Gauss method, also called method sequential elimination unknowns is as follows. Using elementary transformations, a system of linear equations is brought to such a form that its matrix of coefficients turns out to be trapezoidal (the same as triangular or stepped) or close to trapezoidal (direct stroke of the Gaussian method, hereinafter - simply straight stroke). An example of such a system and its solution is in the figure above.
In such a system, the last equation contains only one variable and its value can be unambiguously found. The value of this variable is then substituted into the previous equation ( reverse stroke Gauss method , then just the reverse), from which the previous variable is found, and so on.
In a trapezoidal (triangular) system, as we see, the third equation no longer contains variables y And x, and the second equation is the variable x .
After the matrix of the system has taken a trapezoidal shape, it is no longer difficult to understand the issue of compatibility of the system, determine the number of solutions and find the solutions themselves.
Advantages of the method:
- when solving systems linear equations with the number of equations and unknowns more than three, the Gauss method is not as cumbersome as the Cramer method, since solving with the Gauss method requires fewer calculations;
- Using the Gauss method, you can solve indefinite systems of linear equations, that is, having common decision(and we will look at them in this lesson), but using Cramer’s method, we can only state that the system is uncertain;
- you can solve systems of linear equations in which the number of unknowns is not equal to the number of equations (we will also analyze them in this lesson);
- The method is based on elementary (school) methods - the method of substituting unknowns and the method of adding equations, which we touched on in the corresponding article.
In order for everyone to understand the simplicity with which trapezoidal (triangular, step) systems of linear equations are solved, we present a solution to such a system using reverse motion. A quick solution to this system was shown in the picture at the beginning of the lesson.
Example 1. Solve a system of linear equations using inverse:
Solution. In this trapezoidal system the variable z can be uniquely found from the third equation. We substitute its value into the second equation and get the value of the variable y:
Now we know the values of two variables - z And y. We substitute them into the first equation and get the value of the variable x:
From the previous steps we write out the solution to the system of equations:
To obtain such a trapezoidal system of linear equations, which we solved very simply, it is necessary to use a forward stroke associated with elementary transformations of the system of linear equations. It's also not very difficult.
Elementary transformations of a system of linear equations
Repeating the school method of algebraically adding the equations of a system, we found out that to one of the equations of the system we can add another equation of the system, and each of the equations can be multiplied by some numbers. As a result, we obtain a system of linear equations equivalent to this one. In it, one equation already contained only one variable, substituting the value of which into other equations, we come to a solution. Such addition is one of the types of elementary transformation of the system. When using the Gaussian method, we can use several types of transformations.
The animation above shows how the system of equations gradually turns into a trapezoidal one. That is, the one that you saw in the very first animation and convinced yourself that it is easy to find the values of all unknowns from it. How to perform such a transformation and, of course, examples will be discussed further.
When solving systems of linear equations with any number of equations and unknowns in the system of equations and in the extended matrix of the system Can:
- rearrange lines (this was mentioned at the very beginning of this article);
- if other transformations result in equal or proportional rows, they can be deleted, except for one;
- remove “zero” rows where all coefficients are equal to zero;
- multiply or divide any string by a certain number;
- to any line add another line, multiplied by a certain number.
As a result of the transformations, we obtain a system of linear equations equivalent to this one.
Algorithm and examples of solving a system of linear equations with a square matrix of the system using the Gauss method
Let us first consider solving systems of linear equations in which the number of unknowns is equal to the number of equations. The matrix of such a system is square, that is, the number of rows in it is equal to the number of columns.
Example 2. Solve a system of linear equations using the Gauss method
When solving systems of linear equations using school methods, we multiplied one of the equations term by term, so that the coefficients of the first variable in the two equations were opposite numbers. When adding equations, this variable is eliminated. The Gauss method works similarly.
To simplify appearance solutions let's create an extended matrix of the system:
In this matrix, the coefficients of the unknowns are located on the left before the vertical line, and the free terms are located on the right after the vertical line.
