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 collection 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

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 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 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, we 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 do 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 line multiplied by 3. To the third line we add the first line 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, where is a nonzero number, is obtained, 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 (inconsistent).

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 the rank of the matrix and the Kronecker-Capelli theorem.

Please 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. Add the first line multiplied by -1 to the fourth line.

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 stepped 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:

The only solution of the system here does not smell like "usual". 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 (that is, to prove that the 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:

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

We will 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 the 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 variables (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 can 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 using 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 break your head.

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

Answer: Common decision: , particular solutions: , .

I shouldn't have remembered about the blacks here ... ... because all sorts of sadistic motives got into my head and I remembered the famous photo-toad, in which the KKK Klan members 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, one should calmly bring the extended 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.

To investigate a system of linear agebraic equations (SLAE) for compatibility means to find out whether this system has solutions or not. Well, if there are solutions, then indicate how many of them.

We need information from the topic "System of linear algebraic equations. Basic terms. Matrix notation". In particular, we need such concepts as the matrix of the system and the extended matrix of the system, since it is on them that the formulation of the Kronecker-Capelli theorem is based. As usual, the matrix of the system will be denoted by the letter $ A $, and the extended matrix of the system by the letter $ \ widetilde (A) $.

Kronecker-Capelli theorem

The system of linear algebraic equations is consistent if and only if the rank of the matrix of the system is equal to the rank of the extended matrix of the system, i.e. $ \ rang A = \ rang \ widetilde (A) $.

Let me remind you that a system is called joint if it has at least one solution. The Kronecker-Capelli theorem says the following: if $ \ rang A = \ rang \ widetilde (A) $, then there is a solution; if $ \ rang A \ neq \ rang \ widetilde (A) $, then this SLAE has no solutions (inconsistent). The answer to the question about the number of these solutions is given by a corollary to the Kronecker-Capelli theorem. In the formulation of the corollary, the letter $ n $ is used, which is equal to the number of variables of the given SLAE.

Corollary from the Kronecker-Capelli theorem

1. If $ \ rang A \ neq \ rang \ widetilde (A) $, then the SLAE is inconsistent (has no solutions).
2. If $ \ rang A = \ rang \ widetilde (A)< n$, то СЛАУ является неопределённой (имеет бесконечное количество решений).
3. If $ \ rang A = \ rang \ widetilde (A) = n $, then the SLAE is definite (has exactly one solution).

Note that the above theorem and its corollary do not indicate how to find the solution to the SLAE. With their help, you can only find out whether these solutions exist or not, and if they exist, then how many.

Example # 1

Explore SLAE $ \ left \ (\ begin (aligned) & -3x_1 + 9x_2-7x_3 = 17; \\ & -x_1 + 2x_2-4x_3 = 9; \\ & 4x_1-2x_2 + 19x_3 = -42. \ End (aligned ) \ right. $ for compatibility If the SLAE is compatible, indicate the number of solutions.

To find out the existence of solutions to a given SLAE, we use the Kronecker-Capelli theorem. We need the matrix of the system $ A $ and the extended matrix of the system $ \ widetilde (A) $, we write them down:

$$ A = \ left (\ begin (array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \ end (array) \ right); \; \ widetilde (A) = \ left (\ begin (array) (ccc | c) -3 & 9 & -7 & 17 \\ -1 & 2 & -4 & 9 \\ 4 & -2 & 19 & -42 \ end (array) \ right). $$

Find $ \ rang A $ and $ \ rang \ widetilde (A) $. There are many ways to do this, some of which are listed in the Matrix Rank section. Usually, two methods are used to study such systems: "Calculation of the rank of a matrix by definition" or "Calculation of the rank of a matrix by the method of elementary transformations".

Method number 1. Calculation of ranks by definition.

