We continue to deal with systems linear equations. So far, we have considered systems that have a unique solution. Such systems can be solved in any way: substitution method("school") by Cramer's formulas, matrix method, Gauss method. However, two more cases are widespread in practice when:

1) the system is inconsistent (has no solutions);

2) the system has infinitely many solutions.

For these systems, the most universal of all solution methods is used - Gauss method. In fact, the "school" method will also lead to the answer, but in higher mathematics it is customary to use the Gaussian method of successive elimination of unknowns. Those who are not familiar with the Gauss method algorithm, please study the lesson first Gauss method

The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. First, consider a couple of examples where the system has no solutions (inconsistent).

Example 1

What immediately catches your eye in this system? The number of equations is less than the number of variables. There is a theorem that says: “If the number of equations in the system is less than the number of variables, then the system is either inconsistent or has infinitely many solutions. And it remains only to find out.

The beginning of the solution is quite ordinary - we write the extended matrix of the system and, using elementary transformations, we bring it to a stepwise form:

(1). On the upper left step, we need to get (+1) or (-1). There are no such numbers in the first column, so rearranging the rows will not work. The unit will have to be organized independently, and this can be done in several ways. We did so. To the first line we add the third line, multiplied by (-1).

(2). Now we get two zeros in the first column. To the second line, add the first line, multiplied by 3. To the third line, add the first, multiplied by 5.

(3). After the transformation is done, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the desired one (-1) on the second step. Divide the third line by (-3).

(4). Add the second line to the third line. Probably, everyone paid attention to the bad line, which turned out as a result of elementary transformations:

. It is clear that this cannot be so.

Indeed, we rewrite the resulting matrix

back to the system of linear equations:

If as a result of elementary transformations a string of the form , Whereλ is a non-zero number, then the system is inconsistent (has no solutions).

How to record the end of a task? You need to write down the phrase:

“As a result of elementary transformations, a string of the form is obtained, where λ ≠ 0 ". Answer: "The system has no solutions (inconsistent)."

Please note that in this case there is no reverse move of the Gaussian algorithm, there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is an example for independent decision. Complete Solution and the answer at the end of the lesson.

Again, we remind you that your solution process may differ from our solution process, the Gauss method does not set an unambiguous algorithm, you need to guess the procedure and the actions themselves in each case yourself.

Another one technical feature solutions: elementary transformations can be stopped At once, as soon as a line like , where λ ≠ 0 . Consider a conditional example: suppose that after the first transformation we get a matrix

.

This matrix has not yet been reduced to a stepped form, but there is no need for further elementary transformations, since a line of the form has appeared, where λ ≠ 0 . It should be immediately answered that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift to the student, because a short solution is obtained, sometimes literally in 2-3 steps. But everything in this world is balanced, and the problem in which the system has infinitely many solutions is just longer.

Example 3:

Solve a system of linear equations

There are 4 equations and 4 unknowns, so the system can either have a single solution, or have no solutions, or have infinitely many solutions. Whatever it was, but the Gauss method in any case will lead us to the answer. This is its versatility.

The beginning is again standard. We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

That's all, and you were afraid.

(1). Please note that all the numbers in the first column are divisible by 2, so on the upper left step we are also satisfied with a deuce. To the second line we add the first line, multiplied by (-4). To the third line we add the first line, multiplied by (-2). To the fourth line we add the first line, multiplied by (-1).

Attention! Many may be tempted from the fourth line subtract first line. This can be done, but it is not necessary, experience shows that the probability of an error in calculations increases several times. We just add: to the fourth line we add the first line, multiplied by (-1) - exactly!

(2). The last three lines are proportional, two of them can be deleted. Here again it is necessary to show increased attention, but are the lines really proportional? For reinsurance, it will not be superfluous to multiply the second row by (-1), and divide the fourth row by 2, resulting in three identical rows. And only after that remove two of them. As a result of elementary transformations, the extended matrix of the system is reduced to a stepped form:

When completing a task in a notebook, it is advisable to make the same notes in pencil for clarity.

We rewrite the corresponding system of equations:

The “usual” only solution of the system does not smell here. Bad line where λ ≠ 0, also no. Hence, this is the third remaining case - the system has infinitely many solutions.

The infinite set of solutions of the system is briefly written in the form of the so-called general system solution.

We will find the general solution of the system using the reverse motion of the Gauss method. For systems of equations with an infinite set of solutions, new concepts appear: "basic variables" And "free variables". First, let's define what variables we have basic, and what variables - free. It is not necessary to explain in detail the terms of linear algebra, it is enough to remember that there are such basis variables And free variables.

