Gaussian method solution algorithm. Gaussian method (sequential elimination of unknowns)
Today we are looking at the Gauss method for solving systems of linear algebraic equations. You can read about what these systems are in the previous article devoted to solving the same SLAEs using the Cramer method. The Gauss method does not require any specific knowledge, you only need attentiveness and consistency. Despite the fact that, from a mathematical point of view, school training is sufficient to apply it, students often find it difficult to master this method. In this article we will try to reduce them to nothing!
Gauss method
M Gaussian method– the most universal method for solving SLAEs (with the exception of very large systems). Unlike previously discussed, it is suitable not only for systems with only decision, but also for systems that have an infinite number of solutions. There are three possible options here.
- The system has a unique solution (the determinant of the main matrix of the system is not equal to zero);
- The system has an infinite number of solutions;
- There are no solutions, the system is incompatible.
So we have a system (let it have one solution) and we are going to solve it using the Gaussian method. How it works?
The Gauss method consists of two stages - forward and inverse.
Direct stroke of the Gaussian method
First, let's write down the extended matrix of the system. To do this, add a column of free members to the main matrix.
The whole essence of the Gauss method is to bring, through elementary transformations, this matrix to a stepped (or as they also say triangular) appearance. In this form, there should be only zeros under (or above) the main diagonal of the matrix.
What you can do:
- You can rearrange the rows of the matrix;
- If there are equal (or proportional) rows in a matrix, you can remove all but one of them;
- You can multiply or divide a string by any number (except zero);
- Null rows are removed;
- You can append a string multiplied by a number other than zero to a string.
Reverse Gaussian method
After we transform the system in this way, one unknown Xn becomes known, and you can find all the remaining unknowns in reverse order, substituting the already known x's into the equations of the system, up to the first.
When the Internet is always at hand, you can solve a system of equations using the Gaussian method online. You just need to enter the coefficients into the online calculator. But you must admit, it is much more pleasant to realize that the example has not been solved computer program, but with your own brain.
An example of solving a system of equations using the Gauss method
And now - an example so that everything becomes clear and understandable. Let the system be given linear equations, and you need to solve it using the Gaussian method:
First we write the extended matrix:
Now let's do the transformations. We remember that we need to achieve a triangular appearance of the matrix. Let's multiply the 1st line by (3). Multiply the 2nd line by (-1). Add the 2nd line to the 1st and get:
Then multiply the 3rd line by (-1). Let's add the 3rd line to the 2nd:
Let's multiply the 1st line by (6). Let's multiply the 2nd line by (13). Let's add the 2nd line to the 1st:
Voila - the system is brought to the appropriate form. It remains to find the unknowns:
The system in this example has a unique solution. We will consider solving systems with an infinite number of solutions in a separate article. Perhaps at first you will not know where to start transforming the matrix, but after appropriate practice you will get the hang of it and will crack SLAEs using the Gaussian method like nuts. And if you suddenly come across a SLAU that turns out to be too tough nut to crack, contact our authors! you can by leaving a request in the Correspondence Office. Together we will solve any problem!
Carl Friedrich Gauss, the greatest mathematician for a long time hesitated, choosing between philosophy and mathematics. Perhaps it was precisely this mindset that allowed him to make such a noticeable “legacy” in world science. In particular, by creating the "Gauss Method" ...
For almost 4 years, articles on this site concerned school education, mainly from the side of philosophy, the principles of (mis)understanding introduced into the consciousness of children. The time is coming for more specifics, examples and methods... I believe that this is exactly the approach to the familiar, confusing and important areas of life gives better results.
We people are designed in such a way that no matter how much we talk about abstract thinking, But understanding Always happens through examples. If there are no examples, then it is impossible to grasp the principles... Just as it is impossible to get to the top of a mountain except by walking the entire slope from the foot.
Same with school: for now living stories It is not enough that we instinctively continue to regard it as a place where children are taught to understand.
For example, teaching the Gaussian method...
Gauss method in 5th grade school
I’ll make a reservation right away: the Gauss method has a much wider application, for example, when solving systems of linear equations. What we will talk about takes place in 5th grade. This started, having understood which, it is much easier to understand the more “advanced options”. In this article we are talking about Gauss's method (method) for finding the sum of a series
Here is an example that I brought from school younger son, attending 5th grade at a 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 Karl (an outdated method, not used in schools these days) because he
instead of sequentially adding numbers from 1 to 100, 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 executed in front of an astonished public. So that others would be discouraged from thinking.
What did little Gauss do? developed number sense? Noticed some feature number series with a constant step (arithmetic progression). AND exactly this later made him a great scientist, those who know how to notice, having feeling, instinct of understanding.
This is why mathematics is valuable, developing ability to see general in particular - abstract thinking. Therefore, most parents and employers instinctively consider mathematics an important discipline ...
“Then you need to learn mathematics, because it puts your mind in order.
M.V.Lomonosov".
However, the followers of those who flogged future geniuses with rods turned the Method into something the opposite. As my supervisor said 35 years ago: “The question has been learned.” Or as my youngest son said yesterday about Gauss’s method: “Maybe it’s not worth making a big science out of this, huh?”
The consequences of the creativity of "scientists" are visible in the level of the current school mathematics, the level of her teaching and understanding of “The Queen of Sciences” by the majority.
However, let's continue...
Methods for explaining the Gauss method in 5th grade school
A mathematics teacher at a Moscow gymnasium, explaining the Gauss method according to Vilenkin, complicated the task.
What if the difference (step) of an arithmetic progression is not one, but another number? For example, 20.
The problem he gave to the fifth graders:
20+40+60+80+ ... +460+480+500
Before getting acquainted with the gymnasium method, let’s take a look at the Internet: how do school teachers and math tutors do it?..
