Solving arbitrary and homogeneous systems of linear equations. How to find a nontrivial and fundamental solution to a system of linear homogeneous equations

System m linear equations c n called unknowns system of linear homogeneous equations if all free terms are equal to zero. Such a system looks like:

Where and ij (i = 1, 2, …, m; j = 1, 2, …, n) - given numbers; x i– unknown.

Linear system homogeneous equations always joint, because r(A) = r(). It always has at least zero ( trivial) solution (0; 0; …; 0).

Let us consider under what conditions homogeneous systems have non-zero solutions.

Theorem 1. A system of linear homogeneous equations has nonzero solutions if and only if the rank of its main matrix is r less number unknown n, i.e. r < n.

□ 1). Let a system of linear homogeneous equations have a nonzero solution. Since the rank cannot exceed the size of the matrix, then, obviously, r ≤ n. Let r = n. Then one of the minor sizes n n different from zero. Therefore, the corresponding system of linear equations has only decision: . . . This means that there are no other solutions other than trivial ones. So if there is non-trivial solution, That r < n.

2). Let r < n. Then the homogeneous system, being consistent, is uncertain. So she has infinite set decisions, i.e. has non-zero solutions. ■

Consider a homogeneous system n linear equations c n unknown:

(2)

Theorem 2. Homogeneous system n linear equations c n unknowns (2) has non-zero solutions if and only if its determinant equal to zero: = 0.

□ If system (2) has a non-zero solution, then = 0. Because when the system has only a single zero solution. If = 0, then the rank r the main matrix of the system is less than the number of unknowns, i.e. r < n. And, therefore, the system has an infinite number of solutions, i.e. has non-zero solutions. ■

Let us denote the solution of system (1) X 1 = k 1 , X 2 = k 2 , …, x n = k n as a string .

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

1. If the line is a solution to system (1), then the line is a solution to system (1).

2. If the lines And - solutions of system (1), then for any values With 1 and With 2 their linear combination is also a solution to system (1).

The validity of these properties can be verified by directly substituting them into the equations of the system.

From the formulated properties it follows that any linear combination of solutions to a system of linear homogeneous equations is also a solution to this system.

System of linearly independent solutions e 1 , e 2 , …, e r called fundamental, if each solution of system (1) is a linear combination of these solutions e 1 , e 2 , …, e r.

Theorem 3. If rank r coefficient matrices for system variables linear homogeneous equations (1) are less than the number of variables n, then any fundamental system solutions to system (1) consists of n–r decisions.

That's why common decision system of linear homogeneous equations (1) has the form:

Where e 1 , e 2 , …, e r– any fundamental system of solutions to system (9), With 1 , With 2 , …, with p – arbitrary numbers, R = n–r.

Theorem 4. General solution of the system m linear equations c n unknowns is equal to the sum of the general solution of the corresponding system of linear homogeneous equations (1) and an arbitrary particular solution of this system (1).

Example. Solve the system

Solution. For this system m = n= 3. Determinant

by Theorem 2, the system has only a trivial solution: x = y = z = 0.

Example. 1) Find general and particular solutions of the system

2) Find the fundamental system of solutions.

Solution. 1) For this system m = n= 3. Determinant

by Theorem 2, the system has nonzero solutions.

Since there is only one independent equation in the system

x + y – 4z = 0,

then from it we will express x =4z- y. Where do we get an infinite number of solutions: (4 z- y, y, z) – this is the general solution of the system.

At z= 1, y= -1, we get one particular solution: (5, -1, 1). Putting z= 3, y= 2, we get the second particular solution: (10, 2, 3), etc.

2) In the general solution (4 z- y, y, z) variables y And z are free, and the variable X- dependent on them. In order to find a fundamental system of solutions, we assign free variable values: at first y = 1, z= 0, then y = 0, z= 1. We obtain partial solutions (-1, 1, 0), (4, 0, 1), which form the fundamental system of solutions.

