Systems solution linear equations Gaussian method. Suppose we need to find a solution to the system from n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.
The essence of the Gauss method consists in the successive exclusion of unknown variables: first, the x 1 from all equations of the system, starting from the second, then x2 of all equations, starting with the third, and so on, until only the unknown variable remains in the last equation x n. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called direct Gauss method. After the completion of the forward move of the Gauss method, from the last equation we find x n, using this value from the penultimate equation is calculated xn-1, and so on, from the first equation is found x 1. The process of calculating unknown variables when moving from the last equation of the system to the first is called reverse Gauss method.
Let us briefly describe the algorithm for eliminating unknown variables.
We will assume that , since we can always achieve this by rearranging the equations of the system. Eliminate the unknown variable x 1 from all equations of the system, starting from the second. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, to n-th add the first equation, multiplied by . The system of equations after such transformations will take the form 
where , a 
.
We would arrive at the same result if we expressed x 1 through other unknown variables in the first equation of the system and the resulting expression was substituted into all other equations. So the variable x 1 excluded from all equations, starting with the second.
Next, we act similarly, but only with a part of the resulting system, which is marked in the figure 
To do this, add the second multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, to n-th add the second equation, multiplied by . The system of equations after such transformations will take the form 
where , a 
. So the variable x2 excluded from all equations, starting with the third.
Next, we proceed to the elimination of the unknown x 3, while we act similarly with the part of the system marked in the figure 
So we continue the direct course of the Gauss method until the system takes the form 
From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as , using the obtained value x n find xn-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Example.
Solve System of Linear Equations 
Gaussian method.
In this article, the method is considered as a way to solve. 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 infinitely many solutions. Or they don't have it at all.
What does Gauss mean?
First you need to write down our system of equations in It looks like this. The system is taken:
The coefficients are written in the form of a table, and on the right in a separate column - free members. The column with free members is separated for convenience. The matrix that includes this column is called extended.
Further, the main matrix with coefficients must be reduced to the upper triangular shape. This is the main point of solving the system by the Gauss method. Simply put, after certain manipulations, the matrix should look like this, so that there are only zeros in its lower left part:
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 Gauss method in the most general terms. And what happens if suddenly the system does not have a solution? Or are there an infinite number of them? To answer these and many more questions, it is necessary to consider separately all the elements used in the solution by the Gauss method.
Matrices, their properties
There is no hidden meaning in the matrix. It's just a convenient way to record data for later operations. Even schoolchildren should not be afraid of them.
The matrix is always rectangular, because it is more convenient. Even in the Gauss method, where everything comes down to building a matrix triangular, a rectangle appears in the entry, only with zeros in the place where there are no numbers. Zeros can be omitted, but they are implied.
The matrix has a size. Its "width" is the number of rows (m), its "length" is the number of columns (n). Then the size of the matrix A (capital Latin letters are usually used for their designation) 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 the matrix A can be denoted by the number of its row and column: a xy ; x - row number, changes , y - column number, changes .
B is not the main point of the solution. In principle, all operations can be performed directly with the equations themselves, but the notation will turn out to be much more cumbersome, and it will be much easier to get confused in it.
Determinant
The matrix also has a determinant. This is a very important feature. Finding out its meaning now is not worth it, 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 of 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 located at the intersection of the selected columns and rows will form a new square matrix. If the determinant of such a matrix is a number other than zero, then it is called the basis minor of the original rectangular matrix.
Before proceeding with the solution of the system of equations by the Gauss method, it does not 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 there are 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 determinant, which is different from zero (if we recall the basis minor, we can say that the rank of a matrix is the order of the basis minor).
According to how things are with the rank, SLAE can be divided into:
- Joint. At of joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended one (with a column of free members). Such systems have a solution, but not necessarily one, therefore, joint systems are additionally divided into:
- - certain- having a unique 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;
- - indefinite - with an infinite number of solutions. The rank of matrices for such systems is less than the number of unknowns.
- Incompatible. At such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.
The Gauss method is good in that it allows one to obtain either an unambiguous proof of the inconsistency of the system (without calculating the determinants of large matrices) or a general solution for a system with an infinite number of solutions.
Elementary transformations
Before proceeding directly to the solution of the system, it is possible to 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 above elementary transformations are valid only for matrices, the source of which was precisely the SLAE. Here is a list of these transformations:
- String permutation. It is obvious that if we change the order of the equations in the system record, then this will not affect the solution in any way. Consequently, it is also possible to interchange rows in the matrix of this system, not forgetting, of course, about the column of free members.
- Multiplying all elements of a string by some factor. Very useful! It can be used to shorten big numbers in the matrix or remove zeros. The set of solutions, as usual, will not change, and it will become more convenient to perform further operations. The main thing is that the coefficient is not equal to zero.