Basic variables always "sit" strictly on the steps of the matrix. In this example, the base variables are x 1 and x 3 .

Free variables are everything remaining variables that did not get a step. In our case, there are two: x 2 and x 4 - free variables.

Now you need Allbasis variables express only throughfree variables. The reverse move of the Gaussian algorithm traditionally works from the bottom up. From the second equation of the system, we express the basic variable x 3:

Now look at the first equation: . First, we substitute the found expression into it:

It remains to express the basic variable x 1 through free variables x 2 and x 4:

The result is what you need - All basis variables ( x 1 and x 3) expressed only through free variables ( x 2 and x 4):

Actually, the general solution is ready:

.

How to write down the general solution? First of all, free variables are written into the general solution “on their own” and strictly in their places. In this case, the free variables x 2 and x 4 should be written in the second and fourth positions:

.

The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:

From the general solution of the system, one can find infinitely many private decisions. It's very simple. free variables x 2 and x 4 are called so because they can be given any final values. The most popular values ​​are zero values, since this is the easiest way to obtain a particular solution.

Substituting ( x 2 = 0; x 4 = 0) into the general solution, we get one of the particular solutions:

, or is a particular solution corresponding to free variables with values ​​( x 2 = 0; x 4 = 0).

Ones are another sweet couple, let's substitute ( x 2 = 1 and x 4 = 1) into the general solution:

, i.e. (-1; 1; 1; 1) is another particular solution.

It is easy to see that the system of equations has infinitely many solutions since we can give free variables any values.

Each a particular solution must satisfy to each system equation. This is the basis for a “quick” check of the correctness of the solution. Take, for example, a particular solution (-1; 1; 1; 1) and substitute it into the left side of each equation in the original system:

Everything has to come together. And with any particular solution you get, everything should also converge.

Strictly speaking, the verification of a particular solution sometimes deceives, i.e. some particular solution can satisfy each equation of the system, and the general solution itself is actually found incorrectly. Therefore, first of all, the verification of the general solution is more thorough and reliable.

How to check the resulting general solution ?

It's not difficult, but it requires quite a long transformation. We need to take expressions basic variables, in this case and , and substitute them into the left side of each equation of the system.

To the left side of the first equation of the system:

The right side of the original first equation of the system is obtained.

To the left side of the second equation of the system:

The right side of the original second equation of the system is obtained.

And further - to the left parts of the third and fourth equations of the system. This check is longer, but it guarantees the 100% correctness of the overall solution. In addition, in some tasks it is required to check the general solution.

Example 4:

Solve the system using the Gauss method. Find a general solution and two private ones. Check the overall solution.

This is a do-it-yourself example. Here, by the way, again the number of equations is less than the number of unknowns, which means that it is immediately clear that the system will be either inconsistent or with an infinite number of solutions.

Example 5:

Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution

Solution: Let us write down the extended matrix of the system and, with the help of elementary transformations, bring it to a stepped form:

(1). Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.

(2). To the third line we add the second line, multiplied by (-5). To the fourth line we add the second line, multiplied by (-7).

(3). The third and fourth lines are the same, we delete one of them. Here is such a beauty:

Basis variables sit on steps, so they are base variables.

There is only one free variable, which did not get a step: .

(4). Reverse move. We express the basic variables in terms of the free variable:

From the third equation:

Consider the second equation and substitute the found expression into it:

, , ,

Consider the first equation and substitute the found expressions and into it:

Thus, the general solution with one free variable x 4:

Once again, how did it happen? free variable x 4 sits alone in its rightful fourth place. The resulting expressions for the basic variables , , are also in their places.

Let us immediately check the general solution.

We substitute the basic variables , , into the left side of each equation of the system:

The corresponding right-hand sides of the equations are obtained, thus, the correct general solution is found.

Now from the found general solution we get two particular solutions. All variables are expressed here through a single free variable x 4 . You don't need to break your head.

Let x 4 = 0, then is the first particular solution.

Let x 4 = 1, then is another particular solution.

Answer: Common decision: . Private Solutions:

And .

Example 6:

Find the general solution of the system of linear equations.

We have already checked the general solution, the answer can be trusted. Your course of action may differ from our course of action. The main thing is that the general solutions coincide. Probably, many people noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to fiddle with ordinary fractions. In practice, this is true, cases where there are no fractions are much less common. Be prepared mentally, and most importantly, technically.

