Gauss method perfect for solving linear systems algebraic equations(SLAU). It has a number of advantages compared to other methods:
- firstly, there is no need to first examine the system of equations for consistency;
- secondly, the Gauss method can 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-singular, 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 is equal to zero;
- thirdly, the Gaussian method leads to results with a relatively small number of computational operations.
Brief overview of the article.
First, we give the necessary definitions and introduce notations.
Next, we will 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 is the sequential elimination of unknown variables. Therefore, the Gaussian method is also called the method of sequential elimination of unknowns. We will show detailed solutions of several examples.
In conclusion, we will consider the solution by the Gauss method of systems of linear algebraic equations, the main matrix of which is either rectangular or singular. The solution to such systems has some features, which we will examine in detail using examples.
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Basic definitions and notations.
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), and are free terms.
If 
, then the system of linear algebraic equations is called homogeneous, otherwise - heterogeneous.
The set of values of unknown variables at which all equations of the system become identities is called decision of the SLAU.
If there is at least one solution to a system of linear algebraic equations, then it is called joint, otherwise - incompatible.
If a SLAE has a unique solution, then it is called certain. If there is more than one solution, then the system is called uncertain.
They say that the system is written in coordinate form, if it has the form
.
This system in matrix form records has the form , where 
- the main matrix of the SLAE, - the matrix of the column of unknown variables, - the matrix of free terms.
If we add a matrix-column of free terms to matrix A as the (n+1)th column, we get the so-called extended matrix systems of linear equations. Typically, an extended matrix is denoted by the letter T, and the column of free terms is separated by a vertical line from the remaining columns, that is, 
The square matrix A is called degenerate, if its determinant is zero. If , then matrix A is called non-degenerate.
The following point should be noted.
If we perform with a system of linear algebraic equations the following actions
- swap two equations,
- multiply both sides of any equation by an arbitrary and non-zero real (or complex) number k,
- to both sides of any equation add the corresponding parts of another equation, multiplied by an arbitrary number k,
then you get an equivalent system that has the same solutions (or, just like the original one, has no solutions).
For an extended matrix of a system of linear algebraic equations, these actions will mean carrying out elementary transformations with the rows:
- swapping two lines,
- multiplying all elements of any row of matrix T by a nonzero number k,
- adding to the elements of any row of a 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 non-singular, using 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 that.
Note that adding to the left side of the second equation left side first, and to the right side - the right one, 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 =1 into the first and third equations of the system: 
If we multiply both sides of the third equation of the system by -1 and add them to the corresponding parts of the first equation, we get rid of the unknown variable x 3 and can find x 2: 
We substitute the resulting value x 2 = 2 into the third equation and find the remaining unknown variable x 3: 
Others would have done differently.
Let us resolve 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 for x 2 and substitute the result obtained into the third equation to eliminate the unknown variable x 2 from it: 
From the third equation of the system it is clear 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 Gaussian method. When we expressed the unknown variables (first x 1, at the next stage x 2) and substituted them into the remaining equations of the system, we thereby excluded them. We carried out elimination until there was only one unknown variable left in the last equation. The process of sequentially eliminating unknowns is called direct Gaussian method. After finishing forward stroke we now have the opportunity to calculate the unknown variable in the last equation. With its help, we find the next unknown variable from the penultimate equation, and so on. The process of sequentially finding unknown variables while moving from the last equation to the first is called in 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 makes it possible to eliminate the unknown variable x 1 from the second and third equations of the system: 
Nuances with the elimination of unknown variables using the Gaussian 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 for x 1 in order to eliminate this unknown variable from the remaining 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 is always 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 resolve the first equation for x 1 and exclude it from the remaining equations of the system (although x 1 is no longer present in the second equation).
We hope you get the gist.
Let's describe Gaussian method algorithm.
Suppose we need to solve a system of n linear algebraic equations with n unknowns variables of the form 
, and let the determinant of its main matrix be different from zero.
We will assume that , since we can always achieve this by rearranging the equations of the system. Let's eliminate the unknown variable x 1 from all equations of the system, starting with the second. To do this, to the second equation of the system we add the first, multiplied by , to the third equation we add the first, multiplied by , and so on, to the nth equation we add the first, multiplied by . The system of equations after such transformations will take the form 
where , and 
.
