System of linear equations of the 3rd order by the Gauss method. Gauss method (successive exclusion of unknowns)
Carl Friedrich Gauss, the greatest mathematician for a long time hesitated between philosophy and mathematics. Perhaps it was precisely such a mindset that allowed him to "leave" so noticeably in world science. In particular, by creating the "Gauss Method" ...
For almost 4 years, the articles of this site have dealt with school education, mainly from the side of philosophy, the principles of (mis)understanding, introduced into the minds of children. The time is coming for more specifics, examples and methods ... I believe that this is the approach to the familiar, confusing and important areas of life gives the best results.
We humans are so arranged that no matter how much you talk about abstract thinking, but understanding always happens through examples. If there are no examples, then it is impossible to catch the principles ... How impossible it is to be on the top of a mountain otherwise than by going through its entire slope from the foot.
Same with school: for now living stories not enough we instinctively continue to regard it as a place where children are taught to understand.
For example, teaching the Gauss method...
Gauss method in the 5th grade of the school
I will make a reservation right away: the Gauss method has a much wider application, for example, when solving systems linear equations . What we are going to talk about takes place in the 5th grade. This start, having understood which, it is much easier to understand more "advanced options". In this article we are talking about method (method) of Gauss when finding the sum of a series
Here is an example that I brought from school younger son attending the 5th grade of the Moscow gymnasium.
School demonstration of the Gauss method
Math teacher using interactive whiteboard ( modern methods training) showed the children a presentation of the history of the "creation of the method" by little Gauss.
The school teacher whipped little Carl (an outdated method, now not used in schools) for being,
instead of sequentially adding numbers from 1 to 100 to find their sum noticed that pairs of numbers equally spaced from the edges of an arithmetic progression add up to the same number. for example, 100 and 1, 99 and 2. Having counted the number of such pairs, little Gauss almost instantly solved the problem proposed by the teacher. For which he was subjected to execution in front of an astonished public. To the rest to think was disrespectful.
What did little Gauss do developed number sense? Noticed some feature number series with a constant step (arithmetic progression). AND exactly this made him later a great scientist, able to notice, possessing feeling, instinct of understanding.
This is the value of mathematics, which develops ability to see general in particular - abstract thinking. Therefore, most parents and employers instinctively consider mathematics an important discipline ...
“Mathematics should be taught later, so that it puts the mind in order.
M.V. Lomonosov".
However, the followers of those who flogged future geniuses turned the Method into something opposite. As my supervisor said 35 years ago: "They learned the question." Or, as my youngest son said yesterday about the Gauss method: "Maybe it's not worth making a big science out of this, huh?"
The consequences of the creativity of the "scientists" are visible in the level of current school mathematics, the level of its teaching and understanding of the "Queen of Sciences" by the majority.
However, let's continue...
Methods for explaining the Gauss method in the 5th grade of the school
A mathematics teacher at a Moscow gymnasium, explaining the Gauss method in Vilenkin's way, complicated the task.
What if the difference (step) of an arithmetic progression is not one, but another number? For example, 20.
The task he gave the fifth graders:
20+40+60+80+ ... +460+480+500
Before getting acquainted with the gymnasium method, let's look at the Web: how do school teachers - math tutors do it? ..
Gauss Method: Explanation #1
A well-known tutor on his YOUTUBE channel gives the following reasoning:
"let's write the numbers from 1 to 100 like this:
first a series of numbers from 1 to 50, and strictly below it another series of numbers from 50 to 100, but in reverse order"
1, 2, 3, ... 48, 49, 50
100, 99, 98 ... 53, 52, 51
"Please note: the sum of each pair of numbers from the top and bottom rows is the same and equals 101! Let's count the number of pairs, it is 50 and multiply the sum of one pair by the number of pairs! Voila: The answer is ready!".
"If you couldn't understand, don't be upset!" the teacher repeated three times during the explanation. "You will pass this method in the 9th grade!"
Gauss Method: Explanation #2
Another tutor, less well-known (judging by the number of views) uses more scientific approach, offering a solution algorithm of 5 points that must be performed sequentially.
For the uninitiated: 5 is one of the Fibonacci numbers traditionally considered magical. The 5-step method is always more scientific than the 6-step method, for example. ... And this is hardly an accident, most likely, the Author is a hidden adherent of the Fibonacci theory
Given an arithmetic progression: 4, 10, 16 ... 244, 250, 256 .
Algorithm for finding the sum of numbers in a series using the Gauss method:
4, 10, 16 ... 244, 250, 256
256, 250, 244 ... 16, 10, 4
At the same time, you need to remember about plus one rule : one must be added to the resulting quotient: otherwise we will get a result that is one less than the true number of pairs: 42 + 1 = 43.
