Let two inequalities be given f 1(x) > g 1(x) and f 2(x) > g 2(x). System of inequalities is is a conjunction of these inequalities . The system is written like this:

The solution to this system X, which turns each of the inequalities into a true numerical inequality. The set of solutions to a system of inequalities is the intersection of the sets of solutions to inequalities that form the given system.

Inequality | x| < a, where a> 0, is equivalent to a system or a double inequality -- a < x < a.

Set of inequalities f 1(x) > g 1(x) and f 2(x) > g 2(x) is yourself disjunction of these inequalities .

The set is written like this:

The solution of this set is any value of the variable X, which turns into a true numerical inequality at least one of the inequalities in the set. The set of solutions to a set is the union of the sets of solutions to inequalities that form the set.

Inequality | x| > a, where a> 0, is equivalent to the set

Task. Find a set of solutions to the system of inequalities:

Decision. Let's find the solution sets for each of the system's inequalities, and then find their intersection. Let us transform each of the inequalities to the form x > a or x < a.

Û Û

Û Û Û

X> -7 is a numerical interval (-7; ¥), and the set of solutions to the inequality X < 7 - промежуток (-¥; 7). Найдем их пересечение: (-7; ¥) Ç (-¥; 7) = (-7; 7). Таким образом, множеством решений данной системы является промежуток (-7; 7).

Task. Solve inequality | x+ 3| £4.

Decision. This inequality is equivalent to the double inequality -4 £ x+ 3 £ 4. Solving it, we find that -7 £ x£ 1, i.e. XО [-7; one].

Task. Find set of population solutions

Decision. Let us first find the solution sets for each of the population inequalities, and then their union.

We transform each of the population inequalities, replacing it with an equivalent one: Û Û Û

The set of solutions to the inequality X> 2 is the numerical interval (2; ¥), and the set of solutions of the inequality X> 1 - interval (1; ¥). Let us find their union: (2; ¥) È (1; ¥) = (1; ¥). Therefore, the set of solutions of the collection is the numerical interval (1; ¥).

Task. Solve inequality | x+ 3| > 5.

Decision. This inequality is equivalent to the set of inequalities:

Thus, the solution of the resulting set is the numerical interval (-¥; -8) È (2; ¥).

Exercises for independent work

1. Find the truth sets of the following conjunctions of inequalities and draw them on a real line:

a) ( X> 3) u ( X> 5); G) ( X³ -7) u ( X³ -9);

b) ( X < 3) Ù (X < 5); д) (X> 4) u ( X£ -2);

in) ( X³ -4) u ( X£ -2); e) ( X³ -6) u ( X < 11).

2. Solve systems of inequalities:

a) b)

in) G)

3. Find sets of solutions to inequalities:

a) | x - 6| < 13; в) |3x- 6| £0;

b) |5 - 2 x| £3; d) |3 x - 8| < - 1.

4. Find the truth sets of the following disjunctions of inequalities:

a) ( X> -9) Ú ( X> 1) Ú ( X> 6); G) ( X < 2) Ú (X > 8);

§one. Systems of linear equations.

view system

called a system m linear equations with n unknown.

Here
- unknown, - coefficients for unknowns,
- free members of the equations.

If all free terms of the equations are equal to zero, the system is called homogeneous.Decision system is called a set of numbers
, when substituting them into the system instead of unknowns, all equations turn into identities. The system is called joint if it has at least one solution. A joint system with a unique solution is called certain. The two systems are called equivalent if the sets of their solutions are the same.

System (1) can be represented in matrix form using the equation

(2)

.

§2. Compatibility of systems of linear equations.

We call the extended matrix of system (1) the matrix

Kronecker - Capelli theorem. System (1) is consistent if and only if the rank of the system matrix is ​​equal to the rank of the extended matrix:

.

§3. Systems solutionn linear equations withn unknown.

Consider an inhomogeneous system n linear equations with n unknown:

(3)

Cramer's theorem.If the main determinant of the system (3)
, then the system has a unique solution determined by the formulas:

those.
,

where - the determinant obtained from the determinant replacement th column to the column of free members.

If a
, and at least one of ≠0, then the system has no solutions.

If a
, then the system has infinitely many solutions.

System (3) can be solved using its matrix notation (2). If the rank of the matrix BUT equals n, i.e.
, then the matrix BUT has an inverse
. Multiplying the matrix equation
to matrix
on the left, we get:

.

The last equality expresses a way to solve systems of linear equations using an inverse matrix.

Example. Solve the system of equations using the inverse matrix.