- Delete rows with proportional coefficients. This partly follows from the previous paragraph. If two or more rows in the matrix have proportional coefficients, then when multiplying / dividing one of the rows by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and you can remove the extra ones, leaving only one.
- Removing the null line. If in the course of transformations a string is obtained somewhere in which all elements, including the free member, are zero, then such a string 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 obscure 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 disassembling 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
Suppose you need to add the first to the second, multiplied by the coefficient "-2".
a" 21 \u003d a 21 + -2 × a 11
a" 22 \u003d a 22 + -2 × a 12
a" 2n \u003d a 2n + -2 × a 1n
Then in the matrix the second row is replaced with a new one, and the first one remains unchanged.
a 11 a 12 ... a 1n | b1
a" 21 a" 22 ... a" 2n | b 2
It should be noted that the multiplication factor can be chosen in such a way that, as a result of the addition of two strings, one of the elements of the new string is equal to zero. Therefore, it is possible to obtain an equation in the 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 already contain two less unknowns. And if each time we turn to zero one coefficient for all rows that are lower than the original one, then we can, like steps, 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 down like this:
The main matrix is compiled from the coefficients of the system. A column of free members is added to the extended matrix and separated by a bar for convenience.
- 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, in each step of the algorithm, the 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 equal to zero. Now we need to forget about line number one and execute the same algorithm starting from the second line:
- coefficient k \u003d (-a 32 / a 22);
- the second modified line is added to the "current" line;
- the result of the addition is substituted in 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. The bottom line contains 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 row 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 for 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 turn out that in the reduced triangular matrix there are no rows with one element-the coefficient of the equation, and one - a free member. There are only strings 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 - these are those that stand "on the edge" of the rows in the stepped matrix. The rest are free. In the general solution, the basic variables are written in terms of the free ones.
For convenience, the matrix is first rewritten back into a system of equations. Then in the last of them, where exactly only one basic variable remained, it remains on one side, and everything else is transferred to the other. This is done for each equation with one basic variable. Then, in the rest of the equations, where possible, instead of the basic variable, the expression obtained for it is substituted. If the result is again an expression containing only one basic variable, it is expressed from there again, 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 particular case calculate the values of the basic variables. There are infinitely many particular solutions.
Solution with specific examples
Here is the system of equations.
For convenience, it is better to immediately create its matrix
It is known that when solving by the Gauss 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 in place of the first row.
second line: k = (-a 21 / a 11) = (-3/1) = -3
a" 21 \u003d a 21 + k × a 11 \u003d 3 + (-3) × 1 \u003d 0
a" 22 \u003d a 22 + k × a 12 \u003d -1 + (-3) × 2 \u003d -7
a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11
b "2 \u003d b 2 + k × b 1 \u003d 12 + (-3) × 12 \u003d -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 \u003d b 3 + k × b 1 \u003d 3 + (-5) × 12 \u003d -57
Now, in order not to get confused, it is necessary to write down the matrix with the intermediate results of the transformations.
It is obvious that such a matrix can be made more convenient for perception with the help of some 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 row 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 alone the first line and work with the second and third. The task is to add the second row to the third row, multiplied by such a factor that the element a 32 becomes equal to zero.
k = (-a 32 / a 22) = (-3/7) = -3/7 common fraction, and only then, when the answers are received, decide whether to round up and translate into another form of record)
a" 32 = a 32 + k × a 22 = 3 + (-3/7) × 7 = 3 + (-3) = 0
a" 33 \u003d a 33 + k × a 23 \u003d 6 + (-3/7) × 11 \u003d -9/7
b "3 \u003d b 3 + k × b 2 \u003d 19 + (-3/7) × 24 \u003d -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 by the Gauss method are not required. What can be done here is to remove the overall coefficient "-1/7" from the third line.
Now everything is beautiful. The point is small - 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 Gauss method. Equation (3) contains the value of z:
y = (24 - 11×(61/9))/7 = -65/9
And the first equation allows you to find x:
x = (12 - 4z - 2y)/1 = 12 - 4x(61/9) - 2x(-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 response is written in the following form:
x 1 \u003d -2/3, y \u003d -65/9, z \u003d 61/9.
An example of an indefinite system
The variant of solving a certain system by the Gauss method has been analyzed, now it is necessary to consider the case if the system is indefinite, 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 matrix of the system is already exactly less than this number, because the number of rows is m = 4, that is greatest order determinant-square - 4. This means that there are an infinite number of solutions, and we must look for its general form. The Gauss method for linear equations makes it possible to do this.
First, as usual, the augmented 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 it is. Fourth line: k = (-a 4 1 /a 11) = -5
Multiplying the elements of the first row by each of their coefficients in turn and adding them to the desired rows, we obtain a matrix of the following form:
As you can see, the second, third and fourth rows consist of elements that are proportional to each other. The second and fourth are generally the same, so one of them can be removed immediately, and the rest multiplied by the coefficient "-1" and get line number 3. And again, leave one of two identical lines.
