home biology tutor Is the statement true homogeneous system of linear equations. Homogeneous systems of equations. Work on lectures on topics

Is the statement true homogeneous system of linear equations. Homogeneous systems of equations. Work on lectures on topics

Solution of linear systems algebraic equations(SLAE) is undoubtedly the most important topic of the linear algebra course. A huge number of problems from all branches of mathematics are reduced to solving systems linear equations. These factors explain the reason for creating this article. The material of the article is selected and structured so that with its help you can

choose the optimal method for solving your system of linear algebraic equations,
study the theory of the chosen method,
solve your system of linear equations, having considered in detail the solutions of typical examples and problems.

Brief description of the material of the article.

Let's give it all first necessary definitions, concepts, and introduce notation.

Next, we consider methods for solving systems of linear algebraic equations in which the number of equations is equal to the number of unknown variables and which have a unique solution. First, let us dwell on Cramer's method, and secondly, we will show matrix method solving such systems of equations, thirdly, we will analyze the Gauss method (the method of successive elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in various ways.

After that, we turn to solving systems of linear algebraic equations of a general form, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is degenerate. We formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (in the case of their compatibility) using the concept of the basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions of the examples.

Be sure to dwell on the structure of the general solution of homogeneous and inhomogeneous systems of linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how to write common decision SLAE with the help of vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.

In conclusion, we consider systems of equations that are reduced to linear ones, as well as various problems, in the solution of which SLAEs arise.

Page navigation.

Definitions, concepts, designations.

We will consider systems of p linear algebraic equations with n unknown variables (p may be equal to n ) of the form

Unknown variables, - coefficients (some real or complex numbers), - free members (also real or complex numbers).

This form of SLAE is called coordinate.

AT matrix form this system of equations has the form ,
where - the main matrix of the system, - the matrix-column of unknown variables, - the matrix-column of free members.

If we add to the matrix A as the (n + 1)-th column the matrix-column of free terms, then we get the so-called expanded matrix systems of linear equations. Usually, the augmented matrix is denoted by the letter T, and the column of free members is separated by a vertical line from the rest of the columns, that is,

By solving a system of linear algebraic equations called a set of values of unknown variables , which turns all the equations of the system into identities. The matrix equation for the given values of the unknown variables also turns into an identity.

If a system of equations has at least one solution, then it is called joint.

If the system of equations has no solutions, then it is called incompatible.

If a SLAE has a unique solution, then it is called certain; if there is more than one solution, then - uncertain.

If the free terms of all equations of the system are equal to zero , then the system is called homogeneous, otherwise - heterogeneous.

Solution of elementary systems of linear algebraic equations.

If the number of system equations is equal to the number of unknown variables and the determinant of its main matrix is not equal to zero, then we will call such SLAEs elementary. Such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.

We began to study such SLAEs in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the addition method, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are essentially modifications of the Gauss method.

The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's sort them out.

Solving systems of linear equations by Cramer's method.

Let us need to solve a system of linear algebraic equations

in which the number of equations is equal to the number of unknown variables and the determinant of the main matrix of the system is different from zero, that is, .

Let be the determinant of the main matrix of the system, and are determinants of matrices that are obtained from A by replacing 1st, 2nd, …, nth column respectively to the column of free members:

With such notation, the unknown variables are calculated by the formulas of Cramer's method as . This is how the solution of a system of linear algebraic equations is found by the Cramer method.

Example.

Cramer method .

Decision.

The main matrix of the system has the form . Calculate its determinant (if necessary, see the article):

Since the determinant of the main matrix of the system is different from zero, the system has a unique solution that can be found by Cramer's method.

Compose and calculate the necessary determinants (the determinant is obtained by replacing the first column in matrix A with a column of free members, the determinant - by replacing the second column with a column of free members, - by replacing the third column of matrix A with a column of free members):

Finding unknown variables using formulas :

Answer:

The main disadvantage of Cramer's method (if it can be called a disadvantage) is the complexity of calculating the determinants when the number of system equations is more than three.

Solving systems of linear algebraic equations by the matrix method (using the inverse matrix).

Let the system of linear algebraic equations be given in matrix form , where the matrix A has dimension n by n and its determinant is nonzero.