Decision. Matrix
non-degenerate, because
, so there is an inverse matrix. Let's calculate the inverse matrix:
.

,

Exercise. Solve the system by Cramer's method.

§4. Solution of arbitrary systems of linear equations.

Let an inhomogeneous system of linear equations of the form (1) be given.

Let us assume that the system is consistent, i.e. the condition of the Kronecker-Capelli theorem is fulfilled:
. If the rank of the matrix
(to the number of unknowns), then the system has a unique solution. If a
, then the system has infinitely many solutions. Let's explain.

Let the rank of the matrix r(A)= r< n. Insofar as
, then there exists some nonzero minor of order r. Let's call it the basic minor. The unknowns whose coefficients form the basic minor are called basic variables. The remaining unknowns are called free variables. We rearrange the equations and renumber the variables so that this minor is located in the upper left corner of the system matrix:

.

First r rows are linearly independent, the rest are expressed through them. Therefore, these lines (equations) can be discarded. We get:

Let's give free variables arbitrary numerical values: . We leave only the basic variables on the left side, and move the free variables to the right side.

Got a system r linear equations with r unknown, whose determinant is different from 0. It has a unique solution.

This system is called the general solution of the system of linear equations (1). Otherwise: the expression of basic variables in terms of free ones is called common solution systems. From it you can get an infinite number private decisions, giving free variables arbitrary values. A particular solution obtained from a general one at zero values ​​of the free variables is called basic solution. The number of different basic solutions does not exceed
. A basic solution with non-negative components is called pivotal system solution.

Example.

,r=2.

Variables
- basic,
- free.

Let's add the equations; express
through
:

- common decision.

- private solution
.

- basic solution, basic.

§5. Gauss method.

The Gauss method is a universal method for studying and solving arbitrary systems of linear equations. It consists in bringing the system to a diagonal (or triangular) form by sequential elimination of unknowns using elementary transformations that do not violate the equivalence of systems. A variable is considered excluded if it is contained in only one equation of the system with a coefficient of 1.

Elementary transformations systems are:

Multiplying an equation by a non-zero number;

Adding an equation multiplied by any number with another equation;

Rearrangement of equations;

Dropping the equation 0 = 0.

Elementary transformations can be performed not on equations, but on extended matrices of the resulting equivalent systems.

Example.

Decision. We write the extended matrix of the system:

.

Performing elementary transformations, we bring the left side of the matrix to the unit form: we will create units on the main diagonal, and zeros outside it.

Comment. If, when performing elementary transformations, an equation of the form 0 = k(where to0), then the system is inconsistent.

The solution of systems of linear equations by the method of successive elimination of unknowns can be formalized in the form tables.

The left column of the table contains information about the excluded (basic) variables. The remaining columns contain the coefficients of the unknowns and the free terms of the equations.

The expanded matrix of the system is written into the source table. Next, proceed to the implementation of the Jordan transformations:

1. Choose a variable , which will become the basis. The corresponding column is called the key column. Choose an equation in which this variable will remain, being excluded from other equations. The corresponding table row is called the key row. Coefficient The , standing at the intersection of the key row and the key column, is called the key.

2. Elements of the key string are divided by the key element.

3. The key column is filled with zeros.

4. The remaining elements are calculated according to the rectangle rule. They make up a rectangle, at opposite vertices of which there are a key element and a recalculated element; from the product of the elements on the diagonal of the rectangle with the key element, the product of the elements of another diagonal is subtracted, the resulting difference is divided by the key element.

Example. Find the general solution and the basic solution of the system of equations:

Decision.

General solution of the system:

Basic solution:
.

A one-time substitution transformation allows one to go from one basis of the system to another: instead of one of the main variables, one of the free variables is introduced into the basis. To do this, a key element is selected in the free variable column and transformations are performed according to the above algorithm.

§6. Finding support solutions

The reference solution of a system of linear equations is a basic solution that does not contain negative components.

The support solutions of the system are found by the Gauss method under the following conditions.

1. In the original system, all free terms must be non-negative:
.

2. The key element is chosen among positive coefficients.

3. If the variable introduced into the basis has several positive coefficients, then the key string is the one in which the ratio of the free term to the positive coefficient is the smallest.

Remark 1. If, in the process of eliminating the unknowns, an equation appears in which all coefficients are nonpositive, and the free term
, then the system has no non-negative solutions.

Remark 2. If there is not a single positive element in the columns of coefficients for free variables, then the transition to another reference solution is impossible.

Example.

If a problem has fewer than three variables, it's not a problem; if more than eight, it is undecidable. Enon.

