FIC FSR and general system solution. Fundamental system of solutions

We will continue to grind the equipment elementary transformations on the uniform system linear equations .
According to the first paragraphs, the material may seem boring and ordinary, but this impression is deceptive. In addition to further working out technical techniques there will be many new information, so please try not to neglect the examples of this article.

What is a homogeneous system of linear equations?

The answer suggests itself. The system of linear equations is homogeneous if free dick eVERY System equations equal to zero.. For example:

It is quite clear that the homogeneous system is always coordinatedThat is, always has a solution. And, above all, the so-called eye rushes trivial decision . Trivial, for those who do not understand the meaning of the adjective, which means that the limit. Not academic, of course, but then it is intelligible \u003d) ... what to go around and about, let's find out if this system has any other solutions:

Example 1.

Decision: To solve a homogeneous system you need to record system matrix And with the help of elementary transformations, lead it to a stepped form. Please note that there is no need to record a vertical line and a zero column of free members - because they do not do with zeros, they will remain zeros:

(1) The second line added the first string multiplied by -2. To the third line added the first string multiplied by -3.

(2) To the third line added the second string multiplied by -1.

Sharing a third line to 3 does not make much sense.

As a result of elementary transformations, an equivalent homogeneous system was obtained. , and applying reverse Gauss method, it is easy to make sure that the solution is unique.

Answer:

We formulate an obvious criterion: A homogeneous system of linear equations has only a trivial solution, if a rank matrix system (In this case, 3) is equal to the number of variables (in this case - 3 pcs.).

Preheat and tighten your radio to the wave of elementary transformations:

Example 2.

Solve a homogeneous system of linear equations

To finally consolidate the algorithm, we will analyze the final task:

Example 7.

Solve a homogeneous system, write the answer in vector form.

Decision: We write the system matrix and with the help of elementary transformations we give it to a step type:

(1) The first line changed the sign. Once again, focusing on a repeatedly encountered reception, which allows you to significantly simplify the following action.

(1) The 2nd and 3rd rows added the first string. To the 4th line added the first string multiplied by 2.

(3) The last three lines are proportional, two of them removed.

As a result, a standard stepped matrix was obtained, and the solution continues at the rolled track:

- basic variables;
- free variables.

Express the basic variables through free variables. From the 2nd equation:

- Substitute in the 1st equation:

In this way, common decision:

Since there are three free variables in the example of the example, the fundamental system contains three vectors.

We substitute the top three values In general solution and we obtain the vector whose coordinates satisfy each equation of a homogeneous system. And again I repeat, that it is extremely desirable to check each resulting vector - time will take not so much, and it will make one hundred percent from errors.

For triple values Find vector

And finally, for the top three We get the third vector:

Answer:, where

Those who wish to avoid fractional values \u200b\u200bcan consider the top three and get an answer in equivalent:

By the word about frauds. Let's look at the matrix obtained in the task And we ask a question - is it possible to simplify the further decision? After all, here we first expressed through the fraught basic variable, then through the fraction of the basic variable, and, I must say, the process was not the easiest and not most pleasant.

Second solution solution:

The idea is to try select other basic variables. Let's look at the matrix and notice two units in the third column. So why not get zero at the top? Let's draw another elementary transformation:

You can order detailed solution Your task !!!

To understand what is fundamental system of solutions You can watch a video lesson for the same example clicking. Now let's turn to the description of the whole necessary work. This will help you in more detail in the essence of this issue.

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

Take for example such a system of linear equations:

Find the solution of this linear system of equations. To start us it is necessary to write a matrix of the system coefficients.

We transform this matrix to triangular. I rewrite the first string unchanged. And all the elements that stand under $ a_ (11) $, you need to make zeros. To make zero in the place of the element $ A_ (21) $, it is necessary to subtract the first from the second line, and write the difference in the second line. To make zero in the place of the $ a_ (31) $ element, it is necessary to make the first and difference in the third line in the third line. To make zero to the place of the element $ A_ (41) $, it is necessary from the fourth line to subtract the first multiplied by 2 and the difference to write in the fourth string. To make zero in the place of the element $ A_ (31) $, it is necessary from the fifth line to make the first multiplied by 2 and the difference to write in the fifth line.