We would have arrived at the same result if we had 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 proceed in a similar way, but only with part of the resulting system, which is marked in the figure 
To do this, to the third equation of the system we add the second, multiplied by , to the fourth equation we add the second, multiplied by , and so on, to the nth equation we add the second, multiplied by . The system of equations after such transformations will take the form 
where , and 
. Thus, the variable x 2 is excluded from all equations, starting from the third.
Next, we proceed to eliminating the unknown x 3, while we act similarly with the part of the system marked in the figure 
So we continue the direct progression of the Gaussian method until the system takes the form 
From this moment we begin the reverse of the Gaussian 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 look at the algorithm using an example.
Example.
Gauss method.
Solution.
The coefficient a 11 is different from zero, so let’s proceed to the direct progression of the Gaussian method, that is, to the exclusion of the unknown variable x 1 from all equations of the system except the first. To do this, to the left and right sides of the second, third and fourth equations, add the left and right sides of the first equation, multiplied by , respectively. 
And : 
The unknown variable x 1 has been eliminated, let's move on to eliminating x 2 . To the left and right sides of the third and fourth equations of the system we add the left and right sides of the second equation, multiplied by respectively 
And 
:
To complete the forward progression of the Gaussian method, we need to eliminate the unknown variable x 3 from the last equation of the system. Let us add to the left and right sides of the fourth equation, respectively, the left and right side third equation multiplied by 
:
You can begin the reverse of the Gaussian method.
From the last equation we have 
,
from the third equation we get,
from the second,
from the first one.
To check, you can substitute the obtained values of the unknown variables into the original system of equations. All equations turn into identities, which indicates that the solution using the Gauss method was found correctly.
Answer:
Now let’s give a solution to the same example using the Gaussian method in matrix notation.
Example.
Find the solution to the system of equations Gauss method.
Solution.
The extended matrix of the system has the form 
. At the top of each column are the unknown variables that correspond to the elements of the matrix.
The direct approach of the Gaussian method here involves reducing the extended matrix of the system to a trapezoidal form using elementary transformations. This process is similar to the elimination of unknown variables that we did with the system in coordinate form. Now you will see this.
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 lines we add the corresponding elements of the first line multiplied by , and accordingly: 
Next, we transform the resulting matrix so that in the second column all elements, starting from the third, become zero. This would correspond to eliminating the unknown variable x 2 . To do this, to the elements of the third and fourth rows we add the corresponding elements of the first row of the matrix, multiplied by respectively 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 a system of linear equations 
which was obtained earlier after a forward move.
It's time to turn back. In matrix notation, the inverse of the Gaussian method involves transforming the resulting matrix such that the matrix marked in the figure 
became diagonal, that is, took the form 
where are some numbers.
These transformations are similar to the forward transformations of the Gaussian 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 lines the corresponding elements of the last line, multiplied by 
, on and on 
respectively: 
Now add to the elements of the second and first lines the corresponding elements of the third line, multiplied by and by, respectively: 
At the last step of the reverse Gaussian method, to the elements of the first row we add the corresponding elements of the second row, multiplied by: 
The resulting matrix corresponds to the system of equations 
, from where 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 completely incorrect results. We recommend not rounding decimals. Better from decimals move on to ordinary fractions.
Example.
Solve a system of three equations using the Gauss 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: 
Let us exclude the unknown x from the second and third equations of the system: 
In the resulting system, the unknown variable y is absent in the second equation, but y is present in the third equation, therefore, let’s swap the second and third equations: 
This completes the direct progression of the Gauss method (there is no need to exclude y from the third equation, since this unknown variable no longer exists).
Let's start the reverse move.
From the last equation we find 
,
from the penultimate 
from the first equation we have 
Answer:
X = 10, y = 5, z = -20.
Solving 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 singular, using the Gauss method.
Systems of equations, the main matrix of which is rectangular or square singular, 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 incompatibility 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 going into detail about some situations that may arise.
Let's move on to the most important stage.
So, let us assume that the system of linear algebraic equations, after completing the forward progression of the Gauss method, takes the form 
and not a single equation was reduced to (in this case we would conclude that the system is incompatible). A logical question arises: “What to do next”?