Problems with parameters are found in all variants of the USE, since when solving them, it is most clearly revealed how deep and informal the knowledge of the graduate is. The difficulties encountered by students in the performance of such tasks are caused not only by their relative complexity, but also by the fact that insufficient attention is paid to them in textbooks. In variants of KIMs in mathematics, there are two types of assignments with parameters. First: "for each value of the parameter, solve the equation, inequality, or system." Second: "find all values ​​of the parameter, for each of which the solutions of the inequality, equation or system satisfy the given conditions." Accordingly, the answers in these two types of problems differ in essence. In the first case, all possible values ​​of the parameter are listed in the answer, and solutions to the equation are written for each of these values. The second one lists all parameter values ​​under which the conditions of the problem are met. Recording the answer is an essential stage of the solution, it is very important not to forget to reflect all the stages of the decision in the answer. This needs to be brought to the attention of students.
The appendix to the lesson contains additional material on the topic "Solving systems of linear equations with parameters", which will help in preparing students for the final certification.

Lesson Objectives:

• systematization of students' knowledge;
• development of skills to apply graphical representations in solving systems of equations;
• formation of the ability to solve systems of linear equations containing parameters;
• implementation of operational control and self-control of students;
• development of research and cognitive activity of schoolchildren, the ability to evaluate the results obtained.

The lesson is designed for two teaching hours.

During the classes

1. Organizing time

Message topics, goals and objectives of the lesson.

1. Updating the basic knowledge of students

Checking homework. As homework, students were asked to solve each of the three systems of linear equations

a) b) in)

graphically and analytically; draw a conclusion about the number of solutions obtained for each case

The conclusions made by the students are heard and analyzed. The results of the work under the guidance of the teacher are summarized in notebooks.

In general, a system of two linear equations with two unknowns can be represented as: .

To solve a given system of equations graphically means to find the coordinates of the intersection points of the graphs of these equations or to prove that there are none. The graph of each equation of this system on the plane is some straight line.

There are three cases of mutual arrangement of two straight lines on a plane:

<Рисунок1>;

<Рисунок2>;

<Рисунок3>.

For each case, it is useful to draw a picture.

1. Learning new material

Today in the lesson we will learn how to solve systems of linear equations containing parameters. We will call a parameter an independent variable, the value of which in the problem is considered to be a given fixed or arbitrary real number, or a number belonging to a predetermined set. To solve a system of equations with a parameter means to establish a correspondence that allows for any value of the parameter to find the corresponding set of solutions to the system.

The solution of a problem with a parameter depends on the question posed in it. If you just need to solve a system of equations for different values ​​of a parameter or investigate it, then you need to give a reasonable answer for any parameter value or for a parameter value that belongs to a set specified in advance in the problem. If it is necessary to find the values ​​of the parameter that satisfy certain conditions, then a complete study is not required, and the solution of the system is limited to finding these particular values ​​of the parameter.

Example 1 For each parameter value, we solve the system of equations

Decision.

1. The system has a unique solution if

In this case we have

1. If a = 0, then the system takes the form

The system is inconsistent, i.e. has no solutions.

1. If then the system can be written in the form

It is obvious that in this case the system has infinitely many solutions of the form x = t; where t is any real number.

Answer:

Example 2

• has a unique solution;
• has many solutions;
• has no solutions?

Decision.

Answer:

Example 3 Let us find the sum of parameters a and b for which the system

has an infinite number of solutions.

Decision. The system has an infinite number of solutions if

That is, if a = 12, b = 36; a + b = 12 + 36 = 48.

Answer: 48.

1. Consolidation of what was learned in the course of solving problems

1. No. 15.24(a) . For each parameter value, solve the system of equations

1. #15.25(a) For each parameter value, solve the system of equations

1. For what values ​​of the parameter a the system of equations

a) has no solutions; b) has infinitely many solutions.

Answer: for a = 2 there are no solutions, for a = -2 there are an infinite number of solutions

1. Practical work in groups

The class is divided into groups of 4-5 people. Each group includes students with different levels of mathematical training. Each group receives a card with a task. You can invite all groups to solve one system of equations, and draw up the solution. The group that completed the task correctly first presents its solution; the rest hand over the decision to the teacher.

Card. Solve System of Linear Equations

for all values ​​of the parameter a.

Answer: when the system has a unique solution ; when there are no solutions; for a = -1 there are infinitely many solutions of the form, (t; 1- t) where t R

If the class is strong, groups can be offered different systems of equations, a list of which is in Appendix 1. Then each group presents its solution to the class.