By definition, rank is the highest order of the matrix minors, among which there is at least one nonzero. Usually, the study starts with first-order minors, but here it is more convenient to start immediately calculating the third-order minor of the matrix $ A $. The elements of the third order minor are at the intersection of three rows and three columns of the matrix under consideration. Since the matrix $ A $ contains only 3 rows and 3 columns, the third order minor of the matrix $ A $ is the determinant of the matrix $ A $, i.e. $ \ Delta A $. To calculate the determinant, let's apply formula # 2 from the topic "Formulas for calculating determinants of the second and third orders":

$$ \ Delta A = \ left | \ begin (array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \ end (array) \ right | = -21. $$

So, there is a third order minor of the matrix $ A $, which is not equal to zero. It is impossible to compose a minor of the fourth order, since it requires 4 rows and 4 columns, and in the $ A $ matrix there are only 3 rows and 3 columns. So, the highest order of the minors of the matrix $ A $, among which there is at least one non-zero, is equal to 3. Therefore, $ \ rang A = 3 $.

We also need to find $ \ rang \ widetilde (A) $. Let's take a look at the structure of the matrix $ \ widetilde (A) $. The matrix $ \ widetilde (A) $ contains the elements of the matrix $ A $, and we found out that $ \ Delta A \ neq 0 $. Therefore, the matrix $ \ widetilde (A) $ has a third-order minor that is not zero. We cannot compose the fourth order minors of the matrix $ \ widetilde (A) $, so we conclude: $ \ rang \ widetilde (A) = 3 $.

Since $ \ rang A = \ rang \ widetilde (A) $, according to the Kronecker-Capelli theorem, the system is consistent, i.e. has a solution (at least one). To indicate the number of solutions, let's take into account that our SLAE contains 3 unknowns: $ x_1 $, $ x_2 $ and $ x_3 $. Since the number of unknowns is $ n = 3 $, we conclude: $ \ rang A = \ rang \ widetilde (A) = n $, therefore, according to the corollary from the Kronecker-Capelli theorem, the system is definite, i.e. has only one solution.

The problem has been solved. What are the disadvantages and advantages of this method? First, let's talk about the pros. First, we only needed to find one determinant. After that, we immediately made a conclusion about the number of solutions. Usually, in standard standard calculations, systems of equations are given that contain three unknowns and have a unique solution. For such systems, this method is even very convenient, because we know in advance that there is a solution (otherwise there would be no example in a typical calculation). Those. we just have to show the existence of a solution in the fastest way. Secondly, the calculated value of the determinant of the matrix of the system (i.e., $ \ Delta A $) will come in handy after: when we begin to solve the given system by the Cramer method or using the inverse matrix.

However, the method of calculating the rank is by definition undesirable if the matrix of the system $ A $ is rectangular. In this case, it is better to use the second method, which will be discussed below. In addition, if $ \ Delta A = 0 $, then we cannot say anything about the number of solutions of a given inhomogeneous SLAE. Maybe SLAE has an infinite number of solutions, or maybe none. If $ \ Delta A = 0 $, then additional research is required, which is often cumbersome.

Summarizing what has been said, I note that the first method is good for those SLAEs in which the matrix of the system is square. In this case, the SLAE itself contains three or four unknowns and is taken from standard typical calculations or control works.

Method number 2. Calculation of the rank by the method of elementary transformations.

This method is described in detail in the related topic. We will calculate the rank of the matrix $ \ widetilde (A) $. Why exactly matrices $ \ widetilde (A) $ and not $ A $? The fact is that the matrix $ A $ is a part of the matrix $ \ widetilde (A) $, therefore, calculating the rank of the matrix $ \ widetilde (A) $, we will simultaneously find the rank of the matrix $ A $.

