However, in practice, two more cases are widespread:

- The system is incompatible (has no solutions);
- The system is compatible and has infinitely many solutions.

Note : The term "interoperability" implies that the system has at least some solution. In a number of tasks, it is required to first investigate the system for compatibility, how to do this - see the article about rank of matrices.

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

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

Example 1

What immediately catches the 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 incompatible or has infinitely many solutions. And it remains only to find out.

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

(1) On the top left rung, we need to get +1 or –1. There are no such numbers in the first column, so rearranging the rows will give nothing. The unit will have to be organized independently, and this can be done in several ways. I did this: 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 we add the first row multiplied by 3. To the third line we add the first row multiplied by 5.

(3) After the performed transformation, it is always advisable to look, and is it possible to simplify the resulting lines? Can. Divide the second row by 2, at the same time getting the desired –1 on the second step. Divide the third row by –3.

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

Probably everyone paid attention to the bad line that 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 nonzero number, then the system is incompatible (has no solutions).

How do I record the ending of an assignment? Let's draw with white chalk: “as a result of elementary transformations, a line of the form, where” was obtained and give the answer: the system has no solutions (incompatible).

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

Note that there is no Gauss backtracking here - there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is an example for a do-it-yourself solution. Complete solution and answer at the end of the tutorial. Again, I remind you that your decision course may differ from my decision course, the Gauss algorithm does not have a strong "rigidity".

Another technical feature of the solution: elementary transformations can be stopped immediately, as soon as a line of the form appeared, where. Consider a conditional example: suppose that after the very first transformation the matrix is ​​obtained ... 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 appeared, where. You should immediately answer that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift, since 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. Be that as it may, but the Gauss method will lead us to the answer anyway. This is its versatility.

The beginning is again standard. Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

That's all, and you were afraid.

(1) Note that all the numbers in the first column are divisible by 2, so we are happy with two on the top left step. To the second line, add the first line multiplied by –4. To the third line, add the first line multiplied by –2. To the fourth line, 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: 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 you need to show increased attention, but are the lines really proportional? To be on the safe side (especially for a teapot) it will not be superfluous to multiply the second line by –1, and divide the fourth line by 2, resulting in three identical lines. And only then delete two of them.

As a result of elementary transformations, the expanded matrix of the system is reduced to a stepwise form:

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

Let's rewrite the corresponding system of equations:

It doesn't smell like a "usual" only solution of the system. 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 do I find the rank of a matrix? But for now, we are analyzing the basics:

The infinite number of solutions to the system are briefly written in the form of the so-called overall system solution .

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

First we need to determine which 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 basic 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 rung. In our case, there are two of them: - free variables.

Now you need all basic variables to express only through free variables.

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

Now let's 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:

In the end, we got what we need - all basic variable (s) are expressed only through free variables:

Actually, the general solution is ready:

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

The obtained expressions for the basic variables and, obviously, you need to write in the first and third positions:

Giving free variables arbitrary values, you can find infinitely many private solutions... The most popular values ​​are zeros, since the particular solution is the easiest. Let's substitute in the general solution:

- a private solution.

Units are another sweet couple, let's substitute them in the general solution:

- one more 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 the particular solution must satisfy to each equation of the system. This is the basis for the "quick" check of the correctness of the solution. Take, for example, a particular solution and plug it into the left side of each equation in the original system:

Everything should fit together. And with any particular decision you receive - everything should also agree.

But, strictly speaking, checking a particular solution sometimes deceives, i.e. some particular solution may satisfy each equation of the system, but the general solution itself is actually found incorrectly.

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

It's easy, but pretty dreary. You need to take expressions basic variables, in this case and, and substitute them in the left side of each equation of the system.

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

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


The right-hand side of the original equation is obtained.

Example 4

Solve the system by the Gaussian method. Find a general solution and two particular ones. Check the general solution.

This is an example for a do-it-yourself solution. 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 be either incompatible or with an infinite set of solutions. What is important in the decision process itself? Attention, attention again... Complete solution and answer at the end of the tutorial.

And a couple more examples to consolidate 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

Solution: Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise 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) Add the second row multiplied by –5 to the third line. To the fourth line, add the second line multiplied by –7.
(3) The third and fourth lines are the same, we delete one of them.

Here is such beauty:

Basic variables sit on rungs, therefore basic variables.
There is only one free variable that did not get a step here:

Reverse:
Let's express the basic variables in terms of a 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 handy.

So the general solution is:

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

Let's check the general solution right away. Work for blacks, but I have already done it, so catch =)

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

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

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

Let, then - a private solution.
Let, then - one more particular solution.

Answer: Common decision: , particular solutions: , .

I shouldn't have remembered about blacks here ... ... because all sorts of sadistic motives got into my head and I remembered the well-known foto-toad, in which the Klans in white robes are running across the field after a black football player. I sit, smiling quietly. You know how distracting….

A lot of math is harmful, so a similar final example for your own solution.

Example 6

Find the general solution of a system of linear equations.

I have already checked the general solution, the answer can be trusted. Your decision course may differ from my decision course, the main thing is that the general decisions 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 when there are no fractions are much less common. Be prepared mentally, and most importantly, technically.

I will dwell on some of the 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 an A 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, you should calmly bring the expanded matrix of the system to a stepwise form according to the standard algorithm. Such a system can be inconsistent, it can have infinitely many solutions, and, oddly enough, it can have a single solution.

Theorem. The system of linear equations is consistent only if the rank of the extended matrix is ​​equal to the rank of the matrix of the system itself.

