Homogeneous dialing system. Homogeneous systems of equations

Homogeneous systems linear algebraic equations

As part of the lessons Gaussian method And Incompatible systems/systems with a common solution we considered heterogeneous systems linear equations , Where free member(which is usually on the right) at least one from the equations was different from zero.
And now, after a good warm-up with matrix rank, we will continue to polish the technique elementary transformations on homogeneous system of linear equations.
Based on the first paragraphs, the material may seem boring and mediocre, but this impression is deceptive. In addition to further development of technical techniques, there will be many new information, so please try not to neglect the examples in this article.

What is a homogeneous system of linear equations?

The answer suggests itself. A system of linear equations is homogeneous if the free term everyone system equations equal to zero. For example:

It is absolutely clear that a homogeneous system is always consistent, that is, it always has a solution. And, first of all, what catches your eye is the so-called trivial solution . Trivial, for those who do not understand the meaning of the adjective at all, means without a show-off. Not academic, of course, but intelligible =) ...Why beat around the bush, let's find out if this system has any other solutions:

Example 1

Solution: to solve a homogeneous system it is necessary to write system matrix and with the help of elementary transformations bring it to stepped view. Please note that here there is no need to write down the vertical bar and the zero column of free terms - after all, no matter what you do with zeros, they will remain zeros:

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

(2) The second line was added to the third line, multiplied by –1.

Dividing the third line by 3 doesn't make much sense.

As a result of elementary transformations, an equivalent homogeneous system is obtained , and, using the inverse of the Gaussian method, it is easy to verify that the solution is unique.

Answer:

Let us formulate an obvious criterion: a homogeneous system of linear equations has just a trivial solution, If system matrix rank(V in this case 3) equal to the number of variables (in this case – 3 pieces).

Let's warm up and tune our radio to the wave of elementary transformations:

Example 2

Solve a homogeneous system of linear equations

From the article How to find the rank of a matrix? remember rational technique parallel reduction of matrix numbers. Otherwise, you will have to cut large, and often biting fish. Approximate sample completing the assignment at the end of the lesson.

Zeros are good and convenient, but in practice the case is much more common when the rows of the system matrix linearly dependent. And then the inevitable appearance general solution:

Example 3

Solve a homogeneous system of linear equations

Solution: let’s write down the matrix of the system and, using elementary transformations, bring it to a stepwise form. The first action is aimed not only at obtaining a single value, but also at decreasing the numbers in the first column:

(1) A third line was added to the first line, multiplied by –1. The third line was added to the second line, multiplied by –2. At the top left I got a unit with a “minus”, which is often much more convenient for further transformations.

(2) The first two lines are the same, one of them was deleted. Honestly, I didn’t customize the solution - that’s how it turned out. If you perform transformations in a template way, then linear dependence lines would have been revealed a little later.

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

(4) The sign of the first line was changed.

As a result of elementary transformations, an equivalent system was obtained:

The algorithm works exactly the same as for heterogeneous systems. The variables “sitting on the steps” are the main ones, the variable that did not get a “step” is free.

Let's express the basic variables through a free variable:

Answer: general solution:

The trivial solution is included in general formula, and it is unnecessary to write it down separately.

The check is also carried out according to the usual scheme: the resulting general solution must be substituted into left side each equation of the system and obtain a legal zero for all substitutions.

It would be possible to finish this quietly and peacefully, but the solution to a homogeneous system of equations often needs to be represented V vector form by using fundamental system of solutions. Please forget about it for now analytical geometry, since now we will talk about vectors in the general algebraic sense, which I opened a little in the article about matrix rank. There is no need to gloss over the terminology, everything is quite simple.

The Gaussian method has a number of disadvantages: it is impossible to know whether the system is consistent or not until all the transformations necessary in the Gaussian method have been carried out; Gauss's method is not suitable for systems with letter coefficients.

Let's consider other methods for solving systems of linear equations. These methods use the concept of matrix rank and reduce the solution to any joint system to the solution of a system to which Cramer's rule applies.