\ begin (aligned) & \ widetilde (A) = \ left (\ begin (array) (ccc | c) -3 & 9 & -7 & 17 \\ -1 & 2 & -4 & 9 \\ 4 & - 2 & 19 & -42 \ end (array) \ right) \ rightarrow \ left | \ text (swap first and second lines) \ right | \ rightarrow \\ & \ rightarrow \ left (\ begin (array) (ccc | c) -1 & 2 & -4 & 9 \\ -3 & 9 & -7 & 17 \\ 4 & -2 & 19 & - 42 \ end (array) \ right) \ begin (array) (l) \ phantom (0) \\ r_2-3r_1 \\ r_3 + 4r_1 \ end (array) \ rightarrow \ left (\ begin (array) (ccc | c) -1 & 2 & -4 & 9 \\ 0 & 3 & 5 & -10 \\ 0 & 6 & 3 & -6 \ end (array) \ right) \ begin (array) (l) \ phantom (0 ) \\ \ phantom (0) \\ r_3-2r_2 \ end (array) \ rightarrow \\ & \ rightarrow \ left (\ begin (array) (ccc | c) -1 & 2 & -4 & 9 \\ 0 & 3 & 5 & -10 \\ 0 & 0 & -7 & 14 \ end (array) \ right) \ end (aligned)

We have transformed the matrix $ \ widetilde (A) $ into a stepped form. The resulting step matrix has three nonzero rows, so its rank is 3. Consequently, the rank of the matrix $ \ widetilde (A) $ is 3, that is, $ \ rang \ widetilde (A) = 3 $. Making transformations with the elements of the matrix $ \ widetilde (A) $, we simultaneously transformed the elements of the matrix $ A $, located up to the line. Matrix $ A $ is also stepped: $ \ left (\ begin (array) (ccc) -1 & 2 & -4 \\ 0 & 3 & 5 \\ 0 & 0 & -7 \ end (array) \ right ) $. Conclusion: the rank of the matrix $ A $ is also equal to 3, i.e. $ \ rang A = 3 $.

Since $ \ rang A = \ rang \ widetilde (A) $, then, according to the Kronecker-Capelli theorem, the system is consistent, i.e. has a solution. To indicate the number of solutions, let's take into account that our SLAE contains 3 unknowns: $ x_1 $, $ x_2 $ and $ x_3 $. Since the number of unknowns is $ n = 3 $, we conclude: $ \ rang A = \ rang \ widetilde (A) = n $, therefore, according to the corollary from the Kronecker-Capelli theorem, the system is defined, i.e. has only one solution.

What are the advantages of the second method? The main advantage is its versatility. It doesn't matter to us at all whether the matrix of the system is square or not. In addition, we actually carried out transformations of the forward course of the Gauss method. There are only a couple of actions left, and we could get the solution of this SLAE. To be honest, I like the second method more than the first, but the choice is a matter of taste.

Answer: The given SLAE is consistent and defined.

Example No. 2

Explore SLAE $ \ left \ (\ begin (aligned) & x_1-x_2 + 2x_3 = -1; \\ & -x_1 + 2x_2-3x_3 = 3; \\ & 2x_1-x_2 + 3x_3 = 2; \\ & 3x_1- 2x_2 + 5x_3 = 1; \\ & 2x_1-3x_2 + 5x_3 = -4. \ End (aligned) \ right. $ For compatibility.

We will find the ranks of the matrix of the system and the extended matrix of the system by the method of elementary transformations. Extended system matrix: $ \ widetilde (A) = \ left (\ begin (array) (ccc | c) 1 & -1 & 2 & -1 \\ -1 & 2 & -3 & 3 \\ 2 & -1 & 3 & 2 \\ 3 & -2 & 5 & 1 \\ 2 & -3 & 5 & -4 \ end (array) \ right) $. Find the required ranks by transforming the extended matrix of the system:

