements are equal to zero! This is the next example.

Example 3. Solve the system of linear equations by Cramer's method:

.

Solution. We find the determinant of the system:

Look carefully at the system of equations and at the determinant of the system and repeat the answer to the question in which cases one or more elements of the determinant are equal to zero. So, the determinant is not equal to zero, therefore, the system is definite. To find its solution, we calculate the determinants for unknowns

According to Cramer's formulas, we find:

So, the solution to the system is (2; -1; 1).

To check the solutions of the 3 X 3 and 4 X 4 systems of equations, you can use the online calculator that solves the Cramer method.

Back to the top of the page

We continue to solve systems by Cramer's method together

As already mentioned, if the determinant of the system is equal to zero, and the determinants for the unknowns are not equal to zero, the system is inconsistent, that is, it has no solutions. Let us illustrate with the following example.

Example 6. Solve the system of linear equations by Cramer's method:

Solution. We find the determinant of the system:

The determinant of the system is equal to zero, therefore, the system of linear equations is either inconsistent and definite, or inconsistent, that is, it has no solutions. To make it more precise, we calculate the determinants for unknowns

Determinants for unknowns are not equal to zero, therefore, the system is inconsistent, that is, it has no solutions.

To check the solutions of the 3 X 3 and 4 X 4 systems of equations, you can use the online calculator that solves the Cramer method.

In problems on systems of linear equations, there are also those where, in addition to the letters denoting variables, there are also other letters. These letters represent a certain number, most often a real number. In practice, search problems lead to such equations and systems of equations general properties any phenomena and objects. That is, have you invented any new material or a device, and to describe its properties that are common regardless of the size or number of an instance, you need to solve a system of linear equations, where instead of some coefficients of variables - letters. You don't have to go far for examples.

The next example is for a similar task, only the number of equations, variables, and letters denoting some real number increases.

Example 8. Solve the system of linear equations by Cramer's method:

Solution. We find the determinant of the system:

Find determinants for unknowns

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. Determinant composed of coefficients for independent system variables(1.5) is called the main determinant of the system. We will denote it by the Greek letter D. Thus,

. (1.6)

If the main determinant is arbitrary ( j-th) column, replace with the column of free terms of system (1.5), then we can get another n auxiliary determinants:

(j = 1, 2, …, n). (1.7)

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

(1.8)

Example 1.5. Using Cramer's method to solve the system of equations

.

Let us calculate the main determinant of the system:

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

In this way,

Matrix operations

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

. (1.9)

Example 1.6. .

Addition of matrices.

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 addition of matrices has the properties of associativity and commutativity.

Example 1.7. .

Matrix multiplication.

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

2

Thus, when multiplying the matrix A dimensions m´ n on the matrix V dimensions n´ k we get the matrix WITH dimensions m´ k... Moreover, the elements of the matrix WITH are calculated using the following formulas:

Task 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, since the number of columns in the matrix B does not match the number of rows in the matrix A.

Inverse matrix. Matrix solution of systems of linear equations

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

where through I denoted identity matrix of the same order as the matrix A:

.

In order to square matrix has an inverse it is necessary and sufficient for its determinant to be nonzero. The inverse matrix is ​​found by the formula:

, (1.13)

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

Example 1.9. Find Inverse Matrix A - 1 to the matrix

.

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

.

Find det A = | A| = 1 × 3 × 8 + 2 × 5 × 3 + 2 × 4 × 3 - 3 × 3 × 3 - 1 × 5 × 4 - 2 × 2 × 8 = 24 + 30 + 24 - 27 - 20 - 32 = - 1. Since the determinant of the original matrix is ​​nonzero, the inverse matrix exists.

1) Find the algebraic complements A ij:

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

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

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

where

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

, where

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.

Task 1.10. Solve a system of linear equations

using the inverse matrix.

