Solve a homogeneous system of linear algebraic equations examples. Fundamental decision system (specific example)

Systems of linear homogeneous equations- has the form ∑a k i x i = 0. where m > n or m A homogeneous system of linear equations is always consistent, since rangA = rangB. It obviously has a solution consisting of zeros, which is called trivial.

Purpose of the service. The online calculator is designed to find a non-trivial and fundamental solution to the SLAE. The resulting solution is saved in a Word file (see example solution).

Instructions. Select matrix dimension:

Properties of systems of linear homogeneous equations

In order for the system to have non-trivial solutions, it is necessary and sufficient that the rank of its matrix be less than the number of unknowns.

Theorem. A system in the case m=n has a nontrivial solution if and only if the determinant of this system is equal to zero.

Theorem. Any linear combination of solutions to a system is also a solution to that system.
Definition. The set of solutions to a system of linear homogeneous equations is called fundamental system of solutions, if this set consists of linearly independent solutions and any solution to the system is a linear combination of these solutions.

Theorem. If the rank r of the system matrix is ​​less than the number n of unknowns, then there exists a fundamental system of solutions consisting of (n-r) solutions.

Algorithm for solving systems of linear homogeneous equations

  1. Finding the rank of the matrix.
  2. We select the basic minor. We distinguish dependent (basic) and free unknowns.
  3. We cross out those equations of the system whose coefficients are not included in the basis minor, since they are consequences of the others (according to the theorem on the basis minor).
  4. We move the terms of the equations containing free unknowns to the right side. As a result, we obtain a system of r equations with r unknowns, equivalent to the given one, the determinant of which is nonzero.
  5. We solve the resulting system by eliminating unknowns. We find relationships expressing dependent variables through free ones.
  6. If the rank of the matrix is ​​not equal to the number of variables, then we find the fundamental solution of the system.
  7. In the case rang = n we have a trivial solution.

Example. Find the basis of the system of vectors (a 1, a 2,...,a m), rank and express the vectors based on the base. If a 1 =(0,0,1,-1), and 2 =(1,1,2,0), and 3 =(1,1,1,1), and 4 =(3,2,1 ,4), and 5 =(2,1,0,3).
Let's write down the main matrix of the system:


Multiply the 3rd line by (-3). Let's add the 4th line to the 3rd:
0 0 1 -1
0 0 -1 1
0 -1 -2 1
3 2 1 4
2 1 0 3

Multiply the 4th line by (-2). Let's multiply the 5th line by (3). Let's add the 5th line to the 4th:
Let's add the 2nd line to the 1st:
Let's find the rank of the matrix.
The system with the coefficients of this matrix is ​​equivalent to the original system and has the form:
- x 3 = - x 4
- x 2 - 2x 3 = - x 4
2x 1 + x 2 = - 3x 4
Using the method of eliminating unknowns, we find a nontrivial solution:
We obtained relations expressing the dependent variables x 1 , x 2 , x 3 through the free ones x 4 , that is, we found a general solution:
x 3 = x 4
x 2 = - x 4
x 1 = - x 4

A homogeneous system is always consistent and has a trivial solution
. For a nontrivial solution to exist, it is necessary that the rank of the matrix was less than the number of unknowns:

.

Fundamental system of solutions homogeneous system
call a system of solutions in the form of column vectors
, which correspond to the canonical basis, i.e. basis in which arbitrary constants
are alternately set equal to one, while the rest are set to zero.

Then the general solution of the homogeneous system has the form:

Where
- arbitrary constants. In other words, the overall solution is a linear combination of the fundamental system of solutions.

Thus, basic solutions can be obtained from the general solution if the free unknowns are given the value of one in turn, setting all others equal to zero.

Example. Let's find a solution to the system

Let's accept , then we get a solution in the form:

Let us now construct a fundamental system of solutions:

.

The general solution will be written as:

Solutions of a system of homogeneous linear equations have the following properties:

In other words, any linear combination of solutions to a homogeneous system is again a solution.

