Solution using the Gauss method. Return of the Gauss method
Today we understand with the Gauss method for solving linear systems algebraic equations. What kind of systems is, you can read in the previous article dedicated to the solution of the same Slava method of Cramer. The Gauss method does not require any specific knowledge, only care and sequence are needed. Despite the fact that in terms of mathematics, school training is enough for its application, students have the development of this method often causes complexity. In this article, let's try to reduce them!
Gauss method
M. etod Gaussa - Most. universal method solutions Slava (except for well large Systems). In contrast to the previously discussed, it is suitable not only for systems having only decision, but also for systems that have infinite set solutions. Here are three options.
- The system has a single solution (the determinant of the main matrix of the system is not equal to zero);
- The system has infinite set solutions;
- No solutions, the system is incomplete.
So, we have a system (even if she has one solution), and we are going to solve it by Gauss method. How it works?
The Gauss method consists of two stages - direct and reverse.
Direct Gaussian Method
First write the expanded system matrix. To do this, in the main matrix add a column of free members.
The whole essence of the Gauss method is that by elementary transformations, bring this matrix to the step (or as they also say triangular). In this form under (or above), the main diagonal of the matrix must be alone zeros.
What can I do:
- You can rearrange the strings of the matrix by places;
- If there are identical (or proportional) lines in the matrix, you can delete them all but one;
- You can multiply or divide the string to any number (except zero);
- Zero lines are removed;
- You can add a string line multiplied by a number other than zero.
Return of the Gauss method
After we transform the system in this way, one unknown XN. becomes known, and you can reverse order Find all the remaining unknowns, substituting already known ices in the system equation, up to the first.
When the Internet is always at hand, you can solve the system of equations by Gauss online.It is enough to drive the coefficients to the online calculator. But you agree, much more pleasant to realize that an example is decided not to a computer program, but your own brain.
An example of solving the system of equations by Gauss method
And now - an example so that everything becomes clear and understandable. Let the system be given linear equationsAnd you need to solve it by Gauss:
First, write down an extended matrix:
Now we will deal with transformations. We remember that we need to achieve the triangular species of the matrix. I multiply the 1st line on (3). Multiply 2nd string on (-1). Add a 2nd line to the 1st and get:
Then I multiply the 3rd line on (-1). Add a 3rd line to the 2nd:
Multiply 1st line on (6). Multiply 2nd string on (13). Add a 2nd string to the 1st:
Voila - the system is given to the appropriate form. It remains to find unknowns:
System B. this example It has a single decision. Solution solutions with infinite set solutions We will look at in a separate article. Perhaps first you will not know where to start converting the matrix, but after appropriate practice, you will leave the hand and click the Gauss method as nuts. And if you suddenly come across the Slava, which will be too strong a nut, contact our authors! You can, leaving an application in the nameball. Together we will solve any task!
We continue to consider systems of linear equations. This lesson is the third on the topic. If you are vague imagine that such a system of linear equations in general, feel like a kettle, I recommend starting with the osses on the page further useful to study the lesson.
Gauss method is easy! Why? The famous German mathematician Johann Karl Friedrich Gauss still received recognition of the greatest mathematics of all times, genius and even the nickname "King of Mathematics". And everything is ingenious, as you know - just!By the way, not only suckers get on money, but also genius - the portrait of Gaussa was concerned about the bill at 10 daiffers (before the introduction of the euro), and until now, Gauss mysteriously smiles in Germans from ordinary postage stamps.
The Gauss method is simple that there is enough knowledge of the fifth grader for his development. You need to be able to fold and multiply!It is not by chance that the method of consistent exclusion of unknown teachers are often considered on school mathematical facultatives. Paradox, but students of the Gauss method causes the greatest difficulties. Nothing amazing is the whole thing in the technique, and I will try in an affordable form to tell about the method algorithm.
First, some systematize knowledge of the systems of linear equations. The system of linear equations can:
1) Have the only solution. 2) have infinitely many solutions. 3) not to have solutions (be non-stop).