Let us dwell on the features of the solution that were not found in the solved examples. The general solution of the system may sometimes include a constant (or constants).

For example, the general solution: . Here one of the basic variables is equal to a constant number: . There is nothing exotic in this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.

Rarely, but there are systems in which number of equations more quantity variables. However, the Gauss method works under the most severe conditions. You should calmly bring the extended matrix of the system to a stepped form according to the standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a unique solution.

We repeat in our advice - in order to feel comfortable when solving a system using the Gauss method, you should fill your hand and solve at least a dozen systems.

Solutions and answers:

Example 2:

Solution:Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepped form.

Performed elementary transformations:

(1) The first and third lines have been swapped.

(2) The first line was added to the second line, multiplied by (-6). The first line was added to the third line, multiplied by (-7).

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

As a result of elementary transformations, a string of the form, Where λ ≠ 0 .So the system is inconsistent.Answer: there are no solutions.

Example 4:

Solution:We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

Conversions performed:

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

There is no unit for the second step , and transformation (2) is aimed at obtaining it.

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

(3). The second and third rows were swapped (the resulting -1 was moved to the second step)

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

(5). The sign of the first two lines was changed (multiplied by -1), the third line was divided by 14.

Reverse move:

(1). Here are the basic variables (which are on steps), and are free variables (who did not get the step).

(2). We express the basic variables in terms of free variables:

From the third equation: .

(3). Consider the second equation:, particular solutions:

Answer: Common decision:

Complex numbers

In this section, we will introduce the concept complex number, consider algebraic, trigonometric And show form complex number. We will also learn how to perform operations with complex numbers: addition, subtraction, multiplication, division, exponentiation and root extraction.

For development complex numbers no special knowledge is required from the course of higher mathematics, and the material is available even to a schoolchild. Enough to be able to algebraic actions with "ordinary" numbers, and remember trigonometry.

First, let's remember the "ordinary" Numbers. In mathematics they are called set of real numbers and are marked with the letter R, or R (thick). All real numbers sit on the familiar number line:

The company of real numbers is very colorful - here are integers, and fractions, and irrational numbers. In this case, each point of the numerical axis necessarily corresponds to some real number.

System of m linear equations with n unknowns called a system of the form

Where aij And b i (i=1,…,m; b=1,…,n) are some known numbers, and x 1 ,…,x n- unknown. In the notation of the coefficients aij first index i denotes the number of the equation, and the second j is the number of the unknown at which this coefficient stands.

The coefficients for the unknowns will be written in the form of a matrix , which we will call system matrix.

The numbers on the right sides of the equations b 1 ,…,b m called free members.

Aggregate n numbers c 1 ,…,c n called decision of this system, if each equation of the system becomes an equality after substituting numbers into it c 1 ,…,c n instead of the corresponding unknowns x 1 ,…,x n.

Our task will be to find solutions to the system. In this case, three situations may arise:

A system of linear equations that has at least one solution is called joint. Otherwise, i.e. if the system has no solutions, then it is called incompatible.

Consider ways to find solutions to the system.

MATRIX METHOD FOR SOLVING SYSTEMS OF LINEAR EQUATIONS

Matrices make it possible to briefly write down a system of linear equations. Let a system of 3 equations with three unknowns be given:

Consider the matrix of the system and matrix columns of unknown and free members

Let's find the product

those. as a result of the product, we obtain the left-hand sides of the equations of this system. Then, using the definition of matrix equality, this system can be written as

or shorter A∙X=B.

Here matrices A And B are known, and the matrix X unknown. She needs to be found, because. its elements are the solution of this system. This equation is called matrix equation.

Let the matrix determinant be different from zero | A| ≠ 0. Then the matrix equation is solved as follows. Multiply both sides of the equation on the left by the matrix A-1, the inverse of the matrix A: . Because the A -1 A = E And E∙X=X, then we obtain the solution of the matrix equation in the form X = A -1 B .

Note that since the inverse matrix can only be found for square matrices, the matrix method can only solve those systems in which the number of equations is the same as the number of unknowns. However, the matrix notation of the system is also possible in the case when the number of equations is not equal to the number of unknowns, then the matrix A is not square and therefore it is impossible to find a solution to the system in the form X = A -1 B.

Examples. Solve systems of equations.

CRAMER'S RULE

Consider a system of 3 linear equations with three unknowns:

Third-order determinant corresponding to the matrix of the system, i.e. composed of coefficients at unknowns,

called system determinant.