Solution. Let us write the system in matrix form:,

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

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

We find the solution of the system by the formula (1.15):

In this way,

Solving systems of linear equations by the method of ordinary Jordan exceptions

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

(1.16)

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

When solving such problems, the method of eliminating unknowns, well known from the school course, is used, which is also called the method of ordinary Jordan exceptions. The 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 variable less 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 satisfied 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 contradictory equations did not arise, then one of the variables remaining in it is found from the last equation. If there is only one variable left 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". 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 decision will be unique 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 (a new solution corresponds to each set of parameters). Formulas that allow you to find a solution to a 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 reducing similar terms in the second and third equations, we arrive at the system:

Let us express y from the second equation and substitute it into the first equation:

Let us 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 where we find y:

.

Then we substitute in the first memorized equation from where we find x:

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

. (1.17)

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

.

Let's remember the first equation

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

We suggest the readers 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.

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

. (1.18)

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

.

Let's remember the first equation and give similar terms in the second and third equations. We come to the system:

By 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 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), choosing an arbitrary value of the parameter t:

(1.19)
So 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 exceptions seems cumbersome. However, it is not. It is enough to derive an algorithm for recalculating the coefficients of the system at one step in general view and formulate the solution to the problem in the form of special Jordan tables.

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

, (1.20)
where x j- independent (sought) variables, a ij- constant coefficients
(i = 1, 2,…, m; j = 1, 2,…, n). Right side 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.

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

We get the following system:

. (1.21)

From s th equality of system (1.21), we subsequently find the variable x s(after the rest of the variables have been found). S-th line is remembered and further 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 the expression of the variable x s through the rest of the variables will look like this:

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

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

After bringing similar 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 Jordan exceptions is formalized 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		a ij		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		a mn

Jordan table 1.1 contains the left header column, in which the right-hand sides of the system (1.20) are written, and the upper header row, in which the independent variables are written.

The rest of the table elements form the main matrix of the coefficients of the system (1.20). If we multiply the matrix A to the matrix consisting of the elements of the upper heading row, then you get the matrix consisting of the elements of the left header column. That is, in essence, a Jordan table is a matrix notation of 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 =	b r 1	b r 2		b rj		b rs		b rn
………………………………………………………………….
y n =	b m 1	b m 2		b mj		b ms		b mn

Permissive element a rs we will highlight it in bold. Recall that for one step of Jordan exceptions to take place, the resolving element must be nonzero. The table row containing the permitting element is called the permitting row. The column containing the permitting element is called the permitting column. When moving from this table to the next table, one variable ( x s) from the upper header row of the table is moved to the left header column and, conversely, one of the free members of the system ( y r) from the left head column of the table is moved to the top head row.

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

1. The permitting element is replaced with a reciprocal:

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

3. The rest of the elements of the resolving column are divided into the resolving element:

4. Elements that are not included in the resolving line and resolving column are recalculated using 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 th and r-th lines and j th and s-th columns (the resolving row, the resolving column, and the row and column at the intersection of which the element being recalculated is located). More precisely, when memorizing the formula the following diagram can be used:

-21 -26 -13 -37

Taking the first step of Jordan exceptions, any element of Table 1.3, located in the columns x 1 ,…, x 5 (all specified elements are nonzero). You should not only select the resolving element in the last column, because you want to find independent variables x 1 ,…, x 5 . We choose, for example, the coefficient 1 at variable x 3 in the third row of Table 1.3 (the enabling element is shown in bold). When going to table 1.4, the variable x The 3 from the top heading row is swapped with the left head column constant 0 (third row). In this case, the variable x 3 is expressed in terms of the remaining variables.

String x 3 (Table 1.4) can, after remembering, be excluded from Table 1.4. The third column with zero in the upper heading line is also excluded from table 1.4. The fact is that regardless of the coefficients of this column b i 3 all the corresponding terms of each equation 0 b i 3 systems will be zero. Therefore, the indicated coefficients can be omitted. Eliminating one variable x 3 and remembering one of the equations, we come to the system, corresponding 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 a 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 stored lines, we find the remaining variables:

Thus, the system has countless solutions. Variable x 5, you can assign arbitrary values. This variable acts as a parameter x 5 = t. We have 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

By giving the parameter t different meanings, we get countless solutions to the original system. So, for example, the solution to the system is the following set of variables (- 3; - 1; - 2; 4; 0).