Report of the group that first correctly completed the task

Participants voice and explain their version of the solution and answer questions that have arisen from representatives of other groups.

1. Independent work

Option 1

Option 2

1. Lesson summary

Solving systems of linear equations with parameters can be compared to a study that includes three main conditions. The teacher asks the students to formulate them.

When deciding, keep in mind:

1. in order for the system to have a unique solution, it is necessary that the lines corresponding to the equation of the system intersect, i.e. it is necessary to fulfill the condition;
2. to have no solutions, the lines must be parallel, i.e. the condition was met
3. and, finally, for the system to have infinitely many solutions, the lines must coincide, i.e. condition was met.

The teacher evaluates the work in the lesson of the class as a whole and sets marks for the lesson for individual students. After checking independent work, each student will receive an assessment for the lesson.

1. Homework

For what values ​​of the parameter b the system of equations

• has infinitely many solutions;
• has no solutions?

The graphs of the functions y = 4x + b and y = kx + 6 are symmetrical about the y-axis.

• Find b and k,
• find the coordinates of the point of intersection of these graphs.

Solve the system of equations for all values ​​of m and n.

Solve a system of linear equations for all values ​​of the parameter a (any choice).

Literature

1. Algebra and the beginning of mathematical analysis: textbook. for 11 cells. general education institutions: basic and profile. levels / S. M. Nikolsky, M. K. Potapov, N. N. Reshetnikov, A. V. Shevkin - M .: Education, 2008.
2. Mathematics: Grade 9: Preparation for the state final certification / M. N. Korchagina, V. V. Korchagin - M .: Eksmo, 2008.
3. Getting ready for university. Mathematics. Part 2 state technol. un-t; Institute of modern technol. and economy; Compiled by: S. N. Gorshkova, L. M. Danovich, N. A. Naumova, A.V. Martynenko, I.A. Palshchikov. – Krasnodar, 2006.
4. Collection of problems in mathematics for preparatory courses TUSUR: Study guide / Z. M. Goldstein, G. A. Kornievskaya, G. A. Korotchenko, S. N. Kudinov. – Tomsk: Tomsk. State. University of Control Systems and Radioelectronics, 1998.
5. Mathematics: an intensive course of preparation for the exam / O. Yu. Cherkasov, A.G. Yakushev. - M .: Rolf, Iris-press, 1998.

We continue to deal with systems of linear equations. So far, we have considered systems that have a unique solution. Such systems can be solved in any way: substitution method("school") by Cramer's formulas, matrix method, Gauss method. However, two more cases are widespread in practice when:

1) the system is inconsistent (has no solutions);

2) the system has infinitely many solutions.

For these systems, the most universal of all solution methods is used - Gauss method. In fact, the "school" method will also lead to the answer, but in higher mathematics it is customary to use the Gaussian method of successive elimination of unknowns. Those who are not familiar with the Gauss method algorithm, please study the lesson first Gauss method

The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. First, consider a couple of examples where the system has no solutions (inconsistent).

Example 1

What immediately catches your eye in this system? The number of equations is less than the number of variables. There is a theorem that says: “If the number of equations in the system is less than the number of variables, then the system is either inconsistent or has infinitely many solutions. And it remains only to find out.

The beginning of the solution is quite ordinary - we write the extended matrix of the system and, using elementary transformations, we bring it to a stepwise form:

(one). On the upper left step, we need to get (+1) or (-1). There are no such numbers in the first column, so rearranging the rows will not work. The unit will have to be organized independently, and this can be done in several ways. We did so. To the first line we add the third line, multiplied by (-1).

(2). Now we get two zeros in the first column. To the second line, add the first line, multiplied by 3. To the third line, add the first, multiplied by 5.

(3). After the transformation is done, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the desired one (-1) on the second step. Divide the third line by (-3).

(4). Add the second line to the third line. Probably, everyone paid attention to the bad line, which turned out as a result of elementary transformations:

. It is clear that this cannot be so.

Indeed, we rewrite the resulting matrix

back to the system of linear equations:

If as a result of elementary transformations a string of the form , whereλ is a non-zero number, then the system is inconsistent (has no solutions).

How to record the end of a task? You need to write down the phrase:

“As a result of elementary transformations, a string of the form is obtained, where λ ≠ 0 ". Answer: "The system has no solutions (inconsistent)."

Please note that in this case there is no reverse move of the Gaussian algorithm, there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is a do-it-yourself example. Full solution and answer at the end of the lesson.

Again, we remind you that your solution path may differ from our solution path, the Gauss method does not set an unambiguous algorithm, you must guess the procedure and the actions themselves in each case yourself.

