Joint slough. Solution of systems of linear algebraic equations of general form. Gaussian method for solving slough

However, in practice, two more cases are widespread:

- The system is incompatible (has no solutions);
- The system is compatible and has infinitely many solutions.

Note : The term "interoperability" implies that the system has at least some solution. In a number of tasks, it is required to first investigate the system for compatibility, how to do this - see the article about rank of matrices.

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

The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. Let's first consider a couple of examples when 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. If the number of equations is less than the number of variables, then we can immediately say that the system is either incompatible or has infinitely many solutions. And it remains only to find out.

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

(1) On the top left rung, we need to get +1 or –1. There are no such numbers in the first column, so rearranging the rows will do nothing. The unit will have to be organized independently, and this can be done in several ways. I did this: 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 we add the first line multiplied by 3. To the third line we add the first line multiplied by 5.

(3) After the performed transformation, it is always advisable to look, and is it possible to simplify the resulting lines? Can. Divide the second row by 2, at the same time getting the desired –1 on the second step. Divide the third row by –3.

(4) Add the second line to the third line.

Probably, everyone paid attention to the bad line that 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 linear equations:

If, as a result of elementary transformations, a string of the form, where is a nonzero number, is obtained, then the system is incompatible (has no solutions).

How do I record the ending of an assignment? Let's draw with white chalk: "as a result of elementary transformations, a line of the form, where" was obtained and give the answer: the system has no solutions (inconsistent).

If, according to the condition, it is required to RESEARCH the system for compatibility, then it is necessary to issue a solution in a more solid style with the involvement of the concept the rank of the matrix and the Kronecker-Capelli theorem.

Please note that there is no Gauss backtracking here - 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, I remind you that your decision course may differ from my decision course, the Gauss algorithm does not have a strong "rigidity".

Another one technical feature solutions: elementary transformations can be stopped immediately, as soon as a line of the form appeared, where. Consider a conditional example: suppose that after the very first transformation the matrix is ​​obtained ... The 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 appeared, where. You should immediately answer that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift, since 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. Be that as it may, but the Gauss method will lead us to the answer anyway. This is its versatility.

The beginning is again standard. Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

That's all, and you were afraid.

(1) Note that all the numbers in the first column are divisible by 2, so we are happy with two on the top left step. To the second line, add the first line multiplied by –4. To the third line, add the first line multiplied by –2. To the fourth line, 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. Just add: To the fourth line, add the first line multiplied by -1 - exactly!

(2) The last three lines are proportional, two of them can be deleted.

Here again you need to show increased attention, but are the lines really proportional? To be on the safe side (especially for a teapot) it will not be superfluous to multiply the second line by –1, and divide the fourth line by 2, resulting in three identical lines. And only then delete two of them.

As a result of elementary transformations, the expanded matrix of the system is reduced to a stepped form:

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

Let's rewrite the corresponding system of equations:

The only solution of the system here does not smell like "usual". There is no bad line either. This means that this is the third remaining case - the system has infinitely many solutions. Sometimes, by condition, it is necessary to investigate the compatibility of the system (that is, to prove that the solution exists at all), you can read about this in the last paragraph of the article How do I find the rank of a matrix? But for now, we are analyzing the basics:

An infinite number of system solutions are briefly written in the form of the so-called overall system solution .

We find the general solution of the system using the reverse move of the Gauss method.

First we need to determine which variables we have basic and which variables free... It is not necessary to bother with the terms of linear algebra, it is enough to remember that there are such basic variables and free variables.

Basic variables always "sit" strictly on the steps of the matrix.
In this example, the basic variables are and

Free variables are everything remaining variables that did not get a rung. In our case, there are two of them: - free variables.

Now you need all basic variables to express only through free variables.

The reverse of the Gaussian algorithm traditionally works from the bottom up.
From the second equation of the system, we express the basic variable:

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

It remains to express the basic variable in terms of free variables:

In the end, we got what we need - all basic variables (s) are expressed only through free variables:

Actually, the general solution is ready:

How to write down the general solution correctly?
Free variables are written into the general solution "by themselves" and strictly in their places. In this case, free variables should be written in the second and fourth positions:
.

The obtained expressions for the basic variables and, obviously, you need to write in the first and third positions:

Giving free variables arbitrary values, you can find infinitely many private solutions... The most popular values ​​are zeros, since the particular solution is the easiest. Let's substitute in the general solution:

- a private solution.

Units are another sweet couple, let's substitute them in the general solution:

- one more 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 the particular solution must satisfy to each equation of the system. This is the basis for the "quick" check of the correctness of the solution. Take, for example, a particular solution and plug it into the left side of each equation in the original system:

Everything should fit together. And with any particular decision you receive - everything should also agree.