Gaussian method: explanation No. 1
A well-known tutor on his YOUTUBE channel gives the following reasoning:
"Let's write the numbers from 1 to 100 as follows:
first a series of numbers from 1 to 50, and strictly below it another series of numbers from 50 to 100, but in the 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 take this method in 9th grade!"
Gaussian method: explanation No. 2
Another tutor, less well-known (judging by the number of views), takes a more scientific approach, offering a solution algorithm of 5 points that must be completed sequentially.
For the uninitiated, 5 is one of the Fibonacci numbers traditionally considered magical. A 5 step method is always more scientific than a 6 step method, for example. ...And this is hardly an accident, most likely, the Author is a hidden supporter of the Fibonacci theory
Dana 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 plus one rule : we must add one to the resulting quotient: otherwise we will get a result that is less by one than the true number of pairs: 42 + 1 = 43.
This is the required sum of the arithmetic progression from 4 to 256 with a difference of 6!
Gauss method: explanation in 5th grade at a Moscow gymnasium
Here's how to solve the problem of finding the sum of a series:
20+40+60+ ... +460+480+500
in the 5th grade of a Moscow gymnasium, Vilenkin’s textbook (according to my son).
After showing the presentation, the math teacher showed a couple of examples using the Gaussian method and gave the class a task of finding the sum of the numbers in a series in increments of 20.
This required the following:
As you can see, this is more compact and effective technique: 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 “method” would be turned into by his followers German teacher, who flogged Karl with rods. He would have seen the symbolism, 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: did 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?
IN simplification. IN observing and grasping simple patterns of numbers. IN turning dry school arithmetic into interesting and exciting activity , activating in the brain the desire to continue, rather than blocking high-cost mental activity.
Is it possible to use one of the given “modifications of Gauss’s method” to calculate the sum of the numbers of an arithmetic progression almost instantly? According to the “algorithms”, little Karl would be guaranteed to avoid spanking, develop an aversion to mathematics and suppress his creative impulses in the bud.
Why did the tutor so persistently advise fifth-graders “not to be afraid of misunderstanding” of the method, convincing them that they would solve “such” problems as early as 9th grade? Psychologically illiterate action. It was a good move to note: "See? You already in 5th grade you can solve problems that you will complete only in 4 years! What a great fellow you are!”
To use the Gaussian method, a level of class 3 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 are “out of touch” to explain the simplest things in normal human language, not to mention mathematical... They are unable to get people interested in mathematics and completely discourage even those who are “capable.”
Or, as my son commented: “making 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 and answers
"Look, here are the numbers from 1 to 100. What do you see?"
The point is not what exactly the child sees. The trick is to get him to look.
"How can you put them together?" The son realized 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 stopped being afraid to look, or as I say: “moved the task”. This is the beginning of the journey to understanding
“Which is easier: adding, for example, 5 and 6 or 5 and 95?” A leading question... But any training comes down to “guiding” a person to the “answer” - in any way acceptable to him.
At this stage, guesses may already arise about how to “save” on calculations.
All we did was hint: the “frontal, linear” method of counting is not the only possible one. If a child understands this, then later he will come up with many more such methods, because it's interesting!!! And he will definitely avoid “misunderstanding” mathematics and will not feel disgusted with it. He got the win!
If child discovered that adding pairs of numbers that add up to a hundred is a piece of cake, then "arithmetic progression with difference 1"- a rather dreary and uninteresting thing for a child - suddenly found life for him . Order emerged from chaos, and this always causes enthusiasm: that's how we are made!
A question to answer: why, after the insight a child has received, should he again be forced into the framework of dry algorithms, which are also functionally useless in this case?!
Why force stupid rewrites? sequence numbers in a notebook: so that even the capable do not have a single chance of understanding? Statistically, of course, but mass education is geared towards “statistics”...
Where did the zero go?
And yet, adding numbers that add up to 100 is much more acceptable to the mind than those that add up to 101...
The "Gauss School Method" requires exactly this: mindlessly fold pairs of numbers equidistant from the center of the progression, Despite everything.
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 is much easier to transform a series of numbers starting with 1 into a series starting with 0. The sum will not change, will it? You need to stop “thinking in textbooks” and start looking... And see that pairs with a sum of 101 can be completely replaced by pairs with a sum of 100!
0 + 100, 1 + 99, 2 + 98 ... 49 + 51
How to abolish the "plus 1 rule"?
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?
I look at the sequence:
1, 2, 3, .. 8, 9, 10
and when you’re completely tired, then move on to a simpler row:
1, 2, 3, 4, 5
and I figure: if you subtract one from 5, you get 4, but I’m absolutely clear I 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.
What the hell are the rules?..
So that in a couple or three years you can fill all the space between your forehead and the back of your head and stop thinking? How to earn your bread and butter? After all, we are moving in even ranks into the era of the digital economy!
More about Gauss’s school method: “why make science out of this?..”
It was not for nothing that I posted a screenshot from my son’s notebook...
"What happened in class?"
“Well, I counted right away, raised my hand, but she didn’t ask. Therefore, while the others were counting, I began to do homework 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 it. She said: “Wrong, you need to count as I showed!”
“It’s good that she didn’t give a bad grade. And she made me write in their notebook “the course of the solution” in their own way. Why make a big science out of this?..”
The main crime of a math teacher
Hardly after that incident Karl Gauss experienced high feeling respect towards the school mathematics teacher. But if he knew how followers of that teacher will distort the very essence of the method... he would roar with indignation and, through the World Intellectual Property Organization WIPO, achieve a ban on the use of his good name in school textbooks!..
In what the main mistake of the school approach? Or, as I put it - a crime school teachers mathematicians against children?
Algorithm of misunderstanding
What do school methodologists do, the vast majority of whom don’t know how to think?
They 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, rather than the stupidity of the school system.