Example 1. Find a general solution next system linear equations using a fundamental system of solutions to the reduced homogeneous system and a particular solution to the inhomogeneous system.

1. Making a matrix A and extended system matrix (1)

2. Explore the system (1) for togetherness. To do this, we find the ranks of the matrices A and https://pandia.ru/text/78/176/images/image006_90.gif" width="17" height="26 src=">). If it turns out that , then the system (1) incompatible. If we get that , then this system is consistent and we will solve it. (The compatibility study is based on the Kronecker-Capelli theorem).

a. We find rA.

To find rA, we will consider sequentially non-zero minors of the first, second, etc. orders of the matrix A and the minors surrounding them.

M1=1≠0 (take 1 from the left top corner matrices A).

We border M1 the second row and second column of this matrix. . We continue to border M1 the second line and the third column..gif" width="37" height="20 src=">. Now we border the non-zero minor M2′ second order.

We have: (since the first two columns are the same)

(since the second and third lines are proportional).

We see that rA=2, a is the basis minor of the matrix A.

b. We find.

Fairly basic minor M2′ matrices A border with a column of free terms and all rows (we have only the last row).

. It follows that M3′′ remains the basic minor of the matrix https://pandia.ru/text/78/176/images/image019_33.gif" width="168 height=75" height="75"> (2)

Because M2′- basis minor of the matrix A systems (2) , then this system is equivalent to the system (3) , consisting of the first two equations of the system (2) (for M2′ is in the first two rows of matrix A).

(3)

Since the basic minor https://pandia.ru/text/78/176/images/image021_29.gif" width="153" height="51"> (4)

In this system there are two free unknowns ( x2 And x4 ). That's why FSR systems (4) consists of two solutions. To find them, we assign free unknowns in (4) values ​​first x2=1 , x4=0 , and then - x2=0 , x4=1 .

At x2=1 , x4=0 we get:

.

This system already has the only thing solution (it can be found using Cramer's rule or any other method). Subtracting the first from the second equation, we get:

Her solution will be x1= -1 , x3=0 . Given the values x2 And x4 , which we gave, we get the first fundamental solution systems (2) : .

Now we believe in (4) x2=0 , x4=1 . We get:

.

We solve this system using Cramer’s theorem:

.

We obtain the second fundamental solution of the system (2) : .

Solutions β1 , β2 and make up FSR systems (2) . Then its general solution will be

γ= C1 β1+С2β2=С1(‑1, 1, 0, 0)+С2(5, 0, 4, 1)=(‑С1+5С2, С1, 4С2, С2)

Here C1 , C2 – arbitrary constants.

4. Let's find one private solution heterogeneous system(1) . As in paragraph 3 , instead of the system (1) Let's consider an equivalent system (5) , consisting of the first two equations of the system (1) .

(5)

Let us move the free unknowns to the right sides x2 And x4.

(6)

Let's give free unknowns x2 And x4 arbitrary values, for example, x2=2 , x4=1 and put them in (6) . Let's get the system

This system has a unique solution (since its determinant M2′0). Solving it (using Cramer’s theorem or Gauss’s method), we obtain x1=3 , x3=3 . Given the values ​​of the free unknowns x2 And x4 , we get particular solution of an inhomogeneous system(1)α1=(3,2,3,1).

5. Now all that remains is to write it down general solution α of an inhomogeneous system(1) : it is equal to the sum private solution this system and general solution of its reduced homogeneous system (2) :

α=α1+γ=(3, 2, 3, 1)+(‑С1+5С2, С1, 4С2, С2).

This means: (7)

6. Examination. To check if you solved the system correctly (1) , we need a general solution (7) substitute in (1) . If each equation turns into the identity ( C1 And C2 must be destroyed), then the solution is found correctly.

We'll substitute (7) for example, only the last equation of the system (1) (x1 + x2 + x3 ‑9 x4 =‑1) .

We get: (3–С1+5С2)+(2+С1)+(3+4С2)–9(1+С2)=–1

(С1–С1)+(5С2+4С2–9С2)+(3+2+3–9)=–1

Where –1=–1. We got an identity. We do this with all the other equations of the system (1) .