$$ \ left (\ begin (array) (ccc | c) 1 & -1 & 2 & -1 \\ -1 & 2 & -3 & 3 \\ 2 & -3 & 5 & -4 \\ 3 & -2 & 5 & 1 \\ 2 & -1 & 3 & 2 \ end (array) \ right) \ begin (array) (l) \ phantom (0) \\ r_2 + r_1 \\ r_3-2r_1 \\ r_4 -3r_1 \\ r_5-2r_1 \ end (array) \ rightarrow \ left (\ begin (array) (ccc | c) 1 & -1 & 2 & -1 \\ 0 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \\ 0 & 1 & -1 & 4 \\ 0 & 1 & -1 & 4 \ end (array) \ right) \ begin (array) (l) \ phantom (0) \\ \ phantom (0) \\ r_3-r_2 \\ r_4-r_2 \\ r_5 + r_2 \ end (array) \ rightarrow \\ $$ $$ \ rightarrow \ left (\ begin (array) (ccc | c) 1 & -1 & 2 & -1 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \ end (array) \ right) \ begin (array) (l) \ phantom (0) \\\ phantom (0) \\\ phantom (0) \\ r_4-r_3 \\\ phantom (0) \ end (array) \ rightarrow \ left (\ begin (array) (ccc | c) 1 & -1 & 2 & -1 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \ end (array) \ right) $$

The expanded system matrix is ​​stepped. The rank of a stepped matrix is ​​equal to the number of its nonzero rows, therefore $ \ rang \ widetilde (A) = 3 $. The matrix $ A $ (to the point) is also reduced to a stepped form, and its rank is equal to 2, $ \ rang (A) = 2 $.

Since $ \ rang A \ neq \ rang \ widetilde (A) $, according to the Kronecker-Capelli theorem, the system is inconsistent (i.e., has no solutions).

Answer: The system is inconsistent.

Example No. 3

Explore SLAE $ \ left \ (\ begin (aligned) & 2x_1 + 7x_3-5x_4 + 11x_5 = 42; \\ & x_1-2x_2 + 3x_3 + 2x_5 = 17; \\ & -3x_1 + 9x_2-11x_3-7x_5 = -64 ; \\ & -5x_1 + 17x_2-16x_3-5x_4-4x_5 = -90; \\ & 7x_1-17x_2 + 23x_3 + 15x_5 = 132. \ End (aligned) \ right. $ For compatibility.

We bring the extended matrix of the system to a stepwise form:

$$ \ left (\ begin (array) (ccccc | c) 2 & 0 & 7 & -5 & 11 & 42 \\ 1 & -2 & 3 & 0 & 2 & 17 \\ -3 & 9 & -11 & 0 & -7 & -64 \\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \ end (array) \ right) \ overset (r_1 \ leftrightarrow (r_3)) (\ rightarrow) $$ $$ \ rightarrow \ left (\ begin (array) (ccccc | c) 1 & -2 & 3 & 0 & 2 & 17 \\ 2 & 0 & 7 & -5 & 11 & 42 \\ -3 & 9 & -11 & 0 & -7 & -64 \\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \ end (array) \ right) \ begin (array) (l) \ phantom (0) \\ r_2-2r_1 \\ r_3 + 3r_1 \\ r_4 + 5r_1 \\ r_5-7r_1 \ end ( array) \ rightarrow \ left (\ begin (array) (ccccc | c) 1 & -2 & 3 & 0 & 2 & 17 \\ 0 & 4 & 1 & -5 & 7 & 8 \\ 0 & 3 & - 2 & 0 & -1 & -13 \\ 0 & 7 & -1 & -5 & 6 & -5 \\ 0 & -3 & 2 & 0 & 1 & 13 \ end (array) \ right) \ begin ( array) (l) \ phantom (0) \\ \ phantom (0) \\ 4r_3 + 3r_2 \\ 4r_4-7r_2 \\ 4r_5 + 3r_2 \ end (array) \ rightarrow $$ \ rightarrow \ left (\ begin (array) (ccccc | c) 1 & -2 & 3 & 0 & 2 & 17 \\ 0 & 4 & 1 & -5 & 7 & 8 \\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & 11 & -15 & 25 & 76 \ end (array) \ right) \ begin (array) (l) \ phantom (0) \\ \ phantom (0) \\\ phantom (0) \\ r_4 -r_3 \\ r_5 + r_2 \ end (array) \ rightarrow \ left (\ begin (array) (ccccc | c) 1 & -2 & 3 & 0 & 2 & 17 \\ 0 & 4 & 1 & -5 & 7 & 8 \\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \ end (array) \ right) $$