Illustrations:

Rice. 1 Classification of systems of linear equations

Rice. 2 Study of systems of linear equations

Presentations:

· Solution SLAE_matrix method

· Solution SLAE_Cramer method

· Solution SLAE_Gauss method

· Solution packages mathematical problems Mathematica, MathCad: search for analytical and numerical solution systems of linear equations

Control questions :

1. Define a linear equation

2. What type of system does it look like? m linear equations with n unknown?

3. What is called solving systems of linear equations?

4. What systems are called equivalent?

5. Which system is called incompatible?

6. What system is called joint?

7. Which system is called definite?

8. Which system is called indefinite

9. List the elementary transformations of systems of linear equations

10. List the elementary transformations of matrices

11. Formulate a theorem on the application of elementary transformations to a system of linear equations

12. What systems can be solved matrix method?

13. What systems can be solved by Cramer's method?

14. What systems can be solved by the Gauss method?

15. List 3 possible cases, arising when solving systems of linear equations using the Gauss method

16. Describe the matrix method for solving systems of linear equations

17. Describe Cramer’s method for solving systems of linear equations

18. Describe Gauss’s method for solving systems of linear equations

19. What systems can be solved using inverse matrix?

20. List 3 possible cases that arise when solving systems of linear equations using the Cramer method

Literature:

1. Higher mathematics for economists: Textbook for universities / N.Sh. Kremer, B.A. Putko, I.M. Trishin, M.N. Friedman. Ed. N.Sh. Kremer. – M.: UNITY, 2005. – 471 p.

2. General course Higher mathematics for economists: Textbook. / Ed. IN AND. Ermakova. –M.: INFRA-M, 2006. – 655 p.

3. Collection of problems in higher mathematics for economists: Tutorial/ Edited by V.I. Ermakova. M.: INFRA-M, 2006. – 574 p.

4. Gmurman V. E. Guide to solving problems in probability theory and magmatic statistics. - M.: graduate School, 2005. – 400 p.

5. Gmurman. V.E Probability theory and math statistics. - M.: Higher School, 2005.

6. Danko P.E., Popov A.G., Kozhevnikova T.Ya. Higher mathematics in exercises and problems. Part 1, 2. – M.: Onyx 21st century: Peace and Education, 2005. – 304 p. Part 1; – 416 p. Part 2.

7. Mathematics in economics: Textbook: In 2 parts / A.S. Solodovnikov, V.A. Babaytsev, A.V. Brailov, I.G. Shandara. – M.: Finance and Statistics, 2006.

8. Shipachev V.S. Higher mathematics: Textbook for students. universities - M.: Higher School, 2007. - 479 p.

Related information.

To understand what it is fundamental decision system you can watch a video tutorial for the same example by clicking. Now let's move on to the description of the whole necessary work. This will help you understand the essence of this issue in more detail.

How to find the fundamental system of solutions to a linear equation?

Let's take for example the following system of linear equations:

Let's find a solution to this linear system equations To begin with, we you need to write out the coefficient matrix of the system.

Let's transform this matrix to a triangular one. We rewrite the first line without changes. And all the elements that are under $a_(11)$ must be made zeros. To make a zero in place of the element $a_(21)$, you need to subtract the first from the second line, and write the difference in the second line. To make a zero in place of the element $a_(31)$, you need to subtract the first from the third line and write the difference in the third line. To make a zero in place of the element $a_(41)$, you need to subtract the first multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(31)$, you need to subtract the first multiplied by 2 from the fifth line and write the difference in the fifth line.

We rewrite the first and second lines without changes. And all the elements that are under $a_(22)$ must be made zeros. To make a zero in place of the element $a_(32)$, you need to subtract the second one multiplied by 2 from the third line and write the difference in the third line. To make a zero in place of the element $a_(42)$, you need to subtract the second multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(52)$, you need to subtract the second multiplied by 3 from the fifth line and write the difference in the fifth line.

We see that the last three lines are the same, so if you subtract the third from the fourth and fifth, they will become zero.

According to this matrix write down new system equations.

We see that we have only three linearly independent equations, and five unknowns, so the fundamental system of solutions will consist of two vectors. So we we need to move the last two unknowns to the right.

Now, we begin to express those unknowns that are on the left side through those that are on the right side. We start with the last equation, first we express $x_3$, then we substitute the resulting result into the second equation and express $x_2$, and then into the first equation and here we express $x_1$. Thus, we expressed all the unknowns that are on the left side through the unknowns that are on the right side.

Then instead of $x_4$ and $x_5$, we can substitute any numbers and find $x_1$, $x_2$ and $x_3$. Each five of these numbers will be the roots of our original system of equations. To find the vectors that are included in FSR we need to substitute 1 instead of $x_4$, and substitute 0 instead of $x_5$, find $x_1$, $x_2$ and $x_3$, and then vice versa $x_4=0$ and $x_5=1$.