Since , then the matrix A is invertible, that is, there is an inverse matrix . If we multiply both parts of the equality by on the left, then we get a formula for finding the column matrix of unknown variables. So we got the solution of the system of linear algebraic equations by the matrix method.

Example.

Solve System of Linear Equations matrix method.

Decision.

Let's rewrite the system of equations in matrix form:

As

then the SLAE can be solved by the matrix method. Using the inverse matrix, the solution to this system can be found as .

Let's build an inverse matrix using a matrix of algebraic complements of the elements of matrix A (if necessary, see the article):

It remains to calculate - the matrix of unknown variables by multiplying the inverse matrix on the matrix-column of free members (if necessary, see the article):

Answer:

or in another notation x 1 = 4, x 2 = 0, x 3 = -1.

The main problem in finding solutions to systems of linear algebraic equations by the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than the third.

Solving systems of linear equations by the Gauss method.

Suppose we need to find a solution to a system of n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.

The essence of the Gauss method consists in the successive exclusion of unknown variables: first, x 1 is excluded from all equations of the system, starting from the second, then x 2 is excluded from all equations, starting from the third, and so on, until only the unknown variable x n remains in the last equation. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called direct Gauss method. After the completion of the forward run of the Gaussian method, x n is found from the last equation, x n-1 is calculated from the penultimate equation using this value, and so on, x 1 is found from the first equation. The process of calculating unknown variables when moving from the last equation of the system to the first is called reverse Gauss method.

Let us briefly describe the algorithm for eliminating unknown variables.

We will assume that , since we can always achieve this by rearranging the equations of the system. We exclude the unknown variable x 1 from all equations of the system, starting from the second one. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, add the first multiplied by to the nth equation. The system of equations after such transformations will take the form

where , a .

We would come to the same result if we expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.

Next, we act similarly, but only with a part of the resulting system, which is marked in the figure

To do this, add the second multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, add the second multiplied by to the nth equation. The system of equations after such transformations will take the form

where , a . Thus, the variable x 2 is excluded from all equations, starting from the third.

Next, we proceed to the elimination of the unknown x 3, while acting similarly with the part of the system marked in the figure

So we continue the direct course of the Gauss method until the system takes the form

From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as , using the obtained value x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Example.

Solve System of Linear Equations Gaussian method.

Decision.

Let's exclude the unknown variable x 1 from the second and third equations of the system. To do this, to both parts of the second and third equations, we add the corresponding parts of the first equation, multiplied by and by, respectively:

Now we exclude x 2 from the third equation by adding to its left and right parts the left and right parts of the second equation, multiplied by:

On this, the forward course of the Gauss method is completed, we begin the reverse course.

From the last equation of the resulting system of equations, we find x 3:

From the second equation we get .

From the first equation we find the remaining unknown variable and this completes the reverse course of the Gauss method.

Answer:

X 1 \u003d 4, x 2 \u003d 0, x 3 \u003d -1.

Solving systems of linear algebraic equations of general form.

In the general case, the number of equations of the system p does not coincide with the number of unknown variables n:

Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations whose main matrix is square and degenerate.

Kronecker-Capelli theorem.

Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when SLAE is compatible, and when it is incompatible, gives Kronecker–Capelli theorem:
for a system of p equations with n unknowns (p can be equal to n ) to be consistent it is necessary and sufficient that the rank of the main matrix of the system is equal to the rank of the extended matrix, that is, Rank(A)=Rank(T) .

Let us consider the application of the Kronecker-Cappelli theorem for determining the compatibility of a system of linear equations as an example.

Example.

Find out if the system of linear equations has solutions.

Decision.

. Let us use the method of bordering minors. Minor of the second order different from zero. Let's go over the third-order minors surrounding it:

Since all bordering third-order minors are equal to zero, the rank of the main matrix is two.

In turn, the rank of the augmented matrix is equal to three, since the minor of the third order

different from zero.

Thus, Rang(A) , therefore, according to the Kronecker-Capelli theorem, we can conclude that the original system of linear equations is inconsistent.

Answer:

There is no solution system.

So, we have learned to establish the inconsistency of the system using the Kronecker-Capelli theorem.

But how to find the solution of the SLAE if its compatibility is established?