This is the desired sum of the arithmetic progression from 4 to 256 with a difference of 6!
Gauss method: explanation in the 5th grade of the Moscow gymnasium
And here is how it was required to solve the problem of finding the sum of a series:
20+40+60+ ... +460+480+500
in the 5th grade of the Moscow gymnasium, Vilenkin's textbook (according to my son).
After showing the presentation, the math teacher showed a couple of Gaussian examples and gave the class the task of finding the sum of the numbers in a series with a step of 20.
This required the following:
As you can see, it is more compact and effective technique: the number 3 is also a member of the Fibonacci sequence
My comments on the school version of the Gauss method
The great mathematician would definitely have chosen philosophy if he had foreseen what his followers would turn his "method" into. German teacher who flogged Karl with rods. He would have seen the symbolism and the dialectical spiral and the undying stupidity of the "teachers" trying to measure the harmony of living mathematical thought with the algebra of misunderstanding ....
By the way, do you know. that our education system is rooted in the German school of the 18th and 19th centuries?
But Gauss chose mathematics.
What is the essence of his method?
V simplification. V observation and capture simple patterns of numbers. V turning dry school arithmetic into interesting and fun activity , activating the desire to continue in the brain, and not blocking high-cost mental activity.
Is it possible to calculate the sum of the numbers of an arithmetic progression with one of the above "modifications of the Gauss method" instantly? According to the "algorithms", little Karl would have been guaranteed to avoid spanking, cultivate an aversion to mathematics and suppress his creative impulses in the bud.
Why did the tutor so insistently advise the fifth-graders "not to be afraid of misunderstanding" of the method, convincing them that they would solve "such" problems already in the 9th grade? Psychologically illiterate action. It was a good idea to note: "See? You already in the 5th grade you can solve problems that you will pass only in 4 years! What good fellows you are!"
To use the Gaussian method, level 3 of the class is sufficient when normal children already know how to add, multiply and divide 2-3 digit numbers. Problems arise due to the inability of adult teachers who "do not enter" how to explain the simplest things in a normal human language, not just mathematical ... They are not able to interest mathematics and completely discourage even the "capable".
Or, as my son commented, "make a big science out of it."
Gauss method, my explanations
My wife and I explained this "method" to our child, it seems, even before school ...
Simplicity instead of complexity or a game of questions - answers
""Look, here are the numbers from 1 to 100. What do you see?"
It's not about what the child sees. The trick is to make him look.
"How can you put them together?" The son caught that such questions are not asked "just like that" and you need to look at the question "somehow differently, differently than he usually does"
It doesn't matter if the child sees the solution right away, it's unlikely. It is important that he ceased to be afraid to look, or as I say: "moved the task". This is the beginning of the path to understanding
"Which is easier: add, for example, 5 and 6 or 5 and 95?" A leading question... But after all, any training comes down to "guiding" a person to an "answer" - in any way acceptable to him.
At this stage, there may already be guesses about how to "save" on calculations.
All we have done is hint: the "frontal, linear" counting method is not the only one possible. If the child has truncated this, then later he will invent many more such methods, because it's interesting!!! And he will definitely avoid "misunderstanding" of mathematics, will not feel disgust for it. He got the win!
If baby discovered that adding pairs of numbers that add up to a hundred is a trifling task, then "arithmetic progression with difference 1"- a rather dreary and uninteresting thing for a child - suddenly gave life to him . Out of chaos came order, and this is always enthusiastic: that's the way we are!
A quick question: why, after a child’s insight, should they again be driven into the framework of dry algorithms, which are also functionally useless in this case?!
Why make stupid rewrite sequence numbers in a notebook: so that even the capable would not have a single chance for understanding? Statistically, of course, but mass education is focused on "statistics" ...
Where did zero go?
And yet, adding up numbers that add up to 100 is much more acceptable to the mind than giving 101 ...
The "school Gauss method" requires exactly this: mindlessly fold equidistant from the center of the progression of a pair of numbers, no matter what.
What if you look?
Still, zero greatest invention humanity, which is more than 2,000 years old. And math teachers continue to ignore him.
It's much easier to convert a series of numbers starting at 1 into a series starting at 0. The sum won't change, will it? You need to stop "thinking in textbooks" and start looking ... And to see that pairs with sum 101 can be completely replaced by pairs with sum 100!
0 + 100, 1 + 99, 2 + 98 ... 49 + 51
How to abolish the "rule plus 1"?
To be honest, I first heard about such a rule from that YouTube tutor ...