For the convenience of dividing coefficients for variables (to obtain division by unity) Let's swap the first and second rows of the system matrix. We obtain a system equivalent to this one, since in a system of linear equations the equations can be interchanged:
Using the new first equation eliminate the variable x from the second and all subsequent equations. To do this, to the second row of the matrix we add the first row multiplied by (in our case by ), to the third row - the first row multiplied by (in our case by ).
This is possible because
If there were more than three equations in our system, then we would have to add to all subsequent equations the first line, multiplied by the ratio of the corresponding coefficients, taken with a minus sign.
As a result, we obtain a matrix equivalent to this system new system equations in which all equations, starting from the second do not contain a variable x :
To simplify the second line of the resulting system, multiply it by and again obtain the matrix of a system of equations equivalent to this system:
Now, keeping the first equation of the resulting system unchanged, using the second equation we eliminate the variable y from all subsequent equations. To do this, to the third row of the system matrix we add the second row, multiplied by (in our case by ).
If there were more than three equations in our system, then we would have to add a second line to all subsequent equations, multiplied by the ratio of the corresponding coefficients taken with a minus sign.
As a result, we again obtain the matrix of a system equivalent to this system of linear equations:
We have obtained an equivalent trapezoidal system of linear equations:
If the number of equations and variables is greater than in our example, then the process of sequentially eliminating variables continues until the system matrix becomes trapezoidal, as in our demo example.
We will find the solution “from the end” - the reverse move. For this from the last equation we determine z:
.
Substituting this value into the previous equation, we'll find y:
From the first equation we'll find x:
Answer: the solution to this system of equations is 
.
: in this case the same answer will be given if the system has a unique solution. If the system has an infinite number of solutions, then this will be the answer, and this is the subject of the fifth part of this lesson.
Solve a system of linear equations using the Gaussian method yourself, and then look at the solution
Here again we have an example of a consistent and definite system of linear equations, in which the number of equations is equal to the number of unknowns. The difference from our demo example from the algorithm is that there are already four equations and four unknowns.
Example 4. Solve a system of linear equations using the Gauss method:
Now you need to use the second equation to eliminate the variable from subsequent equations. Let's carry out preparatory work. To make it more convenient with the ratio of coefficients, you need to get one in the second column of the second row. To do this, subtract the third from the second line, and multiply the resulting second line by -1.
Let us now carry out the actual elimination of the variable from the third and fourth equations. To do this, add the second line, multiplied by , to the third line, and the second, multiplied by , to the fourth line.
Now, using the third equation, we eliminate the variable from the fourth equation. To do this, add the third line to the fourth line, multiplied by . We obtain an extended trapezoidal matrix.
We obtained a system of equations to which the given system is equivalent:
Consequently, the resulting and given systems are compatible and definite. We find the final solution “from the end”. From the fourth equation we can directly express the value of the variable “x-four”:
We substitute this value into the third equation of the system and get
,
,
Finally, value substitution
The first equation gives
,
where do we find “x first”:
Answer: this system of equations has a unique solution 
.
You can also check the solution of the system on a calculator using Cramer's method: in this case, the same answer will be given if the system has a unique solution.
Solving applied problems using the Gauss method using the example of a problem on alloys
Systems of linear equations are used to model real objects in the physical world. Let's solve one of these problems - alloys. Similar problems are problems on mixtures, the cost or share of individual goods in a group of goods, and the like.
Example 5. Three pieces of alloy have a total mass of 150 kg. The first alloy contains 60% copper, the second - 30%, the third - 10%. Moreover, in the second and third alloys taken together there is 28.4 kg less copper than in the first alloy, and in the third alloy there is 6.2 kg less copper than in the second. Find the mass of each piece of the alloy.
Solution. We compose a system of linear equations:
We multiply the second and third equations by 10, we obtain an equivalent system of linear equations:
We create an extended matrix of the system:
Attention, straight ahead. By adding (in our case, subtracting) one row multiplied by a number (we apply it twice), the following transformations occur with the extended matrix of the system:
The direct move is over. We obtained an expanded trapezoidal matrix.
We apply the reverse move. We find the solution from the end. We see that.