First and second string rewrite unchanged. And all the elements that cost under $ A_ (22) $, you need to make zeros. To make zero in the place of the $ a_ (32) $ element, it is necessary to subtract the second line multiplied by 2 and write the difference in the third line. What to make zero in the place of the element $ A_ (42) $, it is necessary from the fourth line to subtract the second multiplied by 2 and write the difference in the fourth line. To make zero in the place of the element $ A_ (52) $, it is necessary from the fifth line to subtract the second multiplied by 3 and the difference is written in the fifth line.

We see that the last three lines are the sameTherefore, if from the fourth and fifth subtract the third, then they will be zero.

On this matrix record new system equations.

We see that the linearly independent equations of us, only three, and unknown five, therefore the fundamental system of solutions will consist of two vectors. So, us we must transfer the last two unknown to the right.

Now, we begin to express those unknowns that they stand in the left side through those that stand in the right part. We begin with the last equation, first we will express $ x_3 $, then we substitute the resulting result in the second equation and express $ x_2 $, and then in the first equation and here we will express $ x_1 $. Thus, we are all unknown that they stand in the left side, expressed through unknowns that they stand in the right part.

After that, instead of $ x_4 $ and $ x_5 $, we can substitute any numbers and find $ x_1 $, $ x_2 $ and $ x_3 $. Each such a fifth of numbers will be the roots of our original system of equations. What would be the vectors that come in FSR We need to substitute 1 instead of $ x_4, and instead $ x_5 $ substitute 0, to find $ x_1 $, $ x_2 $ and $ x_3 $, and then the opposite of $ x_4 \u003d 0 $ and $ x_5 \u003d 1 $.

System of linear equations in which all free members are zero, called uniform :

Any homogeneous system is always co-developed because it always possesses zero (trivial ) Decision. The question arises, under what conditions the homogeneous system will have a nontrivial solution.

Theorem 5.2. The homogeneous system has a nontrivial solution then and only if the rank of the main matrix is \u200b\u200bless than the number of its unknown.

Corollary. The square homogeneous system has a nontrivial solution if and only if the determinant of the main matrix of the system is not zero.

Example 5.6. Determine the values \u200b\u200bof the parameter L, in which the system has non-trivial solutions, and find these solutions:

Decision. This system will have a nontrivial solution when the determinant of the main matrix is \u200b\u200bzero:

Thus, the system is nontrivial, when l \u003d 3 or l \u003d 2. When L \u003d 3, the rank of the main matrix of the system is 1. Then leaving only one equation and believing that y.=a. and z.=b., get x \u003d b-a.

With L \u003d 2, the rank of the main matrix of the system is 2. Then, choosing as a basic minor:

we get a simplified system

From here we find that x \u003d Z./4, y \u003d z/ 2. Believed z.=4a., get

The set of all solutions of a homogeneous system has a very important linear property : if the columns X. 1 and X. 2 - solutions of the homogeneous system AX \u003d 0, then all their linear combination A. X. 1 + B. X. 2 will also solve this system. Indeed, since AX. 1 = 0 and AX. 2 = 0 T. A.(A. X. 1 + B. X. 2) \u003d a AX. 1 + B. AX. 2 \u003d A · 0 + b · 0 \u003d 0. It is precisely because of this property, if the linear system has more than one solution, then these solutions will be infinitely a lot.

Linear independent columns E. 1 , E. 2 , E K.which are solutions of a homogeneous system called fundamental system solutions Uniform system of linear equations, if the general solution of this system can be written in the form of a linear combination of these columns:

If a homogeneous system has n. variables, and the rank of the main system matrix is \u200b\u200bequal r.T. k. = n-R..

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

Decision. We will find the rank of the main system matrix:

Thus, the set of solutions of this system of equations forms a linear subspace of dimension n - R.\u003d 5 - 2 \u003d 3. Select as a basic minor

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

Believed x. 3 = a., x. 4 = b., x. 5 = c.Find

, .

Believed a.= 1, b \u003d C.\u003d 0, we get the first basic solution; believed b.= 1, a \u003d C.\u003d 0, we obtain the second basis solution; believed c.= 1, a \u003d B.\u003d 0, we get the third basic solution. As a result, a normal fundamental system of solutions will take

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

X. = aE 1 + bE. 2 + cE 3. à.

Note some properties of solutions of a non-uniform system of linear equations AX \u003d B. and their relationship of the corresponding homogeneous system of equations AX \u003d 0.