Gauss method - the most powerful and universal tool for finding a solution any Systems of linear equations. How do we remember cramer rule and matrix method Understandable in cases where the system has infinitely many solutions or inconsistent. A method consistent exception Unknown anywaywill lead us to the answer! In this lesson, we again consider the Gauss method for the case number 1 (the only solution of the system), an article is assigned under the situation of paragraphs No. 2-3. I note that the algorithm of the method itself in all three cases works equally.
Return to K. simplest system From lesson How to solve a system of linear equations? and solving it method Gauss.
At the first stage you need to record extended system matrix:. What principle the coefficients are recorded, I think everyone can see. The vertical feature inside the matrix does not carry any mathematical meaning - it is just a drawing for the convenience of design.
reference : i recommend remember terms linear algebra. System matrix - This is a matrix, compiled only from coefficients at unknown, in this example, a system matrix: . Expanded system matrix - This is the same matrix of the system plus a column of free members, in this case: . Any of the matrices can be called simply matrix for brevity.
After the extended system matrix is \u200b\u200brecorded, it is necessary to perform some actions that are also called elementary transformations.
There are the following elementary transformations:
1) Strings Matrians can rearrange places. For example, in the matrix under consideration, you can painlessly rearrange the first and second lines: 
2) if there is a matrix (or appeared) proportional (as a special case - the same) lines, then it follows delete From the matrix all these lines besides one. Consider, for example, the matrix 
. In this matrix, the last three lines are proportional, so it is enough to leave only one of them: 
.
3) If a zero string appeared in the matrix during the conversion, it should also delete. I will not draw, it is clear, the zero line is a string in which some zeros.
4) the matrix string can be multiply (divided) for any number non-zero. Consider, for example, the matrix. Here it is advisable to divide the first string to -3, and the second line is to multiply by 2: 
. This action is very useful because it simplifies further conversion of the matrix.
5) This transformation causes the greatest difficulties, but in fact there is nothing complicated either. To the matrix string can add another string multiplied by the numberdifferent from zero. Consider our matrix from practical example:. At first I will write a conversion very detailed. We multiply the first line to -2: 
, I. to the second line add the first string multiplied by -2: 
. Now the first line can be divided "back" to -2 :. As you can see a string that adds Lie – not changed. Always The string is changing to which added ÜT.
In practice, it is so detailed, of course, do not paint, but they write in short: 
Once again: to the second line added the first string multiplied by -2. The string is usually orally or on a draft, while the mental course of calculations is approximately like that:
"I rewrite the matrix and rewrite the first string: 
»
"First first column. At the bottom I need to get zero. Therefore, the unit at the top multiply on -2: and I add the first to the second line: 2 + (-2) \u003d 0. I write down the result in the second string: 
»
"Now the second column. Top -1 multiply on -2 :. To the second line I add the first: 1 + 2 \u003d 3. I write down the result in the second string: 
»
"And the third column. Top -5 multiply on -2 :. To the second line I add the first: -7 + 10 \u003d 3. I write down the result in the second line: 
»
Please understand this example carefully and disperse in a consecutive calculation algorithm if you understand this, then the Gauss method is practically "in your pocket." But, of course, we still work on this transformation.
Elementary transformations do not change the solution of the system of equations
! ATTENTION: Considered manipulations can not useIf you are prompted to task where the matrices are given by "alone." For example, at "classic" actions with matrices Something to rearrange inside the matrices in no case! Let's go back to our system. It is almost disassembled around the bones.
We write an extended system matrix and with the help of elementary transformations we give it to standard:
(1) The second line added the first string multiplied by -2. And again: Why do first line multiply on -2? In order to take zero below, and therefore get rid of one variable in the second line.
(2) We divide the second string by 3.
The purpose of elementary transformations
–
Lead a matrix to a step above: 
. In the design of the task, it is directly drawn up with a simple pencil "staircase", and also rub the numbers with circles that are located on the "steps". The term "stepped appearance" itself is not quite theoretical, in scientific and educational literature he is often called trapezoid species or triangular view.