This is what happens: parents blame their children, and teachers... do the same for children who “don’t understand mathematics!”
Are you smart?
What did little Karl do?
A completely unconventional approach to a formulaic task. This is the essence of His approach. This the main thing that should be taught in 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 Gauss's method is to
What, if the number of elements of the series is odd, as in the problem that was assigned to my son?..
The "catch" is that in this case you should find an “extra” number in the series and add it to the sum of the pairs. In our example this number is 260.
How to detect? Copying all pairs of numbers into a notebook!(This is why the teacher made the kids do this stupid job of trying to teach "creativity" using the Gaussian method... And this is why such a "method" is practically inapplicable to large data series, AND this is why it is not the Gaussian method.)
A little creativity in the school routine...
The son acted differently.
(20 + 500, 40 + 480 ...).
0+500, 20+480, 40+460 ...
Not difficult, 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 method: 5, which does not allow the approach to be criticized for being unscientific.
Obviously this approach is simpler, faster and more universal, in the style of the Method. But... the teacher not only did not praise, but also forced me to rewrite it “in the correct way” (see screenshot). That is, she made a desperate attempt to stifle the creative impulse and the ability to understand mathematics at the root! Apparently, so that she could later be hired as a tutor... She attacked the wrong person...
Everything that I described so long and tediously can be explained to a normal child in a maximum of half an hour. Along with examples.
And in such a way that he will never forget it.
And it will be step towards understanding...not just mathematicians.
Admit it: how many times in your life have you added using the Gaussian method? And I never did!
But instinct of understanding, which develops (or is extinguished) 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 have quietly entered under the strict leadership 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 effect.
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 must be done “in accordance with and on the basis of” and everything must be documented. Step aside - stood in line to be fired. Let’s not be hypocrites: the salaries of Moscow teachers are very good... If they fire you, where to go?..
Therefore this site not about education. He's about individual education, only possible way get out of the crowd generation Z ...
We continue to consider systems of linear equations. This lesson is the third on the topic. If you have a vague idea of what a system of linear equations is in general, if you feel like a teapot, then I recommend starting with the basics on the page Next, it is useful to study the lesson.
The Gaussian method is easy! Why? The famous German mathematician Johann Carl Friedrich Gauss, during his lifetime, received recognition as the greatest mathematician of all time, a genius, and even the nickname “King of Mathematics.” And everything ingenious, as you know, is simple! By the way, not only suckers get money, but also geniuses - Gauss’s portrait was on the 10 Deutschmark banknote (before the introduction of the euro), and Gauss still smiles mysteriously at Germans from ordinary postage stamps.
The Gauss method is simple in that the KNOWLEDGE OF A FIFTH-GRADE STUDENT IS ENOUGH to master it. You must know how to add and multiply! It is no coincidence that teachers often consider the method of sequential exclusion of unknowns in school mathematics electives. It’s a paradox, but students find the Gaussian method the most difficult. Nothing surprising - it’s all about the methodology, and I will try to talk about the algorithm of the method in an accessible form.
First, let's systematize a little knowledge about systems of linear equations. A system of linear equations can:
1) Have a unique solution. 2) Have infinitely many solutions. 3) Have no solutions (be non-joint).
The Gauss method is the most powerful and universal tool for finding a solution any systems of linear equations. As we remember, Cramer's rule and matrix method are unsuitable in cases where the system has infinitely many solutions or is inconsistent. A method sequential elimination unknown Anyway will lead us to the answer! In this lesson, we will again consider the Gauss method for case No. 1 (the only solution to the system), an article is devoted to the situations of points No. 2-3. I note that the algorithm of the method itself works the same in all three cases.
Let's go back to the simplest system from class How to solve a system of linear equations? and solve it using the Gaussian method.
The first step is to write down extended system matrix: . I think everyone can see by what principle the coefficients are written. The vertical line inside the matrix does not have any mathematical meaning - it is simply a strikethrough for ease of design.
Reference : I recommend you remember terms linear algebra. System Matrix is a matrix composed only of coefficients for unknowns, in this example the matrix of the system: . Extended System Matrix – this is the same matrix of the system plus a column of free terms, in this case: . For brevity, any of the matrices can be simply called a matrix.
After the extended system matrix is written, it is necessary to perform some actions with it, which are also called elementary transformations.
The following elementary transformations exist:
1) Strings matrices Can rearrange in some places. For example, in the matrix under consideration, you can painlessly rearrange the first and second rows: 
2) If there are (or have appeared) proportional (as a special case - identical) rows in the matrix, then you should delete All these rows are from the matrix except one. Consider, for example, the matrix 
. In this matrix, the last three rows are proportional, so it is enough to leave only one of them: 
.
3) If a zero row appears in the matrix during transformations, then it should also be delete. I won’t draw, of course, the zero line is the line in which all zeros.
4) The matrix row can be multiply (divide) to any number non-zero. Consider, for example, the matrix . Here it is advisable to divide the first line by –3, and multiply the second line by 2: 
. This action is very useful because it simplifies further transformations of the matrix.
5) This transformation causes the most difficulties, but in fact there is nothing complicated either. To a row of a matrix you can add another string multiplied by a number, different from zero. Consider our matrix of practical example: . First I'll describe the transformation in great detail. Multiply the first line by –2: 
, And to the second line we add the first line multiplied by –2: 
. Now the first line can be divided “back” by –2: . As you can see, the line that is ADDED LI – hasn't changed. Always the line TO WHICH IS ADDED changes UT.