Let us write down the unknown variables that come first in all equations of the resulting system: 
In our example these are x 1, x 4 and x 5. On the left sides of the equations of the system we leave only those terms that contain the written unknown variables x 1, x 4 and x 5, the remaining terms are transferred to the right side of the equations with the opposite sign: 
Let's give the unknown variables that are on the right sides of the equations arbitrary values, where 
- arbitrary numbers: 
After this, the right-hand sides of all equations of our SLAE contain numbers and we can proceed to the reverse of the Gaussian method.
From the last equation of the system we have, from the penultimate equation we find, from the first equation we get 
The solution to a system of equations is a set of values of unknown variables 
Giving Numbers different values, we will obtain different solutions to the system 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.
Decide homogeneous system linear algebraic equations 
Gauss method.
Solution.
Let us exclude the unknown variable x from the second and third equations of the system. To do this, to the left and right sides of the second equation, we add, respectively, the left and right sides of the first equation, multiplied by , and to the left and right sides of the third equation, we add the left and right sides of the first equation, multiplied by: 
Now let’s exclude y from the third equation of the resulting system of equations: 
The resulting SLAE is equivalent to the system 
.
We leave on the left side of the system equations only the terms containing the unknown variables x and y, and move the terms with the unknown variable z to the right side: 
Let a system of linear algebraic equations be given that needs to be solved (find such values of the unknowns xi that turn each equation of the system into an equality).
We know that a system of linear algebraic equations can:
1) Have no solutions (be incompatible).
2) Have infinitely many solutions.
3) Have a single solution.
As we remember, Cramer's rule and matrix method are unsuitable in cases where the system has infinitely many solutions or is inconsistent. Gauss method – the most powerful and versatile tool for finding solutions to any system of linear equations, which in every case will lead us to the answer! The algorithm of the method itself in all three cases works the same. If the Cramer and matrix methods require knowledge of determinants, then to apply the Gauss method you only need knowledge arithmetic operations, which makes it accessible even to primary school students.
Augmented matrix transformations ( this is the matrix of the system - a matrix composed only of the coefficients of the unknowns, plus a column of free terms) systems of linear algebraic equations in the Gauss method:
1) With troki matrices Can rearrange in some places.
2) if proportional ones appeared (or exist) in the matrix (as special case– identical) lines, then it follows delete All these rows are from the matrix except one.
3) if a zero row appears in the matrix during transformations, then it should also be delete.
4) a row of the matrix can be multiply (divide) to any number other than zero.
5) to a matrix row you can add another string multiplied by a number, different from zero.
In the Gauss method, elementary transformations do not change the solution of the system of equations.
The Gauss method consists of two stages:
- “Direct move” - using elementary transformations, bring the extended matrix of a system of linear algebraic equations to a “triangular” step form: the elements of the extended matrix located below the main diagonal are equal to zero (top-down move). For example, to this type:
To do this, perform the following steps:
1) Let us consider the first equation of a system of linear algebraic equations and the coefficient for x 1 is equal to K. The second, third, etc. we transform the equations as follows: we divide each equation (coefficients of the unknowns, including free terms) by the coefficient of the unknown x 1 in each equation, and multiply by K. After this, we subtract the first from the second equation (coefficients of unknowns and free terms). For x 1 in the second equation we obtain the coefficient 0. From the third transformed equation we subtract the first equation until all equations except the first, for unknown x 1, have a coefficient 0.
2) Let's move on to the next equation. Let this be the second equation and the coefficient for x 2 equal to M. We proceed with all “lower” equations as described above. Thus, “under” the unknown x 2 there will be zeros in all equations.
3) Move on to the next equation and so on until one last unknown and the transformed free term remain.
- The “reverse move” of the Gauss method is to obtain a solution to a system of linear algebraic equations (the “bottom-up” move).
Example.
From the last “lower” equation we obtain one first solution - the unknown x n. To do this, we solve the elementary equation A * x n = B. In the example given above, x 3 = 4. We substitute the found value into the “upper” next equation and solve it with respect to the next unknown. For example, x 2 – 4 = 1, i.e. x 2 = 5. And so on until we find all the unknowns.
Let's solve the system of linear equations using the Gauss method, as some authors advise:
Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:
We look at the upper left “step”. We should have a unit 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. Let's do this:
1 step
. 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 action: multiply the first line by –1 (change its sign). . The first line, multiplied by 5, was added to the second line. The first line, multiplied by 3, was added to the third line.