Systems of linear equations

Consistent r (A) = r () inconsistent r (A) ≠ r ().

Thus, systems of linear equations have either an infinite set of solutions, or one solution, or have no solutions at all.

End of work -

This topic belongs to the section:

Elementary matrix transformations. Cramer method. Vector definition

Two elements of the permutation form an inversion if in the permutation notation the larger element precedes the smaller one .. there are n different permutations of the nth degree out of n numbers, we will prove this .. the permutation is called even if the total number of inversions is an even number and, accordingly, odd if ..

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All topics in this section:

Kronecker-Capelli theorem
Consider a system of linear equations with n unknowns: Let's compose the matrix and the extended matrix

The concept of a homogeneous system of linear equations
A system of linear equations in which all free terms are equal to 0, i.e. the system of the form is called homogeneous

Property of solutions to a homogeneous SLO
A linear combination of solutions to a homogeneous system of equations is itself a solution to this system. x = and y =

Relationship between solutions of homogeneous and inhomogeneous systems of linear equations
Consider both systems: I and

Axiomatic approach to the definition of linear space
Earlier, the concept of an n-dimensional vector space was introduced as a set of ordered systems of n-real numbers, for which the operations of addition and multiplication by a real number were introduced.

Corollaries from the axioms
1. Uniqueness of the zero vector 2. Uniqueness of the opposite vector

Proof of Consequences
1. Suppose that. - to zero

Basis. Dimension. Coordinates
Definition 1. A basis of a linear space L is a system of elements belonging to L that satisfies two conditions: 1) the system

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1 1 The number of solutions to the system of equations Graphical dynamic method To find the number of solutions to the system of equations containing a parameter, the following technique is useful.We build graphs of each of the equations for a certain fixed value of the parameter and find the number of common points of the constructed graphs Each common point is one of the solutions of the system Next, mentally change parameter and represent how the graph of the equation with the parameter is transformed, how the common points of the graphs appear and disappear Such a study requires a developed imagination To train the imagination, consider a number of typical problems Let us call special values ​​of the parameter those values ​​at which the number of solutions changes These values ​​correspond to situations when the graphs of decisions touch each other or the corner point of one of the graphs falls on another graph.As a rule, when passing through a singular point, the number of solutions changes by two, and at such a point itself it differs by one from the number of solutions with a small change in n parameter Let us consider the problems in which it is required to find the number of solutions of the system of equations, one of which depends on the parameter a, and the other does not depend on the Variables in the systems x and y The numbers xi, yi, r are considered given constants In the course of each solution, we plot the graphs of both equations. how the graph of the equation with the parameter changes when the value of the parameter changes.Then we draw a conclusion about the number of solutions (common points of the constructed graphs) In the interactive figure, the graph of the equation without the parameter is shown in blue, and the dynamic graph of the equation with the parameter is shown in red To study the topic (tasks 1 7 ) use the InMA file 11, 5 Number of system solutions with the parameter For research (task 8) use the GInMA file Number of system solutions with the parameter (x x0) + (y y0) = r; 1 Find the number of solutions of the system (x x1) + y = a (x x0) + (y y 0) = r; Find the number of solutions of the system y = kx + a (x x0) + (y y0) = r; 3 Find the number of solutions of the system y = ax + y1 (x x0) + (y y0) = r; 4 Find the number of solutions of the system (x x1) + y = a (x x0) + y y0 = r; 5 Find the number of solutions of the system (x x0) + (y y0) = a (x x0) + (y y0) = r; 6 Find the number of solutions of the system y = x a + y1 x x0 + y y0 = r; 7 Find the number of solutions of the system (x x0) + (y y0) = a f (x, y) = 0; g (x, y, a) = 0 8 Find the number of solutions of the BB system Shelomovsky Thematic sets, cmdru /

2 1 Graphs of equations smooth curves (x x0) + (y y0) = r; 1 Task Find the number of solutions of the system (x x1) + y = a Solution: The graph of the first equation is a circle of radius r centered at point O (x0; y0) The graph of the second equation is a circle of radius a centered on the abscissa axis at point A (x1 ; 0) The center of the circle is motionless, the radius determines the parameter When the modulus of the parameter increases, the circle "swells" Special values ​​of the parameter are those values ​​at which the number of roots changes, that is, the values ​​of the parameter at which the circle of the second graph touches the circle of the first Condition of tangency of circles modulus of the sum or difference the radii of the circles is equal to the center-to-center distance: a ± r = AO a = ± AO ± r Research: By changing the value of the variables and the parameter, find the number of solutions of the system. when the common axis of the circles is vertical In general, use the Pythagorean triangles For example, x0 x1 = 3, y0 = ± 4 Typically, as with small mo Since two non-coinciding circles can have at most two common points, the number of solutions in the general case is not more than two At the points of tangency, the number of solutions is equal to one, with intermediate values ​​of the parameter two parameter at which three different points (x 1) + (y y0) = 9; are solutions to the system of equations (x x1) + y = a (x x0) + (y y0) = r; Task Find the number of solutions of the system y = kx + a Solution: The graph of the first equation is a circle of radius r centered at the point O (x0; y0) The graph of the second equation is a family of parallel lines passing through points A (0; a) and having a constant slope The tangent of the angle of inclination of straight lines is equal to k When increasing the parameter, the straight lines move up. Special values ​​of the parameter are those values ​​at which the number of roots changes, that is, the values ​​of the parameter at which the straight lines touch the circle. cmdru /