Solving systems of linear equations using the Gauss method

Solving systems of linear equations has interested mathematicians for several centuries. The first results were obtained in the 18th century. In 1750, G. Kramer (1704–1752) published his works on the determinants of square matrices and proposed an algorithm for finding the inverse matrix. In 1809, Gauss outlined a new solution method known as the method of elimination.

The Gauss method, or the method of sequential elimination of unknowns, consists in the fact that, using elementary transformations, a system of equations is reduced to an equivalent system of a step (or triangular) form. Such systems make it possible to sequentially find all unknowns in a certain order.

Let us assume that in system (1)
(which is always possible).

(1)

Multiplying the first equation one by one by the so-called suitable numbers

and adding the result of multiplication with the corresponding equations of the system, we obtain an equivalent system in which in all equations except the first there will be no unknown X 1

(2)

Let us now multiply the second equation of system (2) by suitable numbers, assuming that

,

and adding it with the lower ones, we eliminate the variable from all equations, starting from the third.

Continuing this process, after
step we get:

(3)

If at least one of the numbers
is not equal to zero, then the corresponding equality is contradictory and system (1) is inconsistent. Conversely, for any joint number system
are equal to zero. Number is nothing more than the rank of the matrix of system (1).

The transition from system (1) to (3) is called straight forward Gauss method, and finding the unknowns from (3) – in reverse .

Comment : It is more convenient to carry out transformations not with the equations themselves, but with the extended matrix of the system (1).

Example. Let's find a solution to the system

.

Let's write the extended matrix of the system:

.

Let's add the first one to lines 2,3,4, multiplied by (-2), (-3), (-2) respectively:

.

Let's swap rows 2 and 3, then in the resulting matrix add row 2 to row 4, multiplied by :

.

Add to line 4 line 3 multiplied by
:

.

It's obvious that
, therefore, the system is consistent. From the resulting system of equations

we find the solution by reverse substitution:

,
,
,
.

Example 2. Find a solution to the system:

.

It is obvious that the system is incompatible, because
, A
.

Advantages of the Gauss method :

    Less labor intensive than Cramer's method.

    Unambiguously establishes the compatibility of the system and allows you to find a solution.

    Makes it possible to determine the rank of any matrices.

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 the general form:

It is obvious that every homogeneous system is consistent and has a zero (trivial) solution. Therefore, when applied to homogeneous systems of linear equations, one often has to look for an answer to the question of the existence of nonzero 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 this unique 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 zero.

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

Kronecker-Capelli theorem: An SLU is consistent if and only if the rank of the system matrix is ​​equal to the rank of the extended matrix of this system. A system ur is called consistent if it has at least one solution.

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 terms are equal to 0. A 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 is any fundamental system of solutions, c1, c2,...,ck are 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 there is a system of linearly independent vectors in it, and any system of a larger number of vectors is linearly dependent. The number is called dimension (number of dimensions) linear space and is denoted by . In other words, the dimension of a space is the maximum number of linearly independent vectors of this space. If such a number exists, then the space is called finite-dimensional. If, for any natural number n, there is a system in space 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 a linearly dependent system (since this system consists of vectors of n-dimensional space). Using the property of 7 linearly dependent and linearly independent vectors, we obtain the conclusion of the theorem.

6.3. HOMOGENEOUS SYSTEMS OF LINEAR EQUATIONS

Let now in system (6.1).

A homogeneous system is always consistent. Solution () is called zero, or trivial.

A homogeneous system (6.1) has a nonzero solution if and only if its rank ( ) is less than the number of unknowns. In particular, a homogeneous system in which the number of equations is equal to the number of unknowns has a nonzero solution if and only if its determinant is zero.

Because this time everything, instead of formulas (6.6) we obtain the following:

(6.7)

Formulas (6.7) contain any solution of the homogeneous system (6.1).

1. The set of all solutions to the homogeneous system of linear equations (6.1) forms a linear space.

2. Linear spaceRall solutions of the homogeneous system of linear equations (6.1) withnunknowns and the rank of the main matrix equal tor, has dimensionn–r.