From the second equation we find
From the third equation -
You can also check the solution of the system on a calculator using Cramer's method: in this case, the same answer will be given if the system has a unique solution.
The simplicity of Gauss's method is evidenced by the fact that it took the German mathematician Carl Friedrich Gauss only 15 minutes to invent it. In addition to the method named after him, the saying “We should not confuse what seems incredible and unnatural to us with the absolutely impossible” is known from the works of Gauss - a kind of brief instruction on making discoveries.
In many applied problems there may not be a third constraint, that is, a third equation, then you have to solve a system of two equations with three unknowns using the Gaussian method, or, conversely, there are fewer unknowns than equations. We will now begin to solve such systems of equations.
Using the Gaussian method, you can determine whether any system is compatible or incompatible n linear equations with n variables.
The Gauss method and systems of linear equations with an infinite number of solutions
The next example is a consistent but indeterminate system of linear equations, that is, having an infinite number of solutions.
After performing transformations in the extended matrix of the system (rearranging rows, multiplying and dividing rows by a certain number, adding another to one row), rows of the form could appear
If in all equations having the form
Free terms are equal to zero, this means that the system is indefinite, that is, it has an infinite number of solutions, and equations of this type are “superfluous” and we exclude them from the system.
Example 6.
Solution. Let's create an extended matrix of the system. Then, using the first equation, we eliminate the variable from subsequent equations. To do this, add to the second, third and fourth lines the first, multiplied by :
Now let's add the second line to the third and fourth.
As a result, we arrive at the system
The last two equations turned into equations of the form. These equations are satisfied for any value of the unknowns and can be discarded.
To satisfy the second equation, we can choose arbitrary values for and , then the value for will be determined uniquely: 
. From the first equation the value for is also found uniquely: 
.
Both the given and the last systems are consistent, but uncertain, and the formulas
for arbitrary and give us all solutions of a given system.
Gauss method and systems of linear equations without solutions
The next example is an inconsistent system of linear equations, that is, one that has no solutions. The answer to such problems is formulated this way: the system has no solutions.
As already mentioned in connection with the first example, after performing transformations, rows of the form could appear in the extended matrix of the system
corresponding to an equation of the form
If among them there is at least one equation with a nonzero free term (i.e. ), then this system of equations is inconsistent, that is, it has no solutions and its solution is complete.
Example 7. Solve the system of linear equations using the Gauss method:
Solution. We compose an extended matrix of the system. Using the first equation, we exclude the variable from subsequent equations. To do this, add the first line multiplied by to the second line, the first line multiplied by the third line, and the first line multiplied by the fourth line.
Now you need to use the second equation to eliminate the variable from subsequent equations. To obtain integer ratios of coefficients, we swap the second and third rows of the extended matrix of the system.
To exclude the third and fourth equations, add the second one multiplied by , to the third line, and the second multiplied by , to the fourth line.
Now, using the third equation, we eliminate the variable from the fourth equation. To do this, add the third line to the fourth line, multiplied by .
The given system is therefore equivalent to the following:
The resulting system is inconsistent, since its last equation cannot be satisfied by any values of the unknowns. Therefore, this system has no solutions.
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 Gauss method, where everything comes down to constructing a matrix of a triangular form, a rectangle appears in the entry, only with zeros in the place where there are no numbers. Zeros may not be written, but they are implied.
The matrix has a size. Its “width” is the number of rows (m), “length” is the number of columns (n). Then the size of the matrix A (capital Latin letters are usually used to denote them) will be denoted as A m×n. If m=n, then this matrix is square, and m=n is its order. Accordingly, any element of matrix A can be denoted by its row and column numbers: a xy ; x - row number, changes, y - column number, changes.
B is not the main point of the decision. In principle, all operations can be performed directly with the equations themselves, but the notation will be much more cumbersome, and it will be much easier to get confused in it.
Determinant
The matrix also has a determinant. This is a 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 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 non-zero number, it is called the basis minor of the original rectangular matrix.
Before you start solving a system of equations using the Gaussian method, it doesn’t hurt to calculate the determinant. If it turns out to be zero, then we can immediately say that the matrix has either an infinite number of solutions or none at all. In such a sad case, you need to go further and find out about the rank of the matrix.