One more technical feature of the solution: elementary transformations can be stopped At once, as soon as a line like , where λ ≠ 0 . Consider a conditional example: suppose that after the first transformation we get a matrix

.

This matrix has not yet been reduced to a stepped form, but there is no need for further elementary transformations, since a line of the form has appeared, where λ ≠ 0 . It should be immediately answered that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift to the student, due to the fact that a short solution is obtained, sometimes literally in 2-3 steps. But everything in this world is balanced, and the problem in which the system has infinitely many solutions is just longer.

Example 3:

Solve a system of linear equations

There are 4 equations and 4 unknowns, so the system can either have a single solution, or have no solutions, or have infinitely many solutions. Whatever it was, but the Gauss method in any case will lead us to the answer. This is its versatility.

The beginning is again standard. We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

That's all, and you were afraid.

(one). Please note that all the numbers in the first column are divisible by 2, so on the upper left step we are also satisfied with a deuce. To the second line we add the first line, multiplied by (-4). To the third line we add the first line, multiplied by (-2). To the fourth line we add the first line, multiplied by (-1).

Attention! Many may be tempted from the fourth line subtract first line. This can be done, but it is not necessary, experience shows that the probability of an error in calculations increases several times. We just add: to the fourth line we add the first line, multiplied by (-1) - exactly!

(2). The last three lines are proportional, two of them can be deleted. Here again it is necessary to show increased attention, but are the lines really proportional? For reinsurance, it will not be superfluous to multiply the second row by (-1), and divide the fourth row by 2, resulting in three identical rows. And only after that remove two of them. As a result of elementary transformations, the extended matrix of the system is reduced to a stepped form:

When completing a task in a notebook, it is advisable to make the same notes in pencil for clarity.

We rewrite the corresponding system of equations:

The “usual” only solution of the system does not smell here. Bad line where λ ≠ 0, also no. Hence, this is the third remaining case - the system has infinitely many solutions.

The infinite set of solutions of the system is briefly written in the form of the so-called general system solution.

We will find the general solution of the system using the reverse motion of the Gauss method. For systems of equations with an infinite set of solutions, new concepts appear: "basic variables" and "free variables". First, let's define what variables we have basic, and what variables - free. It is not necessary to explain in detail the terms of linear algebra, it is enough to remember that there are such basis variables and free variables.

Basic variables always "sit" strictly on the steps of the matrix. In this example, the base variables are x 1 and x 3 .

Free variables are everything remaining variables that did not get a step. In our case, there are two: x 2 and x 4 - free variables.

Now you need allbasis variables express only throughfree variables. The reverse move of the Gaussian algorithm traditionally works from the bottom up. From the second equation of the system, we express the basic variable x 3:

Now look at the first equation: . First, we substitute the found expression into it:

It remains to express the basic variable x 1 through free variables x 2 and x 4:

The result is what you need - all basis variables ( x 1 and x 3) expressed only through free variables ( x 2 and x 4):

Actually, the general solution is ready:

.

How to write down the general solution? First of all, free variables are written into the general solution “on their own” and strictly in their places. In this case, the free variables x 2 and x 4 should be written in the second and fourth positions:

.

The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:

From the general solution of the system, one can find infinitely many private decisions. It's very simple. free variables x 2 and x 4 are called so because they can be given any final values. The most popular values ​​are zero values, since this is the easiest way to obtain a particular solution.

Substituting ( x 2 = 0; x 4 = 0) into the general solution, we get one of the particular solutions:

, or is a particular solution corresponding to free variables with values ​​( x 2 = 0; x 4 = 0).

Ones are another sweet couple, let's substitute ( x 2 = 1 and x 4 = 1) into the general solution:

, i.e. (-1; 1; 1; 1) is another particular solution.

It is easy to see that the system of equations has infinitely many solutions since we can give free variables any values.

Each a particular solution must satisfy to each system equation. This is the basis for a “quick” check of the correctness of the solution. Take, for example, a particular solution (-1; 1; 1; 1) and substitute it into the left side of each equation in the original system:

Everything has to come together. And with any particular solution you get, everything should also converge.

Strictly speaking, the verification of a particular solution sometimes deceives, i.e. some particular solution can satisfy each equation of the system, and the general solution itself is actually found incorrectly. Therefore, first of all, the verification of the general solution is more thorough and reliable.

How to check the resulting general solution ?

It's not difficult, but it requires quite a long transformation. We need to take expressions basic variables, in this case and , and substitute them into the left side of each equation of the system.