It turned out such a matrix. The system has not yet been written down, it is necessary here to determine the basic variables - standing at the coefficients a 11 \u003d 1 and a 22 \u003d 1, and free - all the rest.
The second equation has only one basic variable - x 2 . Hence, it can be expressed from there, writing through the variables x 3 , x 4 , x 5 , which are free.
We substitute the resulting expression into the first equation.
It turned out 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 you can write the answer in a general form.
You can also specify one of the particular solutions of the system. For such cases, as a rule, zeros are chosen as values for free variables. Then the answer will be:
16, 23, 0, 0, 0.
An example of an incompatible system
The solution of inconsistent systems of equations by the Gauss method is the fastest. It ends as soon as at one of the stages an equation is obtained that has no solution. That is, the stage with the calculation of the roots, which is quite long and dreary, disappears. 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 stepped form:
k 1 \u003d -2k 2 \u003d -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
having no solution. Therefore, the system is inconsistent, and the answer is the empty set.
Advantages and disadvantages of the method
If you choose which method to solve SLAE on paper with a pen, then the method that was considered in this article looks the most attractive. In elementary transformations, it is much more difficult to get confused than it happens if you have to manually look 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 a mistake, it is more expedient 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, in fact, 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 shove the method into is spreadsheets, for example, Excel. Again, any SLAE entered in 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, matrix multiplication (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 much faster to determine the rank of a matrix and, therefore, to establish its compatibility or inconsistency.
The Gauss method was proposed by the famous German mathematician Carl Friedrich Gauss (1777 - 1855) and is one of the most universal methods for solving SLAEs. The essence of this method lies in the fact that by means of successive eliminations of unknowns, the given system is transformed into a stepped (in particular, triangular) system equivalent to the given one. In the practical solution of the problem, the extended matrix of the system with the help of elementary transformations over its rows is reduced to a stepped form. Then all the unknowns are sequentially found, starting from the bottom up.
The principle of the Gauss method
The Gaussian method includes direct (reducing the extended matrix to a stepped form, that is, obtaining zeros under the main diagonal) and reverse (obtaining zeros above the main diagonal of the expanded matrix) moves. The forward move is called the Gauss method, the reverse - the Gauss-Jordan method, which differs from the first one only in the sequence of excluding variables.
The Gauss method is ideal for solving systems containing more than three linear equations, for solving systems of equations that are not quadratic (which cannot be said about the Cramer method and the matrix method). That is, the Gauss method is the most universal method for finding a solution to any system of linear equations; it works in the case when the system has infinitely many solutions or is inconsistent.
Examples of solving systems of equations
Example
Exercise. Solve SLAE by Gauss method.
Decision. We write out the extended matrix of the system and, using elementary transformations over its rows, we bring this matrix to a stepped form (forward move) and then perform the reverse move of the Gauss method (we make zeros above the main diagonal). First, change the first and second row so that the element equals 1 (we do this to simplify the calculations):
Divide all elements of the third row by two (or, which is the same, multiply by):
From the third line we subtract the second, multiplied by 3:
Multiplying the third row by , we get:
Let us now carry out the reverse course of the Gauss method (the Gassout-Jordan method), that is, we will make zeros above the main diagonal. Let's start with the elements of the third column. It is necessary to reset the element , for this we subtract the third from the second line.
Carl Friedrich Gauss, the greatest mathematician, hesitated for a long time, choosing between philosophy and mathematics. Perhaps it was precisely such a mindset that allowed him to "leave" so noticeably in world science. In particular, by creating the "Gauss Method" ...
For almost 4 years, the articles of this site have dealt with school education, mainly from the side of philosophy, the principles of (mis)understanding, introduced into the minds of children. The time is coming for more specifics, examples and methods ... I believe that this is the approach to the familiar, confusing and important areas of life gives the best results.
We humans are so arranged that no matter how much you talk about abstract thinking, but understanding always happens through examples. If there are no examples, then it is impossible to catch the principles ... How impossible it is to be on the top of a mountain otherwise than by going through its entire slope from the foot.
Same with school: for now living stories not enough we instinctively continue to regard it as a place where children are taught to understand.
For example, teaching the Gauss method...
Gauss method in the 5th grade of the school
I will make a reservation right away: the Gauss method has much more wide application, for example, when solving systems of linear equations. What we are going to talk about takes place in the 5th grade. This is start, having understood which, it is much easier to understand more "advanced options". In this article we are talking about method (method) of Gauss when finding the sum of a series
Here is an example that I brought from school younger son attending the 5th grade of the Moscow gymnasium.
School demonstration of the Gauss method
Math teacher using interactive whiteboard (modern methods training) showed the children a presentation of the history of the "creation of the method" by little Gauss.