To do this, we need the concept of the basis minor of a matrix and the theorem on the rank of a matrix.

The highest order minor of the matrix A, other than zero, is called basic.

It follows from the definition of the basis minor that its order is equal to the rank of the matrix. For a non-zero matrix A, there can be several basic minors; there is always one basic minor.

For example, consider the matrix .

All third-order minors of this matrix are equal to zero, since the elements of the third row of this matrix are the sum of the corresponding elements of the first and second rows.

The following minors of the second order are basic, since they are nonzero

Minors are not basic, since they are equal to zero.

Matrix rank theorem.

If the rank of a matrix of order p by n is r, then all elements of the rows (and columns) of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding elements of the rows (and columns) that form the basis minor.

What does the matrix rank theorem give us?

If, by the Kronecker-Capelli theorem, we have established the compatibility of the system, then we choose any basic minor of the main matrix of the system (its order is equal to r), and exclude from the system all equations that do not form the chosen basic minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still redundant (according to the matrix rank theorem, they are a linear combination of the remaining equations).

As a result, after discarding the excessive equations of the system, two cases are possible.

If the number of equations r in the resulting system is equal to the number of unknown variables, then it will be definite and the only solution can be found by the Cramer method, the matrix method or the Gauss method.

Example.

Decision.

Rank of the main matrix of the system is equal to two, since the minor of the second order different from zero. Extended matrix rank is also equal to two, since the only minor of the third order is equal to zero

and the minor of the second order considered above is different from zero. Based on the Kronecker-Capelli theorem, one can assert the compatibility of the original system of linear equations, since Rank(A)=Rank(T)=2 .

As a basis minor, we take . It is formed by the coefficients of the first and second equations:

The third equation of the system does not participate in the formation of the basic minor, so we exclude it from the system based on the matrix rank theorem:

So we got elementary system linear algebraic equations. Let's solve it by Cramer's method:

Answer:

x 1 \u003d 1, x 2 \u003d 2.

If the number of equations r in the resulting SLAE less than number unknown variables n, then on the left-hand side of the equations we leave the terms that form the basic minor, and transfer the remaining terms to the right-hand side of the equations of the system with the opposite sign.

The unknown variables (there are r of them) remaining on the left-hand sides of the equations are called main.

Unknown variables (there are n - r of them) that ended up on the right side are called free.

Now we assume that the free unknown variables can take arbitrary values, while the r main unknown variables will be expressed in terms of the free unknown variables in a unique way. Their expression can be found by solving the resulting SLAE by the Cramer method, the matrix method, or the Gauss method.

Let's take an example.

Example.

Solve System of Linear Algebraic Equations .

Decision.

Find the rank of the main matrix of the system by the bordering minors method. Let us take a 1 1 = 1 as a non-zero first-order minor. Let's start searching for a non-zero second-order minor surrounding this minor:

So we found a non-zero minor of the second order. Let's start searching for a non-zero bordering minor of the third order:

Thus, the rank of the main matrix is three. The rank of the augmented matrix is also equal to three, that is, the system is consistent.

The found non-zero minor of the third order will be taken as the basic one.

For clarity, we show the elements that form the basis minor:

We leave the terms participating in the basic minor on the left side of the equations of the system, and transfer the rest with opposite signs to the right sides:

We give free unknown variables x 2 and x 5 arbitrary values, that is, we take , where are arbitrary numbers. In this case, the SLAE takes the form

We solve the obtained elementary system of linear algebraic equations by the Cramer method:

Hence, .

In the answer, do not forget to indicate free unknown variables.

Answer:

Where are arbitrary numbers.

Summarize.

To solve a system of linear algebraic equations of a general form, we first find out its compatibility using the Kronecker-Capelli theorem. If the rank of the main matrix is not equal to the rank of the extended matrix, then we conclude that the system is inconsistent.

If the rank of the main matrix is equal to the rank of the extended matrix, then we choose the basic minor and discard the equations of the system that do not participate in the formation of the chosen basic minor.

If the order of the basis minor is equal to the number unknown variables, then the SLAE has a unique solution that can be found by any method known to us.

If the order of the basis minor is less than the number of unknown variables, then on the left side of the equations of the system we leave the terms with the main unknown variables, transfer the remaining terms to the right sides and assign arbitrary values to the free unknown variables. From the resulting system of linear equations, we find the main unknowns method variables Cramer, matrix method or Gauss method.