To the left side of the first equation of the system:

The right side of the original first equation of the system is obtained.

To the left side of the second equation of the system:

The right side of the original second equation of the system is obtained.

And further - to the left parts of the third and fourth equations of the system. This check is longer, but it guarantees the 100% correctness of the overall solution. In addition, in some tasks it is required to check the general solution.

Example 4:

Solve the system using the Gauss method. Find a general solution and two private ones. Check the overall solution.

This is a do-it-yourself example. Here, by the way, again the number of equations is less than the number of unknowns, which means that it is immediately clear that the system will either be inconsistent or have an infinite number of solutions.

Example 5:

Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution

Decision: Let us write down the extended matrix of the system and, with the help of elementary transformations, bring it to a stepped form:

(one). Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.

(2). To the third line we add the second line, multiplied by (-5). To the fourth line we add the second line, multiplied by (-7).

(3). The third and fourth lines are the same, we delete one of them. Here is such a beauty:

Basis variables sit on steps, so they are base variables.

There is only one free variable, which did not get a step: .

(4). Reverse move. We express the basic variables in terms of the free variable:

From the third equation:

Consider the second equation and substitute the found expression into it:

, , ,

Consider the first equation and substitute the found expressions and into it:

Thus, the general solution with one free variable x 4:

Once again, how did it happen? free variable x 4 sits alone in its rightful fourth place. The resulting expressions for the basic variables , , are also in their places.

Let us immediately check the general solution.

We substitute the basic variables , , into the left side of each equation of the system:

The corresponding right-hand sides of the equations are obtained, thus, the correct general solution is found.

Now from the found general solution we get two particular solutions. All variables are expressed here through a single free variable x 4 . You don't need to break your head.

Let be x 4 = 0, then is the first particular solution.

Let be x 4 = 1, then is another particular solution.

Answer: Common decision: . Private Solutions:

and .

Example 6:

Find the general solution of the system of linear equations.

We have already checked the general solution, the answer can be trusted. Your course of action may differ from our course of action. The main thing is that the general solutions coincide. Probably, many have noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to fiddle with ordinary fractions. In practice, this is true, cases where there are no fractions are much less common. Be prepared mentally, and most importantly, technically.

Let us dwell on the features of the solution that were not found in the solved examples. The general solution of the system may sometimes include a constant (or constants).

For example, the general solution: . Here one of the basic variables is equal to a constant number: . There is nothing exotic in this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables. However, the Gauss method works under the most severe conditions. You should calmly bring the extended matrix of the system to a stepped form according to the standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a unique solution.

We repeat in our advice - in order to feel comfortable when solving a system using the Gauss method, you should fill your hand and solve at least a dozen systems.

Solutions and answers:

Example 2:

Decision:Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepped form.

Performed elementary transformations:

(1) The first and third lines have been swapped.

(2) The first line was added to the second line, multiplied by (-6). The first line was added to the third line, multiplied by (-7).

(3) The second line was added to the third line, multiplied by (-1).

As a result of elementary transformations, a string of the form, where λ ≠ 0 .So the system is inconsistent.Answer: there are no solutions.

Example 4:

Decision:We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

Conversions performed:

(one). The first line multiplied by 2 was added to the second line. The first line multiplied by 3 was added to the third line.

There is no unit for the second step , and transformation (2) is aimed at obtaining it.

(2). The second line was added to the third line, multiplied by -3.

(3). The second and third rows were swapped (the resulting -1 was moved to the second step)

(4). The second line was added to the third line, multiplied by 3.

(5). The sign of the first two lines was changed (multiplied by -1), the third line was divided by 14.

Reverse move:

(one). Here are the basic variables (which are on steps), and are free variables (who did not get the step).

(2). We express the basic variables in terms of free variables:

From the third equation: .

(3). Consider the second equation:, particular solutions:

Answer: Common decision:

Complex numbers

In this section, we will introduce the concept complex number, consider algebraic, trigonometric and show form complex number. We will also learn how to perform operations with complex numbers: addition, subtraction, multiplication, division, exponentiation and root extraction.

To master complex numbers, you do not need any special knowledge from the course of higher mathematics, and the material is available even to a schoolboy. It is enough to be able to perform algebraic operations with "ordinary" numbers, and remember trigonometry.

First, let's remember the "ordinary" Numbers. In mathematics they are called set of real numbers and are marked with the letter R, or R (thick). All real numbers sit on the familiar number line:

The company of real numbers is very colorful - here are integers, and fractions, and irrational numbers. In this case, each point of the numerical axis necessarily corresponds to some real number.