Step 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.
Step 4 . The third line was added to the second line, multiplied by 2.
Step 5 . The third line was divided by 3.
A sign that indicates an error in calculations (more rarely, a typo) is a “bad” bottom line. That is, if we got something like (0 0 11 |23) below, and, accordingly, 11x 3 = 23, x 3 = 23/11, then with a high degree of probability we can say that an error was made during elementary transformations.
Let’s do the reverse; in the design of examples, the system itself is often not rewritten, but the equations are “taken directly from the given matrix.” The reverse move, I remind you, works from the bottom up. In this example, the result was a gift:
x 3 = 1
x 2 = 3
x 1 + x 2 – x 3 = 1, therefore x 1 + 3 – 1 = 1, x 1 = –1
Answer:x 1 = –1, x 2 = 3, x 3 = 1.
Let's solve the same system using the proposed algorithm. We get
4 2 –1 1
5 3 –2 2
3 2 –3 0
Divide the second equation by 5, and the third by 3. We get:
4 2 –1 1
1 0.6 –0.4 0.4
1 0.66 –1 0
Multiplying the second and third equations by 4, we get:
4 2 –1 1
4 2,4 –1.6 1.6
4 2.64 –4 0
Subtract the first equation from the second and third equations, we have:
4 2 –1 1
0 0.4 –0.6 0.6
0 0.64 –3 –1
Divide the third equation by 0.64:
4 2 –1 1
0 0.4 –0.6 0.6
0 1 –4.6875 –1.5625
Multiply the third equation by 0.4
4 2 –1 1
0 0.4 –0.6 0.6
0 0.4 –1.875 –0.625
Subtracting the second from the third equation, we obtain a “stepped” extended matrix:
4 2 –1 1
0 0.4 –0.6 0.6
0 0 –1.275 –1.225
Thus, since the error accumulated during the calculations, we obtain x 3 = 0.96 or approximately 1.
x 2 = 3 and x 1 = –1.
By solving in this way, you will never get confused in the calculations and, despite the calculation errors, you will get the result.
This method of solving a system of linear algebraic equations is easy to program and does not take into account specific features coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.
I wish you success! See you in class! Tutor.
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Let the system be given, ∆≠0. (1)Gauss method is a method of sequentially eliminating 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 obtained sequentially (in reverse). Let's consider one of the computational schemes. This circuit is called a single division circuit. So let's look at this diagram. Let a 11 ≠0 (leading element) divide the first equation by a 11. We get
(2)
Using equation (2), it is easy to eliminate the unknowns x 1 from the remaining equations of the system (to do this, it is enough to subtract equation (2) from each equation, previously multiplied by the corresponding coefficient for x 1), that is, in 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” onto the first column and the first (transformed) row.
Following this, leaving the first equation alone, we perform a similar transformation over the remaining equations of the system obtained in the first step: we select from among them the equation with the leading element and, with its help, exclude x 2 from the remaining equations (step 2).
After n steps, instead of (1), we obtain an equivalent system 
(3)
Thus, at the first stage we obtain a triangular system (3). This stage is called forward stroke.
At the second stage (reverse), we find sequentially from (3) the values x n, x n -1, ..., x 1.
Let us denote the resulting solution as x 0 . Then the difference ε=b-A x 0 called residual.
If ε=0, then the found solution x 0 is correct.
Calculations using the Gaussian method are performed in two stages:
- The first stage is called the forward method. At the first stage, the original system is converted to a triangular form.
- The second stage is called the reverse stroke. At the second stage, a triangular system equivalent to the original one is solved.
At each step, the leading element was assumed to be nonzero. If this is not the case, then any other element can be used as a leading element, as if rearranging the equations of the system.
Purpose of the Gauss method
The Gauss method is designed for solving systems of linear equations. Refers to direct solution methods.Types of Gaussian method
- Classical Gaussian method;
- Modifications of the Gauss method. One of the modifications of the Gaussian method is a scheme with the choice of the main element. A feature of the Gauss method with the choice of the main element is such a rearrangement of the equations so that at the kth step the leading element turns out to be the largest element in the kth column.