What do I still do when I need to determine the number of members of a series?
Looking at the sequence:
1, 2, 3, .. 8, 9, 10
and when completely tired, then on a simpler row:
1, 2, 3, 4, 5
and I figure: if you subtract one from 5, you get 4, but I'm quite clear see 5 numbers! Therefore, you need to add one! Number sense developed in primary school, suggests: even if there are a whole Google of members of the series (10 to the hundredth power), the pattern will remain the same.
Fuck the rules?..
So that in a couple of - three years to fill all the space between the forehead and the back of the head and stop thinking? How about earning bread and butter? After all, we are moving in even ranks into the era of the digital economy!
More about the school method of Gauss: "why make science out of this? .."
It was not in vain that I posted a screenshot from my son's notebook...
"What was there in the lesson?"
“Well, I immediately counted, raised my hand, but she didn’t ask. Therefore, while the others were counting, I began to do DZ in Russian so as not to waste time. Then, when the others finished writing (???), she called me to the board. I said the answer."
"That's right, show me how you solved it," said the teacher. I showed. She said: "Wrong, you need to count as I showed!"
“It’s good that I didn’t put a deuce. And I made me write the “decision process” in their own way in a notebook. Why make a big science out of this? ..”
The main crime of a math teacher
hardly after that occasion Carl Gauss experienced a high sense of respect for the school teacher of mathematics. But if he knew how followers of that teacher pervert the essence of the method... he would have roared with indignation and, through the World Intellectual Property Organization WIPO, achieved a ban on the use of his honest name in school textbooks! ..
What main mistake school approach? Or, as I put it, the crime of school mathematics teachers against children?
Misunderstanding algorithm
What do school methodologists do, the vast majority of whom do not know how to think?
Create methods and algorithms (see). This a defensive reaction that protects teachers from criticism ("Everything is done according to ..."), and children from understanding. And thus - from the desire to criticize teachers!(The second derivative of bureaucratic "wisdom", a scientific approach to the problem). A person who does not grasp the meaning will rather blame his own misunderstanding, and not the stupidity of the school system.
What is happening: parents blame the children, and teachers ... the same for children who "do not understand mathematics! ..
Are you savvy?
What did little Carl do?
Absolutely unconventionally approached a template task. This is the quintessence of His approach. This the main thing that should be taught at school is to think not with textbooks, but with your head. Of course, there is also an instrumental component that can be used ... in search of simpler and effective methods accounts.
Gauss method according to Vilenkin
In school they teach that the Gauss method is to
what, if the number of elements in the row is odd, as in the task that was assigned to the son? ..
The "trick" is that in this case you should find the "extra" number of the series and add it to the sum of the pairs. In our example, this number is 260.
How to discover? Rewriting all pairs of numbers in a notebook!(That's why the teacher made the kids do this stupid job, trying to teach "creativity" using the Gaussian method... And that's why such a "method" is practically inapplicable to large data series, And that's why it is not a Gaussian method).
A little creativity in the school routine...
The son acted differently.
(20 + 500, 40 + 480 ...).
0+500, 20+480, 40+460 ...
Easy, right?
But in practice it becomes even easier, which allows you to carve out 2-3 minutes for remote sensing in Russian, while the rest are "counting". In addition, it retains the number of steps of the methodology: 5, which does not allow criticizing the approach for being unscientific.
Obviously this approach is simpler, faster and more versatile, in the style of the Method. But... the teacher not only didn't praise, but also forced me to rewrite it "in the right way" (see screenshot). That is, she made a desperate attempt to stifle the creative impulse and the ability to understand mathematics in the bud! Apparently, in order to later get hired as a tutor ... She attacked the wrong one ...
Everything that I have described so long and tediously can be explained to a normal child in a maximum of half an hour. Along with examples.
And so that he will never forget it.
And it will step towards understanding...not just mathematics.
Admit it: how many times in your life have you added using the Gauss method? And I never!
But instinct of understanding, which develops (or extinguishes) in the process of learning mathematical methods at school ... Oh! .. This is truly an irreplaceable thing!
Especially in the age of universal digitalization, which we quietly entered under the strict guidance of the Party and the Government.
A few words in defense of teachers...
It is unfair and wrong to place all responsibility for this style of teaching solely on school teachers. The system is in operation.
Some teachers understand the absurdity of what is happening, but what to do? Law on Education, Federal State Educational Standards, methods, technological maps lessons... Everything should be done "according to and based on" and everything should be documented. Step aside - stood in line for dismissal. Let's not be hypocrites: the salary of Moscow teachers is very good... If they get fired, where should they go?..