3 3 Solving the resulting equation, we find the coordinates of two points of contact: kr x = x0 ±; x0 x 1 + k = kk (y y0) + (y y0) = rry y0 y = y0 1+ k Substituting the obtained expressions into the equation of a straight line, we find the value of the parameter at singular points: a = y 0 kx0 ± r 1 + k Research : Changing the value of the variables and the parameter, find the number of solutions of the system. It is desirable to start the study with the simplest case k = 0, when the straight lines are parallel to the abscissa axis. Then consider the cases when the root is extracted (for example, k = 3), pay attention to the popular case k = 1 For small and for large values ​​of the parameter there are no solutions Since a straight line and a circle can have no more than two common points, the number of solutions is no more than two For the values ​​of the parameter corresponding to tangency, the number of solutions is equal to one, with intermediate values ​​of the parameter two. Creative task It is known that this system of equations has at most one solution Find the value of the parameter for which the system of equations has a solution: (x) + (y 3) = r; y = x + a (x x0) + (y y0) = r; 3 Find the number of solutions of the system y = ax + y1 Solution: The graph of the first equation is a circle of radius r centered at the point O (x0; y0) The graph of the second equation is a family of straight lines passing through the point A (0; y1) The tangent of the angle of inclination of straight lines ( a) determines the value of the parameter.When the parameter increases, the angle between the graph and the positive direction of the abscissa axis increases. , then any possible straight line intersects the circle at two points We find the condition of tangency by equating the tangents of the angle of inclination of the circle and the straight line Solving the resulting equation, we find the coordinates of two points of tangency: VV Shelomovsky Thematic sets, cmdru /

4 4 ar x = x0 ±; x0 x 1 + a = aa (y y0) + (y y0) = rry y0 y = y0 1+ a Substituting the obtained expressions into the equation of the straight line, we find the value of the parameter at (y1 y 0) r singular points If x0 = 0, then special values ​​of the parameter a = ± r If y0 = y1, x0 r, then the special values ​​of the parameter a = ± (y1 y 0) rr x0 If х0 = ± r, then the circle touches the vertical line passing through the point r (y1 y 0) А (0; у1) and the value of the parameter a = In other cases, x0 (y1 y 0) a = x0 (y 0 y1) ± r (x0 + (y 0 y1) r) r x0 Research: By changing the value of the variables and the parameter, find the number of solutions of the system It is desirable to start the study from the simplest case y0 = y1, x0< r, когда точка А(0; у1) внутри окружности и число решений всегда равно двум Рассмотрите случай х0 = r, когда число решений легко найти (х0 = r =, y0 = 3, y1 =) Затем рассмотрите случаи, когда корень хорошо извлекается (например, х0 = 3, y0 = 4, r =, y1 =) Поскольку прямая и окружность могут иметь не более двух общих точек, число решений не более двух При значениях параметра, соответствующих касанию, число решений равно единице, при остальных значениях параметра нулю или двум (x + 3) + (y 5) = r ; при всех y = ax + 1 Творческое задание Известно, что система уравнений значениях параметра, кроме одного, имеет два решения Найдите то значение параметра, при котором система уравнений имеет единственное решение (x x0) + (y y0) = r ; 4 Задание Найдите число решений системы (x x1) + y = a Решение: В ходе решения строим графики каждого из уравнений и исследуем число общих точек построенных графиков График первого уравнения это пара окружностей одинакового радиуса r Центры окружностей O и Q имеют одинаковую ординату y0 и ВВ Шеломовский Тематические комплекты, cmdru/

5 5 identical in magnitude, but different in sign of the abscissa ± x0 Graphs are shown in blue and purple The graph of the second equation is a circle of radius a centered on the abscissa axis at point A (x1; 0) Special values ​​of the parameter those values ​​at which the number of roots changes , that is, the values ​​of the parameter at which the circle of the second graph touches the circles of the first one.Conditions of touching the sum or difference of the radii of the circles is equal to the center-to-center distance: a ± r = AO, and ± r = AQ Research: By changing the value of the variables and parameter, find the number of solutions of the system Use integers values ​​for one center-to-center distance (for example, x0 = 6, y0 = 3, r = 3, x1 =) Typically, for small in modulus and large values ​​of the parameter, there are no solutions At the points of tangency, the number of roots is odd, at other points the number of roots is even ( x 6) + (yy 0) = r; Creative task It is known that the system of equations for (x x1) + y = a some value of the parameter has exactly two solutions. For this value of the parameter, the graphs touch Find this value of the parameter (x x0) + y y0 = r; 5 Find the number of solutions of the system (x x0) + (y y0) = a Solution: The graph of the first equation consists of a pair of parabolas that are joined at y = y0 Equations of parabolas y = y0 ± (r (x x0)) They have a horizontal axis of symmetry y = y0, vertical axis of symmetry х = х0 Center of symmetry point (x0, y0) The second graph is a circle of radius a, the center of which is located in the center of symmetry of parabolas The number of roots changes at such a value of the parameter at which the circle of the second graph touches the vertices of the parabolas At the point of tangency: x = x0, y = y0 ± r = y = y0 ± a, which means that a = ± r The number of roots changes at such a value of the parameter at which the inner tangency of the circle of the second graph with parabolas occurs. To find this value, go from a system of equations to an equation with one variable: (yy 0) = a (x x0) = (r (x x0)) This is a quadratic equation for (xx 0) It has one root if the discriminant is zero: VV Shelomovsky Thematic sets, cmdru /