General solution of the heterogeneous system equal to the sum of the overall solution of the corresponding homogeneous system AX \u003d 0 and an arbitrary private solution of the inhomogeneous system. Indeed, let Y. 0 arbitrary private solution of the heterogeneous system, i.e. AY. 0 = B., I. Y. - general solution of the heterogeneous system, i.e. Ay \u003d B.. Responted one equality from the other, we get
A.(Y-Y. 0) \u003d 0, i.e. Y - Y. 0 There is a general solution of the corresponding homogeneous system AX.\u003d 0. Hence, Y - Y. 0 = X., or Y \u003d y. 0 + X.. Q.E.D.

Let the heterogeneous system be viewed AX \u003d B 1 + B. 2 . Then the general solution of such a system can be written as x \u003d x 1 + X. 2 , where AX 1 = B. 1 and AX. 2 = B. 2. This property expresses a universal property at all any linear systems (algebraic, differential, functional, etc.). In physics, this property is called superposition principle, in electrical and radio engineering - principle of overlay. For example, in the theory of linear electrical circuits, the current in any circuit can be obtained as an algebraic amount of currents caused by each source of energy separately.

Solution of linear systems algebraic equations (Slava) undoubtedly is the most important topic Course linear algebra. A huge number of tasks from all sections of mathematics is reduced to solving systems of linear equations. These factors explain the reason for creating this article. The article article is selected and structured so that with it you can

pick up optimal method solutions to your system of linear algebraic equations,
explore the theory of the selected method,
solve your system of linear equations, examined in detail disassembled solutions of characteristic examples and tasks.

Brief description of the material of the article.

First, we will give all the necessary definitions, concepts and introduce notation.

Next, we consider the methods of solving systems of linear algebraic equations, in which the number of equations is equal to the number of unknown variables and which have only decision. First, we will focus on the Cramer method, secondly, we will show the matrix method of solving such systems of equations, thirdly, we will analyze the Gauss method (method consistent exception unknown variables). To secure the theory, it will necessarily solve several slows in various ways.

After that, we turn to solve systems of linear algebraic equations general viewin 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 theorem of the Krocecker - Capelli, which allows you to establish a compatibility of Slava. We will analyze the solution of systems (in the case of their compatibility) with the help of the concept of basic minor of the matrix. We will also consider the Gauss method and describe in detail the solutions of examples.

We will definitely focus on the structure of the overall solution of homogeneous and inhomogeneous systems of linear algebraic equations. We give the concept of a fundamental solution system and show how the general solution is written to the Slava using the vectors of the fundamental solutions system. For a better understanding we will analyze several examples.

In conclusion, consider the system of equations reduced to linear, as well as various tasks, when solving a slope.

Navigating page.

Definitions, concepts, notation.

We will consider systems from p linear algebraic equations with n unknown variables (P may be equal to n)

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

Such a form of wrote is called coordinate.

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

If you add to the matrix and add a matrix-column-column of free members, then we get the so-called extended matrix Systems of linear equations. Typically, the expanded matrix is \u200b\u200bdenoted by the letter T, and the column of free members is separated by the vertical line from the remaining columns, that is,

By solving the system of linear algebraic equations Call a set of values \u200b\u200bof unknown variables, adding all equations of the system in identities. Matrix equation With these values \u200b\u200bof unknown variables also addresses identity.

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

If the system of solutions does not have, then it is called non-stop.

If the only solution has a single decision, then it is called defined; If solutions are more than one, then - uncertain.

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

The 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 \u200b\u200bnot zero, then such a slope will be called elementary. Such systems of equations have a single solution, and in the case of a homogeneous system, all unknown variables are zero.

We started to study in high school such a skull. When they were solved, we took some kind of equation, expressed one unknown variable through others and substituted it into the remaining equations, followed the following equation, expressed the following unknown variable and substituted into other equations and so on. Or used the method of addition, that is, two or more equations folded to exclude some unknown variables. We will not stop in detail on these methods, as they are essentially modifications of the Gauss method.

The main methods of solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. We will analyze them.

Solution of systems of linear equations by the Cramer 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 - the determinant of the main matrix of the system, and - determinants of matrices that are obtained from a replacement 1st, 2nd, ..., N-wow Column, respectively, on the column of free members:

With such notation, unknown variables are calculated using the formulas of the Cramer method as . So there is a solution to the system of linear algebraic equations by the Cramer method.

Example.

Cramer method .

Decision.

The main matrix of the system has the form . We 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 single solution that can be found by the Cramer method.