Therefore this site not about education. He is about individual education, only possible way get out of the crowd Generation Z ...
Solving systems of linear equations by the Gauss method. Suppose we need to find a solution to the system from n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.
The essence of the Gauss method consists in the successive exclusion of unknown variables: first, the x 1 from all equations of the system, starting from the second, then x2 of all equations, starting with the third, and so on, until only the unknown variable remains in the last equation x n. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called direct Gauss method. After the completion of the forward move of the Gauss method, from the last equation we find x n, using this value from the penultimate equation is calculated xn-1, and so on, from the first equation is found x 1. The process of calculating unknown variables when moving from the last equation of the system to the first is called reverse Gauss method.
Let us briefly describe the algorithm for eliminating unknown variables.
We will assume that , since we can always achieve this by rearranging the equations of the system. Eliminate the unknown variable x 1 from all equations of the system, starting from the second. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, to n-th add the first equation, multiplied by . The system of equations after such transformations will take the form 
where , a 
.
We would arrive at the same result if we expressed x 1 through other unknown variables in the first equation of the system and the resulting expression was substituted into all other equations. So the variable x 1 excluded from all equations, starting with the second.
Next, we act similarly, but only with a part of the resulting system, which is marked in the figure 
To do this, add the second multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, to n-th add the second equation, multiplied by . The system of equations after such transformations will take the form 
where , a 
. So the variable x2 excluded from all equations, starting with the third.
Next, we proceed to the elimination of the unknown x 3, while we act similarly with the part of the system marked in the figure 
So we continue the direct course of the Gauss method until the system takes the form 
From now on we start reverse stroke Gauss method: calculate x n from the last equation as , using the obtained value x n find xn-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Example.
Solve System of Linear Equations 
Gaussian method.
Gauss method is a method of successive elimination of unknowns.
The essence of the Gauss method is to transform (1) to a system with a triangular matrix , from which the values of all unknowns are then sequentially (reversely) obtained. Let's consider one of the computational schemes. This circuit is called the single division circuit. So let's take a look at this diagram. Let a 11 ≠0 (leading element) divide by a 11 the first equation. Get
(2)
Using equation (2), it is easy to exclude the unknowns x 1 from the remaining equations of the system (for this, it is enough to subtract equation (2) from each equation preliminarily multiplied by the corresponding coefficient at x 1), that is, at the first step we obtain
.
In other words, at step 1, each element of subsequent rows, starting from the second, is equal to the difference between the original element and the product of its “projection” on the first column and the first (transformed) row.
After that, leaving the first equation alone, over the rest of the equations of the system obtained at the first step, we will perform a similar transformation: we choose from among them an equation with a leading element and use it to exclude x 2 from the remaining equations (step 2).
After n steps, instead of (1) we get an equivalent system 
(3)
Thus, at the first stage, we will obtain a triangular system (3). This step is called forward.
At the second stage (reverse move) we sequentially find from (3) the values x n , x n -1 , …, x 1 .
Let's denote the obtained solution as x 0 . Then the difference ε=b-A x 0 is called residual.
If ε=0, then the found solution x 0 is correct.
Calculations by the Gauss method are performed in two stages:
- The first stage is called the direct course of the method. At the first stage, the original system is converted to a triangular form.
- The second stage is called reverse. At the second stage, a triangular system equivalent to the original one is solved.
At each step, it was assumed that the leading element is different from zero. If this is not the case, then any other element can be used as a leader, as if rearranging the equations of the system.
Purpose of the Gauss method
The Gauss method is intended for solving systems of linear equations. Refers to direct methods of solution.Types of Gauss method
- Classical Gauss method;
- Modifications of the Gauss method. One of the modifications of the Gaussian method is the circuit with the choice of the main element. A feature of the Gauss method with the choice of the main element is such a permutation of the equations so that at the k-th step the leading element is the largest element in the k-th column.
- Jordan-Gauss method;
Illustrate the difference Jordan-Gauss method from the Gauss method on examples.
Gauss solution example
Let's solve the system: 
For the convenience of calculations, we swap the lines: 
Multiply the 2nd row by (2). Add the 3rd line to the 2nd 
Multiply the 2nd row by (-1). Add the 2nd row to the 1st 
From the 1st line we express x 3:
From the 2nd line we express x 2:
From the 3rd line we express x 1:
An example of a solution by the Jordan-Gauss method
We will solve the same SLAE using the Jordano-Gauss method. 
We will sequentially choose the resolving element of the RE, which lies on the main diagonal of the matrix.
The enabling element is equal to (1).