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

• pick up optimal method solutions to your system of linear algebraic equations,
• study the theory of the chosen method,
• solve your system of linear equations by reviewing detailed solutions typical examples and tasks.

Brief description of the article material.

First let's give everything necessary definitions, concepts and introduce notations.

Next, we will 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. Firstly, we will focus on Cramer’s method, secondly, we will show the matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of sequential elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in different ways.

After this, we will move on to solving systems of linear algebraic equations general view, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is singular. Let us formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (if they are compatible) using the concept of a basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions to the examples.

We will definitely dwell on the structure of the general solution of homogeneous and heterogeneous systems linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how the general solution of a SLAE is written using the vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.

In conclusion, we will consider systems of equations that can be reduced to linear ones, as well as various tasks, 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 can be equal to n) of the form

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

This form of recording SLAE is called coordinate.

IN matrix form writing this system of equations has the form,
Where - main matrix of the system, - column matrix of unknown variables, - column matrix free members.

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

Solving a system of linear algebraic equations called a set of values ​​of unknown variables that turns all equations of the system into identities. The matrix equation for given values ​​of the unknown variables also becomes an identity.

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

If a system of equations has no solutions, then it is called non-joint.

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.

Solving elementary systems of linear algebraic equations.

If the number of equations of a system is equal to the number of unknown variables and the determinant of its main matrix is ​​not equal to zero, then such SLAEs will be called 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 following 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 using Cramer's method.

Suppose we 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 - determinants of matrices that are obtained from A by replacement 1st, 2nd, …, nth column respectively to the column of free members:

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

Example.

Cramer's method .

Solution.

The main matrix of the system has the form . Let's calculate its determinant (if necessary, see the article):

Since the determinant of the main matrix of the system is nonzero, the system has a unique solution that can be found by Cramer’s method.

Let's compose and calculate the necessary determinants (we obtain the determinant by replacing the first column in matrix A with a column of free terms, the determinant by replacing the second column with a column of free terms, and by replacing the third column of matrix A with a column of free terms):

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 determinants when the number of equations in the system is more than three.

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

Let a 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 , matrix A is invertible, that is, there is an inverse matrix. If we multiply both sides of the equality by the left, we get a formula for finding a matrix-column of unknown variables. This is how we obtained a solution to a system of linear algebraic equations using the matrix method.

Example.

Solve system of linear equations matrix method.

Solution.

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

Because

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

Let's construct the inverse matrix using the matrix from algebraic additions elements of matrix A (if necessary, see article):

It remains to calculate the matrix of unknown variables by multiplying the inverse matrix to a 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 when finding solutions to systems of linear algebraic equations using the matrix method is the complexity of finding the inverse matrix, especially for square matrices order higher than third.

Solving systems of linear equations using 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 of sequentially eliminating 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. This process of transforming system equations to sequentially eliminate unknown variables is called direct Gaussian method. After completing the forward stroke of the Gaussian method, x n is found from the last equation, using this value from the penultimate equation, x n-1 is calculated, 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 inverse of the Gaussian 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. Let's eliminate the unknown variable x 1 from all equations of the system, starting with the second. To do this, to the second equation of the system we add the first, multiplied by , to the third equation we add the first, multiplied by , and so on, to the nth equation we add the first, multiplied by . The system of equations after such transformations will take the form

where and .

We would have arrived at the same result if we had 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 proceed in a similar way, but only with part of the resulting system, which is marked in the figure

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

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

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

So we continue the direct progression of the Gaussian method until the system takes the form

From this moment we begin the reverse of the Gaussian method: we calculate x n from the last equation as , using the obtained value of 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 Gauss method.

Solution.

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

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

This completes the forward stroke of the Gauss method; we begin the reverse stroke.

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 thereby complete the reverse of the Gauss method.

Answer:

X 1 = 4, x 2 = 0, x 3 = -1.

Solving systems of linear algebraic equations of general form.

IN 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 singular.

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 inconsistent is given by Kronecker–Capelli theorem:
In order 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 be equal to the rank of the extended matrix, that is, Rank(A)=Rank(T).

Let us consider, as an example, the application of the Kronecker–Capelli theorem to determine the compatibility of a system of linear equations.

Example.