6 6 D = (r 0,5) (ra) = 0, a = ± r 1 4 The number of roots changes at such a value of the parameter at which the intersection of the circle and the parabola occurs at the break points of the first graph, that is, at y = y0 Research : Changing the value of the variables and the parameter, find the number of solutions to the system Use the values ​​r = 1, 4 and 9 Note that the parameters x0 and y0 do not affect the answer to the problem For small in magnitude and large values ​​of the parameter, there are no solutions x x0 + y y0 = r; 6 Find the number of solutions of the system (x x0) + (y y0) = a Solution: The graph of the first equation is a square inclined at an angle of 45 to the coordinate axes, the length of half of the diagonal of which is r The second graph is a circle of radius a, the center of which is located in the center symmetry of the square The number of roots changes at the value of the parameter at which the circle passes through the vertices of the square In this case, y = y0, a = ± r The number of roots changes at the value of the parameter at which the inner tangency of the circle with the sides of the square occurs To find this value, go from a system of equations to an equation with one variable: (yy 0) = a (x x0) = (rx x0) This is a quadratic equation for xx 0 It has one root if the discriminant is zero In this case, a = ± r The radius of the circle in this case refers to the radius in the previous case, as sin 45: 1 VV Shelomovsky Thematic kits, cmdru /

7 7 (x x0) + (y y0) = r; 7 Find the number of solutions of the system y = xa + y1 The graph of the first equation is a circle with the center O (x0; y0) The graph of the second equation consists of two rays with a common origin this is “bird, wings up”, the top of the graph is located at point A (a; y1) The number of roots changes at the value of the parameter at which the "wing" of the second graph touches the circle or the top of the graph lies on this circle. this wing touches the circle at points (xk; yk) such that r yk = y0 The tangency condition yk = xk a + y1 a = xk yka + y1 = x0 y0 + y1 ± r Since the "wing" is a ray going up , the condition is added that the ordinate of the vertex should be no more than the ordinate of the point of contact, that is, у1 уk y0 у1 ± r Similarly, we write down the conditions of contact with the "left wing" If the vertex of the graph lies on a circle, then its coordinates satisfy the equation of the circle: (a x0) + (y1 y0) = r no solutions to the system, that is, the number of common points of the graphs At singular points, the number of roots is odd, at other points the number of roots is even (x) + (yy 0) = r, Creative task It is known that the system of equations for y = xa + y1, some value parameter has three solutions Find this value of the parameter if it is known that the ordinates of the two solutions coincide f (x, y) = 0; g (x, y, a) = 0 8 Find the number of solutions of the system. Define the functions yourself according to the model and investigate the number of solutions VV Shelomovsky Thematic sets, cmdru /

8 8 VV Shelomovsky Thematic sets, cmdru /

9 9 Tasks C5 (Semyonov Yashchenko) Option 1 Find all the values ​​of a, for each of which the set of solutions to the inequality 4 x 1 x + 3 a 3 is a segment 3 a 4 x Thinking Perform transformations xb 1, 1 xb 1, 4 x 1 x + 3 axb 3 =, b = 3 a 3 a 4 xx (x) 0, (x +1) b 1 0 The boundary lines of the plane x 3a are: x = 0, x =, x = 3a, x = ± 3 aa = (x + 1) 1 4 If 0 x, then b< 4x, b (x +1) 1 Так как 4x >(x +1) 1, then b (x +1) 1 If 0> x then b> 4x, (x +1) 1 b There is a solution for 1 b For example, x = 1 If x>, then b> 4x, (x +1) 1 b Since 4x< (x +1) 1, то (x +1) 1 b Значит, решения таковы Если 3а >8, then x [3 a + 1 1.0] [, 3 a +1 1] If 3a = 8, then x [4.0] x [3 a +1 1.0] [3 a + 1 1,] If 0< 3а < 8, то Если 3а = 0, то х [,0) (0, ] Если 1< 3а < 0, то х [ 3 a +1 1, 3 a+1 1] [ 0, ] Если 1 = 3а, то х 1 } Если 1 >3a, then x Solution Let 1 3a Then x = 1 satisfies the inequality, 4 x 1 x + 3 a 16 + 3 a 3 a 3 = 3 =, a contradiction, this is a number outside the segment 3 a 4 x 3 a + 4 3 a +4 Let 1> 3а Then xb 1, 4 x 1 x + 3 axb 3 =, b = 3 a< 1 3 a 4 x 1 x b 1, x (x) 0, (x +1) b 1 0 Числа из промежутка 0 х удовлетворяют обоим неравенствам Если x >, then the first inequality is not satisfied VV Shelomovsky Thematic sets, cmdru /