NE \u003d SE - (A * B) / RE
RE - enabling element (1), A and B - matrix elements forming a rectangle with elements of STE and RE.
Let's present the calculation of each element in the form of a table:
| x 1 | x2 | x 3 | B |
| 1 / 1 = 1 | 2 / 1 = 2 | -2 / 1 = -2 | 1 / 1 = 1 |
The enabling element is equal to (3).
In place of the resolving element, we get 1, and in the column itself we write zeros.
All other elements of the matrix, including the elements of column B, are determined by the rectangle rule.
To do this, select four numbers that are located at the vertices of the rectangle and always include the enabling element of the RE.
| x 1 | x2 | x 3 | B |
| 0 / 3 = 0 | 3 / 3 = 1 | 1 / 3 = 0.33 | 4 / 3 = 1.33 |
The enabling element is (-4).
In place of the resolving element, we get 1, and in the column itself we write zeros.
All other elements of the matrix, including the elements of column B, are determined by the rectangle rule.
To do this, select four numbers that are located at the vertices of the rectangle and always include the enabling element of the RE.
Let's present the calculation of each element in the form of a table:
| x 1 | x2 | x 3 | B |
| 0 / -4 = 0 | 0 / -4 = 0 | -4 / -4 = 1 | -4 / -4 = 1 |
Answer: x 1 = 1, x 2 = 1, x 3 = 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 online implementation of the Gauss method.Using the Gauss method
Application of the Gauss method in game theory
In game theory, when finding the maximin optimal strategy of a player, 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 particular solution to a differential equation, first find the derivatives of the corresponding degree for the written particular solution (y=f(A,B,C,D)), which are substituted into the original 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 Jordano-Gauss method in linear programming
In linear programming, in particular, in the simplex method, to transform a simplex table at each iteration, the rectangle rule is used, which uses the Jordan-Gauss method.Gauss method perfect for solving linear systems algebraic equations(SLAU). It has several advantages over other methods:
- firstly, there is no need to pre-investigate the system of equations for compatibility;
- secondly, the Gauss method can be used to solve not only SLAEs in which the number of equations coincides with the number of unknown variables and the main matrix of the system is non-degenerate, but also systems of equations in which the number of equations does not coincide with the number of unknown variables or the determinant of the main matrix zero;
- thirdly, the Gauss method leads to a result with a relatively small number of computational operations.
Brief review of the article.
First, we give the necessary definitions and introduce some notation.
Next, we describe the algorithm of the Gauss method for the simplest case, that is, for systems of linear algebraic equations, the number of equations in which coincides with the number of unknown variables and the determinant of the main matrix of the system is not equal to zero. When solving such systems of equations, the essence of the Gauss method is most clearly visible, which consists in the successive elimination of unknown variables. Therefore, the Gaussian method is also called the method of successive elimination of unknowns. Let's show detailed solutions a few examples.
In conclusion, we consider the Gaussian solution of systems of linear algebraic equations whose main matrix is either rectangular or degenerate. The solution of such systems has some features, which we will analyze in detail using examples.
Page navigation.
Basic definitions and notation.
Consider a system of p linear equations with n unknowns (p can be equal to n ):
Where are unknown variables, are numbers (real or complex), are free members.
If 
, then the system of linear algebraic equations is called homogeneous, otherwise - heterogeneous.
The set of values of unknown variables, in which all equations of the system turn into identities, is called SLAU decision.
If there is at least one solution to a system of linear algebraic equations, then it is called joint, otherwise - incompatible.
If the SLAE has only decision, then it is called certain. If there is more than one solution, then the system is called uncertain.
The system is said to be written in coordinate form if it has the form
.
This system in matrix form records has the form , where 
- the main matrix of SLAE, - the matrix of the column of unknown variables, - the matrix of free members.
If we add to the matrix A as the (n + 1)-th column the matrix-column of free terms, then we get the so-called expanded matrix systems of linear equations. Usually, the augmented matrix is denoted by the letter T, and the column of free members is separated by a vertical line from the rest of the columns, that is, 
The square matrix A is called degenerate if its determinant is zero. If , then the matrix A is called non-degenerate.
The following point should be noted.
If the following actions are performed with a system of linear algebraic equations
- swap two equations,
- multiply both sides of any equation by an arbitrary and non-zero real (or complex) number k,
- to both parts of any equation add the corresponding parts of the other equation, multiplied by an arbitrary number k,
then we get an equivalent system that has the same solutions (or, like the original one, has no solutions).
For an extended matrix of a system of linear algebraic equations, these actions will mean elementary transformations with rows:
- swapping two strings
- multiplication of all elements of any row of the matrix T by a non-zero number k ,
- adding to the elements of any row of the matrix the corresponding elements of another row, multiplied by an arbitrary number k .