System classification
There is such a thing as the rank of a matrix. This is the maximum order of its non-zero determinant (if we remember about the basis minor, we can say that the rank of a matrix is the order of the basis minor).
Based on the situation with rank, SLAE can be divided into:
- Joint. U In joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended matrix (with a column of free terms). Such systems have a solution, but not necessarily one, therefore, additionally joint systems are divided into:
- - certain- having a single solution. In certain systems, the rank of the matrix and the number of unknowns (or the number of columns, which is the same thing) are equal;
- - undefined - with an infinite number of solutions. The rank of matrices in such systems is less than the number of unknowns.
- Incompatible. U In such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.
The Gauss method is good because during the solution it allows one to obtain either an unambiguous proof of the inconsistency of the system (without calculating the determinants of large matrices), or a solution in general form for a system with an infinite number of solutions.
Elementary transformations
Before proceeding directly to solving the system, you can make it less cumbersome and more convenient for calculations. This is achieved through elementary transformations - such that their implementation does not change the final answer in any way. It should be noted that some of the given elementary transformations are valid only for matrices, the source of which was the SLAE. Here is a list of these transformations:
- Rearranging lines. Obviously, if you change the order of the equations in the system record, this will not affect the solution in any way. Consequently, rows in the matrix of this system can also be swapped, not forgetting, of course, the column of free terms.
- Multiplying all elements of a string by a certain coefficient. Very helpful! It can be used to shorten big numbers in the matrix or remove zeros. Many decisions, as usual, will not change, but further operations will become more convenient. The main thing is that the coefficient is not equal to zero.
- Removing rows with proportional factors. This partly follows from previous paragraph. If two or more rows in a matrix have proportional coefficients, then when one of the rows is multiplied/divided by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and the extra ones can be removed, leaving only one.
- Removing a null line. If, during the transformation, a row is obtained somewhere in which all elements, including the free term, are zero, then such a row can be called zero and thrown out of the matrix.
- Adding to the elements of one row the elements of another (in the corresponding columns), multiplied by a certain coefficient. The most unobvious and most important transformation of all. It is worth dwelling on it in more detail.
Adding a string multiplied by a factor
For ease of understanding, it is worth breaking down this process step by step. Two rows are taken from the matrix:
a 11 a 12 ... a 1n | b1
a 21 a 22 ... a 2n | b 2
Let's say you need to add the first to the second, multiplied by the coefficient "-2".
a" 21 = a 21 + -2×a 11
a" 22 = a 22 + -2×a 12
a" 2n = a 2n + -2×a 1n
Then the second row in the matrix is replaced with a new one, and the first remains unchanged.
a 11 a 12 ... a 1n | b1
a" 21 a" 22 ... a" 2n | b 2
It should be noted that the multiplication coefficient can be selected in such a way that, as a result of adding two rows, one of the elements of the new row is equal to zero. Therefore, it is possible to obtain an equation in a system where there will be one less unknown. And if you get two such equations, then the operation can be done again and get an equation that will contain two fewer unknowns. And if each time you turn one coefficient of all rows that are below the original one to zero, then you can, like stairs, go down to the very bottom of the matrix and get an equation with one unknown. This is called solving the system using the Gaussian method.
In general
Let there be a system. It has m equations and n unknown roots. You can write it as follows:
The main matrix is compiled from the system coefficients. A column of free terms is added to the extended matrix and, for convenience, separated by a line.
- the first row of the matrix is multiplied by the coefficient k = (-a 21 /a 11);
- the first modified row and the second row of the matrix are added;
- instead of the second row, the result of the addition from the previous paragraph is inserted into the matrix;
- now the first coefficient in the new second row is a 11 × (-a 21 /a 11) + a 21 = -a 21 + a 21 = 0.