10 10 If 0> x, then b (x +1) 1, the second inequality is not satisfied Answer: 1> 3а Option 3 Find all values ​​of a, for each of which the equation a +7 xx + x +5 has at least one root = a + 3 x 4 a +1 Thinking Let f (a, x) = a +7 xx + x +5 a 3 x 4 a + 1 Singular point of the function х + 1 = 0 If х = 1, then the equation has the form a +10 a 1 a = 0 It is easy to find four solutions It is necessary to prove that the original function is always greater than this Solution Let f (a, x) = a + 7 xx + x +5 a 3 x 4 a + 1 Equation f (a, x) = 0 Then f (a, 1) = a +10 a 1 a = 0 Difference f (a, x) f (a, 1) = 7 x +1 +5 (x + x +5) + 3 4 a 3 x 4 a + 1 3 (xa 4 ax 1) 0 Hence, the equation f (a, x) = 0 has roots only if f (a, 1) 0 The equation f (a, 1) = 0 has four roots a 1 =, a =, a 3 =, a 4 = Function f (a, 1) 0 (not positive) for a For example, if a = 10, that is, the root For other values ​​of a x = f (a, x) f (a, 1)> 0 No roots Answer: [5 15, 5+ 15] Option 5 Find all values ​​of a, for each of which it has at least one root ur a +11 x + +3 x + 4 x + 13 = 5 a + xa + Use the function f (a,) = a +9 5 a 4 a = 0 and the inequality f (a, x) f (a,) (x + + ax a +) 0 Answer: [,] Option 9 Find the number of roots of the equation x + 4x 5 3a = x + a 1 Thinking We consider the following (obvious) statement known Let the functions f (x) and g (x) be given on some interval Let the derivative of one is greater on the interval than the other Let the difference of the values ​​of the functions at the left end have one sign, at the right another Then the equation f (x) = g (x) has exactly one root on the interval Solution Denote f (x, a) = 3а + x + a, g (x) = x + 4x Equation f (x, a) = g (x) VV Shelomovsky Thematic kits, cmdru /

11 11 The singular points of the function g (x) are the minima at x = 1 and x = 5 and the maximum at x = Values ​​g (1) = g (5) = 1, g () = 10 The function has an axis of symmetry x = 3 At large modulo x values, the quadratic function g (x) is greater than the linear function f (x, a) The slope of the function outside the segment [5,1] is determined by the derivative (x + 4x 5) "= x for x> 1 Function g (x) for x > 1 monotonically increases with a coefficient greater than 6 By virtue of symmetry, the function g (x) decreases monotonically with a coefficient greater than 6 for x< 5 Наклон g(x) равен 1 только на промежутке (5, 1) При этом производная (x 4x + 5)" = x 4 = 1 Значит, в точке x = 5 наклон равен 1 Функция f(x, a) = 3а + x + a монотонно убывает с коэффициентом 1 при x + а < 0 и монотонно возрастает с коэффициентом 1 при x + а >0 Values ​​at a number of points f (a, a) = 3а, f (5, a) = 3а + 5 a, f (, a) = 3а + a, f (1, a) = 3а + 1+ a Graphs f (x, a) and g (x) touch if their slopes are equal Touching is possible at x = 5 Moreover, g (x) = 39/4 f (x, a) = 4а + x = 39/4, 4a = 49 / 4, a = 49/16 Analyze the roots of the equation f (x, a) = g (x) If a<, f(5, a) = а +5 < 1, f(1, a) = а 1 < 5 f(x, a) < g(x), так как в промежутке 5 < x < 1 f(x, a) < 1 < g(x) Если x >1, g (x) increases faster than f (x, a), that is, everywhere f (x, a)< g(x) Если x < 5, g(x) убывает быстрее, чем f(x, a), то есть всюду f(x, a) < g(x) Других корней нет Если a =, f(5, a) = 1, f(1, a) = 5 f(5,) = g(5) Один корень х = 5 Во всех других точках f(x, a) < g(x), как и в предыдущем случае Если < a < 0, f(5, a) = а +5 >1, f (1, a) = 4a + 1< 1f(, a) = а + < 10 При x >f (x, a)< g(x), корней нет При x < f(1,a) >1 For x< 5 быстро убывающая g(x) пересекает медленно убывающую левую ветвь f(x,а), один корень При 5 < x < возрастающая g(x) пересекает убывающую f(x,а), один корень, всего корней два, один при x < 5, второй при 5 < x < Если a = 0, f(5, a) = 5, f(1, a) = 1 f(1, a) = g(1), один корень х = 1 Как и раньше, один корень при x < 5, один корень при 5 < x < Всего корней три Если 0 < a < 3, корней 4, два на левой ветке f(х, a) при x <, два на правой при x >If a = 3, f (3, 3) = 8 = g (3), f (, 3) = 10 = g (), roots 4, one two on the left branch f (x, a) for x< 5, один в вершине f(х, 3) при x = 3, один в вершине g(x) при x =, один при x >1 If 3< a < 49/16, корней 4, один на левой ветке f(х, a) при x < 5, два на правой ветви g(x) при 3 < x <, один при x >1 If a = 49/16, then the number of roots is 3, one on the left branch f (x, a) for x< 5, один в точке касания при x = 5, один при x >1 If a> 49/16, then the number of roots, one on the left branch f (x, a) for x< 5, один на правой при x >1 Answer: no roots for a< ; один корень при a =, два корня при < a < 0 или 49/16 < a, три корня при a = 0 или а = 49/16, четыре корня при 0 < a < 49/16 ВВ Шеломовский Тематические комплекты, cmdru/