We compose three more determinants as follows: we replace successively 1, 2 and 3 columns in the determinant D with a column of free members

Then we can prove the following result.

Theorem (Cramer's rule). If the determinant of the system is Δ ≠ 0, then the system under consideration has one and only one solution, and

Proof. So, consider a system of 3 equations with three unknowns. Multiply the 1st equation of the system by the algebraic complement A 11 element a 11, 2nd equation - on A21 and 3rd - on A 31:

Let's add these equations:

Consider each of the brackets and the right side of this equation. By the theorem on the expansion of the determinant in terms of the elements of the 1st column

Similarly, it can be shown that and .

Finally, it is easy to see that

Thus, we get the equality: .

Hence, .

The equalities and are derived similarly, whence the assertion of the theorem follows.

Thus, we note that if the determinant of the system is Δ ≠ 0, then the system has a unique solution and vice versa. If the determinant of the system is equal to zero, then the system either has an infinite set of solutions or has no solutions, i.e. incompatible.

Examples. Solve a system of equations

GAUSS METHOD

The previously considered methods can be used to solve only those systems in which the number of equations coincides with the number of unknowns, and the determinant of the system must be different from zero. The Gaussian method is more universal and is suitable for systems with any number of equations. It consists in the successive elimination of unknowns from the equations of the system.

Consider again a system of three equations with three unknowns:

.

We leave the first equation unchanged, and from the 2nd and 3rd we exclude the terms containing x 1. To do this, we divide the second equation by A 21 and multiply by - A 11 and then add with the 1st equation. Similarly, we divide the third equation into A 31 and multiply by - A 11 and then add it to the first one. As a result, the original system will take the form:

Now, from the last equation, we eliminate the term containing x2. To do this, divide the third equation by , multiply by and add it to the second. Then we will have a system of equations:

Hence from the last equation it is easy to find x 3, then from the 2nd equation x2 and finally from the 1st - x 1.

When using the Gaussian method, the equations can be interchanged if necessary.

Often, instead of writing a new system of equations, they limit themselves to writing out the extended matrix of the system:

and then bring it to a triangular or diagonal form using elementary transformations.

TO elementary transformations matrices include the following transformations:

1. permutation of rows or columns;
2. multiplying a string by a non-zero number;
3. adding to one line other lines.

Examples: Solve systems of equations using the Gauss method.

Thus, the system has an infinite number of solutions.

Section 5. ELEMENTS OF LINEAR ALGEBRA

Systems of linear equations

Basic concepts

A system of linear algebraic equations, containing T equations and P unknowns, is called a system of the form

where are the numbers A ij , i=
, j= called coefficients systems, numbers b i - free members. To be found number X P .

It is convenient to write such a system in a compact matrix form
.

Here A is the coefficient matrix of the system, called main matrix:

,

-column vector of unknowns X j , is a column vector of free members b i .

Extended the matrix of the system is the matrix system, supplemented by a column of free terms

.

Decision system is called P unknown values X 1 = with 1 , X 2 = with 2 , ..., X P = with P , upon substitution of which all equations of the system turn into true equalities. Any solution of the system can be written as a matrix-column .

The system of equations is called joint if it has at least one solution, and incompatible if it has no solution.

The joint system is called certain if it has a unique solution, and uncertain if it has more than one solution. In the latter case, each of its solutions is called private decision systems. The set of all particular solutions is called general solution.

Solve the system it means finding out whether it is compatible or not. If the system is consistent, then find its general solution.

The two systems are called equivalent(equivalent) if they have the same general solution. In other words, systems are equivalent if every solution to one of them is a solution to the other, and vice versa.

Equivalent systems are obtained, in particular, when elementary transformations system, provided that the transformations are performed only on the rows of the matrix.

The system of linear equations is called homogeneous if all free terms are equal to zero:

A homogeneous system is always consistent, since X 1 =x 2 =…=x P =0 is the solution to the system. This solution is called zero or trivial.

Solving systems of linear equations

Let an arbitrary system be given T linear equations with P unknown

Theorem 1(Kronecker-Cappelli). A system of linear algebraic equations is consistent if and only if the rank of the extended matrix is ​​equal to the rank of the main matrix.

Theorem 2. If the rank of a consistent system is equal to the number of unknowns, then the system has a unique solution.

Theorem 3. If the rank of a consistent system is less than the number of unknowns, then the system has an infinite number of solutions.

EXAMPLE Examine the system for compatibility

Solution.
,r(A)=1;
, r()=2,
.