Now we can proceed to the description of the Gauss method.
Solving systems of linear algebraic equations, in which the number of equations is equal to the number of unknowns and the main matrix of the system is nondegenerate, by the Gauss method.
What would we do at school if we were given the task of finding a solution to a system of equations 
.
Some would do so.
Note that by adding the left side of the first equation to the left side of the second equation, and the right side to the right side, you can get rid of the unknown variables x 2 and x 3 and immediately find x 1: 
We substitute the found value x 1 \u003d 1 into the first and third equations of the system: 
If we multiply both parts of the third equation of the system by -1 and add them to the corresponding parts of the first equation, then we get rid of the unknown variable x 3 and can find x 2: 
We substitute the obtained value x 2 \u003d 2 into the third equation and find the remaining unknown variable x 3: 
Others would have done otherwise.
Let's solve the first equation of the system with respect to the unknown variable x 1 and substitute the resulting expression into the second and third equations of the system in order to exclude this variable from them: 
Now let's solve the second equation of the system with respect to x 2 and substitute the result obtained into the third equation in order to exclude the unknown variable x 2 from it: 
It can be seen from the third equation of the system that x 3 =3. From the second equation we find 
, and from the first equation we get .
Familiar solutions, right?
The most interesting thing here is that the second solution method is essentially the method of sequential elimination of unknowns, that is, the Gauss method. When we expressed unknown variables (first x 1 , next x 2 ) and substituted them into the rest of the equations of the system, we thereby excluded them. We carried out the exception until the moment when the last equation left only one unknown variable. The process of sequential elimination of unknowns is called direct Gauss method. After the forward move is completed, we have the opportunity to calculate the unknown variable in the last equation. With its help, from the penultimate equation, we find the next unknown variable, and so on. The process of successively finding unknown variables while moving from the last equation to the first is called reverse Gauss method.
It should be noted that when we express x 1 in terms of x 2 and x 3 in the first equation, and then substitute the resulting expression into the second and third equations, the following actions lead to the same result:
Indeed, such a procedure also allows us to exclude the unknown variable x 1 from the second and third equations of the system: 
Nuances with the elimination of unknown variables by the Gauss method arise when the equations of the system do not contain some variables.
For example, in SLAU 
in the first equation, there is no unknown variable x 1 (in other words, the coefficient in front of it is zero). Therefore, we cannot solve the first equation of the system with respect to x 1 in order to exclude this unknown variable from the rest of the equations. The way out of this situation is to swap the equations of the system. Since we are considering systems of linear equations whose determinants of the main matrices are different from zero, there always exists an equation in which the variable we need is present, and we can rearrange this equation to the position we need. For our example, it is enough to swap the first and second equations of the system 
, then you can solve the first equation for x 1 and exclude it from the rest of the equations of the system (although x 1 is already absent in the second equation).
We hope you get the gist.
Let's describe Gauss method algorithm.
Let us need to solve a system of n linear algebraic equations with n unknown variables of the form 
, and let the determinant of its main matrix be nonzero.
We will assume that , since we can always achieve this by rearranging the equations of the system. We exclude the unknown variable x 1 from all equations of the system, starting from the second one. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, add the first multiplied by to the nth equation. The system of equations after such transformations will take the form 
where , a 
.
We would come to the same result if we expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.
Next, we act similarly, but only with a part of the resulting system, which is marked in the figure 
To do this, add the second multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, add the second multiplied by to the nth equation. The system of equations after such transformations will take the form 
where , a 
. Thus, the variable x 2 is excluded from all equations, starting from the third.
Next, we proceed to the elimination of the unknown x 3, while acting similarly with the part of the system marked in the figure 
So we continue the direct course of the Gauss method until the system takes the form 
From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as , using the obtained value of x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Let's analyze the algorithm with an example.
Example.
Gaussian method.
Solution.
The coefficient a 11 is different from zero, so let's proceed to the direct course of the Gauss method, that is, to the elimination of the unknown variable x 1 from all equations of the system, except for the first one. To do this, to the left and right parts of the second, third and fourth equations, add the left and right parts of the first equation, multiplied by , respectively, 
and : 
The unknown variable x 1 has been eliminated, let's move on to the exclusion x 2 . To the left and right parts of the third and fourth equations of the system, we add the left and right parts of the second equation, multiplied by 
and 
:
To complete the forward course of the Gauss method, we need to exclude the unknown variable x 3 from the last equation of the system. Let us add to the left and right parts of the fourth equation, respectively, the left and right side third equation multiplied by 
:
You can start the reverse course of the Gauss method.