Gauss method for solving systems of linear algebraic equations of general form.

Using the Gauss method, one can solve systems of linear algebraic equations of any kind without their preliminary investigation for compatibility. The process of successive elimination of unknown variables makes it possible to draw a conclusion about both the compatibility and inconsistency of the SLAE, and if a solution exists, it makes it possible to find it.

From the point of view of computational work, the Gaussian method is preferable.

Watch it detailed description and analyzed examples in the article Gauss method for solving systems of linear algebraic equations of general form.

Recording the general solution of homogeneous and inhomogeneous linear algebraic systems using the vectors of the fundamental system of solutions.

In this section, we will focus on joint homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.

Let's deal with homogeneous systems first.

Fundamental decision system A homogeneous system of p linear algebraic equations with n unknown variables is a set of (n – r) linearly independent solutions of this system, where r is the order of the basis minor of the main matrix of the system.

If we designate linearly independent solutions of a homogeneous SLAE as X (1) , X (2) , …, X (n-r) (X (1) , X (2) , …, X (n-r) are matrices columns of dimension n by 1 ) , then the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary constant coefficients С 1 , С 2 , …, С (n-r), that is, .

What does the term general solution of a homogeneous system of linear algebraic equations (oroslau) mean?

The meaning is simple: the formula sets everything possible solutions the original SLAE, in other words, taking any set of values of arbitrary constants С 1 , С 2 , …, С (n-r) , according to the formula we get one of the solutions of the original homogeneous SLAE.

Thus, if we find a fundamental system of solutions, then we can set all solutions of this homogeneous SLAE as .

Let us show the process of constructing a fundamental system of solutions for a homogeneous SLAE.

We choose the basic minor of the original system of linear equations, exclude all other equations from the system, and transfer to the right-hand side of the equations of the system with opposite signs all terms containing free unknown variables. Let's give the free unknown variables the values 1,0,0,…,0 and calculate the main unknowns by solving the resulting elementary system of linear equations in any way, for example, by the Cramer method. Thus, X (1) will be obtained - the first solution of the fundamental system. If we give the free unknowns the values 0,1,0,0,…,0 and calculate the main unknowns, then we get X (2) . Etc. If we give the free unknown variables the values 0,0,…,0,1 and calculate the main unknowns, then we get X (n-r) . This is how it will be built fundamental system solutions of a homogeneous SLAE and its general solution can be written in the form .

For inhomogeneous systems of linear algebraic equations, the general solution is represented as

Let's look at examples.

Example.

Find the fundamental system of solutions and the general solution of a homogeneous system of linear algebraic equations .

Decision.

The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let us find the rank of the main matrix by the method of fringing minors. As a nonzero minor of the first order, we take the element a 1 1 = 9 of the main matrix of the system. Find the bordering non-zero minor of the second order:

A minor of the second order, different from zero, is found. Let's go through the third-order minors bordering it in search of a non-zero one:

All bordering minors of the third order are equal to zero, therefore, the rank of the main and extended matrix is two. Let's take the basic minor. For clarity, we note the elements of the system that form it:

The third equation of the original SLAE does not participate in the formation of the basic minor, therefore, it can be excluded:

We leave the terms containing the main unknowns on the right-hand sides of the equations, and transfer the terms with free unknowns to the right-hand sides:

Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basic minor is two. To find X (1), we give the free unknown variables the values x 2 \u003d 1, x 4 \u003d 0, then we find the main unknowns from the system of equations
.

2.4.1. Definition. Let an inhomogeneous system of linear equations be given

Consider a homogeneous system

for which the matrix of coefficients coincides with the matrix of coefficients of system (2.4.1). Then system (2.4.2) is called reduced homogeneous system (2.4.1).

2.4.2. Theorem. The general solution of an inhomogeneous system is equal to the sum of some particular solution of the inhomogeneous system and the general solution of the reduced homogeneous system.

Thus, to find the general solution of the inhomogeneous system (2.4.1), it suffices:

1) Examine it for compatibility. In case of compatibility:

2) Find the general solution of this homogeneous system.

3) Find any particular solution to the original (non-homogeneous) one.