In practice, of course, they don’t write it in such detail, but write it briefly: 
Once again: to the second line added the first line multiplied by –2. A line is usually multiplied orally or on a draft, with the mental calculation process going something like this:
“I rewrite the matrix and rewrite the first line: 
»
“First column. At the bottom I need to get zero. Therefore, I multiply the one at the top by –2: , and add the first one to the second line: 2 + (–2) = 0. I write the result in the second line: 
»
“Now the second column. At the top, I multiply -1 by -2: . I add the first to the second line: 1 + 2 = 3. I write the result in the second line: 
»
“And the third column. At the top I multiply -5 by -2: . I add the first to the second line: –7 + 10 = 3. I write the result in the second line: 
»
Please carefully understand this example and understand the sequential calculation algorithm, if you understand this, then the Gaussian method is practically in your pocket. But, of course, we will still work on this transformation.
Elementary transformations do not change the solution of the system of equations
! ATTENTION: considered manipulations can not use, if you are offered a task where the matrices are given “by themselves.” For example, with “classical” operations with matrices Under no circumstances should you rearrange anything inside the matrices! Let's return to our system. It is practically taken to pieces.
Let us write down the extended matrix of the system and, using elementary transformations, reduce it to stepped view:
(1) The first line was added to the second line, multiplied by –2. And again: why do we multiply the first line by –2? In order to get zero at the bottom, which means getting rid of one variable in the second line.
(2) Divide the second line by 3.
The purpose of elementary transformations
–
reduce the matrix to stepwise form: 
. In the design of the task, they just mark out the “stairs” with a simple pencil, and also circle the numbers that are located on the “steps”. The term “stepped view” itself is not entirely theoretical, in scientific and educational literature it is often called trapezoidal view or triangular view.
As a result of elementary transformations, we obtained equivalent original system of equations:
Now the system needs to be “unwinded” in the opposite direction - from bottom to top, this process is called inverse of the Gaussian method.
In the lower equation we already have a ready-made result: .
Let's consider the first equation of the system and substitute the already known value of “y” into it:
Let's consider the most common situation, when the Gaussian method requires solving a system of three linear equations with three unknowns.
Example 1
Solve the system of equations using the Gauss method: 
Let's write the extended matrix of the system: 
Now I will immediately draw the result that we will come to during the solution: 
And I repeat, our goal is to bring the matrix to a stepwise form using elementary transformations. Where to start?
First, look at the top left number: 
Should almost always be here unit. Generally speaking, –1 (and sometimes other numbers) will do, but somehow it has traditionally happened that one is usually placed there. How to organize a unit? We look at the first column - we have a finished unit! Transformation one: swap the first and third lines: 
Now the first line will remain unchanged until the end of the solution. Now fine.
Unit in left top corner organized. Now you need to get zeros in these places: 
We get zeros using a “difficult” transformation. First we deal with the second line (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 line by –2: (–2, –4, 2, –18). And we consistently carry out (again mentally or on a draft) addition, to the second line we add the first line, already multiplied by –2:
We write the result in the second line: 
We deal with the third line in the same way (3, 2, –5, –1). To get a zero in the first position, you need to the third line add the first line multiplied by –3. Mentally or on a draft, multiply the first line by –3: (–3, –6, 3, –27). AND to the third line we add the first line multiplied by –3:
We write the result in the third line:
In practice, these actions are usually performed orally and written down in one step: 
No need to count everything at once and at the same time. The order of calculations and “writing in” the results consistent and usually it’s like this: first we rewrite the first line, and slowly puff on ourselves - CONSISTENTLY and ATTENTIVELY:
And I have already discussed the mental process of the calculations themselves above.
In this example, this is easy to do; we divide the second line by –5 (since all numbers there are divisible by 5 without a remainder). At the same time, we divide the third line by –2, because the smaller the numbers, the simpler the solution:
At the final stage of elementary transformations, you need to get another zero here: 
For this to the third line we add the second line multiplied by –2:
Try to figure out this action yourself - mentally multiply the second line by –2 and perform the addition.
The last action performed is the hairstyle of the result, divide the third line by 3.
As a result of elementary transformations, an equivalent system of linear equations was obtained: 
Cool.
Now the reverse of the Gaussian method comes into play. The equations “unwind” from bottom to top.
In the third equation we already have a ready result:
Let's look at the second equation: . The meaning of "zet" is already known, thus:
And finally, the first equation: . “Igrek” and “zet” are known, it’s just a matter of little things:
Answer: 
As has already been noted several times, for any system of equations it is possible and necessary to check the solution found, fortunately, this is easy and quick.
Example 2
This is an example for an independent solution, a sample of the final design and an answer at the end of the lesson.
It should be noted that your progress of the decision may not coincide with my decision process, and this is a feature of the Gauss method. But the answers must be the same!
Example 3
Solve a system of linear equations using the Gauss method 
We look at the upper left “step”. We should have one there. The problem is that there are no units in the first column at all, so rearranging the rows will not solve anything. In such cases, the unit must be organized using an elementary transformation. This can usually be done in several ways. I did this: (1) To the first line we add the second line, multiplied by –1. That is, we mentally multiplied the second line by –1 and added the first and second lines, while the second line did not change.
Now at the top left there is “minus one”, which suits us quite well. Anyone who wants to get +1 can perform an additional movement: multiply the first line by –1 (change its sign).
(2) The first line multiplied by 5 was added to the second line. The first line multiplied by 3 was added to the third line.
(3) The first line was multiplied by –1, in principle, this is for beauty. The sign of the third line was also changed and it was moved to second place, so that on the second “step” we had the required unit.
(4) The second line was added to the third line, multiplied by 2.
(5) The third line was divided by 3.
A bad sign that indicates an error in calculations (more rarely, a typo) is a “bad” bottom line. That is, if we got something like , below, and, accordingly, 
, then with a high degree of probability we can say that an error was made during elementary transformations.
We charge the reverse, in the design of examples they often do not rewrite the system itself, but the equations are “taken directly from the given matrix.” The reverse stroke, I remind you, works from bottom to top. Yes, here is a gift:
Answer: 
.