Now the same series of transformations is performed, only the first and third rows are involved. Accordingly, at each step of the algorithm, element a 21 is replaced by a 31. Then everything is repeated for a 41, ... a m1. The result is a matrix where the first element in the rows is zero. Now you need to forget about line number one and perform the same algorithm, starting from line two:
- coefficient k = (-a 32 /a 22);
- the second modified line is added to the “current” line;
- the result of the addition is substituted into the third, fourth, and so on lines, while the first and second remain unchanged;
- in the rows of the matrix the first two elements are already equal to zero.
The algorithm must be repeated until the coefficient k = (-a m,m-1 /a mm) appears. This means that in last time the algorithm was performed only for the lower equation. Now the matrix looks like a triangle, or has a stepped shape. In the bottom line there is the equality a mn × x n = b m. The coefficient and free term are known, and the root is expressed through them: x n = b m /a mn. The resulting root is substituted into the top line to find x n-1 = (b m-1 - a m-1,n ×(b m /a mn))÷a m-1,n-1. And so on by analogy: in each next line there is a new root, and, having reached the “top” of the system, you can find many solutions. It will be the only one.
When there are no solutions
If in one of the matrix rows all elements except the free term are equal to zero, then the equation corresponding to this row looks like 0 = b. It has no solution. And since such an equation is included in the system, then the set of solutions of the entire system is empty, that is, it is degenerate.
When there are an infinite number of solutions
It may happen that in the given triangular matrix there are no rows with one coefficient element of the equation and one free term. There are only lines that, when rewritten, would look like an equation with two or more variables. This means that the system has infinite number decisions. In this case, the answer can be given in the form of a general solution. How to do it?
All variables in the matrix are divided into basic and free. Basic ones are those that stand “on the edge” of the rows in the step matrix. The rest are free. In the general solution, the basic variables are written through free ones.
For convenience, the matrix is first rewritten back into a system of equations. Then in the last of them, where exactly there is only one basic variable left, it remains on one side, and everything else is transferred to the other. This is done for every equation with one basic variable. Then, in the remaining equations, where possible, the expression obtained for it is substituted instead of the basic variable. If the result is again an expression containing only one basic variable, it is again expressed from there, and so on, until each basic variable is written as an expression with free variables. This is the general solution of SLAE.
You can also find the basic solution of the system - give the free variables any values, and then for this specific case calculate the values of the basic variables. There are an infinite number of particular solutions that can be given.
Solution with specific examples
Here is a system of equations.
For convenience, it is better to immediately create its matrix
It is known that when solved by the Gaussian method, the equation corresponding to the first row will remain unchanged at the end of the transformations. Therefore, it will be more profitable if the upper left element of the matrix is the smallest - then the first elements of the remaining rows after the operations will turn to zero. This means that in the compiled matrix it will be advantageous to put the second row in place of the first one.
second line: k = (-a 21 /a 11) = (-3/1) = -3
a" 21 = a 21 + k×a 11 = 3 + (-3)×1 = 0
a" 22 = a 22 + k×a 12 = -1 + (-3)×2 = -7
a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11
b" 2 = b 2 + k×b 1 = 12 + (-3)×12 = -24
third line: k = (-a 3 1 /a 11) = (-5/1) = -5
a" 3 1 = a 3 1 + k×a 11 = 5 + (-5)×1 = 0
a" 3 2 = a 3 2 + k×a 12 = 1 + (-5)×2 = -9
a" 3 3 = a 33 + k×a 13 = 2 + (-5)×4 = -18
b" 3 = b 3 + k×b 1 = 3 + (-5)×12 = -57
Now, in order not to get confused, you need to write down a matrix with the intermediate results of the transformations.
Obviously, such a matrix can be made more convenient for perception using certain operations. For example, you can remove all “minuses” from the second line by multiplying each element by “-1”.
It is also worth noting that in the third line all elements are multiples of three. Then you can shorten the string by this number, multiplying each element by "-1/3" (minus - at the same time, to remove negative values).