4) Having added the particular solution found and the general solution of the given one, find the general solution of the original system.

2.4.3. An exercise. Investigate the system for compatibility and, in the case of compatibility, find its general solution in the form of the sum of the quotient and the general reduced.

Decision. a) To solve the problem, we use the above scheme:

1) We examine the system for compatibility (by the method of bordering minors): The rank of the main matrix is 3 (see the solution of exercise 2.2.5, a), and the non-zero minor of maximum order is composed of elements of the 1st, 2nd, 4th rows and 1st, 3 th, 4th columns. To find the rank of the expanded matrix, we border it with the 3rd row and the 6th column of the expanded matrix: =0. Means, rg A =rg=3, and the system is consistent. In particular, it is equivalent to the system

2) Find a general solution X 0 reduced homogeneous of this system

X 0 ={(-2a - b ; a ; b ; b ; b ) | a , b Î R}

(see the solution of exercise 2.2.5, a)).

3) Find some particular solution x h of the original system . To do this, in system (2.4.3), which is equivalent to the original one, the free unknowns x 2 and x We set 5 equal, for example, to zero (these are the most convenient data):

and solve the resulting system: x 1 =- , x 3 =- , x 4=-5. Thus, (- ; 0; - ; -5; 0) ¾ is a particular solution of the system.

4) We find the general solution X n of the original system :

X n={x h }+X 0 ={(- ; 0; - ; -5; 0)} + {(-2a - b ; a ; b ; b ; b )}=

={(- -2a - b ; a ; - + b ; -5+b ; b )}.

Comment. Compare your answer with the second answer in example 1.2.1 c). To obtain an answer in the first form for 1.2.1 c), we take as basic unknowns x 1 , x 3 , x 5 (the minor for which is also not equal to zero), and as free ¾ x 2 and x 4 .

§3. Some applications.

3.1. On the question of matrix equations. We remind you that matrix equation over the field F is an equation in which some matrix over the field acts as an unknown F .

The simplest matrix equations are equations of the form

AX=B , XA =B (2.5.1)

where A , B ¾ given (known) matrices over field F , a X ¾ such matrices, when substituting which equations (2.5.1) turn into true matrix equalities. In particular, the matrix method of certain systems is reduced to solving a matrix equation.

When the matrices A in equations (2.5.1) are non-degenerate, they have solutions, respectively X =A B and X =BA .

In the case where at least one of the matrices on the left side of equations (2.5.1) is degenerate, this method no longer suitable, since the corresponding inverse matrix A does not exist. In this case, finding solutions to equations (2.5.1) reduces to solving systems.

But first, let's introduce some concepts.

The set of all solutions of the system is called common solution . An individual solution of an indefinite system, let's call it private decision .

3.1.1. Example. Decide matrix equation over the field R.

a) X = ; b) X = ; in) X = .

Decision. a) Since \u003d 0, then the formula X =A B not suitable for solving this equation. If in the work XA =B matrix A has 2 rows, then the matrix X has 2 columns. Number of lines X must match the number of rows B . So X has 2 lines. Thus, X ¾ is some second-order square matrix: X = . Substitute X into the original equation:

Multiplying the matrices on the left side of (2.5.2), we arrive at the equality

Two matrices are equal if and only if they have the same dimensions and their corresponding elements are equal. Therefore (2.5.3) is equivalent to the system

This system is equivalent to the system

Solving it, for example, by the Gauss method, we arrive at a set of solutions (5-2 b , b , -2d , d ), where b , d run independently of each other R. Thus, X = .

b) Similarly to a) we have X = and.

This system is inconsistent (check it out!). Therefore, this matrix equation has no solutions.

c) Denote this equation by AX =B . As A has 3 columns and B has 2 columns then X ¾ some 3´2 matrix: X = . Therefore, we have the following chain of equivalences:

We solve the last system using the Gauss method (we omit the comments)

Thus, we arrive at the system

whose solution is (11+8 z , 14+10z , z , -49+8w , -58+10w ,w ) where z , w run independently of each other R.

Answer: a) X = , b , d Î R.

b) There are no solutions.

in) X = z , w Î R.