The school teacher whipped little Carl (an outdated method, now not used in schools) for being,
instead of sequentially adding numbers from 1 to 100 to find their sum noticed that pairs of numbers equally spaced from the edges of an arithmetic progression add up to the same number. for example, 100 and 1, 99 and 2. Having counted the number of such pairs, little Gauss almost instantly solved the problem proposed by the teacher. For which he was subjected to execution in front of an astonished public. To the rest to think was disrespectful.
What did little Gauss do developed number sense? Noticed some feature number series with a constant step (arithmetic progression). And exactly this made him later a great scientist, able to notice, possessing feeling, instinct of understanding.
This is the value of mathematics, which develops ability to see general in particular - abstract thinking. Therefore, most parents and employers instinctively consider mathematics an important discipline ...
“Mathematics should be taught later, so that it puts the mind in order.
M.V. Lomonosov".
However, the followers of those who flogged future geniuses turned the Method into something opposite. As my supervisor said 35 years ago: "They learned the question." Or, as my youngest son said yesterday about the Gauss method: "Maybe it's not worth making a big science out of this, huh?"
The consequences of the creativity of the "scientists" are visible in the level of current school mathematics, the level of its teaching and understanding of the "Queen of Sciences" by the majority.
However, let's continue...
Methods for explaining the Gauss method in the 5th grade of the school
A mathematics teacher at a Moscow gymnasium, explaining the Gauss method in Vilenkin's way, complicated the task.
What if the difference (step) of an arithmetic progression is not one, but another number? For example, 20.
The task he gave the fifth graders:
20+40+60+80+ ... +460+480+500
Before getting acquainted with the gymnasium method, let's look at the Web: how do school teachers - math tutors do it? ..
Gauss Method: Explanation #1
A well-known tutor on his YOUTUBE channel gives the following reasoning:
"let's write the numbers from 1 to 100 like this:
first a series of numbers from 1 to 50, and strictly below it another series of numbers from 50 to 100, but in reverse order"
1, 2, 3, ... 48, 49, 50
100, 99, 98 ... 53, 52, 51
"Please note: the sum of each pair of numbers from the top and bottom rows is the same and equals 101! Let's count the number of pairs, it is 50 and multiply the sum of one pair by the number of pairs! Voila: The answer is ready!".
"If you couldn't understand, don't be upset!" the teacher repeated three times during the explanation. "You will pass this method in the 9th grade!"
Gauss Method: Explanation #2
Another tutor, less well-known (judging by the number of views) uses more scientific approach, offering a solution algorithm of 5 points that must be performed sequentially.
For the uninitiated: 5 is one of the Fibonacci numbers traditionally considered magical. The 5-step method is always more scientific than the 6-step method, for example. ... And this is hardly an accident, most likely, the Author is a hidden adherent of the Fibonacci theory
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 about plus one rule : one must be added to the resulting quotient: otherwise we will get a result that is one less than the true number of pairs: 42 + 1 = 43.
This is the desired sum of the arithmetic progression from 4 to 256 with a difference of 6!
Gauss method: explanation in the 5th grade of the Moscow gymnasium
And here is how it was required to solve the problem of finding the sum of a series:
20+40+60+ ... +460+480+500
in the 5th grade of the Moscow gymnasium, Vilenkin's textbook (according to my son).
After showing the presentation, the math teacher showed a couple of Gaussian examples and gave the class the task of finding the sum of the numbers in a series with a step of 20.
This required the following:
As you can see, it is more compact and effective technique: the number 3 is also a member of the Fibonacci sequence
My comments on the school version of the Gauss method
The great mathematician would definitely have chosen philosophy if he had foreseen what his followers would turn his "method" into. German teacher who flogged Karl with rods. He would have seen the symbolism and the dialectical spiral and the undying stupidity of the "teachers" trying to measure the harmony of living mathematical thought with the algebra of misunderstanding ....
By the way, do you know. that our education system is rooted in the German school of the 18th and 19th centuries?
But Gauss chose mathematics.
What is the essence of his method?
AT simplification. AT observation and capture simple patterns of numbers. AT turning dry school arithmetic into interesting and fun activity , activating the desire to continue in the brain, and not blocking high-cost mental activity.
Is it possible to calculate the sum of the numbers of an arithmetic progression with one of the above "modifications of the Gauss method" instantly? According to the "algorithms", little Karl would have been guaranteed to avoid spanking, cultivate an aversion to mathematics and suppress his creative impulses in the bud.
Why did the tutor so insistently advise the fifth-graders "not to be afraid of misunderstanding" of the method, convincing them that they would solve "such" problems already in the 9th grade? Psychologically illiterate action. It was a good idea to note: "See? You already in the 5th grade you can solve problems that you will pass only in 4 years! What good fellows you are!"