From the last equation we have 
,
from the third equation we get ,
from the second
from the first.
To check, you can substitute the obtained values of unknown variables into the original system of equations. All equations turn into identities, which means that the solution by the Gauss method was found correctly.
Answer:
And now we will give the solution of the same example by the Gauss method in matrix form.
Example.
Find a solution to the system of equations Gaussian method.
Solution.
The extended matrix of the system has the form 
. Above each column, unknown variables are written, which correspond to the elements of the matrix.
The direct course of the Gauss method here involves bringing the extended matrix of the system to a trapezoidal form using elementary transformations. This process is similar to the exclusion of unknown variables that we did with the system in coordinate form. Now you will be convinced of it.
Let's transform the matrix so that all elements in the first column, starting from the second, become zero. To do this, to the elements of the second, third and fourth rows, add the corresponding elements of the first row multiplied by , and on respectively: 
Next, we transform the resulting matrix so that in the second column, all elements, starting from the third, become zero. This would correspond to excluding the unknown variable x 2 . To do this, add to the elements of the third and fourth rows the corresponding elements of the first row of the matrix, multiplied by and :
It remains to exclude the unknown variable x 3 from the last equation of the system. To do this, to the elements of the last row of the resulting matrix, we add the corresponding elements of the penultimate row, multiplied by :
It should be noted that this matrix corresponds to the system of linear equations 
which was obtained earlier after the direct move.
It's time to turn back. In the matrix form of the notation, the reverse course of the Gauss method involves such a transformation of the resulting matrix so that the matrix marked in the figure 
became diagonal, that is, took the form 
where are some numbers.
These transformations are similar to those of the Gauss method, but are performed not from the first line to the last, but from the last to the first.
Add to the elements of the third, second and first rows the corresponding elements of the last row, multiplied by 
, on and on 
respectively: 
Now let's add to the elements of the second and first rows the corresponding elements of the third row, multiplied by and by, respectively: 
At the last step of the reverse motion of the Gaussian method, we add the corresponding elements of the second row, multiplied by , to the elements of the first row: 
The resulting matrix corresponds to the system of equations 
, from which we find the unknown variables.
Answer:
NOTE.
When using the Gauss method to solve systems of linear algebraic equations, approximate calculations should be avoided, as this can lead to absolutely incorrect results. We recommend that you do not round decimals. Better off decimal fractions go to ordinary fractions.
Example.
Solve System of Three Equations by Gaussian Method 
.
Solution.
Note that in this example, the unknown variables have a different designation (not x 1 , x 2 , x 3 , but x, y, z ). Let's move on to ordinary fractions: 
Eliminate the unknown x from the second and third equations of the system: 
In the resulting system, there is no unknown variable y in the second equation, and y is present in the third equation, therefore, we swap the second and third equations: 
At this point, the direct course of the Gauss method is over (you do not need to exclude y from the third equation, since this unknown variable no longer exists).
Let's go back.
From the last equation we find 
,
from penultimate 
from the first equation we have 
Answer:
X=10, y=5, z=-20.
The solution of systems of linear algebraic equations, in which the number of equations does not coincide with the number of unknowns, or the main matrix of the system is degenerate, by the Gauss method.
Systems of equations whose main matrix is rectangular or square degenerate may have no solutions, may have a single solution, or may have an infinite number of solutions.
Now we will understand how the Gauss method allows us to establish the compatibility or inconsistency of a system of linear equations, and in the case of its compatibility, determine all solutions (or one single solution).
In principle, the process of eliminating unknown variables in the case of such SLAEs remains the same. However, it is worth dwelling in detail on some situations that may arise.
Let's move on to the most important step.
So, let us assume that the system of linear algebraic equations after the completion of the forward run of the Gauss method takes the form 
and none of the equations reduced to (in this case, we would conclude that the system is inconsistent). A logical question arises: "What to do next"?
We write out the unknown variables that are in the first place of all the equations of the resulting system: 
In our example, these are x 1 , x 4 and x 5 . In the left parts of the equations of the system, we leave only those terms that contain the written out unknown variables x 1, x 4 and x 5, we transfer the remaining terms to the right side of the equations with the opposite sign: 
Let us assign arbitrary values to the unknown variables that are on the right-hand sides of the equations, where 
- arbitrary numbers: 
After that, the numbers are found in the right parts of all the equations of our SLAE and we can proceed to the reverse course of the Gauss method.
From the last equation of the system we have , from the penultimate equation we find , from the first equation we get 
The solution of the system of equations is the set of values of unknown variables 
Giving numbers various meanings, we will receive various solutions systems of equations. That is, our system of equations has infinitely many solutions.