As a result of elementary transformations received equivalent The initial system of equations:
Now the system needs to be "promoting" in the opposite direction - from the bottom up, this process is called return of the Gauss method.
In the lower equation, we have a ready-made result :.
Consider the first system equation and substitute the already known meaning of "Igarek" in it:
Consider the most common situation when the Gauss method is required to solve the system of three linear equations with three unknown.
Example 1.
Solve the Gauss method of the system of equations: 
We write an extended system matrix: 
Now I immediately draw the result to which we will come during the solution: 
And I repeat, our goal - with the help of elementary transformations, lead the matrix to the stepped form. Where to start actions?
First we look at the left top number: 
Almost always should be here unit. Generally speaking, it will arrange and -1 (and sometimes other numbers), but somehow it has traditionally developed that it is usually placed one. How to organize a unit? We look at the first column - the finished unit we have! Transformation first: We change the first and third lines in places: 
Now the first line will remain unchanged until the end of the decision. Now fine.
Unit in the upper left corner is organized. Now you need to get zeros in these places: 
Zeros get just with the help of a "hard" transformation. First we designate with the second string (2, -1, 3, 13). What needs to be done to get zero in the first position? Need to to the second line add the first line multiplied by -2. Mentally or on a draft multiply the first string to -2: (-2, -4, 2, -18). And consistently carry out (again mentally or on the draft) addition, to the second line add the first string, already multiplied by -2:
The result is recorded in the second line: 
Similarly deal with the third line (3, 2, -5, -1). To get in the first position zero, you need to the third line, add the first string multiplied by -3. Mentally or on a draft multiply the first string to -3: (-3, -6, 3, -27). AND to the third line add the first string multiplied by -3:
The result is written to the third line:
In practice, these actions are usually performed orally and recorded in one step: 
No need to consider everything immediately and at the same time. The procedure for calculations and "fitting" results sequence And usually such: first rewrite the first string, and let yourself swollen - sequentially and CAREFULLY:
I have already considered the mental course of the calculations yourself.
In this example, it is easy to do, we divide the second line to -5 (as there all numbers are divided into 5 without a residue). At the same time we divide the third line on -2, because the smaller the number, the easier it is:
At the final stage of elementary transformations, you need to get another zero here: 
For this to the third line add a second string multiplied by -2:
Try to disassemble this action yourself - mentally multiply the second string on -2 and make addition.
The last action is the hairstyle, we divide the third line by 3.
As a result of elementary transformations, an equivalent source system of linear equations was obtained: 
Cool.
The reverse move of the Gauss method comes into effect. The equations are "unwind" from the bottom up.
In the third equation, we already have a ready-made result:
We look at the second equation :. The value "Zet" is already known, so:
And finally, the first equation :. "Igarek" and "Zet" are known, it's small:
Answer: 
As it has already been repeatedly noted, it is possible for any system of equations and need to check the solution found, good, it is easy and quickly.
Example 2.
This is an example for an independent solution, a sample of a clean design and response at the end of the lesson.
It should be noted that your proceedings may not coincide with my decision of the decision, and this is the feature of the Gauss method.. But now the answers must be equal to the same!
Example 3.
Solve the system of linear equations by Gauss 
We look at the left upper "step". There we must have a unit. The problem is that there are no units in the first column at all, so nothing to solve the permutation of the rows. In such cases, one needs to be organized using an elementary transformation. This can usually be done in several ways. I did this: (1) To the first line add a second string multiplied by -1. That is, mentally multiplied the second line on -1 and completed the addition of the first and second line, while we did not change the second line.
Now on the left at the top of "minus one" that it is quite suitable. Who wants to get +1, can perform additional television: multiply the first string on -1 (change the sign from it).
(2) The second line added the first string multiplied by 5. To the third line added the first string multiplied by 3.
(3) The first string was multiplied by -1, in principle, it is for beauty. The third line also changed the sign and rearranged it in second place, so on the second "step we had the desired unit.
(4) To the third line added a second string multiplied by 2.
(5) The third line was divided into 3.