We will compose and calculate the necessary determinants (We obtain the determinant, replacing in the matrix and the first column on the column of free members, the determinant - replacing the second column on the column of free members, - replacing the third column of the matrix and on the column of free members):

We find unknown variables by formulas :

Answer:

The main disadvantage of the Cramer 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 a reverse matrix).

Let the system of linear algebraic equations are specified in the matrix form, where the matrix A has the dimension n on N and its determinant is different from zero.

Since, then the matrix A is reversible, that is, there is a reverse matrix. If you multiply both parts of equality to the left, we obtain the formula for finding a column-column of unknown variables. So we obtained a solution of a system of linear algebraic equations by the matrix method.

Example.

Decide the system of linear equations matrix method.

Decision.

I rewrite the system of equations in the matrix form:

As

That the slope can be solved by the matrix method. Via reverse matrix Solution of this system can be found as .

We construct an inverse matrix using a matrix from algebraic additions of the elements of the matrix A (if necessary, see the article):

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

Answer:

Or in another record x 1 \u003d 4, x 2 \u003d 0, x 3 \u003d -1.

The main problem when solving solutions of linear algebraic equations, the matrix method is consistent with the inverse matrix, especially for square matrices About the third.

Solving systems of linear equations by Gauss method.

Let us need to find a solution of a system from N linear equations with n unknown variables
The determinant of the main matrix is \u200b\u200bdifferent from zero.

The essence of the Gauss method It consists in the sequential exclusion of unknown variables: first excludes x 1 of all the equations of the system, starting from the second, then x 2 of 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 converting system equations for consistent exclusion of unknown variables is called direct running of the Gauss method. After the removal of the direct movement of the Gauss method from the last equation is X N, with the help of this value from the penultimate equation, x N-1 is calculated, and so on, x 1 is calculated from the first equation. The process of calculating unknown variables when driving from the last equation of the system to the first one is called return of the Gauss method.

Briefly describe an algorithm to exclude unknown variables.

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

where, A. .

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

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

To do this, we add the second, multiplied by, to the fourth equation to the fourth equation, the second, multiplied by, and so on, to the N-th equation, add the second, multiplied by. 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, proceed to the exclusion of an unknown X 3, while acting similarly to the part of the system marked in the figure

So we continue the direct move of the Gauss method while the system does not take

From that moment, we begin the reverse course of the Gauss method: Calculate the X N from the last equation, as using the resulting X N, we find X N-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Example.

Decide the system of linear equations Gauss method.

Decision.

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

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

On this, the direct move of the Gauss method is finished, we begin the opposite.

From the last equation of the obtained system of equations, we find X 3:

From the second equation we get.

From the first equation, we find the remaining unknown variable and these are completing the reverse move 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 the system P equations does not coincide with the number of unknown variables N:

Such a slope may not have solutions, have a single decision or have infinitely many solutions. This statement also refers to the systems of equations, the main matrix of which is square and degenerate.

The theorem of the Kronkera - Capelli.

Before finding a solution of a system of linear equations, it is necessary to establish its compatibility. The answer to the question when the Slava is together, and when incomplete, gives koncheker theorem - Capelli:
In order for the system from P equations with n unknown (P may be equal to N), it is necessary and enough that the rank of the main matrix of the system was equal to the rank of an extended matrix, that is, RANK (A) \u003d RANK (T).

Consider on the example the use of the theorem of the Krakeker - Capelli to determine the compilation of the system of linear equations.

Example.

Find out whether the system of linear equations has solutions.

Decision.

. We use the method of bustling minor. Minor of second order Different from zero. We will overcome the third-order minors from the forefront:

Since all third-order fundamental minors are zero, the rank of the main matrix is \u200b\u200btwo.

In turn, the rank of an extended matrix equal to three, as minor of the third order

Different from zero.

In this way, Rang (A), therefore, on the Krakecker theorem - Capelli, it can be concluded that the initial system of linear equations is incomplete.

Answer:

The system of solutions has no.

So, we learned how to establish the incompleteness of the system using the Klekeker - Capelli theorem.

But how to find a solution to the Slava, if its compatibility is installed?

To do this, we need the concept of the base minor of the matrix and the theorem on the ring of the matrix.

Minor of the highest order of the matrix A, different from zero, is called basis.

From the definition of the basic minor it follows that its order is equal to the margin of the matrix. For a nonzero matrix, but there may be several basic minorors, one basic minor is always.