3.2. On the question of permutability of matrices. In general, the product of matrices is nonpermutable, that is, if A and B such that AB and BA defined, then, generally speaking, AB ¹ BA . But the identity matrix example E shows that commutability is also possible AE =EA for any matrix A , if only AE and EA were determined.

In this subsection, we consider problems of finding the set of all matrices that commute with a given one. Thus,

Unknown x 1 , y 2 and z 3 can take any value: x 1 =a , y 2 =b , z 3 =g . Then

Thus, X = .

Answer. a) X d ¾ any number.

b) X ¾ set of matrices of the form , where a , b and g ¾ any numbers.

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; the Gaussian method is not suitable for systems with letter coefficients.

Consider other methods for solving systems of linear equations. These methods use the concept of the rank of a matrix and reduce the solution of 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 the fundamental system of solutions of the reduced homogeneous system and a particular solution of the inhomogeneous system.

1. We make a matrix A and the augmented matrix of the system (1)

2. Explore the system (1) for compatibility. 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 consistency study is based on the Kronecker-Capelli theorem).

a. We find rA.

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

M1=1≠0 (1 is taken from the upper left corner of the matrix BUT).

Bordering 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 М2′ second order.

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

(because the second and third lines are proportional).

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

b. We find .

Sufficiently basic minor М2′ matrices A border with a column of free members and all lines (we have only the last line).

. It follows from this that М3′′ remains the basis minor of the matrix https://pandia.ru/text/78/176/images/image019_33.gif" width="168 height=75" height="75"> (2)

As М2′- 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 М2′ is in the first two rows of matrix A).

(3)

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

In this system, two free unknowns ( x2 and x4 ). So FSR systems (4) consists of two solutions. To find them, we assign free unknowns to (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 by Cramer's rule or by any other method). Subtracting the first equation from the second equation, we get:

Her decision will be x1= -1 , x3=0 . Given the values x2 and x4 , which we have given, we obtain the first fundamental solution of the system (2) : .

Now we put 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 are arbitrary constants.

4. Find one private decision heterogeneous system(1) . As in paragraph 3 , instead of the system (1) consider the equivalent system (5) , consisting of the first two equations of the system (1) .

(5)

We transfer the free unknowns to the right-hand sides x2 and x4.

(6)

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

This system has a unique solution (because its determinant М2′0). Solving it (using the Cramer theorem or the Gauss 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 it remains to write general solution α of an inhomogeneous system(1) : it is equal to the sum private decision this system and general solution of its reduced homogeneous system (2) :

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

It means: (7)

6. Examination. To check if you have solved the system correctly (1) , we need a general solution (7) substitute in (1) . If each equation becomes an identity ( C1 and C2 should be destroyed), then the solution is found correctly.

We will substitute (7) for example, only in 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 other equations of the system (1) .

Comment. Verification is usually quite cumbersome. We can recommend the following "partial verification": in the overall solution of the system (1) assign some values to arbitrary constants and substitute the resulting particular solution only into the discarded equations (i.e., into those equations from (1) that are not included in (5) ). If you get identities, then more likely, solution of the system (1) found correctly (but such a check does not give a full 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 main unknowns in terms of free ones.

Decision. 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, which is equivalent to 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 parts as free members.

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Option 2.

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Option 4.

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Option 5.

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Option 6.

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A system of linear equations in which all free terms are equal to zero is called homogeneous :

Any homogeneous system is always consistent, since it always has zero (trivial ) solution. The question arises under what conditions a homogeneous system will have a non-trivial solution.

Theorem 5.2.A homogeneous system has a non-trivial solution if and only if the rank of the underlying matrix is less than the number of its unknowns.

Consequence. A square homogeneous system has a non-trivial solution if and only if the determinant of the main matrix of the system is not equal to zero.

Example 5.6. Determine the values of the parameter l for which the system has nontrivial solutions and find these solutions:

Decision. This system will have a non-trivial solution when the determinant of the main matrix is equal to zero:

Thus, the system is nontrivial when l=3 or l=2. For l=3, the rank of the main matrix of the system is 1. Then, leaving only one equation and assuming that y=a and z=b, we get x=b-a, i.e.