12 1 Option 10 Find all values ​​of the parameter a, for each of which the equation has two roots 4x 3x x + a = 9 x 3 Solution Denote f (x, a) = 4x 3x x + a, g (x) = 9 x 3 The singular point of the function g (x) is x = 3 The function decreases monotonically with a coefficient of 9 at x< 3 и монотонно возрастает с коэффициентом 9 при x >3 The function f (x, a) is piecewise linear with coefficients 8, 6, or 0 This means that it does not decrease in x, its growth rate is less than that of the right branch of the function 9 x 3 f (3, a) = a The graph of this the expression is a polyline with vertices (1, 1), (3, 3), (6, 1) The values ​​of the function are positive for a (4, 18) From the found it follows If f (3, a)< 0, уравнение не может иметь корней, так как g(x) >f (x, a) If f (3, a) = 0, the equation has exactly one root x = 3 For other xes g (x)> f (x, a) If f (3, a)> 0, the equation has exactly two roots, one for x< 3, когда пересекаются убывающая ветвь g(x) и монотонно не убывающая f(x, a) Другой при x >3, when the rapidly increasing branch g (x) intersects the slowly increasing branch f (x, a) Answer: а (4, 18) Option 11 Find all values ​​of the parameter a, for each of which, for any value of the parameter b, it has at least one solution system of equations (1+ 3 x) a + (b 4 b + 5) y =, xy + (b) x y + a + a = 3 Thinking The system has the form (1+ 3 x) a + (1+ (b) ) y =, Conveniently xy + (b) xy = 4 (a + 1) a (1 + 3 x) = 1, We see the solution x = y = 0 and xy = 4 (a +1) the corresponding values ​​of the parameter a = 1 and a = 3 analyze the singular point b = Then (1+ 3 x) a + (1+ (b)) y =, xy + (b) xy = 4 (a + 1) Solution Write the system as Solution x = y = 0 always exists for a = 1 or a = 3 If b =, then the system has the form (1+ 3 x) a +1 y =, or xy = 4 (a +1) (1 + 3 x) a = 1, xy = 4 (a +1) If a> 1 or a< 3 система не имеет решений, так как их не имеет второе уравнение Если 1 < a < 3, из второго уравнения получим, что x >0, from the first we find a = 0 Let a = 0 Then for b = 4 from the first equation we obtain that y = 0 In this case, the second equation has no solution Answer: 1 or 3 BB Shelomovsky Thematic sets, cmdru /

13 13 Option 14 Find all the values ​​of the parameter, for each of which the modulus of the difference between the roots of the equation x 6x a 4a = 0 takes the largest value Solution Write the equation in the form (x 3) = 1 (a) Its solution = 0 because of the periodicity of the functions sine and cosine , the problem can be solved for the segment x = 3 ± 1 (a) The largest difference of the roots is equal at a = Answer: Option 15 Find all the values ​​of the parameter, for each of which the equation (4 4 k) sin t = 1 has at least one solution on the segment [3 π; 5 π] cos t 4 sin t Solution Due to the periodicity of the sine and cosine functions, the problem can be solved for the interval t [π; 15 π], then subtract 4π from each obtained solution. Transform the equation to the form + 4 k sin t cos t = 0 cos t 4 sin t On the segment t [π; 15 π] the sine decreases monotonically from zero to minus one, the cosine monotonically increases from minus one to zero. The denominator vanishes at 4tgt = 1, that is, at sin t = 1 4, cos t = The numerator at t = π is 1, at t = 15π equals 4k If k 0, the numerator is positive and the equation has no roots If k> 0, both variable terms of the numerator decrease, that is, the numerator changes monotonically Hence, the numerator takes zero value exactly once if k 05 and is positive for smaller values k The equation has a root if the numerator is zero, and the denominator is not zero, that is, in the case of 4k = + 4 k sin t cos t + k Answer: k [05, +) \ 1+) Option 18 Find all the values ​​of the parameter, for each of which the system of equations (xa 5) + (y 3 a +5) = 16, (xa) + (y a + 1) = 81 has a unique solution. Thinking Each equation describes a circle. Solution is unique in the case of tangency of circles. Solution The first equation defines a circle centered at point (a + 5, 3a 5) and radius 4 The second equation is circular st centered at the point (a +, a 1) with a radius of 9 BB Shelomovsky Thematic sets, cmdru /

14 14 The system has a unique solution if the circles touch In this case, the distance between the centers is = 13 or 0 4 = 5 The square of the center-to-center distance: ((a + 5) (a +)) + ((3a 5) (a 1)) = aa + 5 If the distance is 5, then a = 0 or a = 1 If the distance is 13, then a = 8 or a = 9 Answer: 8, 0, 1, 9 Option 1 Find all the parameter values ​​for each of which has exactly two non-negative solutions equation 10 0,1 xa 5 x + a = 004 x Solution Perform transformations 5 xa 5 x + a = 5 x Denote t = 5x 1 Due to the monotonicity of the exponential function 5x, each root t 1 generates exactly one root x 0 The equation takes the form ta t + at = 0 If at, then t + 3t + a = 0 there are no roots greater than 1 If t> at /, then tt + 3a = 0 For t> 1, the function increases monotonically, there is only one root If 1 /> t /> a, then t 3t a = 0 For t> 1 the function t 3t decreases monotonically from at t = 1 to 5 at t = 15 and then monotonically increases.Thus, for 5> a there are two roots, for smaller a, there are no roots, for large a, the root is exactly odi n Answer: 5> a Option Find, depending on the parameter, the number of solutions of the system x (a + 1) x + a 3 = y, y (a + 1) y + a 3 = x Thinking The system has the form f (x) = y, f (y) = x, or f (f (х)) = x One of the solutions f (x) = x We find the second solution by subtracting the equations Solution Subtract the second from the first equation We obtain (x + ya) (xy) = 0 Let x = y Substitute in the first equation, transform We get (xa 1) = 4 + a Let x + y = a Substitute in the first equation, transform: (xa) = 3 + a If a<, корней нет Если a =, то x = y = a + 1, единственное решение Если 15 >a>, that is, a pair of solutions x = y = a + 1 ± 4+ a If a = 15, then two solutions: x = y = a, x = y = a + If 15< a то решения x= y =a+ 1± 4+ a, x=a± 3+ a, y= a x Ответ: a < нет решений, а = одно, 15 a >, two solutions, a> 15 four solutions VV Shelomovsky Thematic sets, cmdru /