Find out whether the system of linear equations has solutions.

Solution.

. Let's use the method of bordering minors. Minor of the second order different from zero. Let's look at the third-order minors bordering it:

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

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

different from zero.

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

Answer:

The system has no solutions.

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

But how to find a solution to an SLAE if its compatibility is established?

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

The minor of the highest order of the matrix A, different from zero, is called basic.

From the definition of a basis minor it follows that its order is equal to the rank of the matrix. For a non-zero matrix A there can be several basis minors; there is always one basis 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 second-order minors are basic, since they are non-zero

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 equal to r, then all row (and column) elements of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding row (and column) elements forming the basis minor.

What does the matrix rank theorem tell us?

If, according to the Kronecker–Capelli theorem, we have established the compatibility of the system, then we choose any basis 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 selected basis 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 unnecessary 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.

.

Solution.

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

and the second-order minor considered above is different from zero. Based on the Kronecker–Capelli theorem, we 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 basis minor, so we exclude it from the system based on the theorem on the rank of the matrix:


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


Answer:

x 1 = 1, x 2 = 2.

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

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

Unknown variables (there are n - r pieces) that are on the right sides are called free.

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

Let's look at it with an example.

Example.

Solve a system of linear algebraic equations .

Solution.

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


This is how 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 extended matrix is ​​also equal to three, that is, the system is consistent.

We take the found non-zero minor of the third order as the basis one.

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


We leave the terms involved in the basis minor on the left side of the equations of the system, and transfer the rest from opposite signs to the right sides:


Let's give the free unknown variables x 2 and x 5 arbitrary values, that is, we accept , where are arbitrary numbers. In this case, the SLAE will take the form


Let us solve the resulting elementary system of linear algebraic equations using Cramer’s method:


Hence, .

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

Answer:

Where are arbitrary numbers.

Summarize.

To solve a system of general linear algebraic equations, we first determine 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 incompatible.

If the rank of the main matrix is ​​equal to the rank of the extended matrix, then we select a basis minor and discard the equations of the system that do not participate in the formation of the selected basis minor.

If the order of the basis minor equal to the number unknown variables, then the SLAE has a unique solution, which we find 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 system equations we leave the terms with the main unknown variables, transfer the remaining terms to the right sides and give arbitrary values ​​to the free unknown variables. From the resulting system of linear equations we find the main unknowns variables by method Cramer, matrix method or Gaussian method.

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

The Gauss method can be used to solve systems of linear algebraic equations of any kind without first testing them for consistency. The process of sequential elimination of unknown variables makes it possible to draw a conclusion about both the compatibility and incompatibility of the SLAE, and if a solution exists, it makes it possible to find it.

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

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

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

In this section we will talk about simultaneous homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.

Let us first deal with homogeneous systems.

Fundamental system of solutions homogeneous system of p linear algebraic equations with n unknown variables is a collection 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 denote linearly independent solutions homogeneous SLAE as X (1) , X (2) , …, X (n-r) (X (1) , X (2) , …, X (n-r) are columnar matrices 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 C 1, C 2, ..., C (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 C 1, C 2, ..., C (n-r), according to the formula we will obtain one of the solutions to the original homogeneous SLAE.

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

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

We select the basis minor of the original system of linear equations, exclude all other equations from the system and transfer all terms containing free unknown variables to the right-hand sides of the system equations with opposite signs. 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, using the Cramer method. This will result in X (1) - 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, we get X (2) . And so on. If we assign the values ​​0.0,…,0.1 to the free unknown variables and calculate the main unknowns, we obtain X (n-r) . In this way, a fundamental system of solutions to a homogeneous SLAE will be constructed and its general solution can be written in the form .

For inhomogeneous systems of linear algebraic equations, the general solution is represented in the form , where is the general solution of the corresponding homogeneous system, and is the particular solution of the original inhomogeneous SLAE, which we obtain by giving the free unknowns the values ​​0,0,...,0 and calculating the values ​​of the main unknowns.

Let's look at examples.

Example.

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

Solution.

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

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

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

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

We leave the terms containing the main unknowns on the right sides of the equations, and transfer the terms with free unknowns to the right 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 basis minor is equal to two. To find X (1), we give the free unknown variables the values ​​x 2 = 1, x 4 = 0, then we find the main unknowns from the system of equations
.

Solution find using 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.

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:

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