A bad feature that indicates an error in calculations (less often about typing) is the "bad" bottom line. That is, if we were at the bottom, something like, and, accordingly, 
, with a large probability, it can be argued that an error is made in the course of elementary transformations.
We charge the reverse move, in the design of examples often do not rewrite the system itself, and the equations "take directly from the given matrix". Return, I remind you, works, bottom up. Yes, here the gift turned out:
Answer: 
.
Example 4.
Solve the system of linear equations by Gauss 
This is an example for an independent solution, it is somewhat more complicated. Nothing terrible if someone is confused. Complete solution and sample design at the end of the lesson. Your decision may differ from my decision.
In the last part, consider some features of the Gauss algorithm. The first feature is that sometimes there are no variables in the system equations, for example: 
How to record an extended system matrix? About this moment I already talked at the lesson Cramer rule. Matrix method. In the extended matrix of the system on the site of the missing variables, we put zeros: 
By the way, this is a fairly easy example, since in the first column there is already one zero, and there are fewer elementary transformations.
The second feature consists in this. In all the examples considered, we placed either -1, or +1. Can there be other numbers there? In some cases can. Consider the system: 
.
Here on the left upper "step" we have a two. But we notice the fact that all numbers in the first column are divided into 2 without a residue - and the other twice and six. And the deuce left at the top will suit us! In the first step, you need to perform the following transformations: to add the first string multiplied by -1 to the second line; To the third line, add the first string multiplied by -3. Thus, we get the desired zeros in the first column.
Or one more conditional example: 
. Here Troika on the second "step" is satisfied with us, because 12 (the place where we need to get zero) is divided by 3 without a balance. It is necessary to carry out the following transformation: to the third line, add a second string multiplied by -4, as a result of which the zero we need will be obtained.
Gaussian method is universal, but there is one peculiarity. Confidently learn how to solve the systems by other methods (by the Cramer method, matrix method) You can literally the first time - there is a very rigid algorithm. But in order to confidently feel in the Gauss method, you should "fill your hand", and break at least 5-10 ten systems. Therefore, it is possible to confusion, errors in calculations, and there is nothing unusual or tragic.
Rainy autumn weather outside the window .... Therefore, for all those who wish a more complex example for an independent solution:
Example 5.
To solve the Gauss method of 4 linear equations with four unknowns. 
Such a task in practice is not so rare. I think even a teapot who studied this page in detail, the algorithm for solving such a system is intuitively understood. In principle, everything is the same - just more action.
Cases when the system has no solutions (inconsistent) or has infinitely many solutions, considered in the lesson Invobilities and systems with a general solution. There you can also consolidate the considered algorithm of the Gauss method.
I wish you success!
Solutions and answers:
Example 2:
Decision
:
We write the expanded matrix of the system and with the help of elementary transformations we give it to a stepped form.
Performed elementary transformations:
(1) The second line added the first string multiplied by -2. To the third line added the first string multiplied by -1.
Attention!
Here, there may be a temptation from the third line to subtract the first, I extremely not recommended to deduct - the risk of error is strongly rising. Just fold!
(2) The second line changed the sign (multiplied by -1). Second and third lines changed places.
note
That on the "steps" is satisfied with us not only the unit, but also -1, which is even more convenient.
(3) To the third line added a second string multiplied by 5.
(4) The second line changed the sign (multiplied by -1). The third line was divided into 14.
Return:
Answer
:
.
Example 4:
Decision
:
We write the expanded matrix of the system and with the help of elementary transformations we give it to the step type:
Conversion performed: (1) The second line was added to the second. Thus, the desired unit is organized on the left upper "step". (2) The second line was added to the first string multiplied by 7. To the third line added the first line multiplied by 6.
With the second "step" is worse , "Candidates" on it - numbers 17 and 23, and we need either a single one or -1. Transformation (3) and (4) will be aimed at obtaining the desired unit (3) To the third line added the second, multiplied by -1. (4) The second line added the third, multiplied by -3. The desired thing on the second step is obtained . (5) To the third line added the second, multiplied by 6. (6) The second line was multiplied by -1, the third line was divided into -83.