But, strictly speaking, checking a particular solution sometimes deceives, i.e. some particular solution can satisfy each equation of the system, but the general solution itself is actually found incorrectly.

Therefore, the check of the general solution is more thorough and reliable. How to check the resulting general solution ?

It's easy, but pretty dreary. You need to take expressions basic variables, in this case and, and substitute them in the left side of each equation of the system.

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

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


The right-hand side of the original equation is obtained.

Example 4

Solve the system using the Gaussian method. Find a general solution and two particular ones. Check the general solution.

This is an example for a do-it-yourself solution. 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 incompatible or with an infinite set of solutions. What is important in the decision process itself? Attention, attention again... Complete solution and answer at the end of the tutorial.

And a couple more examples to consolidate the material

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, using elementary transformations, bring it to a stepwise 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) Add the second row multiplied by –5 to the third line. To the fourth line, add the second line multiplied by –7.
(3) The third and fourth lines are the same, we delete one of them.

Here is such beauty:

Basic variables sit on rungs, therefore basic variables.
There is only one free variable that did not get a step here:

Reverse:
Let's express the basic variables in terms of a 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:

Yes, a calculator that counts ordinary fractions is still handy.

So the general solution is:

Once again, how did it come about? The free variable sits alone in its rightful fourth place. The resulting expressions for the basic variables also took their ordinal places.

Let's check the general solution right away. Work for blacks, but I have already done it, so catch =)

We substitute three heroes,, in the left side of each equation of the system:

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

Now from the found common solution we get two particular solutions. The only free variable is the chef here. You don't need to break your head.

Let, then - a private solution.
Let, then - one more particular solution.

Answer: Common decision: , particular solutions: , .

I shouldn't have remembered about blacks here ... ... because all sorts of sadistic motives got into my head and I remembered the well-known photo-toad, in which the Klans in white robes are running across the field after a black football player. I sit, smiling quietly. You know how distracting….

A lot of math is harmful, so a similar final example for your own solution.

Example 6

Find the general solution of a system of linear equations.

I have already checked the general solution, the answer can be trusted. Your decision course may differ from my decision course, the main thing is that the general decisions coincide.

Probably, many have noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to tinker with ordinary fractions... In practice, this is true, cases when there are no fractions are much less common. Be prepared mentally, and most importantly, technically.

I will dwell on some of the features of the solution that were not found in the solved examples.

The general solution of the system can sometimes include a constant (or constants), for example:. 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 an A in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables... The Gauss method works in the most severe conditions, you should calmly bring the extended matrix of the system to a stepwise form according to the standard algorithm. Such a system can be inconsistent, it can have infinitely many solutions, and, oddly enough, it can have a single solution.

where x* - one of the solutions of the inhomogeneous system (2) (for example (4)), (E − A + A) forms the kernel (zero space) of the matrix A.

Let's do the skeletal decomposition of the matrix (E − A + A):

E − A + A = Q S

where Q n × n − r- matrix rank (Q) = n − r, S n − r × n-matrix rank (S) = n − r.

Then (13) can be written in as follows:

x = x * + Q k, ∀ k ∈ R n-r.

where k = Sz.

So, general solution procedure systems of linear equations using a pseudoinverse matrix can be represented as follows:

1. Computing the pseudoinverse A + .
2. We calculate a particular solution of the inhomogeneous system of linear equations (2): x*=A + b.
3. We check the compatibility of the system. To do this, calculate AA + b... If AA + b≠b, then the system is inconsistent. Otherwise, we continue the procedure.
4. We hatch E − A + A.
5. Making skeletal decomposition E − A + A = Q S.
6. Building a solution

x = x * + Q k, ∀ k ∈ R n-r.

Solving a system of linear equations online

The online calculator allows you to find a general solution to a system of linear equations with detailed explanations.

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

where a ij and b i (i=1,…,m; b=1,…,n) Are some known numbers, and x 1, ..., x n- unknown. In the designation of the coefficients a ij first index i denotes the equation number, and the second j- the number of the unknown at which this coefficient stands.

We will write the coefficients for unknowns 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 are called free members.

The aggregate n numbers c 1, ..., c n called decision of the given system, if each equation of the system turns into equality after substitution of 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 inconsistent.

Consider ways to find solutions to the system.

MATRIX METHOD FOR SOLVING SYSTEMS OF LINEAR EQUATIONS

Matrices make it possible to concisely 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 terms

Find the work

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

or shorter A∙X = B.

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

Let the determinant of the matrix be nonzero | A| ≠ 0. Then the matrix equation is solved as follows. We multiply both sides of the equation on the left by the matrix A -1, the inverse of the matrix A:. Insofar as 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 be found only for square matrices, the matrix method can be used to solve only those systems in which the number of equations coincides with the number of unknowns... However, the matrix representation 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 will not be 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.