Thus, r(A) r(), hence the system is inconsistent.

Solution of non-degenerate systems of linear equations. Cramer's formulas

Let the system P linear equations with P unknown

or in matrix form A∙X=B.

The main matrix A of such a system is square. The determinant of this matrix is ​​called system determinant. If the determinant of the system is non-zero, then the system is called non-degenerate.

Let's find the solution of this system of equations in the case of ∆0. multiplying both sides of the equation А∙Х=В on the left by the matrix А  1 , we get А  1 ∙ A∙Х= A  1 ∙B. Since A - 1 ∙ A \u003d E and E ∙ X \u003d X, then X \u003d A - 1 ∙ B. This method of solving the system is called matrix.

From the matrix method follow Cramer's formulas
, where ∆ is the determinant of the main matrix of the system, and ∆ i is the determinant obtained from the determinant ∆ by replacing i th column of coefficients by a column of free members.

EXAMPLE Solve the system

Solution.
, 70,
,
. Means, X 1 =, X 2 =
.

Solution of systems of linear equations by the Gauss method

The Gauss method consists in the successive elimination of unknowns.

Let the system of equations

The Gaussian solution process consists of two steps. At the first stage (forward run), the system is reduced to stepped(in particular, triangular) mind.

Where k≤ n, a ii  0, i= . Odds A ii called main elements of the system.

At the second stage (reverse move), the unknowns from this stepwise system are sequentially determined.

Notes:

If the step system turns out to be triangular, i.e. k= n, then the original system has a unique solution. From the last equation we find X P , from the penultimate equation we find X P  1 , then, going up the system, we find all the other unknowns.

In practice, it is more convenient to work with the extended matrix of the system, performing all elementary transformations on its rows. It is convenient that the coefficient A 11 was equal to 1 (rearrange the equations, or divide by A 11 1).

EXAMPLE Solve the system using the Gauss method

Solution. As a result of elementary transformations over the extended matrix of the system

~
~
~

~

the original system was reduced to a stepwise one:

Therefore, the general solution of the system is: x 2 =5 x 4  13 x 3  3; x 1 =5 x 4  8 x 3  1.

If we put, for example, X 3 =x 4 =0, then we find one of the particular solutions of this system X 1 =  1, x 2 =  3, x 3 =0, x 4 =0.

Systems of homogeneous linear equations

Let the system of linear homogeneous equations be given

Obviously, a homogeneous system is always compatible, it has a zero (trivial) solution.

Theorem 4. In order for a system of homogeneous equations to have a nonzero solution, it is necessary and sufficient that the rank of its main matrix be less than the number of unknowns, i.e. r< n.

Theorem 5. In order for a homogeneous system P linear equations with P unknowns has a nonzero solution, it is necessary and sufficient that the determinant of its main matrix be equal to zero, i.e. ∆=0.

If the system has non-zero solutions, then ∆=0.

EXAMPLE Solve the system

Solution.
,r(A)=2
, n=3. Because r< n, then the system has an infinite number of solutions.

,
. That is, X 1 ==2x 3 , X 2 ==3x 3 - common decision.

Putting X 3 =0, we get one particular solution: X 1 =0, x 2 =0, x 3 =0. Putting X 3 =1, we get the second particular solution: X 1 =2, x 2 =3, x 3 =1 etc.

Questions to control

What is a system of linear algebraic equations?

Explain the following concepts: coefficient, intercept, main and extended matrices.

What are systems of linear equations? Formulate the Kronker-Capelli theorem (on the compatibility of a system of linear equations).

List and explain methods for solving systems of linear equations.

Service assignment. The online calculator is designed to study a system of linear equations. Usually in the condition of the problem it is required to find general and particular solution of the system. When studying systems of linear equations, the following problems are solved:

1. whether the system is collaborative;
2. if the system is consistent, then it is definite or indefinite (the criterion of system compatibility is determined by the theorem);
3. if the system is defined, then how to find its unique solution (the Cramer method, the inverse matrix method or the Jordan-Gauss method are used);
4. if the system is indefinite, then how to describe the set of its solutions.

Classification of systems of linear equations

An arbitrary system of linear equations has the form:
a 1 1 x 1 + a 1 2 x 2 + ... + a 1 n x n = b 1
a 2 1 x 1 + a 2 2 x 2 + ... + a 2 n x n = b 2
...................................................
a m 1 x 1 + a m 2 x 2 + ... + a m n x n = b m

1. Systems of linear inhomogeneous equations (the number of variables is equal to the number of equations, m = n).
2. Arbitrary systems of linear inhomogeneous equations (m > n or m< n).