Return:
Answer :
Example 5:
Decision
:
We write the system matrix and with the help of elementary transformations we give it to the step type:
Conversion performed: (1) The first and second lines changed places. (2) The second line added the first string multiplied by -2. To the third line added the first line multiplied by -2. To the fourth line added the first string multiplied by -3. (3) To the third line added the second, multiplied by 4. To the fourth line added the second, multiplied by -1. (4) The second line changed the sign. The fourth string was divided into 3 and placed instead of the third line. (5) The fourth line added the third line multiplied by -5.
Return:
Answer :
Let the system be given, δ ≠ 0. (one)Gauss method - This is the method of consistent exclusion of unknown.
The essence of the Gauss method consists in converting (1) to a system with a triangular matrix, from which the values \u200b\u200bof all unknowns are then sequentially (reverse). Consider one of the computing schemes. This scheme is called a single division scheme. So, consider this scheme. Let A 11 ≠ 0 (the presenter element) divide the first equation to A 11. Receive
(2)
Using equation (2), it is easy to exclude unknown x 1 from the remaining system equations (for this, it is enough to subtract equation (2) pre-multiplied by the corresponding coefficient at x 1), that is, in the first step we get
.
In other words, on 1 step, each element of subsequent rows, starting with the second, is equal to the difference between the original element and the product of its "projection" to the first column and the first (converted) string.
Following this, leaving the first equation at rest over the rest of the system equations obtained in the first step, we will make a similar transformation: choose an equation with a leading element from their number and exclude from it from the remaining equations x 2 (step 2).
After n steps instead of (1) we get the equivalent system 
(3)
Thus, at the first stage, we get a triangular system (3). This stage is called direct stroke.
In the second stage (reverse stroke) we find sequentially from (3) the values \u200b\u200bof x n, x n -1, ..., x 1.
Denote the solution received for x 0. Then the difference ε \u003d b-a · x 0 called hopeless.
If ε \u003d 0, the solution found x 0 is true.
Calculations according to the Gauss method are performed in two stages:
- The first stage is called the direct progress of the method. At the first stage, the initial system is converted to a triangular form.
- The second stage is called reverse. At the second stage, a triangular system is solved equivalent to the initial one.
At each step it was assumed that the leading element is different from zero. If this is not the case, you can use any other element as the master, as if to rearrange the system equation.
Appointment of Gauss method
The Gauss method is designed to solve systems of linear equations. Refers to direct decision methods.Types of Gauss method
- Classic Gauss method;
- Modifications of the Gauss method. One of the modifications of the Gauss method is a scheme with a choice of the main element. A feature of the Gauss method with a choice of the main element is such a permutation of equations so that the highest element of the K-one column in the K-th step is the leading element.
- Jordan-Gauss method;
We illustrate the difference jordan-Gauss method From the Gauss method on the examples.
Example Solutions by Gauss
Resolving the system: 
For the convenience of computing, change the lines in places: 
Multiply a 2nd line on (2). Add the 3rd line to the 2nd 
Multiply 2nd string on (-1). Add a 2nd line to the 1st 
From the 1st line express x 3:
From the 2nd string, express x 2:
From the 3rd string, express x 1: 
Example Solutions by Jordan-Gauss
The same Slami solves the method of Jordan-Gauss. 
We will sequentially choose the resolution element of the RE, which lies on the main diagonal of the matrix.
The resolution element is equal to (1).
NE \u003d SE - (A * B) / RE
RE is the resolution element (1), A and B - the elements of the matrices forming a rectangle with the elements of STE and RE.
Imagine the calculation of each element in the form of a table:
| x 1 | x 2 | x 3. | B. |
| 1 / 1 = 1 | 2 / 1 = 2 | -2 / 1 = -2 | 1 / 1 = 1 |
 |  |  |  |
The resolution element is equal to (3).
In place of the resolution element, we get 1, and in the column itself, write zeros.
All other elements of the matrix, including column elements B, are determined by the rules of the rectangle.