Any set of (n–r) linearly independent solutions of the homogeneous system (6.1) forms a basis in spaceRall decisions. It is called fundamental a set of solutions to the homogeneous system of equations (6.1). Special emphasis is placed on "normal" fundamental set of solutions of the homogeneous system (6.1):




(6.8)

By definition of the basis, any solution X homogeneous system (6.1) can be represented in the form

(6.9)

Where – arbitrary constants.

Since formula (6.9) contains any solution to the homogeneous system (6.1), it gives common decision this system.

Example.

Example 1. Find a general solution and some fundamental system of solutions for the system

Solution find using a calculator. The solution algorithm is the same as for systems of linear inhomogeneous 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.


The first and second lines are proportional, let’s cross out one of them:

.
Dependent variables – x 2, x 3, x 5, free – x 1, x 4. From the first equation 10x 5 = 0 we find x 5 = 0, then
; .
The general solution is:

We find a fundamental system of solutions, which consists of (n-r) solutions. In our case, n=5, r=3, therefore, the fundamental system of solutions consists of two solutions, and these solutions must be linearly independent. For the rows to be linearly independent, it is necessary and sufficient that the rank of the matrix composed of the elements of the rows be equal to the number of rows, that is, 2. It is enough to give the free unknowns x 1 and x 4 values ​​from the rows of the second-order determinant, nonzero, and calculate x 2 , x 3 , x 5 . The simplest non-zero determinant is .
So the first solution is: , second – .
These two decisions constitute a fundamental decision system. Note that the fundamental system is not unique (you can create as many nonzero determinants as you like).

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 of unknowns. 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:

5 -2 9 -4 -1
1 4 2 2 -5
6 2 11 -2 -6
x 1x 2x 3x 4x 5

Let's reduce the matrix to triangular form. We will work only with rows, since multiplying a matrix row by a number other than zero and adding it to another row for the system means multiplying the equation by the same number and adding it with another equation, which does not change the solution of the system.
Multiply the 2nd line by (-5). Let's add the 2nd line to the 1st:
0 -22 -1 -14 24
1 4 2 2 -5
6 2 11 -2 -6

Let's multiply the 2nd line by (6). Multiply the 3rd line by (-1). Let's add the 3rd line to the 2nd:
Let's find the rank of the matrix.
0 22 1 14 -24
6 2 11 -2 -6
x 1x 2x 3x 4x 5

The selected minor has the highest order (of possible minors) and is non-zero (it is equal to the product of the elements on the reverse diagonal), therefore rang(A) = 2.
This minor is basic. It includes coefficients for the unknowns x 1 , x 2 , which means that the unknowns x 1 , x 2 are dependent (basic), and x 3 , x 4 , x 5 are free.
Let's transform the matrix, leaving only the basis minor on the left.
0 22 14 -1 -24
6 2 -2 -11 -6
x 1x 2x 4x 3x 5

The system with the coefficients of this matrix is ​​equivalent to the original system and has the form:
22x 2 = 14x 4 - x 3 - 24x 5
6x 1 + 2x 2 = - 2x 4 - 11x 3 - 6x 5
Using the method of eliminating unknowns, we find non-trivial solution:
We obtained relations expressing the dependent variables x 1 , x 2 through the free ones x 3 , x 4 , x 5 , that is, we found common decision:
x 2 = 0.64x 4 - 0.0455x 3 - 1.09x 5
x 1 = - 0.55x 4 - 1.82x 3 - 0.64x 5
We find a fundamental system of solutions, which consists of (n-r) solutions.
In our case, n=5, r=2, therefore, the fundamental system of solutions consists of 3 solutions, and these solutions must be linearly independent.
For the rows to be linearly independent, it is necessary and sufficient that the rank of the matrix composed of row elements be equal to the number of rows, that is, 3.
It is enough to give the free unknowns x 3 , x 4 , x 5 values ​​from the lines of the 3rd order determinant, non-zero, and calculate x 1 , x 2 .
The simplest non-zero determinant is the identity matrix.
1 0 0
0 1 0
0 0 1

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

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