Definition. A solution of a system is any collection of numbers c 1 ,c 2 ,...,c n , whose substitution into the system instead of the corresponding unknowns turns each equation of the system into an identity.

Definition. Two systems are said to be equivalent if the solution to the first is the solution to the second and vice versa.

Definition. A system that has at least one solution is called joint. A system that does not have any solution is called inconsistent.

Definition. A system with a unique solution is called certain, and having more than one solution is indefinite.

Algorithm for solving systems of linear equations

1. Find the ranks of the main and extended matrices. If they are not equal, then, by the Kronecker-Capelli theorem, the system is inconsistent, and this is where the study ends.
2. Let rank(A) = rank(B) . We select the basic minor. In this case, all unknown systems of linear equations are divided into two classes. The unknowns, the coefficients of which are included in the basic minor, are called dependent, and the unknowns, the coefficients of which are not included in the basic minor, are called free. Note that the choice of dependent and free unknowns is not always unique.
3. We cross out those equations of the system whose coefficients are not included in the basic minor, since they are consequences of the rest (according to the basic minor theorem).
4. 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.
5. The resulting system is solved in one of the following ways: the Cramer method, the inverse matrix method, or the Jordan-Gauss method. Relations are found that express the dependent variables in terms of the free ones.

Definition. System m equations with n unknowns in general view is written as follows:

Where aij are coefficients, and b i- permanent.

The solutions of the system are n numbers that, when substituted into the system, turn each of its equations into an identity.

Definition. If a system has at least one solution, then it is called compatible. If the system has no solution, then it is called inconsistent.

Definition. A system is called definite if it has only one solution and indefinite if it has more than one.

Definition. For a system of linear equations, the matrix

A = is called the matrix of the system, and the matrix

A*= is called the augmented matrix of the system

Definition. If b 1 , b 2 , …,b m = 0, then the system is said to be homogeneous. Comment. A homogeneous system is always consistent, because always has a zero solution.

Elementary transformations of systems.

1. Adding to both parts of one equation the corresponding parts of the other, multiplied by the same number, not equal to zero.

2. Rearrangement of equations in places.

3. Removal from the system of equations that are identities for all X.

Cramer formulas.

This method is also applicable only in the case of systems of linear equations, where the number of variables coincides with the number of equations.

Theorem. System of n equations with n unknowns

if the determinant of the matrix of the system is not equal to zero, then the system has a unique solution and this solution is found by the formulas: x i = Where D = detA, A D i is the determinant of the matrix obtained from the matrix of the system by changing the column i free members column b i.

D i =

Example. Find a solution to the system of equations:

D \u003d \u003d 5 (4 - 9) + (2 - 12) - (3 - 8) \u003d -25 - 10 + 5 \u003d -30;

D 1 \u003d \u003d (28 - 48) - (42 - 32) \u003d -20 - 10 \u003d -30.

D 2 \u003d\u003d 5 (28 - 48) - (16 - 56) \u003d -100 + 40 \u003d -60.

D 3 \u003d \u003d 5 (32 - 42) + (16 - 56) \u003d -50 - 40 \u003d -90.

Remark 1. If the system is homogeneous, i.e. b i = 0, then for D¹0 the system has a unique zero solution x 1 \u003d x 2 \u003d ... \u003d x n \u003d 0.

Remark 2. At D=0 The system has an infinite number of solutions.

Inverse matrix method.

Matrix method applicable to solving systems of equations where the number of equations is equal to the number of unknowns.

Let the system of equations be given: Let's make matrices:

A= - matrix of coefficients for variables or system matrix;

B = - matrix-column of free members;

X = - matrix - column of unknowns.

Then the system of equations can be written: A×X = B. Multiply on the left both sides of the equality by A -1: A -1 ×A×X = A -1 ×B since A -1 × A \u003d E, That E × X \u003d A -1 × B, then the following formula is true:

X \u003d A -1 × B

Thus, to apply this method, it is necessary to find inverse matrix.

Example. Solve the system of equations:

X = , B = , A =

Find the inverse matrix A -1 .

D = det A = 5(4-9) + 1(2 - 12) - 1(3 - 8) = -25 - 10 +5 = -30≠0 ⇒ inverse matrix exists.