RULE OF CRAMER

Consider a system of 3 linear equations with three unknowns:

Determinant of the third order corresponding to the matrix of the system, i.e. composed of coefficients with unknowns,

called system determinant.

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

Then the following result can be proved.

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, let's consider a system of 3 equations with three unknowns. Let us multiply the 1st equation of the system by the algebraic complement A 11 element a 11, 2nd equation - on A 21 and 3rd - on A 31:

Let's add these equations:

Let's look at 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 obtain the equality:.

Hence, .

The equalities and are derived in a similar way, 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. inconsistent.

Examples. Solve 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 nonzero. Gauss's 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 second and third we exclude the terms containing x 1... For this, we divide the second equation by a 21 and multiply by - a 11 and then add it to the 1st equation. Similarly, we divide the third equation into a 31 and multiply by - a 11, and then add to the first. As a result, the original system will take the form:

Now we exclude from the last equation the term containing x 2... To do this, divide the third equation by, multiply by and add 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 x 2 and finally from the 1st - x 1.

When using the Gaussian method, the equations can be swapped as needed.

Often, instead of writing new system equations are limited 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. rearrangement of rows or columns;
2. multiplying a string by a nonzero number;
3. adding other lines to one line.

Examples: Solve systems of equations by the Gauss method.

Thus, the system has an infinite number of solutions.

Service purpose... The online calculator is designed to study a system of linear equations. Usually, in the problem statement, you need to find general and specific 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 compatible, then it is definite or indefinite (the criterion for the compatibility of the system is determined by the theorem);
3. if the system is definite, then how to find its unique solution (using the Cramer method, the inverse matrix method, or the Jordan-Gauss method);
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 to a system is any set of numbers c 1, c 2, ..., c n, the substitution of which into the system instead of the corresponding unknowns turns each equation of the system into an identity.

Definition... Two systems are called equivalent if the solution of the first is the solution of the second and vice versa.

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

Definition... A system that has 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, according to the Kronecker-Capelli theorem, the system is inconsistent, and this is where the study ends.
2. Let rang (A) = rang (B). Highlight the base minor. In this case, all unknown systems of linear equations are subdivided into two classes. Unknowns, the coefficients of which are included in the basic minor, are called dependent, and unknowns, the coefficients of which did not fall into the basic minor, are called free. Note that the choice of dependent and free unknowns is not always unambiguous.
3. We delete those equations of the system whose coefficients are not included in the basic minor, since they are consequences of the others (by the theorem on the basic minor).
4. The terms of the equations containing free unknowns are transferred to the right-hand side. As a result, we obtain a system of r equations with r unknowns, equivalent to the given one, the determinant of which is nonzero.
5. The resulting system is solved in one of the following ways: Cramer's method, the inverse matrix method, or the Jordan-Gauss method. Relationships are found that express dependent variables in terms of free ones.

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

1) the system is incompatible (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 lead to the answer, but in higher mathematics it is customary to use the Gaussian method of successive elimination of unknowns. For those who are not familiar with the Gaussian 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. Let's first consider a couple of examples when 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 states: "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 completely ordinary - we write down the extended matrix of the system and, using elementary transformations, we bring it to a stepwise form:

(1). On the top left step, we need to get (+1) or (–1). There are no such numbers in the first column, so rearranging the rows will do nothing. The unit will have to be organized independently, and this can be done in several ways. We did this. To the first line add the third line multiplied by (–1).

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

(3). After the performed transformation, it is always advisable to look, and is it possible to simplify the resulting lines? Can. Divide the second row by 2, at the same time getting the desired (–1) on the second step. Divide the third row by (–3).

(4). Add the second line to the third line. Probably, everyone paid attention to the bad line that 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 - a number other than zero, then the system is incompatible (has no solutions).

How do I record the ending of an assignment? You must write down the phrase:

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

Please note that in this case there is no backtracking 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 a do-it-yourself solution. Complete solution and answer at the end of the tutorial.

Again, we remind you that your decision course may differ from our decision course, the Gauss method does not specify an unambiguous algorithm, you have to guess about the order of actions and the actions themselves in each case independently.

Another technical feature of the solution: elementary transformations can be stopped immediately, as soon as a line of the form appeared, where λ ≠ 0 ... Consider a conditional example: suppose that after the very first transformation the matrix is ​​obtained

.

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

When a system of linear equations has no solutions, this is almost a gift to the student, since 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. Be that as it may, but the Gauss method will lead us to the answer anyway. This is its versatility.

The beginning is again standard. Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

That's all, and you were afraid.