To do this, select the four numbers that are located in the vertices of the rectangle and always include the resolution element of the RE.
| x 1 | x 2 | x 3. | B. |
 |  |
||
| 0 / 3 = 0 | 3 / 3 = 1 | 1 / 3 = 0.33 | 4 / 3 = 1.33 |
 |
The resolution element is equal to (-4).
In place of the resolution element, we get 1, and in the column itself, write zeros.
All other elements of the matrix, including column elements B, are determined by the rules of the rectangle.
To do this, select the four numbers that are located in the vertices of the rectangle and always include the resolution element of the RE.
Imagine the calculation of each element in the form of a table:
| x 1 | x 2 | x 3. | B. |
 |  |  |  |
 |  |
||
| 0 / -4 = 0 | 0 / -4 = 0 | -4 / -4 = 1 | -4 / -4 = 1 |
Answer: x 1 \u003d 1, x 2 \u003d 1, x 3 \u003d 1
Implementation of the Gauss method
The Gauss method is implemented in many programming languages, in particular: Pascal, C ++, PHP, Delphi, and there is also an implementation of the Gauss method in online mode.Using Gauss method
Application of the Gauss method in the theory of games
In the game theory, when finding a maximal optimal player strategy, a system of equations is compiled, which is solved by the Gauss method.Application of the Gauss method in solving differential equations
To search for a private solution of the differential equation, the derivatives of the corresponding extent are found for the recorded private solution (Y \u003d F (A, B, C, D)), which are substituted into the initial equation. Next to find variables A, B, C, D A system of equations is compiled, which is solved by the Gauss method.Application of the Jordan-Gauss method in linear programming
In linear programming, in particular in the simplex method for converting the simplex table, the rectangle rule is used on each iteration, which uses the Jordan-Gauss method.Here you can solve the system of linear equations for free. gaussian method online Large sizes in integrated numbers with a very detailed solution. Our calculator can solve online as an ordinary specific and indefinite system of linear equations by Gauss, which has infinite set solutions. In this case, in response, you will receive the dependence of alone variables through other free. You can also check the system of equations for online sharing, using the solution by Gauss method.
About method
When solving a system of linear equations online method Gauss follows the following steps.
- Record the expanded matrix.
- In fact, the solution is divided into direct and the opposite move of the Gauss method. The direct move of the Gauss method is called the leading matrix to a stepped form. The reference of the Gauss method is called the leading matrix to a special step. But in practice it is more convenient to immediately start what is both from above and below the element under consideration. Our calculator uses this approach.
- It is important to note that when solving the Gauss method, the presence in the matrix at least one zero line with nonzero the right part (column of free members) talks about system incompleteness. Decision linear system In this case, there is no.
To best understand the principle of operation of the Gauss algorithm online Enter any example, select "Very detailed solution"And look at his decision online.
Solving systems of linear equations by Gauss method.Let us need to find a solution system from n. Linear Equations S. 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 excluded x 1 from all the equations of the system, starting from the second, then excluded x 2of all the equations, starting from the third, and so on, until the unknown variable will remain in the last equation. x N.. Such a process of converting system equations for consistent exclusion of unknown variables is called direct running of the Gauss method. After completing the direct movement of the Gauss method from the last equation is x N.With this value from the penultimate equation calculated x N-1, and so on, from the first equation is x 1. 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. Let us exclude an unknown variable x 1 Of all the 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 n-OU.equation will add the first multiplied by. The system of equations after such transformations will take the form 
where, A. 
.
To the same result, we would have come if they expressed x 1 Through other unknown variables in the first equation of the system and the resulting expression substituted into all other equations. Thus, variable x 1 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, to the third equation of the system will add the second, multiplied by, to the fourth equation to add the second, multiplied by, and so on, to n-OU.the equation will add the second multiplied by. The system of equations after such transformations will take the form 
where, A. 
. Thus, variable x 2 Excluded from all equations starting from the third.
Next, proceed to the exception 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 on, we begin the reverse course of the Gauss method: Calculate x N. from the last equation how, using the result x N. 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.