M 11 = ; M21 = ; M 31 = ;

M 12 = M 22 = M 32 =

M 13 = M 23 = M 33 =

A -1 = ;

Let's check:

A×A -1 =
=E.

We find the X matrix.

X \u003d \u003d A -1 B \u003d × = .

We got system solutions: x=1; y=2; z = 3.

4. Gauss method.

Let the system m linear equations with n unknown:

Assuming that in the system the coefficient a 11 is different from zero (if this is not the case, then the equation with a non-zero coefficient at x 1). We transform the system as follows: we leave the first equation unchanged, and exclude the unknown from all other equations x 1 using equivalent transformations as described above.

In the resulting system

,

assuming that (which can always be obtained by rearranging the equations or terms inside the equations), we leave the first two equations of the system unchanged, and from the remaining equations, using the second equation, using elementary transformations, we exclude the unknown x 2. In the newly received system

subject to the condition, we leave the first three equations unchanged, and from all the rest, using the third equation, elementary transformations exclude the unknown x 3 .

This process continues until one of three possible cases is realized:

1) if as a result we arrive at a system, one of the equations of which has zero coefficients for all unknowns and a nonzero free term, then the original system is inconsistent;

2) if as a result of transformations we obtain a system with a matrix of triangular coefficients, then the system is compatible and is definite;

3) if a stepwise system of coefficients is obtained (and the condition of paragraph 1 is not satisfied), then the system is consistent and indefinite.

Consider the square system : (1)

This system has a coefficient a 11 is different from zero. If this condition were not met, then in order to obtain it, it would be necessary to rearrange the equations, putting first the equation for which the coefficient at x 1 is not equal to zero.

Let us carry out the following transformations of the system:

1) since a 11 ¹0, we leave the first equation unchanged;

2) instead of the second equation, we write the equation obtained by subtracting the first one multiplied by 4 from the second equation;

3) instead of the third equation, we write the difference between the third and the first, multiplied by 3;

4) instead of the fourth equation, we write the difference between the fourth and the first, multiplied by 5.

Received new system is equivalent to the original one and has in all equations, except for the first one, zero coefficients at x 1 (this was the goal of transformations 1 - 4): (2)

For the above transformation and for all further transformations, one should not completely rewrite the entire system, as has just been done. The initial system can be represented as a matrix

. (3)

Matrix (3) is called expanded matrix for the original system of equations. If we remove the column of free members from the expanded matrix, we get system coefficient matrix, which is sometimes called simply system matrix.

System (2) corresponds to the augmented matrix

.

Let's transform this matrix as follows:

1) we will leave the first two lines unchanged, since the element a 22 is not zero;

2) instead of the third line, we write the difference between the second line and the doubled third;

3) the fourth row is replaced by the difference between the second row doubled and the fourth row multiplied by 5.

The result is a matrix corresponding to a system whose unknown x 1 is excluded from all equations except the first, and the unknown x 2 - from all equations except the first and second:

.

Now we eliminate the unknown x 3 from the fourth equation. To do this, we transform the last matrix as follows:

1) the first three lines will be left unchanged, since a 33 ¹ 0;

2) the fourth line is replaced by the difference between the third, multiplied by 39, and the fourth: .

The resulting matrix corresponds to the system

. (4)

From the last equation of this system we obtain x 4 = 2. Substituting this value into the third equation, we get x 3 = 3. Now it follows from the second equation that x 2 = 1, and from the first - x 1 = -1. It is obvious that the obtained solution is unique (since the value x 4 , then x 3, etc.).

Definition: Let's call a square matrix, which has numbers other than zero on the main diagonal, and zeros under the main diagonal, triangular matrix.

The coefficient matrix of system (4) is a triangular matrix.

Comment: If, with the help of elementary transformations, the matrix of coefficients of a square system can be reduced to a triangular matrix, then the system is consistent and definite.

Consider another example: . (5)

Let us carry out the following transformations of the extended matrix of the system:

1) leave the first line unchanged;

2) instead of the second line, we write the difference between the second line and twice the first;

3) instead of the third line, we write the difference between the third line and triple the first;

4) the fourth row is replaced by the difference between the fourth and the first;

5) the fifth row is replaced by the difference between the fifth row and twice the first one.

As a result of the transformations, we obtain the matrix

.

Leaving the first two rows of this matrix unchanged, we reduce it by elementary transformations to the following form:

.

If now, following the Gauss method, which is also called the method of successive elimination of unknowns, using the third line, bring the coefficients at zero to zero x 3 in the fourth and fifth rows, then after dividing all the elements of the second row by 5 and dividing all the elements of the third row by 2, we get the matrix

.