15 15 Option 4 Find all values ​​of a, for each of which the equation has no roots 7 x 6 + (4 ax) 3 +6 x +8 a = 4 x Think 8a 4x = (4a x), 7x6 = (3x) 3 This means that the equation includes the sum and the sum of cubes of the same expressions This can be used Solution We transform the equation to the form (3 x) 3 + (4 ax) 3+ (3 x + 4 ax) = 0 Expand the sum of cubes (3 x + 4 ax) ( (3 x) 3 x (4 ax) + (4 ax) +) = 0 The second factor is an incomplete square of the difference increased by It is positive Having selected the square in the first factor, we get 1 1 3 (x) + 4 a = This equation does not have roots, if 4 a> 0, a> 3 1 Answer: 1a> 1 Option 8 Find the values ​​of a, for each of which the maximum value of the function xax is not less than one Solution If xa, the function f (x, a) = xax It is maximum for x = 0.5, the maximum is 0.5 a For a< 0,5 наибольшее значение функции 0,5 а 1 при 075 а Если x < a, функция f(x,a) = a x x Она максимальна при x = 0,5, максимум равен a + 05 При a >0.5 is the largest value of the function a + 0.5 1 for a 0.75 Answer: a 0.75 or 075 a Pair of functions Find the range of positive values ​​of a, for each of which there is such b such that the system of equations: y = x4 + a, x = 8y + b has an even number of solutions Solution: From the first equation it follows that y> 0, the second equation can be converted to the form: y =, x (b; +) Excluding y: xbf (x) = xa = 0; f `(x) = 4 x 3 + x b (x b) 3 Each root of the obtained equation generates exactly one solution of the original system At b 0 the function f (x) is monotonically increasing and the equation has exactly one root At negative b< 0 функция f(x) монотонно возрастает от минус бесконечности до f(х1), уменьшается до f(х) и вновь монотонно возрастает при положительных иксах до плюс бесконечности Уравнение может иметь чётное число корней два только если корень совпадает с минимумом или максимумом функции, то есть в точке корня производная равна нулю, то есть f(х1) = g(х1) = 0 Исключая корень из уравнений, найдём: а = (4х1 + х14) Полученная функция имеет максимум при х1 = 1 (а = 3; b = 1,5), поэтому для любого a (0; 3) существуют х1, х х1 и b, при которых число корней равно два Однако при а = 3 х ВВ Шеломовский Тематические комплекты, cmdru/

16 16 = x1, both roots coincide and the equation f (x) = 0 has only one root Derivative f` (x) is positive for x b and for x + It is equal to zero under the condition f `(x) = 0 g (x) = x (xb) + 1 = 0 The last equation can have one or two roots, and only for negative x Let us denote them x1 and x: g (x1) = g (x) = 0 Answer: a (0; 3) BB Shelomovsky Thematic kits, cmdru /

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If the system

a 11 x 1 + a 12 x 2 + ... + a 1n x n = b 1,

a 21 x 1 + a 22 x 2 + ... + a 2n x n = b 2,

a m1 x 1 + a m1 x 2 + ... + a mn x n = b m. (5.1)

turned out to be joint, that is, the matrices of the system A and the matrix of the extended system (with a column of free terms) A | b have the same rank, then two possibilities may appear - a) r = n; b) r< n:

a) if r = n, then we have n independent equations with n unknowns, and the determinant D of this system is nonzero. Such a system has a unique solution obtained by;

b) if r< n, то число независимых уравнений меньше числа неизвестных.

Move the extra unknowns x r + 1, x r + 2, ..., x n, which are usually called free, to the right-hand sides; our system of linear equations will take the form:

a 11 x 1 + a 12 x 2 + ... + a 1r x r = b 1 - a 1, r + 1 x r + 1 -... - a 1n x n,

a 21 x 1 + a 22 x 2 + ... + a 2r x r = b 2 - a 2, r + 1 x r + 1 -... - a 2n x n,

... ... ... ... ... ... ... ... ... ...

a r1 x 1 + a r2 x 2 + ... + a rr x r = b r - a r, r + 1 x r + 1 -... - a rn x n.

It can be solved for x 1, x 2, ..., x r, since the determinant of this system (of the rth order) is nonzero. By assigning arbitrary numerical values ​​to the free unknowns, we obtain, using Cramer's formulas, the corresponding numerical values ​​for x 1, x 2, ..., x r. Thus, for r< n имеем бесчисленное множество решений.

System (5.1) is called homogeneous if all b i = 0, that is, it has the form:

a 11 x 1 + a 12 x 2 + ... + a 1n xn = 0, a 21 x 1 + a 22 x 2 + ... + a 2n xn = 0, (5.5) ... .... .. ... ... ... a m1 x 1 + a m1 x 2 + ... + a mn xn = 0.