We have brought the extended system matrix and the system matrix itself to a stepped form. The rank of the extended matrix of the system is three, the rank of the matrix of the system is also three. Since the system contains $ n = 5 $ unknowns, i.e. $ \ rang \ widetilde (A) = \ rang (A) \ lt (n) $, then, according to the corollary to the Kronecker-Capelli theorem, this system is undefined, i.e. has an infinite number of solutions.

Answer: the system is undefined.

In the second part, we will analyze examples that are often included in typical calculations or tests in higher mathematics: a compatibility study and the solution of a SLAE, depending on the values ​​of the parameters included in it.

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 also 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... Explore 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 countless 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 but.

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. of 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

c) (xe + y "= 1, d) (x" + y "= 2a - 1,

(xy = a; (xy = a - 1?

9.198. Find the number of solutions of the system of equations ((x (+) y ~ = !,

depending on the parameter a.

9.199. How many solutions, depending on a, does the system of equations have:

a) (x "+ y" = 9, b) (x "+ y" +! Ox = 0,

(~ x ~ = y - a; (y = ~ x - a ~?

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

has three solutions? Find these solutions.

9.201. For what values ​​of the parameter p the system of equations

(p + x) (x - p US) = O

has three solutions?

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

a) 1 ~ x ~ +4) y ~ = b, b) 1 x ~ +2 ~ y (= 1, c) (~ y! + x = 4

! ~ y! + xr = 1! ~ y! + xz = b (x + y = b

has four different solutions?

9.208. For what values ​​of the parameter c the system of equations

has eight different solutions?

9.204. Solve the system of equations

where a) 0, and prove that if a is an integer, then for

of each solution (x; y) of this system, the number 1 + xy is the square of an integer.

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

x "+ y" + 2xy - bx - bu + 10 - a = O,

x "+ y" - 2xy - 2x + 2y + a = O

has at least one solution?

Solve the system for the found values ​​of a.

9.206. Find all values ​​of the parameter a for which the system

equations (x "+ (y - 2)" = 1, has at least one solution.

9.207. Find all the values ​​of the parameter a for which the circles x "+ q" = 1 and (x - a) "+ q" = 4 touch.

9.208. Find all the values ​​of the parameter a (a> O) at which the circles x "+ q" = 1 and (x - 3) "+ (q - 4)" = a "touch.

Find the coordinates of the touch point.

9.209. Find all values ​​of a (a> 0) for which the circle

x "+ d" = a "touches the straight line Zx + 4d = 12. Find the coordinates of the point of tangency.

D "- 2x + 4d = 21. Find the coordinates of the intersection points

line and circle.

9.211. At what value of the parameter a will the straight ed = x + 1 be

go through the center of the circle (x - 1) + (d - a) "= 8?

Find the coordinates of the intersection points of the line and the circle.

9 212. It is known that the straight line q = 12x - 9 and the parabola q = ax "have

only one common point. Find the coordinates of this point.

9.213. For what values ​​of b and r (b> 0, r> 0) does the circle

(x - 1) "+ (q - b)" = z "will touch the straight lines q = 0 and q = - x?

Find the coordinates of the touch points.

9.214. Draw on the coordinate plane a set of points with

coordinates (a; b) such that the system of equations

has at least one solution.

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

a (x "+ 1) = q - ~ x ~ + 1,

has the only solution?

9 1O. TEXT TASKS

As a rule, word problems are solved according to the following scheme: select unknowns; make up an equation or a system of equations, and in some problems - an inequality or a system of inequalities; solve the resulting system (sometimes it is enough to find some combination of unknowns from the system, and not solve it in the usual sense).