Each of the last two rows of this matrix corresponds to the equation 0 x 1 +0x 2 +0x 3 +0x 4 +0x 5 = 0. This equation is satisfied by any set of numbers x 1 ,x 2, ¼, x 5 and should be removed from the system. Thus, the system with the augmented matrix just obtained is equivalent to the system with the augmented matrix of the form

. (6)

The last row of this matrix corresponds to the equation
x 3 – 2x 4 + 3x 5 = -4. If unknown x 4 and x 5 give arbitrary values: x 4 = From 1; x 5 = From 2, then from the last equation of the system corresponding to matrix (6), we obtain x 3 = –4 + 2From 1 – 3From 2. Substituting expressions x 3 ,x 4 , and x 5 into the second equation of the same system, we get x 2 = –3 + 2From 1 – 2From 2. Now from the first equation we can get x 1 = 4 – From 1+ From 2. The final solution of the system is represented in the form .

Consider a rectangular matrix A, which has the number of columns m greater than the number of rows n. Such a matrix A let's call stepped.

Obviously, matrix (6) is a step matrix.

If, when applying equivalent transformations to a system of equations, at least one equation is reduced to the form

0x 1 + 0x 2 + ¼0 x n = bj (bj ¹ 0),

then the system is inconsistent or inconsistent, since no set of numbers x 1 , x 2, ¼, x n does not satisfy this equation.

If, when transforming the extended matrix of the system, the matrix of coefficients is reduced to a step form and the system does not turn out to be inconsistent, then the system is consistent and is indefinite, that is, it has infinitely many solutions.

In the latter system, all solutions can be obtained by giving specific numerical values ​​to the parameters From 1 And From 2.

Definition: Those variables whose coefficients are on the main diagonal of the step matrix (this means that these coefficients are nonzero) are called o main. In the example above, these are the unknowns x 1 , x 2 , x 3 . The rest of the variables are called minor. In the example above, these are the variables x 4 , and x 5 . Non-primary variables can be assigned any value or expressed through parameters, as is done in the last example.

Core variables are uniquely expressed in terms of non-core variables.

Definition: If the non-basic variables are given specific numerical values ​​and the main variables are expressed through them, then the resulting solution is called private decision.

Definition: If non-basic variables are expressed in terms of parameters, then a solution is obtained, which is called general solution.

Definition: If all non-primary variables are given zero values, then the resulting solution is called basic.

Comment: The same system can sometimes be reduced to different sets of basic variables. So, for example, you can swap the 3rd and 4th columns in the matrix (6). Then the main variables will be x 1 , x 2 ,x 4 , and minor - x 3 and x 5 .

Definition: If two different sets of basic variables are obtained with different ways of finding a solution to the same system, then these sets necessarily contain the same number of variables, called system rank.

Consider another system that has infinitely many solutions: .

Let us transform the extended matrix of the system using the Gauss method:

.

As you can see, we did not get a step matrix, but the last matrix can be transformed by swapping the third and fourth columns: .

This matrix is ​​already stepwise. The system corresponding to it has two minor variables - x 3 , x 5 and three main - x 1 , x 2 , x 4 . The solution of the original system is presented in the following form:

Here is an example of a system that has no solution:

.

We transform the matrix of the system according to the Gauss method:

.

The last row of the last matrix corresponds to the unsolvable equation 0x1 + 0x2 + 0x3 = 1. Therefore, the original system is inconsistent.

Lecture number 3.

Theme: Vectors. Scalar, vector and mixed product of vectors

1. The concept of a vector. Collinarity, orthogonality and coplanarity of vectors.

2. Linear operation on vectors.

3. Dot product of vectors and its application

4. vector product vectors and its application

5. mixed product vectors and its application

1. The concept of a vector. Collinarity, orthogonality and complanarity of vectors.

Definition: A vector is a line segment with start point A and end point B.

Designation: , ,

Definition: The length or modulus of a vector of a vector is a number, equal to the length segment AB representing a vector.

Definition: A vector is called null if the beginning and end of the vector are the same.

Definition: A vector of unit length is called a unit vector. Definition: Vectors are called collinear if they lie on the same line or on parallel lines. ( || ).

Comment:

1. Collinear vectors can be directed equally or oppositely.

2. Zero vector is considered collinear to any vector.

Definition: Two vectors are said to be equal if they are collinear,

have the same direction and the same length ( = )