To use the Gaussian method, level 3 of the class is sufficient when normal children already know how to add, multiply and divide 2-3 digit numbers. Problems arise due to the inability of adult teachers who "do not enter" how to explain the simplest things in a normal human language, not just mathematical ... They are not able to interest mathematics and completely discourage even the "capable".
Or, as my son commented, "make a big science out of it."
Gauss method, my explanations
My wife and I explained this "method" to our child, it seems, even before school ...
Simplicity instead of complexity or a game of questions - answers
""Look, here are the numbers from 1 to 100. What do you see?"
It's not about what the child sees. The trick is to make him look.
"How can you put them together?" The son caught that such questions are not asked "just like that" and you need to look at the question "somehow differently, differently than he usually does"
It doesn't matter if the child sees the solution right away, it's unlikely. It is important that he ceased to be afraid to look, or as I say: "moved the task". This is the beginning of the path to understanding
"Which is easier: add, for example, 5 and 6 or 5 and 95?" A leading question... But after all, any training comes down to "guiding" a person to an "answer" - in any way acceptable to him.
At this stage, there may already be guesses about how to "save" on calculations.
All we have done is hint: the "frontal, linear" counting method is not the only one possible. If the child has truncated this, then later he will invent many more such methods, because it's interesting!!! And he will definitely avoid "misunderstanding" of mathematics, will not feel disgust for it. He got the win!
If a baby discovered that adding pairs of numbers that add up to a hundred is a trifling task, then "arithmetic progression with difference 1"- a rather dreary and uninteresting thing for a child - suddenly gave life to him . Out of chaos came order, and this is always enthusiastic: that's the way we are!
A question to fill in: why, after the insight received by the child, again drive him into the framework of dry algorithms, moreover, functionally useless in this case ?!
Why make stupid rewrite sequence numbers in a notebook: so that even the capable would not have a single chance for understanding? Statistically, of course, but mass education is focused on "statistics" ...
Where did zero go?
And yet, adding up numbers that add up to 100 is much more acceptable to the mind than giving 101 ...
The "school Gauss method" requires exactly this: mindlessly fold equidistant from the center of the progression of a pair of numbers, no matter what.
What if you look?
Still, zero is the greatest invention of mankind, which is more than 2,000 years old. And math teachers continue to ignore him.
It's much easier to convert a series of numbers starting at 1 into a series starting at 0. The sum won't change, will it? You need to stop "thinking in textbooks" and start looking ... And to see that pairs with sum 101 can be completely replaced by pairs with sum 100!
0 + 100, 1 + 99, 2 + 98 ... 49 + 51
How to abolish the "rule plus 1"?
To be honest, I first heard about such a rule from that YouTube tutor ...
What do I still do when I need to determine the number of members of a series?
Looking at the sequence:
1, 2, 3, .. 8, 9, 10
and when completely tired, then on a simpler row:
1, 2, 3, 4, 5
and I figure: if you subtract one from 5, you get 4, but I'm quite clear see 5 numbers! Therefore, you need to add one! Number sense developed in primary school, suggests: even if there are a whole Google of members of the series (10 to the hundredth power), the pattern will remain the same.
Fuck the rules?..
So that in a couple of - three years to fill all the space between the forehead and the back of the head and stop thinking? How about earning bread and butter? After all, we are moving in even ranks into the era of the digital economy!
More about the school method of Gauss: "why make science out of this? .."
It was not in vain that I posted a screenshot from my son's notebook...
"What was there in the lesson?"
“Well, I immediately counted, raised my hand, but she didn’t ask. Therefore, while the others were counting, I began to do DZ in Russian so as not to waste time. Then, when the others finished writing (???), she called me to the board. I said the answer."
"That's right, show me how you solved it," said the teacher. I showed. She said: "Wrong, you need to count as I showed!"
“It’s good that I didn’t put a deuce. And I made me write the “decision process” in their own way in a notebook. Why make a big science out of this? ..”
The main crime of a math teacher
hardly after that occasion Carl Gauss experienced a high sense of respect for the school teacher of mathematics. But if he knew how followers of that teacher pervert the essence of the method... he would roar in indignation and through the World Organization Intellectual Property Rights WIPO has achieved a ban on the use of his honest name in school textbooks! ..
What the main mistake of the school approach? Or, as I put it, the crime of school mathematics teachers against children?
Misunderstanding algorithm
What do school methodologists do, the vast majority of whom do not know how to think?
Create methods and algorithms (see). This is a defensive reaction that protects teachers from criticism ("Everything is done according to ..."), and children from understanding. And thus - from the desire to criticize teachers!(The second derivative of bureaucratic "wisdom", a scientific approach to the problem). A person who does not grasp the meaning will rather blame his own misunderstanding, and not the stupidity of the school system.
What is happening: parents blame the children, and teachers ... the same for children who "do not understand mathematics! ..
Are you savvy?
What did little Carl do?