For l=2, the rank of the main matrix of the system is 2. Then, choosing as the basic minor:

we get a simplified system

From here we find that x=z/4, y=z/2. Assuming z=4a, we get

The set of all solutions of a homogeneous system has a very important linear property : if X columns 1 and X 2 - solutions of the homogeneous system AX = 0, then any linear combination of them a X 1+b X 2 will also be the solution of this system. Indeed, since AX 1 = 0 and AX 2 = 0 , then A(a X 1+b X 2) = a AX 1+b AX 2 = a · 0 + b · 0 = 0. Due to this property, if a linear system has more than one solution, then there will be infinitely many of these solutions.

Linearly Independent Columns E 1 , E 2 , E k, which are solutions of a homogeneous system, is called fundamental decision system homogeneous system of linear equations if the general solution of this system can be written as a linear combination of these columns:

If a homogeneous system has n variables, and the rank of the main matrix of the system is equal to r, then k = n-r.

Example 5.7. Find the fundamental system of solutions of the following system of linear equations:

Decision. Find the rank of the main matrix of the system:

Thus, the set of solutions of this system of equations forms a linear subspace of dimension n - r= 5 - 2 = 3. We choose as the basic minor

Then, leaving only the basic equations (the rest will be a linear combination of these equations) and the basic variables (we transfer the rest, the so-called free variables to the right), we get a simplified system of equations:

Assuming x 3 = a, x 4 = b, x 5 = c, we find

Assuming a= 1, b=c= 0, we obtain the first basic solution; assuming b= 1, a = c= 0, we obtain the second basic solution; assuming c= 1, a = b= 0, we obtain the third basic solution. As a result, the normal fundamental system of solutions takes the form

Using the fundamental system, the general solution of the homogeneous system can be written as

X = aE 1 + bE 2 + cE 3 . a

Let us note some properties of solutions of the inhomogeneous system of linear equations AX=B and their relationship with the corresponding homogeneous system of equations AX = 0.

General solution of an inhomogeneous systemis equal to the sum of the general solution of the corresponding homogeneous system AX = 0 and an arbitrary particular solution of the inhomogeneous system. Indeed, let Y 0 is an arbitrary particular solution of an inhomogeneous system, i.e. AY 0 = B, and Y is the general solution of an inhomogeneous system, i.e. AY=B. Subtracting one equality from the other, we get
A(Y-Y 0) = 0, i.e. Y-Y 0 is the general solution of the corresponding homogeneous system AX=0. Hence, Y-Y 0 = X, or Y=Y 0 + X. Q.E.D.

Let an inhomogeneous system have the form AX = B 1 + B 2 . Then the general solution of such a system can be written as X = X 1 + X 2 , where AX 1 = B 1 and AX 2 = B 2. This property expresses the universal property of any linear systems(algebraic, differential, functional, etc.). In physics, this property is called superposition principle, in electrical and radio engineering - overlay principle. For example, in the theory of linear electrical circuits, the current in any circuit can be obtained as an algebraic sum of the currents caused by each energy source separately.

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 certainly has a solution consisting of zeros, which is called trivial.

Service assignment. 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 solution example).

Instruction. Select the dimension of the matrix:

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. The system in the case m=n has a non-trivial 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 decision system if this collection consists of linearly independent solutions and any solution of 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 is a fundamental system of solutions consisting of (n-r) solutions.

Algorithm for solving systems of linear homogeneous equations

Find the rank of the matrix.
We select the basic minor. We select dependent (basic) and free unknowns.
We cross out those equations of the system whose coefficients were not included in the basic minor, since they are consequences of the rest (according to the basic minor theorem).
The terms of the equations containing free unknowns will be transferred 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 different from zero.
We solve the resulting system by eliminating the unknowns. We find relations expressing dependent variables in terms of free ones.
If the rank of the matrix is not equal to the number of variables, then we find the fundamental solution of the system.
In the case of 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 in terms of 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).
We write the main matrix of the system:

Multiply the 3rd row 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 row by (-2). Multiply the 5th row by (3). Let's add the 5th line to the 4th:
Let's add the 2nd line to the 1st:
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
2x1 + x2 = - 3x4
By the method of elimination of unknowns, we find a non-trivial solution:
We got relations expressing dependent variables x 1, x 2, x 3 through free x 4, that is, we found a general solution:
x 3 = x 4
x 2 = - x 4
x 1 = - x 4