Example 4
Solve a system of linear equations using the Gauss method 
This is an example for you to solve on your own, it is somewhat more complicated. It's okay if someone gets confused. Full solution and sample design at the end of the lesson. Your solution may be different from my solution.
In the last part we will look at some features of the Gaussian algorithm. The first feature is that sometimes some variables are missing from the system equations, for example: 
How to correctly write the extended system matrix? I already talked about this point in class. Cramer's rule. Matrix method. In the extended matrix of the system, we put zeros in place of missing variables: 
By the way, this is a fairly easy example, since the first column already has one zero, and there are fewer elementary transformations to perform.
The second feature is this. In all the examples considered, we placed either –1 or +1 on the “steps”. Could there be other numbers there? In some cases they can. Consider the system: 
.
Here on the upper left “step” we have a two. But we notice the fact that all the numbers in the first column are divisible by 2 without a remainder - and the other is two and six. And the two at the top left will suit us! In the first step, you need to perform the following transformations: add the first line multiplied by –1 to the second line; to the third line add the first line multiplied by –3. This way we will get the required zeros in the first column.
Or another conventional example: 
. Here the three on the second “step” also suits us, since 12 (the place where we need to get zero) is divisible by 3 without a remainder. It is necessary to carry out the following transformation: add the second line to the third line, multiplied by –4, as a result of which the zero we need will be obtained.
Gauss's method is universal, but there is one peculiarity. You can confidently learn to solve systems using other methods (Cramer’s method, matrix method) literally the first time - they have a very strict algorithm. But in order to feel confident in the Gaussian method, you should “get your teeth into” and solve at least 5-10 ten systems. Therefore, at first there may be confusion and errors in calculations, and there is nothing unusual or tragic about this.
Rainy autumn weather outside the window.... Therefore, for everyone who wants more complex example for independent solution:
Example 5
Solve a system of 4 linear equations with four unknowns using the Gauss method. 
Such a task is not so rare in practice. I think even a teapot who has thoroughly studied this page will understand the algorithm for solving such a system intuitively. Fundamentally, everything is the same - there are just more actions.
Cases when the system has no solutions (inconsistent) or has infinitely many solutions are discussed in the lesson Incompatible systems and systems with a common solution. There you can fix the considered algorithm of the Gaussian method.
I wish you success!
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 line was added to the second line, multiplied by –2. The first line was added to the third line, multiplied by –1.
Attention!
Here you may be tempted to subtract the first from the third line; I highly recommend not subtracting it - the risk of error greatly increases. Just fold it!
(2) The sign of the second line was changed (multiplied by –1). The second and third lines have been swapped.
note
, that on the “steps” we are satisfied not only with one, but also with –1, which is even more convenient.
(3) The second line was added to the third line, multiplied by 5.
(4) The sign of the second line was changed (multiplied by –1). The third line was divided by 14.
Reverse:
Answer
:
.
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) A second line was added to the first line. Thus, the desired unit is organized on the upper left “step”. (2) The first line multiplied by 7 was added to the second line. The first line multiplied by 6 was added to the third line.
With the second “step” everything gets worse , the “candidates” for it are the numbers 17 and 23, and we need either one or –1. Transformations (3) and (4) will be aimed at obtaining the desired unit (3) The second line was added to the third line, multiplied by –1. (4) The third line was added to the second line, multiplied by –3. The required item on the second step has been received. . (5) The second line was added to the third line, multiplied by 6. (6) The second line was multiplied by –1, the third line was divided by -83.
Reverse:
Answer :
Example 5:
Solution
:
Let us write down the matrix of the system and, using elementary transformations, bring it to a stepwise form:
Conversions performed: (1) The first and second lines have been swapped. (2) The first line was added to the second line, multiplied by –2. The first line was added to the third line, multiplied by –2. The first line was added to the fourth line, multiplied by –3. (3) The second line was added to the third line, multiplied by 4. The second line was added to the fourth line, multiplied by –1. (4) The sign of the second line was changed. The fourth line was divided by 3 and placed in place of the third line. (5) The third line was added to the fourth line, multiplied by –5.
Reverse:
Answer :
One of the simplest ways to solve a system of linear equations is a technique based on the calculation of 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 parameters. Its disadvantage is the cumbersomeness of calculations in the case 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 Gaussian method.
Systems of linear equations having the same set of solutions are called equivalent. Obviously, many solutions linear system does not change if any equations are swapped, 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 sequential elimination of unknowns) is that with the help of elementary transformations the system is reduced to an equivalent system of a step type. First, using the 1st equation, we eliminate x 1 of all subsequent equations of the system. Then, using the 2nd equation, we eliminate x 2 from the 3rd and all subsequent equations. This process, called direct Gaussian method, continues until there is only one unknown left on the left side of the last equation x n. After this it is done inverse of the Gaussian method– solving the last equation, we find x n; after that, using this value, from the penultimate equation we calculate x n–1, etc. We find the last one 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 matrix of the system, because, in addition to the main matrix of the system, it includes a column of free terms. The Gaussian method is based on reducing 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 Gaussian method:
Solution. Let's write out the extended matrix of the system and, using the first row, after that we will reset the remaining elements:
we get zeros in the 2nd, 3rd and 4th rows of the first column:
Now we need all 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 it to the 3rd line. However, in order not to deal with fractions, let's create a unit in the 2nd row of the second column and only
Now, to get a triangular matrix, you need to reset the element of the fourth row of the 3rd column; to do 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 rearranging the columns, the corresponding variables change places and this must be remembered; other elementary transformations with columns (addition and multiplication by a number) cannot be performed!
The last simplified matrix corresponds to a system of equations equivalent to the original one:
From here, using the inverse of the Gaussian 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 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 uncertain.