Absolutely unconventionally approached a template task. This is the quintessence of His approach. This is the main thing that should be taught at school is to think not with textbooks, but with your head. Of course, there is also an instrumental component that can be used ... in search of simpler and effective methods accounts.
Gauss method according to Vilenkin
In school they teach that the Gauss method is to
what, if the number of elements in the row is odd, as in the task that was assigned to the son? ..
The "trick" is that in this case you should find the "extra" number of the series and add it to the sum of the pairs. In our example, this number is 260.
How to discover? Rewriting all pairs of numbers in a notebook!(That's why the teacher made the kids do this stupid job, trying to teach "creativity" using the Gaussian method... And that's why such a "method" is practically inapplicable to large data series, And that's why it is not a Gaussian method).
A little creativity in the school routine...
The son acted differently.
(20 + 500, 40 + 480 ...).
0+500, 20+480, 40+460 ...
Easy, right?
But in practice it becomes even easier, which allows you to carve out 2-3 minutes for remote sensing in Russian, while the rest are "counting". In addition, it retains the number of steps of the methodology: 5, which does not allow criticizing the approach for being unscientific.
Obviously this approach is simpler, faster and more versatile, in the style of the Method. But... the teacher not only didn't praise, but also forced me to rewrite it "in the right way" (see screenshot). That is, she made a desperate attempt to stifle the creative impulse and the ability to understand mathematics in the bud! Apparently, in order to later get hired as a tutor ... She attacked the wrong one ...
Everything that I have described so long and tediously can be explained to a normal child in a maximum of half an hour. Along with examples.
And so that he will never forget it.
And it will step towards understanding...not just mathematics.
Admit it: how many times in your life have you added using the Gauss method? And I never!
But instinct of understanding, which develops (or extinguishes) in the process of studying mathematical methods at school ... Oh! .. This is truly an irreplaceable thing!
Especially in the age of universal digitalization, which we quietly entered under the strict guidance of the Party and the Government.
A few words in defense of teachers...
It is unfair and wrong to place all responsibility for this style of teaching solely on school teachers. The system is in operation.
Some teachers understand the absurdity of what is happening, but what to do? The Law on Education, Federal State Educational Standards, methods, lesson cards... Everything should be done "in accordance and on the basis" and everything should be documented. Step aside - stood in line for dismissal. Let's not be hypocrites: the salary of Moscow teachers is very good... If they get fired, where should they go?..
Therefore this site not about education. He is about individual education, only possible way get out of the crowd Generation Z ...
Definition and description of the Gauss method
The Gaussian transform method (also known as the method of sequential elimination of unknown variables from an equation or matrix) for solving systems of linear equations is a classical method for solving a system algebraic equations(SLAU). Also, this classical method is used to solve such problems as obtaining inverse matrices and determining the rank of a matrix.
The transformation using the Gauss method consists in making small (elementary) successive changes in the system of linear algebraic equations, leading to the elimination of variables from it from top to bottom with the formation of a new triangular system of equations, which is equivalent to the original one.
Definition 1
This part of the solution is called the Gaussian forward solution, since the whole process is carried out from top to bottom.
After bringing the original system of equations to a triangular one, all the variables of the system are found from the bottom up (that is, the first variables found are located exactly on the last lines of the system or matrix). This part of the solution is also known as the reverse Gauss solution. Its algorithm consists in the following: first, the variables that are closest to the bottom of the system of equations or a matrix are calculated, then the obtained values are substituted above and thus another variable is found, and so on.
Description of the Gauss method algorithm
The sequence of actions for the general solution of the system of equations by the Gauss method consists in alternately applying the forward and backward strokes to the matrix based on the SLAE. Let the original system of equations have the following form:
$\begin(cases) a_(11) \cdot x_1 +...+ a_(1n) \cdot x_n = b_1 \\ ... \\ a_(m1) \cdot x_1 + a_(mn) \cdot x_n = b_m \end(cases)$
To solve SLAE by the Gauss method, it is necessary to write down the initial system of equations in the form of a matrix:
$A = \begin(pmatrix) a_(11) & … & a_(1n) \\ \vdots & … & \vdots \\ a_(m1) & … & a_(mn) \end(pmatrix)$, $b =\begin(pmatrix) b_1 \\ \vdots \\ b_m \end(pmatrix)$
The matrix $A$ is called the main matrix and represents the coefficients of the variables written in order, and $b$ is called the column of its free terms. The matrix $A$ written through the line with a column of free members is called the augmented matrix:
$A = \begin(array)(ccc|c) a_(11) & … & a_(1n) & b_1 \\ \vdots & … & \vdots & ...\\ a_(m1) & … & a_( mn) & b_m \end(array)$
Now, using elementary transformations over the system of equations (or over the matrix, as it is more convenient), it is necessary to bring it to the following form:
$\begin(cases) α_(1j_(1)) \cdot x_(j_(1)) + α_(1j_(2)) \cdot x_(j_(2))...+ α_(1j_(r)) \cdot x_(j_(r)) +... α_(1j_(n)) \cdot x_(j_(n)) = β_1 \\ α_(2j_(2)) \cdot x_(j_(2)). ..+ α_(2j_(r)) \cdot x_(j_(r)) +... α_(2j_(n)) \cdot x_(j_(n)) = β_2 \\ ...\\ α_( rj_(r)) \cdot x_(j_(r)) +... α_(rj_(n)) \cdot x_(j_(n)) = β_r \\ 0 = β_(r+1) \\ … \ \ 0 = β_m \end(cases)$ (1)
The matrix obtained from the coefficients of the transformed system of equation (1) is called a step matrix, this is how step matrices usually look like:
$A = \begin(array)(ccc|c) a_(11) & a_(12) & a_(13) & b_1 \\ 0 & a_(22) & a_(23) & b_2\\ 0 & 0 & a_(33) & b_3 \end(array)$
These matrices are characterized by the following set of properties:
- All its zero rows come after non-zero ones
- If some row of the matrix with index $k$ is non-zero, then there are fewer zeros in the previous row of the same matrix than in this row with index $k$.