It follows from the Kronecker-Capelli theorem that it is always consistent, since adding a column of zeros cannot increase the rank of a matrix. This, however, can be seen directly - system (5.5) certainly has a zero, or trivial, solution x 1 = x 2 = ... = x n = 0. Let the matrix A of system (5.5) have rank r. If r = n, then the zero solution will be the only solution to system (5.5); at r< n система обладает решениями, отличными от нулевого, и для их разыскания применяют тот же прием, как и в случае произвольной системы уравнений. Всякий ненулевой вектор - столбец X= (x 1 , x 2 ,..., x n) T называется eigenvector of linear transformation (square matrix A ), if there is a number λ such that the equality

The number λ is called eigenvalue of linear transformation (matrices A ), corresponding to vector X. Matrix A is of order n. In mathematical economics, the so-called productive matrices... It is proved that the matrix A is productive if and only if all the eigenvalues ​​of the matrix A are less than one in absolute value. To find the eigenvalues ​​of the matrix A, we rewrite the equality AX = λX in the form (A - λE) X = 0, where E is the nth order unit matrix or in coordinate form:

(a 11 -λ) x 1 + a 12 x 2 + ... + a 1n x n = 0,

a 21 x 1 + (a 22 -λ) x 2 + ... + a 2n x n = 0, (5.6)

... ... ... ... ... ... ... ... ... a n1 x 1 + a n2 x 2 + ... + (a nn -λ) xn = 0 ...

We have obtained a system of linear homogeneous equations that has nonzero solutions if and only if the determinant of this system is equal to zero, i.e.

We got an equation of the nth degree with respect to the unknown λ, which is called characteristic equation of the matrix A, the polynomial is called characteristic polynomial of the matrix A, and its roots are characteristic numbers, or eigenvalues, of the matrix A. To find the eigenmatrices A into the vector equation (A - λE) X = 0 or into the corresponding system of homogeneous equations (5.6), the found values ​​of λ should be substituted and solved in the usual way. Example 2.16... Investigate the system of equations and solve it if it is compatible.

x 1 + x 2 - 2x 3 - x 4 + x 5 = 1, 3x 1 - x 2 + x 3 + 4x 4 + 3x 5 = 4, x 1 + 5x 2 - 9x 3 - 8x 4 + x 5 = 0 ...

Solution. We will find the ranks of the matrices A and A | b by the method of elementary transformations, simultaneously reducing the system to a stepwise form:

Obviously, r (A) = r ( A | b) = 2. The original system is equivalent to the following, reduced to a stepwise form:

x 1 + x 2 - 2x 3 - x 4 + x 5 = 1, - 4x 2 + 7x 3 + 7x 4 = 1.

Since the determinant for unknowns x 1 and x 2 is nonzero, then they can be taken as the main ones and the system can be rewritten as:

x 1 + x 2 = 2x 3 + x 4 - x 5 + 1, - 4x 2 = - 7x 3 - 7x 4 + 1,

Whence x 2 = 7/4 x 3 + 7/4 x 4 -1/4, x 1 = 1/4 x 3 -3/4 x 4 - x 5 + 5/4 - a general solution to a system that has an infinite number of solutions ... Giving to the free unknown x 3, x 4, x 5 specific numerical values, we will receive particular solutions. For example, for x 3 = x 4 = x 5 = 0 x 1 = 5/4, x 2 = - 1/4. Vector C (5/4, - 1/4, 0, 0, 0) is a particular solution of this system. Example 2.17. Examine the system of equations and find a general solution depending on the value of the parameter a.

2x 1 - x 2 + x 3 + x 4 = 1, x 1 + 2x 2 - x 3 + 4x 4 = 2, x 1 + 7x 2 - 4x 3 + 11x 4 = a.

Solution. This system corresponds to the matrix ... We have A ~

therefore, the original system is equivalent to the following:

x 1 + 2x 2 - x 3 + 4x 4 = 2,

5x 2 - 3x 3 + 7x 4 = a-2,

This shows that the system is compatible only for a = 5. The general solution in this case is:

x 2 = 3/5 + 3 / 5x 3 - 7 / 5x 4, x 1 = 4/5 - 1 / 5x 3 - 6 / 5x 4.

Example 2.18. Find out if the vector system is linearly dependent:

a 1 =(1, 1, 4, 2),

a 2 = (1, -1, -2, 4),

a 3 = (0, 2, 6, -2),

a 4 =(-3, -1, 3, 4),

a 5 =(-1, 0, - 4, -7),

Solution. A system of vectors is linearly dependent if there are such numbers x 1, x 2, x 3, x 4, x 5, of which at least one is nonzero
(see item 1.section I) that the vector equality holds:

x 1 a 1 + x 2 a 2 + x 3 a 3 + x 4 a 4 + x 5 a 5 = 0.

In coordinate notation, it is equivalent to the system of equations:

x 1 + x 2 - 3x 4 - x 5 = 0, x 1 - x 2 + 2x 3 - x 4 = 0, 4x 1 - 2x 2 + 6x 3 + 3x 4 - 4x 5 = 0, 2x 1 + 4x 2 - 2x 3 + 4x 4 - 7x 5 = 0.

So, we got a system of linear homogeneous equations. We solve it by eliminating unknowns:

The system is reduced to a stepwise form, equal to 3, which means that the homogeneous system of equations has solutions other than zero (r< n). Определитель при неизвестных x 1, x 2, x 4 is nonzero, so they can be selected as the main ones and the system can be rewritten as:

x 1 + x 2 - 3x 4 = x 5, -2x 2 + 2x 4 = -2x 3 - x 5, - 3x 4 = - x 5.

We have: x 4 = 1/3 x 5, x 2 = 5 / 6x 5 + x 3, x 1 = 7/6 x 5 -x 3. The system has countless solutions; if free unknowns x 3 and x 5 are not equal to zero at the same time, then the principal unknowns are also nonzero. Therefore, the vector equation

x 1 a 1 + x 2 a 2 + x 3 a 3 + x 4 a 4 + x 5 a 5 = 0