For example, consider the matrix .

All the minors of the third order of this matrix are zero, since the elements of the third line of this matrix are the sum of the corresponding elements of the first and second lines.

The basic are the following minors of the second order, as they are different from zero

Minora The basic are not, as they are zero.

The theorem on the rank of the matrix.

If the ring of the order p per n is equal to R, then all the elements of the strings (and columns) of the matrix that do not form the selected base minor are linearly expressed through the corresponding elements of strings (and columns) forming the base minor.

What gives us the theorem on the rank of the matrix?

If, on the theorem of the Kreconeker - Capelli, we set the units of the system, 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 selected base minor. The slope thus obtained will be equivalent to the original, since the discarded equations are still unnecessary (they are the linear combination of the remaining equations in the direction of the matrix's rank theorem).

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

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

Example.

Decision.

Rank main system matrix equal to two, as the second order minor Different from zero. The rank of an extended matrix Also equal to two, since the only minor of the third order is zero

And the first-order minor discussed above is different from zero. Based on the theorem of the Krocecker - Capelli, it is possible to approve the sharing of the original system of linear equations, since Rank (a) \u003d Rank (T) \u003d 2.

As a basic minor, take . It forms the coefficients of the first and second equations:

The third equation of the system is not involved in the formation of a base minor, therefore, we will exclude it from the system based on the theorem on the Ring matrix:

So we obtained an elementary system of linear algebraic equations. By solving it using the crater:

Answer:

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

If the number of R equations in the resulting slope is less than the number of unknown variables N, then in the left parts of the equations, we leave the components that form the base minor, the rest of the components are transferred to the right parts of the system equations with the opposite sign.

Unknown variables (their R pieces) remaining in the left parts of the equations are called basic.

Unknown variables (their N - R pieces), which were in the right parts, are called free.

Now we believe that free unknown variables can make arbitrary values, while R basic unknown variables will be expressed through free unknown variables by the only way. Their expression can be found solving the resulting sample by the drive method, the matrix method or method of Gauss.

We will analyze on the example.

Example.

Decide the system of linear algebraic equations .

Decision.

We find the rank of the main matrix of the system The method of bustling minors. As a nonzero minor of the first order, take a 1 1 \u003d 1. Let's start the search for a second-order non-zero minor, which cuts this Minor:

So we found the nonsense minor of the second order. Let's start the search for nonzero bordering the third order:

Thus, the rank of the main matrix is \u200b\u200bthree. The rank of an extended matrix is \u200b\u200balso equal to three, that is, the system is coordinated.

The founded nonzero minor of the third order will take as a basic one.

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

We leave the components of the system in the left part of the equations involved in the base minor, the rest are transferred with opposite signs to the right parts:

Give the free unknown variables x 2 and x 5 arbitrary values, that is, we will take where - arbitrary numbers. At the same time, the slope will take

The resulting elementary system of linear algebraic equations by solving the control system:

Hence, .

In response, do not forget to specify free unknown variables.

Answer:

Where - arbitrary numbers.

Summarize.

To solve a system of linear algebraic equations of a common type, we first find out its compatibility using the Konpeker's theorem - Capelli. If the rank of the main matrix is \u200b\u200bnot equal to the rank of an extended matrix, then we conclude the incompleteness of the system.

If the rank of the main matrix is \u200b\u200bequal to the rank of an expanded matrix, then we select the base minor and discard the equation of the system that do not participate in the formation of the chosen base minor.

If the order of the basic minor equal to the number Unknown variables, the Slava has a single solution that we find any method known to us.

If the order of the base minor is less than the number of unknown variables, then in the left part of the system equations, we leave the components with the main unknown variables, the remaining components are transferred to the right parts and give free unknown variables arbitrary values. From the resulting system of linear equations, we find the main unknown variables by the manufacturer, the matrix method or method of Gauss.

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

The Gauss method can solve the system of linear algebraic equations of any kind without prior to their research on units. The process of consistent exclusion of unknown variables allows us to conclude both of the compatibility and incompleteness of the Slava, and in the case of the existence of the solution makes it possible to find it.

From the point of view of computational operation, the Gauss method is preferred.

See Him detailed description and disassembled examples in the article of the Gauss method for solving systems of linear algebraic equations of general form.

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

In this section, we will discuss the joint homogeneous and inhomogeneous systems of linear algebraic equations having infinite set solutions.