Example 5.2. Explore the system using the Gaussian method:
Solution. We write out and transform the extended matrix of the system
We write a simplified system of equations:
Here, in the last equation it turns out that 0=4, i.e. contradiction. Consequently, 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, the last line contains only zeros. This means that the number of equations has decreased by one:
Thus, after simplifications, there are two equations left, and four unknowns, i.e. two unknown "extra". Let them be "superfluous", or, as they say, free variables, will x 3 and x 4 . Then
Believing 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, because, giving parameters a And b different meanings, it is possible to describe all possible solutions of the system. a
In this article, the method is considered as a method for solving systems of linear equations (SLAEs). The method is analytical, that is, it allows you to write a solution algorithm in general view, and then substitute values from specific examples there. Unlike the matrix method or Cramer's formulas, when solving a system of linear equations using the Gauss method, you can also work with those that have an infinite number of solutions. Or they don't have it at all.
What does it mean to solve using the Gaussian method?
First, we need to write our system of equations in It looks like this. Take the system:
The coefficients are written in the form of a table, and the free terms are written in a separate column on the right. The column with free terms is separated for convenience. The matrix that includes this column is called extended.
Next, the main matrix with coefficients must be reduced to an upper triangular form. This is the main point of solving the system using the Gaussian method. Simply put, after certain manipulations, the matrix should look so that its lower left part contains only zeros:
Then, if you write the new matrix again as a system of equations, you will notice that the last row already contains the value of one of the roots, which is then substituted into the equation above, another root is found, and so on.
This is a description of the solution by the Gaussian method in the most general terms. What happens if suddenly the system has no solution? Or are there infinitely many of them? To answer these and many other questions, it is necessary to consider separately all the elements used in solving the Gaussian method.
Matrices, their properties
There is no hidden meaning in the matrix. It's simple convenient way recording data for subsequent operations with them. Even schoolchildren do not need to be afraid of them.
The matrix is always rectangular, because it is more convenient. Even in the Gauss method, where everything comes down to constructing a matrix of a triangular form, a rectangle appears in the entry, only with zeros in the place where there are no numbers. Zeros may not be written, but they are implied.
The matrix has a size. Its “width” is the number of rows (m), “length” is the number of columns (n). Then the size of the matrix A (capital Latin letters are usually used to denote them) will be denoted as A m×n. If m=n, then this matrix is square, and m=n is its order. Accordingly, any element of matrix A can be denoted by its row and column numbers: a xy ; x - row number, changes, y - column number, changes.
B is not the main point of the decision. In principle, all operations can be performed directly with the equations themselves, but the notation will be much more cumbersome, and it will be much easier to get confused in it.
Determinant
The matrix also has a determinant. This is very important characteristic. There is no need to find out its meaning now; you can simply show how it is calculated, and then tell what properties of the matrix it determines. The easiest way to find the determinant is through diagonals. Imaginary diagonals are drawn in the matrix; the elements located on each of them are multiplied, and then the resulting products are added: diagonals with a slope to the right - with a plus sign, with a slope to the left - with a minus sign.
It is extremely important to note that the determinant can only be calculated for a square matrix. For a rectangular matrix, you can do the following: choose the smallest from the number of rows and the number of columns (let it be k), and then randomly mark k columns and k rows in the matrix. The elements at the intersection of the selected columns and rows will form a new square matrix. If the determinant of such a matrix is a non-zero number, it is called the basis minor of the original rectangular matrix.
Before you start solving a system of equations using the Gaussian method, it doesn’t hurt to calculate the determinant. If it turns out to be zero, then we can immediately say that the matrix has either an infinite number of solutions or none at all. In such a sad case, you need to go further and find out about the rank of the matrix.
System classification
There is such a thing as the rank of a matrix. This is the maximum order of its non-zero determinant (if we remember about the basis minor, we can say that the rank of a matrix is the order of the basis minor).
Based on the situation with rank, SLAE can be divided into:
- Joint. U In joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended matrix (with a column of free terms). Such systems have a solution, but not necessarily one, therefore, additionally joint systems are divided into:
- - certain- having a single solution. In certain systems, the rank of the matrix and the number of unknowns (or the number of columns, which is the same thing) are equal;
- - undefined - with an infinite number of solutions. The rank of matrices in such systems is less than the number of unknowns.
- Incompatible. U In such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.
The Gauss method is good because during the solution it allows one to obtain either an unambiguous proof of the inconsistency of the system (without calculating the determinants of large matrices), or a solution in general form for a system with an infinite number of solutions.
Elementary transformations
Before proceeding directly to solving the system, you can make it less cumbersome and more convenient for calculations. This is achieved through elementary transformations - such that their implementation does not change the final answer in any way. It should be noted that some of the given elementary transformations are valid only for matrices, the source of which was the SLAE. Here is a list of these transformations:
- Rearranging lines. Obviously, if you change the order of the equations in the system record, this will not affect the solution in any way. Consequently, rows in the matrix of this system can also be swapped, not forgetting, of course, the column of free terms.
- Multiplying all elements of a string by a certain coefficient. Very helpful! It can be used to shorten big numbers in the matrix or remove zeros. Many decisions, as usual, will not change, but further operations will become more convenient. The main thing is that the coefficient is not equal to zero.
- Removing rows with proportional factors. This partly follows from the previous paragraph. If two or more rows in a matrix have proportional coefficients, then when one of the rows is multiplied/divided by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and the extra ones can be removed, leaving only one.
- Removing a null line. If, during the transformation, a row is obtained somewhere in which all elements, including the free term, are zero, then such a row can be called zero and thrown out of the matrix.
- Adding to the elements of one row the elements of another (in the corresponding columns), multiplied by a certain coefficient. The most unobvious and most important transformation of all. It is worth dwelling on it in more detail.