After obtaining the step matrix, it is necessary to substitute the obtained variables into the remaining equations (starting from the end) and obtain the remaining values of the variables.
Basic rules and permitted transformations when using the Gauss method
When simplifying a matrix or a system of equations by this method, only elementary transformations should be used.
Such transformations are operations that can be applied to a matrix or system of equations without changing its meaning:
- permutation of several lines in places,
- adding or subtracting from one line of the matrix another line from it,
- multiplying or dividing a string by a constant that is not equal to zero,
- a line consisting of only zeros, obtained in the process of calculating and simplifying the system, must be deleted,
- You also need to remove unnecessary proportional lines, choosing for the system the only one with coefficients that are more suitable and convenient for further calculations.
All elementary transformations are reversible.
Analysis of the three main cases that arise when solving linear equations using the method of simple Gaussian transformations
There are three cases that arise when using the Gauss method to solve systems:
- When the system is inconsistent, that is, it does not have any solutions
- The system of equations has a solution, and the only one, and the number of non-zero rows and columns in the matrix is equal to each other.
- The system has a certain number or set of possible solutions, and the number of rows in it is less than the number of columns.
Solution outcome with inconsistent system
For this variant, when solving the matrix equation by the Gauss method, it is typical to obtain some line with the impossibility of fulfilling the equality. Therefore, if at least one incorrect equality occurs, the resulting and original systems have no solutions, regardless of the other equations they contain. An example of an inconsistent matrix:
$\begin(array)(ccc|c) 2 & -1 & 3 & 0 \\ 1 & 0 & 2 & 0\\ 0 & 0 & 0 & 1 \end(array)$
An unsatisfied equality appeared in the last line: $0 \cdot x_(31) + 0 \cdot x_(32) + 0 \cdot x_(33) = 1$.
A system of equations that has only one solution
The data of the system after reduction to a stepped matrix and deletion of rows with zeros have the same number of rows and columns in the main matrix. Here the simplest example such a system:
$\begin(cases) x_1 - x_2 = -5 \\ 2 \cdot x_1 + x_2 = -7 \end(cases)$
Let's write it in the form of a matrix:
$\begin(array)(cc|c) 1 & -1 & -5 \\ 2 & 1 & -7 \end(array)$
To bring the first cell of the second row to zero, we multiply the top row by $-2$ and subtract it from the bottom row of the matrix, and leave the top row in its original form, as a result we have the following:
$\begin(array)(cc|c) 1 & -1 & -5 \\ 0 & 3 & 10 \end(array)$
This example can be written as a system:
$\begin(cases) x_1 - x_2 = -5 \\ 3 \cdot x_2 = 10 \end(cases)$
The following value of $x$ comes out of the lower equation: $x_2 = 3 \frac(1)(3)$. Substituting this value into the upper equation: $x_1 – 3 \frac(1)(3)$, we get $x_1 = 1 \frac(2)(3)$.
A system with many possible solutions
This system is characterized by a smaller number of significant rows than the number of columns in it (the rows of the main matrix are taken into account).
Variables in such a system are divided into two types: basic and free. When transforming such a system, the main variables contained in it must be left in the left area before the “=” sign, and the remaining variables should be transferred to the right side of the equality.
Such a system has only a certain general solution.
Let's analyze the following system of equations:
$\begin(cases) 2y_1 + 3y_2 + x_4 = 1 \\ 5y_3 - 4y_4 = 1 \end(cases)$
Let's write it in the form of a matrix:
$\begin(array)(cccc|c) 2 & 3 & 0 & 1 & 1 \\ 0 & 0 & 5 & 4 & 1 \\ \end(array)$
Our task is to find a general solution to the system. For this matrix, the basic variables will be $y_1$ and $y_3$ (for $y_1$ - since it is in the first place, and in the case of $y_3$ - it is located after the zeros).