Answer:
where - arbitrary numbers.
To consolidate the material, we will analyze in detail the solutions of several more examples.
Example.
Solve Homogeneous System of Linear Algebraic Equations 
Gaussian method.
Solution.
Let us exclude the unknown variable x from the second and third equations of the system. To do this, add the left and right parts of the first equation, respectively, to the left and right parts of the second equation, multiplied by , and to the left and right parts of the third equation, the left and right parts of the first equation, multiplied by : 
Now we exclude y from the third equation of the resulting system of equations: 
The resulting SLAE is equivalent to the system 
.
We leave only the terms containing the unknown variables x and y on the left side of the equations of the system, and transfer the terms with the unknown variable z to the right side: 
One of the simplest ways to solve a system of linear equations is a method based on calculating the determinants ( Cramer's rule). Its advantage is that it allows you to immediately record the solution, it is especially convenient in cases where the coefficients of the system are not numbers, but some kind of parameters. Its drawback is the cumbersomeness of calculations in the case a large number equations, moreover, Cramer's rule is not directly applicable to systems in which the number of equations does not coincide with the number of unknowns. In such cases, it is usually used Gauss method.
Systems of linear equations that have the same set of solutions are called equivalent. It is obvious that the set of solutions linear system does not change if any equations are interchanged, or if one of the equations is multiplied by some non-zero number, or if one equation is added to another.
Gauss method (method of successive elimination of unknowns) lies in the fact that, with the help of elementary transformations, the system is reduced to an equivalent stepwise system. First, with the help of the 1st equation, x 1 of all subsequent equations of the system. Then, using the 2nd equation, we eliminate x 2 of the 3rd and all subsequent equations. This process, called direct Gauss method, continues until only one unknown remains on the left side of the last equation x n. After that, it is made Gaussian reverse– solving the last equation, we find x n; after that, using this value, from the penultimate equation we calculate x n-1 etc. Last we find x 1 from the first equation.
It is convenient to carry out Gaussian transformations by performing transformations not with the equations themselves, but with the matrices of their coefficients. Consider the matrix:
called extended system matrix, because in addition to the main matrix of the system, it includes a column of free members. The Gauss method is based on bringing the main matrix of the system to a triangular form (or trapezoidal form in the case of non-square systems) using elementary row transformations (!) of the extended matrix of the system.
Example 5.1. Solve the system using the Gauss method:
Solution. Let's write out the augmented matrix of the system and, using the first row, after that we will set the rest of the elements to zero:
we get zeros in the 2nd, 3rd and 4th rows of the first column:
Now we need all the elements in the second column below the 2nd row to be equal to zero. To do this, you can multiply the second line by -4/7 and add to the 3rd line. However, in order not to deal with fractions, we will create a unit in the 2nd row of the second column and only
Now, to get a triangular matrix, you need to zero out the element of the fourth row of the 3rd column, for this you can multiply the third row by 8/54 and add it to the fourth. However, in order not to deal with fractions, we will swap the 3rd and 4th rows and the 3rd and 4th columns, and only after that we will reset the specified element. Note that when the columns are rearranged, the corresponding variables are swapped, and this must be remembered; other elementary transformations with columns (addition and multiplication by a number) cannot be done!
The last simplified matrix corresponds to a system of equations equivalent to the original one:
From here, using the reverse course of the Gauss method, we find from the fourth equation x 3 = -1; from the third x 4 = -2, from the second x 2 = 2 and from the first equation x 1 = 1. In matrix form, the answer is written as
We have considered the case when the system is definite, i.e. when there is only one solution. Let's see what happens if the system is inconsistent or indeterminate.
Example 5.2. Explore the system using the Gaussian method:
Solution. We write out and transform the augmented matrix of the system
We write a simplified system of equations:
Here, in the last equation, it turned out that 0=4, i.e. contradiction. Therefore, the system has no solution, i.e. she incompatible. à
Example 5.3. Explore and solve the system using the Gaussian method:
Solution. We write out and transform the extended matrix of the system:
As a result of the transformations, only zeros were obtained in the last line. This means that the number of equations has decreased by one:
Thus, after simplifications, two equations remain, and four unknowns, i.e. two unknown "extra". Let "superfluous", or, as they say, free variables, will x 3 and x 4 . Then
Assuming x 3 = 2a and x 4 = b, we get x 2 = 1–a and x 1 = 2b–a; or in matrix form
A solution written in this way is called general, since, by giving the parameters a and b different meanings, you can describe everything possible solutions systems. a