We will understand first with homogeneous systems.

Fundamental system solutions The homogeneous system from p linear algebraic equations with n unknown variables are called a set (n - r) linearly independent solutions of this system, where R is the order of the base minor of the main matrix of the system.

If you designate linearly independent solutions of a homogeneous slope as x (1), x (2), ..., x (nr) (x (1), x (2), ..., x (nr) - these are the matrices of the dimension columns N by 1) , the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary permanent coefficients C 1, C 2, ..., C (N-R), that is,.

What denotes the term general solution of a homogeneous system of linear algebraic equations (Orostal)?

The meaning is simple: the formula sets all possible solutions The original Slava, in other words, taking any set of values \u200b\u200bof arbitrary constants C 1, C 2, ..., C (N-R), according to the formula, we will get one of the solutions of the initial homogeneous slope.

Thus, if we find a fundamental system of solutions, we will be able to ask all solutions to this homogeneous slope as.

Let us show the process of building a fundamental solution system with a homogeneous slope.

We choose the basic minor of the original system of linear equations, we exclude all other equations from the system and transferred to the right parts of the system equations with opposite signs, all terms containing free unknown variables. Let us give a free unknown variable value of 1.0.0, ..., 0 and calculate the main unknown, solving the resulting elementary system of linear equations in any way, for example, by the drive method. So x (1) will be obtained - the first solution of the fundamental system. If you give a free unknown value of 0.1.0.0, ..., 0 and calculate the main unknown, then we obtain X (2). Etc. If the free unknown variables give the value of 0.0, ..., 0.1 and calculate the main unknown, then we obtain X (N-R). This will be built a fundamental system of solutions to a homogeneous slope and its general solution may be recorded.

For inhomogeneous systems of linear algebraic equations, a general solution is represented in the form, where is the general solution of the corresponding homogeneous system, and the private solution of the initial inhomogeneous slope, which we get, giving a free unknown value of 0.0, ..., 0 and calculating the values \u200b\u200bof the main unknowns.

We will analyze on the examples.

Example.

Find a fundamental solutions system and a 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 an extended matrix. We find the rank of the main matrix by the method of bustling minors. As a nonzero minor of the first order, take the element A 1 1 \u003d 9 of the main matrix of the system. We will find the bordering the nonzero minor of the second order:

Minor of second order, different from zero, found. We will overcome the third-order minor foods in search of non-zero:

All third-order focusing minors are zero, therefore, the rank of the main and extended matrix is \u200b\u200btwo. We take the basic minor. We note for clarity the elements of the system that form it:

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

We leave the alignments containing the main unknowns in the right parts of the equations, and we carry the terms with free unknowns into the right parts:

We construct a fundamental system of solutions of the initial homogeneous system of linear equations. Fundamental system Solutions of this Slava consists of two solutions, since the initial slope contains four unknown variables, and the order of its basis minraul is two. To find X (1), let us give a free unknown variable value x 2 \u003d 1, x 4 \u003d 0, then the main unknown to find from the system of equations
.

Uniform system of linear equations over the field

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

Note that a homogeneous system of linear equations, which has only a zero solution, has no fundamental solutions system.

Proposition 3.11. Any two fundamental solutions solutions of a homogeneous system of linear equations consist of same number solutions.

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

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

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

Evidence. If the rank of the main matrix A of a homogeneous system (1) is zero or, then above it was shown that the theorem is correct. Therefore, it is assumed below that believing, we assume that the first columns of the matrix are linearly independent. In this case, the matrix A is linked equivalent to the reduced stepped matrix, and the system (1) is equivalent to the following above-mentioned system of equations:

It is easy to verify that any system of values \u200b\u200bof free system variables (2) corresponds to one and only one solution of the system (2) and, it means, systems (1). In particular, the system of zero values \u200b\u200bcorresponds only to the zero solution of the system (2) and the system (1).

We will give in system (2) to give one of the free variables value equal to 1, and the remaining variables - zero values. As a result, we obtain solutions of the system of equations (2), which we write in the form of strings of the following matrix with:

The system of lines of this matrix is \u200b\u200blinearly independent. In fact, for any scales from equality

equality follows

and, therefore, equality

We prove that the linear shell of the system lines of the matrix co coincides with the set of all solutions of the system (1).

Arbitrary Solution Solution (1). Then vector

is also a solution to the system (1), and