As basic variables, we choose exactly those that are not equal to zero first in the row.
The remaining variables are called free, through them we need to express the basic ones.
Using the so-called reverse move, we disassemble the system from the bottom up, for this we first express $y_3$ from the bottom line of the system:
$5y_3 – 4y_4 = 1$
$5y_3 = 4y_4 + 1$
$y_3 = \frac(4/5)y_4 + \frac(1)(5)$.
Now we substitute the expressed $y_3$ into the upper equation of the system $2y_1 + 3y_2 + y_4 = 1$: $2y_1 + 3y_2 - (\frac(4)(5)y_4 + \frac(1)(5)) + y_4 = 1$
We express $y_1$ in terms of free variables $y_2$ and $y_4$:
$2y_1 + 3y_2 - \frac(4)(5)y_4 - \frac(1)(5) + y_4 = 1$
$2y_1 = 1 - 3y_2 + \frac(4)(5)y_4 + \frac(1)(5) - y_4$
$2y_1 = -3y_2 - \frac(1)(5)y_4 + \frac(6)(5)$
$y_1 = -1.5x_2 – 0.1y_4 + 0.6$
The solution is ready.
Example 1
Solve the slough using the Gaussian method. Examples. An example of solving a system of linear equations given by a 3 by 3 matrix using the Gauss method
$\begin(cases) 4x_1 + 2x_2 - x_3 = 1 \\ 5x_1 + 3x_2 - 2x^3 = 2\\ 3x_1 + 2x_2 - 3x_3 = 0 \end(cases)$
We write our system in the form of an augmented matrix:
$\begin(array)(ccc|c) 4 & 2 & -1 & 1 \\ 5 & 3 & -2 & 2 \\ 3 & 2 & -3 & 0\\ \end(array)$
Now, for convenience and practicality, we need to transform the matrix so that $1$ is in the upper corner of the last column.
To do this, we need to add the line from the middle multiplied by $-1$ to the 1st line, and write the middle line itself as it is, it turns out:
$\begin(array)(ccc|c) -1 & -1 & 1 & -1 \\ 5 & 3 & -2 & 2 \\ 3 & 2 & -3 & 0\\ \end(array)$
$\begin(array)(ccc|c) -1 & -1 & 1 & -1 \\ 0 & -2 & 3 & -3 \\ 0 & -1 & 0 & -3\\ \end(array) $
Multiply the top and last rows by $-1$, and swap the last and middle rows:
$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & -2 & 3 & -3\\ \end(array)$
$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 3 & 3\\ \end(array)$
And split the last line by $3$:
$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1\\ \end(array)$
We obtain the following system of equations, equivalent to the original one:
$\begin(cases) x_1 + x_2 – x_3 = 1\\ x_2 = 3 \\ x_3 = 1 \end(cases)$
From the upper equation, we express $x_1$:
$x1 = 1 + x_3 - x_2 = 1 + 1 - 3 = -1$.
Example 2
An example of solving a system defined using a 4 by 4 matrix using the Gaussian method
$\begin(array)(cccc|c) 2 & 5 & 4 & 1 & 20 \\ 1 & 3 & 2 & 1 & 11 \\ 2 & 10 & 9 & 7 & 40\\ 3 & 8 & 9 & 2 & 37 \\ \end(array)$.
At the beginning, we swap the top lines following it to get $1$ in the upper left corner:
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 2 & 5 & 4 & 1 & 20 \\ 2 & 10 & 9 & 7 & 40\\ 3 & 8 & 9 & 2 & 37 \\ \end(array)$.
Now let's multiply the top line by $-2$ and add to the 2nd and to the 3rd. To the 4th we add the 1st line, multiplied by $-3$:
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & -1 & 0 & -1 & -2 \\ 0 & 4 & 5 & 5 & 18\\ 0 & - 1 & 3 & -1 & 4 \\ \end(array)$
Now to line number 3 we add line 2 multiplied by $4$, and to line 4 we add line 2 multiplied by $-1$.
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & -1 & 0 & -1 & -2 \\ 0 & 0 & 5 & 1 & 10\\ 0 & 0 & 3 & 0 & 6 \\ \end(array)$
Multiply row 2 by $-1$, divide row 4 by $3$ and replace row 3.
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & 0 & 2\\ 0 & 0 & 5 & 1 & 10 \\ \end(array)$
Now we add to the last line the penultimate one, multiplied by $-5$.
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & 0 & 2\\ 0 & 0 & 0 & 1 & 0 \\ \end(array)$
We solve the resulting system of equations:
$\begin(cases) m = 0 \\ g = 2\\ y + m = 2\ \ x + 3y + 2g + m = 11\end(cases)$