However, two more cases are widespread in practice:

– The system is inconsistent (has no solutions);
The system is consistent and has infinitely many solutions.

Note : the term "consistency" implies that the system has at least some solution. In a number of tasks, it is required to preliminarily examine the system for compatibility, how to do this - see the article on matrix rank.

For these systems, the most universal of all solution methods is used - Gauss method. In fact, the "school" method will also lead to the answer, but in higher mathematics it is customary to use the Gaussian method of successive elimination of unknowns. Those who are not familiar with the Gauss method algorithm, please study the lesson first gauss method for dummies.

The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. First, consider a couple of examples where the system has no solutions (inconsistent).

Example 1

What immediately catches your eye in this system? The number of equations is less than the number of variables. If the number of equations is less than the number of variables, then we can immediately say that the system is either inconsistent or has infinitely many solutions. And it remains only to find out.

The beginning of the solution is quite ordinary - we write the extended matrix of the system and, using elementary transformations, we bring it to a stepwise form:

(1) On the upper left step, we need to get +1 or -1. There are no such numbers in the first column, so rearranging the rows will not work. The unit will have to be organized independently, and this can be done in several ways. I did this: To the first line, add the third line, multiplied by -1.

(2) Now we get two zeros in the first column. To the second line we add the first line multiplied by 3. To the third line we add the first line multiplied by 5.

(3) After the transformation is done, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the desired -1 on the second step. Divide the third line by -3.

(4) Add the second line to the third line.

Probably, everyone paid attention to the bad line, which turned out as a result of elementary transformations: . It is clear that this cannot be so. Indeed, we rewrite the resulting matrix back to the system of linear equations:

If, as a result of elementary transformations, a string of the form is obtained, where is a non-zero number, then the system is inconsistent (has no solutions) .

How to record the end of a task? Let's draw with white chalk: "as a result of elementary transformations, a line of the form is obtained, where" and give the answer: the system has no solutions (inconsistent).

If, according to the condition, it is required to EXPLORE the system for compatibility, then it is necessary to issue a solution in a more solid style involving the concept matrix rank and the Kronecker-Capelli theorem.

Please note that there is no reverse motion of the Gaussian algorithm here - there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is a do-it-yourself example. Full solution and answer at the end of the lesson. Again, I remind you that your solution path may differ from my solution path, the Gaussian algorithm does not have a strong “rigidity”.

One more technical feature of the solution: elementary transformations can be stopped At once, as soon as a line like , where . Consider a conditional example: suppose that after the first transformation we get a matrix . The matrix has not yet been reduced to a stepped form, but there is no need for further elementary transformations, since a line of the form has appeared, where . It should be immediately answered that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift, because a short solution is obtained, sometimes literally in 2-3 steps.

But everything in this world is balanced, and the problem in which the system has infinitely many solutions is just longer.

Example 3

Solve a system of linear equations

There are 4 equations and 4 unknowns, so the system can either have a single solution, or have no solutions, or have infinitely many solutions. Whatever it was, but the Gauss method in any case will lead us to the answer. Therein lies its versatility.

The beginning is again standard. We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

That's all, and you were afraid.

(1) Note that all the numbers in the first column are divisible by 2, so a 2 is fine on the top left rung. To the second line we add the first line, multiplied by -4. To the third line we add the first line, multiplied by -2. To the fourth line we add the first line, multiplied by -1.

Attention! Many may be tempted from the fourth line subtract first line. This can be done, but it is not necessary, experience shows that the probability of an error in calculations increases several times. Just add up: To the fourth line, add the first line, multiplied by -1 - exactly!

(2) The last three lines are proportional, two of them can be deleted.

Here again it is necessary to show increased attention, but are the lines really proportional? For reinsurance (especially for a teapot), it would not be superfluous to multiply the second row by -1, and divide the fourth row by 2, resulting in three identical rows. And only after that remove two of them.

As a result of elementary transformations, the extended matrix of the system is reduced to a stepped form:

When completing a task in a notebook, it is advisable to make the same notes in pencil for clarity.

We rewrite the corresponding system of equations:

The “usual” only solution of the system does not smell here. There is no bad line either. This means that this is the third remaining case - the system has infinitely many solutions. Sometimes, by condition, it is necessary to investigate the compatibility of the system (i.e., to prove that a solution exists at all), you can read about this in the last paragraph of the article How to find the rank of a matrix? But for now, let's break down the basics:

The infinite set of solutions of the system is briefly written in the form of the so-called general system solution .