Adding a string multiplied by a factor
For ease of understanding, it is worth breaking down this process step by step. Two rows are taken from the matrix:
a 11 a 12 ... a 1n | b1
a 21 a 22 ... a 2n | b 2
Let's say you need to add the first to the second, multiplied by the coefficient "-2".
a" 21 = a 21 + -2×a 11
a" 22 = a 22 + -2×a 12
a" 2n = a 2n + -2×a 1n
Then the second row in the matrix is replaced with a new one, and the first remains unchanged.
a 11 a 12 ... a 1n | b1
a" 21 a" 22 ... a" 2n | b 2
It should be noted that the multiplication coefficient can be selected in such a way that, as a result of adding two rows, one of the elements of the new row is equal to zero. Therefore, it is possible to obtain an equation in a system where there will be one less unknown. And if you get two such equations, then the operation can be done again and get an equation that will contain two fewer unknowns. And if each time you turn one coefficient of all rows that are below the original one to zero, then you can, like stairs, go down to the very bottom of the matrix and get an equation with one unknown. This is called solving the system using the Gaussian method.
In general
Let there be a system. It has m equations and n unknown roots. You can write it as follows:
The main matrix is compiled from the system coefficients. A column of free terms is added to the extended matrix and, for convenience, separated by a line.
- the first row of the matrix is multiplied by the coefficient k = (-a 21 /a 11);
- the first modified row and the second row of the matrix are added;
- instead of the second row, the result of the addition from the previous paragraph is inserted into the matrix;
- now the first coefficient in the new second row is a 11 × (-a 21 /a 11) + a 21 = -a 21 + a 21 = 0.
Now the same series of transformations is performed, only the first and third rows are involved. Accordingly, at each step of the algorithm, element a 21 is replaced by a 31. Then everything is repeated for a 41, ... a m1. The result is a matrix where the first element in the rows is zero. Now you need to forget about line number one and perform the same algorithm, starting from line two:
- coefficient k = (-a 32 /a 22);
- the second modified line is added to the “current” line;
- the result of the addition is substituted into the third, fourth, and so on lines, while the first and second remain unchanged;
- in the rows of the matrix the first two elements are already equal to zero.
The algorithm must be repeated until the coefficient k = (-a m,m-1 /a mm) appears. This means that in last time the algorithm was performed only for the lower equation. Now the matrix looks like a triangle, or has a stepped shape. In the bottom line there is the equality a mn × x n = b m. The coefficient and free term are known, and the root is expressed through them: x n = b m /a mn. The resulting root is substituted into the top line to find x n-1 = (b m-1 - a m-1,n ×(b m /a mn))÷a m-1,n-1. And so on by analogy: in each next line there is a new root, and, having reached the “top” of the system, you can find many solutions. It will be the only one.
When there are no solutions
If in one of the matrix rows all elements except the free term are equal to zero, then the equation corresponding to this row looks like 0 = b. It has no solution. And since such an equation is included in the system, then the set of solutions of the entire system is empty, that is, it is degenerate.
When there are an infinite number of solutions
It may happen that in the given triangular matrix there are no rows with one coefficient element of the equation and one free term. There are only lines that, when rewritten, would look like an equation with two or more variables. This means that the system has an infinite number of solutions. In this case, the answer can be given in the form of a general solution. How to do it?
All variables in the matrix are divided into basic and free. Basic ones are those that stand “on the edge” of the rows in the step matrix. The rest are free. In the general solution, the basic variables are written through free ones.
For convenience, the matrix is first rewritten back into a system of equations. Then in the last of them, where exactly there is only one basic variable left, it remains on one side, and everything else is transferred to the other. This is done for every equation with one basic variable. Then, in the remaining equations, where possible, the expression obtained for it is substituted instead of the basic variable. If the result is again an expression containing only one basic variable, it is again expressed from there, and so on, until each basic variable is written as an expression with free variables. This is the general solution of SLAE.
You can also find the basic solution of the system - give the free variables any values, and then for this specific case calculate the values of the basic variables. There are an infinite number of particular solutions that can be given.
Solution with specific examples
Here is a system of equations.
For convenience, it is better to immediately create its matrix
It is known that when solved by the Gaussian method, the equation corresponding to the first row will remain unchanged at the end of the transformations. Therefore, it will be more profitable if the upper left element of the matrix is the smallest - then the first elements of the remaining rows after the operations will turn to zero. This means that in the compiled matrix it will be advantageous to put the second row in place of the first one.
second line: k = (-a 21 /a 11) = (-3/1) = -3
a" 21 = a 21 + k×a 11 = 3 + (-3)×1 = 0
a" 22 = a 22 + k×a 12 = -1 + (-3)×2 = -7
a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11
b" 2 = b 2 + k×b 1 = 12 + (-3)×12 = -24
third line: k = (-a 3 1 /a 11) = (-5/1) = -5
a" 3 1 = a 3 1 + k×a 11 = 5 + (-5)×1 = 0
a" 3 2 = a 3 2 + k×a 12 = 1 + (-5)×2 = -9
a" 3 3 = a 33 + k×a 13 = 2 + (-5)×4 = -18
b" 3 = b 3 + k×b 1 = 3 + (-5)×12 = -57
Now, in order not to get confused, you need to write down a matrix with the intermediate results of the transformations.
Obviously, such a matrix can be made more convenient for perception using certain operations. For example, you can remove all “minuses” from the second line by multiplying each element by “-1”.
It is also worth noting that in the third line all elements are multiples of three. Then you can shorten the string by this number, multiplying each element by "-1/3" (minus - at the same time, to remove negative values).