(1). Please note that all the numbers in the first column are divisible by 2, so we are satisfied with the two on the upper left step. To the second line add the first line multiplied by (–4). To the third line, add the first line multiplied by (–2). To the fourth line, 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. Just add: to the fourth line add the first line multiplied by (–1) - exactly!

(2). The last three lines are proportional, two of them can be deleted. Here again you need to show increased attention, but are the lines really proportional? To be on the safe side, it will not be superfluous to multiply the second line by (–1), and divide the fourth line by 2, resulting in three identical lines. And only then delete two of them. As a result of elementary transformations, the expanded matrix of the system is reduced to a stepped form:

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

Let's rewrite the corresponding system of equations:

The only solution of the system here does not smell like "usual". A bad line where λ ≠ 0, also no. This means that this is the third remaining case - the system has infinitely many solutions.

An infinite number of system solutions are briefly written in the form of the so-called overall system solution.

We find the general solution of the system using the reverse move 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 which variables we have basic, and which variables - free... It is not necessary to explain in detail the terms of linear algebra, it is enough to remember that there are such basic variables and free variables.

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

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

Now you need allbasic variables to express only throughfree variables... The reverse 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 let's look at the first equation: ... First, we substitute the found expression into it:

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

In the end, we got what we need - all basic 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 correctly? First of all, free variables are written into the general solution "by themselves" and strictly in their places. In this case, free variables x 2 and x 4 should be written in the second and fourth positions:

.

The obtained expressions for the basic variables and, obviously, you need to write in the first and third positions:

From the general solution of the system, you can find infinitely many private solutions... It’s very simple. Free variables x 2 and x 4 are called so because they can be given any end values... The most popular values ​​are zero, because this is the easiest way to get the 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 at values ​​( x 2 = 0; x 4 = 0).

Units are another sweet couple, substitute ( x 2 = 1 and x 4 = 1) into a 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 the particular solution must satisfy to each equation of the system. This is the basis for the "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 of the original system:

Everything should fit together. And with any particular decision you receive - everything should also agree.

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

How to check the resulting general solution ?

It is not difficult, but it requires a lot of time-consuming transformations. You need to take expressions basic variables, in this case and, and substitute them in the left side of each equation of the system.

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

The right-hand side of the initial first equation of the system is obtained.

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

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

And further - to the left-hand sides of the third and fourth equations of the system. This check takes longer, but it guarantees one hundred percent correctness of the overall solution. In addition, in some tasks, it is precisely the verification of the general solution that is required.

Example 4:

Solve the system using the Gaussian method. Find a general solution and two particular ones. Check the general solution.

This is an example for a do-it-yourself solution. 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 incompatible or with an infinite set 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, using elementary transformations, bring it to a stepwise 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, add the second line multiplied by (–5). To the fourth line, add the second line multiplied by (–7).

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

Basic variables sit on rungs, therefore basic variables.

There is only one free variable that did not get a step here:.

(4). Reverse move. Let's express the basic variables in terms of a 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 for one free variable x 4:

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

Let's check the general solution right away.

We substitute the basic variables,, in 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 common 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 be x 4 = 0, then - the first particular solution.

Let be x 4 = 1, then - one more particular solution.

Answer: Common decision: ... Private solutions:

and .

Example 6:

Find the general solution of a system of linear equations.

We have already checked the general decision, the answer can be trusted. Your course of decision may differ from our course of decision. The main thing is for common decisions to 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 when 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 can sometimes include a constant (or constants).

For example, the general solution is:. 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 an A in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables... However, the Gauss method works under the harshest conditions. It is necessary to calmly reduce the expanded matrix of the system to a stepped form according to the standard algorithm. Such a system can be inconsistent, it can have infinitely many solutions, and, oddly enough, it can have a single solution.

Let's repeat in our advice - in order to feel comfortable when solving a system using the Gaussian 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 stepwise form.

Elementary transformations performed:

(1) The first and third lines are reversed.

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

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

As a result of elementary transformations, a string of the form, where λ ≠ 0 .This means that the system is incompatible.Answer: no solutions.

Example 4:

Solution:Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise 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 one for the second step , and transformation (2) is aimed at obtaining it.

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

(3). The second and third lines were swapped (rearranged the resulting -1 to the second step)

(4). The third line was added to the second 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:

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

(2). Let's 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 get acquainted with the concept complex number, consider algebraic, trigonometric and exemplary form complex number. We will also learn how to perform actions with complex numbers: addition, subtraction, multiplication, division, exponentiation and root extraction.

To master complex numbers, you do not need any special knowledge from the course of higher mathematics, and the material is available even to a student. It is enough to be able to perform algebraic operations with "ordinary" numbers, and remember trigonometry.

First, let's remember the "ordinary" Numbers. In mathematics, they are called many real numbers and denoted by the letter R, or R (thickened). All real numbers sit on the familiar number line:

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