We will find the general solution of the system using the reverse motion of the Gauss method.

First we need to determine what variables we have basic, and which variables free. It is not necessary to bother with the terms of linear algebra, it is enough to remember that there are such basis variables and free variables.

Basic variables always "sit" strictly on the steps of the matrix.
In this example, the basic variables are and

Free variables are everything remaining variables that did not get a step. In our case, there are two of them: – free variables.

Now you need all basis variables express only through free variables.

The reverse move of the Gaussian algorithm traditionally works from the bottom up.
From the second equation of the system, we express the basic variable:

Now look at the first equation: . First, we substitute the found expression into it:

It remains to express the basic variable in terms of free variables:

The result is what you need - all the basis variables ( and ) are expressed only through free variables :

Actually, the general solution is ready:

How to write down the general solution?
Free variables are written into the general solution "on their own" and strictly in their places. In this case, free variables should be written in the second and fourth positions:
.

The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:

Giving free variables arbitrary values, there are infinitely many private decisions. The most popular values ​​are zeros, since the particular solution is the easiest to obtain. Substitute in the general solution:

is a private decision.

Ones are another sweet couple, let's substitute into the general solution:

is another particular solution.

It is easy to see that the system of equations has infinitely many solutions(since we can give free variables any values)

Each a particular solution must satisfy to each system equation. This is the basis for a “quick” check of the correctness of the solution. Take, for example, a particular solution and substitute it into the left side of each equation in the original system:

Everything has to come together. And with any particular solution you get, everything should also converge.

But, strictly speaking, the verification of a particular solution sometimes deceives; some particular solution can satisfy each equation of the system, and the general solution itself is actually found incorrectly.

Therefore, the verification of the general solution is more thorough and reliable. How to check the resulting general solution ?

It's easy, but quite tedious. We need to take expressions basic variables, in this case and , and substitute them into the left side of each equation of the system.

To the left side of the first equation of the system:

To the left side of the second equation of the system:


The right side of the original equation is obtained.

Example 4

Solve the system using the Gauss method. Find a general solution and two private ones. Check the overall solution.

This is a do-it-yourself example. Here, by the way, again the number of equations is less than the number of unknowns, which means that it is immediately clear that the system will either be inconsistent or have an infinite number of solutions. What is important in the decision process itself? Attention, and again attention. Full solution and answer at the end of the lesson.

And a couple more examples to reinforce the material

Example 5

Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution

Decision: Let's write the augmented matrix of the system and with the help of elementary transformations we bring it to the step form:

(1) Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.
(2) To the third line, add the second line, multiplied by -5. To the fourth line we add the second line, multiplied by -7.
(3) The third and fourth lines are the same, we delete one of them.

Here is such a beauty:

Basis variables sit on steps, so they are base variables.
There is only one free variable, which did not get a step:

Reverse move:
We express the basic variables in terms of the free variable:
From the third equation:

Consider the second equation and substitute the found expression into it:

Consider the first equation and substitute the found expressions and into it:

Yes, a calculator that counts ordinary fractions is still convenient.

So the general solution is:

Once again, how did it happen? The free variable sits alone in its rightful fourth place. The resulting expressions for the basic variables , also took their ordinal places.

Let us immediately check the general solution. Work for blacks, but I have already done it, so catch =)

We substitute three heroes , , into the left side of each equation of the system:

The corresponding right-hand sides of the equations are obtained, so the general solution is found correctly.

Now from the found general solution we get two particular solutions. The chef here is the only free variable . You don't need to break your head.

Let then is a private decision.
Let then is another particular solution.

Answer: Common decision: , particular solutions: , .

I shouldn't have mentioned blacks here... ...because all sorts of sadistic motives popped into my head and I remembered the well-known fotozhaba, in which Ku Klux Klansmen in white overalls run across the field after a black football player. I sit and smile quietly. You know how distracting….

A lot of math is harmful, so a similar final example for an independent solution.

Example 6

Find the general solution of the system of linear equations.

I have already checked the general solution, the answer can be trusted. Your solution may differ from my solution, the main thing is that the general solutions match.

Probably, many have noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to fiddle with ordinary fractions. In practice, this is true, cases where there are no fractions are much less common. Be prepared mentally, and most importantly, technically.

I will dwell on some features of the solution that were not found in the solved examples.

The general solution of the system can sometimes include a constant (or constants), for example: . Here one of the basic variables is equal to a constant number: . There is nothing exotic in this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables. The Gauss method works in the most severe conditions; one should calmly bring the extended matrix of the system to a stepped form according to the standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a unique solution.