Comment. The check is usually quite cumbersome. The following “partial check” can be recommended: in the general solution of the system (1) assign some values ​​to arbitrary constants and substitute the resulting partial solution only into the discarded equations (i.e., into those equations from (1) , which were not included in (5) ). If you get identities, then more likely, system solution (1) found correctly (but such a check does not provide a complete guarantee of correctness!). For example, if in (7) put C2=- 1 , C1=1, then we get: x1=-3, x2=3, x3=-1, x4=0. Substituting into the last equation of system (1), we have: - 3+3 - 1 - 9∙0= - 1 , i.e. –1=–1. We got an identity.

Example 2. Find a general solution to a system of linear equations (1) , expressing the basic unknowns in terms of free ones.

Solution. As in example 1, compose matrices A and https://pandia.ru/text/78/176/images/image010_57.gif" width="156" height="50"> of these matrices. Now we leave only those equations of the system (1) , the coefficients of which are included in this basic minor (i.e., we have the first two equations) and consider the system consisting of them, equivalent system (1).

Let us transfer the free unknowns to the right-hand sides of these equations.

system (9) We solve by the Gaussian method, considering the right-hand sides as free terms.

https://pandia.ru/text/78/176/images/image035_21.gif" width="202 height=106" height="106">

Option 2.

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Option 4.

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Option 5.

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Option 6.

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Homogeneous system of linear equations over a field

DEFINITION. The fundamental system of solutions to the system of equations (1) is called a non-empty linear independent system its solutions, the linear span of which coincides with the set of all solutions of system (1).

Note that a homogeneous system of linear equations that has only a zero solution does not have a fundamental system of solutions.

PROPOSAL 3.11. Any two fundamental systems of solutions to a homogeneous system of linear equations consist of the same number decisions.

Proof. In fact, any two fundamental systems of solutions to the homogeneous system of equations (1) are equivalent and linearly independent. Therefore, by Proposition 1.12, their ranks are equal. Consequently, the number of solutions included in one fundamental system is equal to the number of solutions included in any other fundamental system of solutions.

If the main matrix A of the homogeneous system of equations (1) is zero, then any vector from is a solution to system (1); in this case, any collection is linear independent vectors of is a fundamental system of solutions. If the column rank of matrix A is equal to , then system (1) has only one solution - zero; therefore, in this case, the system of equations (1) does not have a fundamental system of solutions.

THEOREM 3.12. If the rank of the main matrix of a homogeneous system of linear equations (1) less number variables , then system (1) has a fundamental solution system consisting of solutions.

Proof. If the rank of the main matrix A of the homogeneous system (1) is equal to zero or , then it was shown above that the theorem is true. Therefore, below it is assumed that Assuming , we will assume that the first columns of matrix A are linearly independent. In this case, matrix A is rowwise equivalent to the reduced stepwise matrix, and system (1) is equivalent to the following reduced stepwise system of equations:

It is easy to check that any system of free values system variables(2) corresponds to one and only one solution to system (2) and, therefore, to system (1). In particular, only the zero solution of system (2) and system (1) corresponds to a system of zero values.

In system (2) we will assign one of the free variables value, equal to 1, and the remaining variables have zero values. As a result, we obtain solutions to the system of equations (2), which we write in the form of rows of the following matrix C:

The row system of this matrix is ​​linearly independent. Indeed, for any scalars from the equality

equality follows

and, therefore, equality

Let us prove that the linear span of the system of rows of the matrix C coincides with the set of all solutions to system (1).

Arbitrary solution of system (1). Then the vector

is also a solution to system (1), and

Solution find using a calculator. The solution algorithm is the same as for systems of linear not homogeneous equations.
Operating only with rows, we find the rank of the matrix, the basis minor; We declare dependent and free unknowns and find a general solution.

Example 2. Find the general solution and fundamental system of solutions of the system
Solution.



,
it follows that the rank of the matrix is ​​3 and equal to the number unknown. This means that the system does not have free unknowns, and therefore has a unique solution - a trivial one.