- Jordano-Gauss method;
Let's illustrate the difference Jordano-Gauss method from the Gaussian method with examples.
Example of a solution using the Gaussian method
Let's solve the system: 
For ease of calculation, let's swap the lines: 
Let's multiply the 2nd line by (2). Add the 3rd line to the 2nd 
Multiply the 2nd line by (-1). Add the 2nd line 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 using the Jordano-Gauss method
Let us solve the same SLAE using the Jordano-Gauss method. 
We will sequentially select the resolving element RE, which lies on the main diagonal of the matrix.
The resolution element is equal to (1).
NE = SE - (A*B)/RE
RE - resolving element (1), A and B - matrix elements forming a rectangle with elements STE and RE.
Let's present the calculation of each element in the form of a table:
| x 1 | x 2 | x 3 | B |
| 1 / 1 = 1 | 2 / 1 = 2 | -2 / 1 = -2 | 1 / 1 = 1 |
 |  |  |  |
The resolving 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 elements of column B, are determined by the rectangle rule.
To do this, we select four numbers that are located at the vertices of the rectangle and always include the resolving element RE.
| x 1 | x 2 | x 3 | B |
 |  |
||
| 0 / 3 = 0 | 3 / 3 = 1 | 1 / 3 = 0.33 | 4 / 3 = 1.33 |
 |
The resolution element is (-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 elements of column B, are determined by the rectangle rule.
To do this, we select four numbers that are located at the vertices of the rectangle and always include the resolving element RE.
Let's present the calculation of each element in the form of a table:
| x 1 | x 2 | x 3 | B |
 |  |  |  |
 |  |
||
| 0 / -4 = 0 | 0 / -4 = 0 | -4 / -4 = 1 | -4 / -4 = 1 |
Answer: x 1 = 1, x 2 = 1, x 3 = 1
Implementation of the Gaussian method
The Gaussian method is implemented in many programming languages, in particular: Pascal, C++, php, Delphi, and there is also an online implementation of the Gaussian method.Using the Gaussian 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 Gaussian method.Application of the Gauss method in solving differential equations
To find a particular solution to a differential equation, first find derivatives of the appropriate degree for the written partial solution (y=f(A,B,C,D)), which are substituted into original equation. Next to find variables A,B,C,D a system of equations is compiled and solved by the Gaussian method.Application of the Jordano-Gauss method in linear programming
IN linear programming, in particular, in the simplex method, the rectangle rule, which uses the Jordano-Gauss method, is used to transform the simplex table at each iteration.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 dealt with school education, mainly from the point of view 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 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
Let me make a reservation right away: the Gauss method has much more wide 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 it would be discouraging for others to think.
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, able 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 friend said 35 years ago scientific adviser: “They have memorized the question.” 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 the “scientists” are visible in the level of current school mathematics, the level of its teaching and the 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 adherent 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 the method” of Gauss 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 gets this right, then later he will come up with many more such methods, 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 is the greatest invention of mankind, 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 Carl Gauss experienced a high sense of respect for his 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 Organization intellectual property WIPO has achieved a ban on the use of its fair name in school textbooks!..
What main mistake school approach? Or, as I put it, a crime of school mathematics teachers 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 is made 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 half an hour maximum. 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 ...
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 outline. 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
None hidden meaning not in the matrix. This is simply a convenient way to record data for subsequent operations with it. Even schoolchildren do not need to be afraid of them.
The matrix is always rectangular, because it is more convenient. Even in the Gaussian method, where everything comes down to constructing a matrix triangular in appearance, the entry contains a rectangle, 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 letters are usually used to denote them) letters) 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 maximum order its determinant, different from zero (if we remember about the basis minor, we can say that the rank of the 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 joint systems are additionally 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 strings. Obviously, if you change the order of the equations in the system record, this will not affect the solution in any way. Consequently, in the matrix of this system it is also possible to swap rows, 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 it 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. Consequently, 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 new second line 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. That's what it is common decision SLAU.
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 can be done here is to remove from the third line overall coefficient "-1/7".
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 highest 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 a matrix of the following form:
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
Solving incompatible systems of equations using the Gauss method is 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 to work with this type of data, 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 with a single command, it is possible to determine the rank of the matrix much more quickly and, therefore, establish its compatibility or incompatibility.