Looks much nicer. Now we need to leave the first line alone and work with the second and third. The task is to add the second line to the third line, multiplied by such a coefficient that the element a 32 becomes equal to zero.
k = (-a 32 /a 22) = (-3/7) = -3/7 (if during some transformations the answer does not turn out to be an integer, it is recommended to maintain the accuracy of the calculations to leave it “as is”, in the form common fraction, and only then, when the answers are received, decide whether to round and convert to another form of recording)
a" 32 = a 32 + k×a 22 = 3 + (-3/7)×7 = 3 + (-3) = 0
a" 33 = a 33 + k×a 23 = 6 + (-3/7)×11 = -9/7
b" 3 = b 3 + k×b 2 = 19 + (-3/7)×24 = -61/7
The matrix is written again with new values.
| 1 | 2 | 4 | 12 |
| 0 | 7 | 11 | 24 |
| 0 | 0 | -9/7 | -61/7 |
As you can see, the resulting matrix already has a stepped form. Therefore, further transformations of the system using the Gaussian method are not required. What you can do here is to remove the overall coefficient "-1/7" from the third line.
Now everything is beautiful. All that’s left to do is write the matrix again in the form of a system of equations and calculate the roots
x + 2y + 4z = 12 (1)
7y + 11z = 24 (2)
The algorithm by which the roots will now be found is called the reverse move in the Gaussian method. Equation (3) contains the z value:
y = (24 - 11×(61/9))/7 = -65/9
And the first equation allows us to find x:
x = (12 - 4z - 2y)/1 = 12 - 4×(61/9) - 2×(-65/9) = -6/9 = -2/3
We have the right to call such a system joint, and even definite, that is, having a unique solution. The answer is written in the following form:
x 1 = -2/3, y = -65/9, z = 61/9.
An example of an uncertain system
The variant of solving a certain system using the Gauss method has been analyzed; now it is necessary to consider the case if the system is uncertain, that is, infinitely many solutions can be found for it.
x 1 + x 2 + x 3 + x 4 + x 5 = 7 (1)
3x 1 + 2x 2 + x 3 + x 4 - 3x 5 = -2 (2)
x 2 + 2x 3 + 2x 4 + 6x 5 = 23 (3)
5x 1 + 4x 2 + 3x 3 + 3x 4 - x 5 = 12 (4)
The very appearance of the system is already alarming, because the number of unknowns is n = 5, and the rank of the system matrix is already exactly less than this number, because the number of rows is m = 4, that is, the largest order of the determinant-square is 4. This means that there are an infinite number of solutions, and you need to look for its general appearance. The Gauss method for linear equations allows you to do this.
First, as usual, an extended matrix is compiled.
Second line: coefficient k = (-a 21 /a 11) = -3. In the third line, the first element is before the transformations, so you don’t need to touch anything, you need to leave it as is. Fourth line: k = (-a 4 1 /a 11) = -5
By multiplying the elements of the first row by each of their coefficients in turn and adding them to the required rows, we obtain the matrix the following type:
As you can see, the second, third and fourth rows consist of elements proportional to each other. The second and fourth are generally identical, so one of them can be removed immediately, and the remaining one can be multiplied by the coefficient “-1” and get line number 3. And again, out of two identical lines, leave one.
The result is a matrix like this. While the system has not yet been written down, it is necessary to determine the basic variables here - those standing at the coefficients a 11 = 1 and a 22 = 1, and free ones - all the rest.
In the second equation there is only one basic variable - x 2. This means that it can be expressed from there by writing it through the variables x 3 , x 4 , x 5 , which are free.
We substitute the resulting expression into the first equation.
The result is an equation in which the only basic variable is x 1 . Let's do the same with it as with x 2.
All basic variables, of which there are two, are expressed in terms of three free ones; now we can write the answer in general form.
You can also specify one of the particular solutions of the system. For such cases, zeros are usually chosen as values for free variables. Then the answer will be:
16, 23, 0, 0, 0.
An example of a non-cooperative system
Solution incompatible systems equations by the Gaussian method - the fastest. It ends immediately as soon as at one of the stages an equation is obtained that has no solution. That is, the stage of calculating the roots, which is quite long and tedious, is eliminated. The following system is considered:
x + y - z = 0 (1)
2x - y - z = -2 (2)
4x + y - 3z = 5 (3)
As usual, the matrix is compiled:
| 1 | 1 | -1 | 0 |
| 2 | -1 | -1 | -2 |
| 4 | 1 | -3 | 5 |
And it is reduced to a stepwise form:
k 1 = -2k 2 = -4
| 1 | 1 | -1 | 0 |
| 0 | -3 | 1 | -2 |
| 0 | 0 | 0 | 7 |
After the first transformation, the third line contains an equation of the form
without a solution. Consequently, the system is inconsistent, and the answer will be the empty set.
Advantages and disadvantages of the method
If you choose which method to solve SLAEs on paper with a pen, then the method that was discussed in this article looks the most attractive. It is much more difficult to get confused in elementary transformations than if you have to manually search for a determinant or some tricky inverse matrix. However, if you use programs for working with data of this type, for example, spreadsheets, then it turns out that such programs already contain algorithms for calculating the main parameters of matrices - determinant, minors, inverse, and so on. And if you are sure that the machine will calculate these values itself and will not make mistakes, it is more advisable to use the matrix method or Cramer’s formulas, because their application begins and ends with the calculation of determinants and inverse matrices.
Application
Since the Gaussian solution is an algorithm, and the matrix is actually a two-dimensional array, it can be used in programming. But since the article positions itself as a guide “for dummies,” it should be said that the easiest place to put the method into is spreadsheets, for example, Excel. Again, any SLAE entered into a table in the form of a matrix will be considered by Excel as a two-dimensional array. And for operations with them there are many nice commands: addition (you can only add matrices of the same size!), multiplication by a number, multiplication of matrices (also with certain restrictions), finding the inverse and transposed matrices and, most importantly, calculating the determinant. If this time-consuming task is replaced by a single command, it is possible to determine the rank of the matrix much more quickly and, therefore, establish its compatibility or incompatibility.