Exercise . Explore and solve a system of linear equations.
Example 4

Exercise . Find the general and particular solutions of each system.
Solution. Let's write down the main matrix of the system:

Task . Find fundamental set solutions of a homogeneous system of linear equations.

The linear equation is called homogeneous, if its free term is equal to zero, and inhomogeneous otherwise. A system consisting of homogeneous equations is called homogeneous and has general view:

It is obvious that every homogeneous system is consistent and has a zero (trivial) solution. Therefore, in relation to homogeneous systems of linear equations, one often has to look for an answer to the question of the existence of non-zero solutions. The answer to this question can be formulated as the following theorem.

Theorem . A homogeneous system of linear equations has a nonzero solution if and only if its rank is less than the number of unknowns .

Proof: Let us assume that a system whose rank is equal has a non-zero solution. Obviously it does not exceed . In case the system has a unique solution. Since a system of homogeneous linear equations always has a zero solution, then the zero solution will be exactly that the only solution. Thus, non-zero solutions are possible only for .

Corollary 1 : A homogeneous system of equations, in which the number of equations is less than the number of unknowns, always has a non-zero solution.

Proof: If a system of equations has , then the rank of the system does not exceed the number of equations, i.e. . Thus, the condition is satisfied and, therefore, the system has a non-zero solution.

Corollary 2 : A homogeneous system of equations with unknowns has a nonzero solution if and only if its determinant is equal to zero.

Proof: Let us assume that a system of linear homogeneous equations, the matrix of which with determinant , has a nonzero solution. Then, according to the proven theorem, and this means that the matrix is ​​singular, i.e. .

Homogeneous system of linear algebraic equations.

A system of m linear equations with n variables is called a system of linear homogeneous equations if all free members are equal to 0. The system of linear homogeneous equations is always consistent, because it always has at least a zero solution. A system of linear homogeneous equations has a non-zero solution if and only if the rank of its matrix of coefficients for variables is less than the number of variables, i.e. for rank A (n. Any linear combination

Lin system solutions. homogeneous. ur-ii is also a solution to this system.

A system of linear independent solutions e1, e2,...,еk is called fundamental if each solution of the system is a linear combination of solutions. Theorem: if the rank r of the matrix of coefficients for the variables of a system of linear homogeneous equations is less than the number of variables n, then every fundamental system of solutions to the system consists of n-r solutions. Therefore, the general solution of the linear system. one-day ur-th has the form: c1e1+c2e2+...+skek, where e1, e2,..., ek – any fundamental system of solutions, c1, c2,..., ck – arbitrary numbers and k=n-r. The general solution of a system of m linear equations with n variables is equal to the sum

of the general solution of the system corresponding to it is homogeneous. linear equations and an arbitrary particular solution of this system.

7. Linear spaces. Subspaces. Basis, dimension. Linear shell. Linear space is called n-dimensional, if it contains a system of linearly independent vectors, and any system of more vectors are linearly dependent. The number is called dimension (number of dimensions) linear space and is designated . In other words, the dimension of space is maximum number linearly independent vectors of this space. If such a number exists, then the space is called finite-dimensional. If for anyone natural number n in space there is a system consisting of linearly independent vectors, then such a space is called infinite-dimensional (written: ). In what follows, unless otherwise stated, finite-dimensional spaces will be considered.

The basis of an n-dimensional linear space is an ordered collection of linearly independent vectors ( basis vectors).

Theorem 8.1 on the expansion of a vector in terms of a basis. If is the basis of an n-dimensional linear space, then any vector can be represented as a linear combination of basis vectors:

V=v1*e1+v2*e2+…+vn+en
and, moreover, in the only way, i.e. the coefficients are determined uniquely. In other words, any vector of space can be expanded into a basis and, moreover, in a unique way.

Indeed, the dimension of space is . The system of vectors is linearly independent (this is a basis). After adding any vector to the basis, we obtain linearly dependent system(since this system consists of vectors n-dimensional space). Using the property of 7 linearly dependent and linearly independent vectors, we obtain the conclusion of the theorem.