Looks much nicer. Now we need to leave the first line alone and work with the second and third. The task is to add the second line to the third line, multiplied by such a coefficient that the element a 32 becomes equal to zero.
k = (-a 32 /a 22) = (-3/7) = -3/7 (if during some transformations the answer does not turn out to be an integer, it is recommended to maintain the accuracy of the calculations to leave it “as is”, in the form common fraction, and only then, when the answers are received, decide whether to round and convert to another form of recording)
a" 32 = a 32 + k×a 22 = 3 + (-3/7)×7 = 3 + (-3) = 0
a" 33 = a 33 + k×a 23 = 6 + (-3/7)×11 = -9/7
b" 3 = b 3 + k×b 2 = 19 + (-3/7)×24 = -61/7
The matrix is written again with new values.
| 1 | 2 | 4 | 12 |
| 0 | 7 | 11 | 24 |
| 0 | 0 | -9/7 | -61/7 |
As you can see, the resulting matrix already has a stepped form. Therefore, further transformations of the system using the Gaussian method are not required. What 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 largest order of the determinant-square is 4. This means that there are an infinite number of solutions, and you need to look for its general appearance. The Gauss method for linear equations allows you to do this.
First, as usual, an extended matrix is compiled.
Second line: coefficient k = (-a 21 /a 11) = -3. In the third line, the first element is before the transformations, so you don’t need to touch anything, you need to leave it as is. Fourth line: k = (-a 4 1 /a 11) = -5
By multiplying the elements of the first row by each of their coefficients in turn and adding them to the required rows, we obtain the matrix the following type:
As you can see, the second, third and fourth rows consist of elements proportional to each other. The second and fourth are generally identical, so one of them can be removed immediately, and the remaining one can be multiplied by the coefficient “-1” and get line number 3. And again, out of two identical lines, leave one.
The result is a matrix like this. While the system has not yet been written down, it is necessary to determine the basic variables here - those standing at the coefficients a 11 = 1 and a 22 = 1, and free ones - all the rest.
In the second equation there is only one basic variable - x 2. This means that it can be expressed from there by writing it through the variables x 3 , x 4 , x 5 , which are free.
We substitute the resulting expression into the first equation.
The result is an equation in which the only basic variable is x 1 . Let's do the same with it as with x 2.
All basic variables, of which there are two, are expressed in terms of three free ones; now we can write the answer in general form.
You can also specify one of the particular solutions of the system. For such cases, zeros are usually chosen as values for free variables. Then the answer will be:
16, 23, 0, 0, 0.
An example of a non-cooperative system
Solution incompatible systems equations by the Gaussian method - the fastest. It ends immediately as soon as at one of the stages an equation is obtained that has no solution. That is, the stage of calculating the roots, which is quite long and tedious, is eliminated. The following system is considered:
x + y - z = 0 (1)
2x - y - z = -2 (2)
4x + y - 3z = 5 (3)
As usual, the matrix is compiled:
| 1 | 1 | -1 | 0 |
| 2 | -1 | -1 | -2 |
| 4 | 1 | -3 | 5 |
And it is reduced to a stepwise form:
k 1 = -2k 2 = -4
| 1 | 1 | -1 | 0 |
| 0 | -3 | 1 | -2 |
| 0 | 0 | 0 | 7 |
After the first transformation, the third line contains an equation of the form
without a solution. Consequently, the system is inconsistent, and the answer will be the empty set.
Advantages and disadvantages of the method
If you choose which method to solve SLAEs on paper with a pen, then the method that was discussed in this article looks the most attractive. It is much more difficult to get confused in elementary transformations than if you have to manually search for a determinant or some tricky inverse matrix. However, if you use programs for working with data of this type, for example, spreadsheets, then it turns out that such programs already contain algorithms for calculating the main parameters of matrices - determinant, minors, inverse, and so on. And if you are sure that the machine will calculate these values itself and will not make a mistake, it is more advisable to use matrix method or Cramer's formulas, because their application begins and ends with the calculation of determinants and inverse matrices.
Application
Since the Gaussian solution is an algorithm, and the matrix is actually a two-dimensional array, it can be used in programming. But since the article positions itself as a guide “for dummies,” it should be said that the easiest place to put the method into is spreadsheets, for example, Excel. Again, any SLAE entered into a table in the form of a matrix will be considered by Excel as a two-dimensional array. And for operations with them there are many nice commands: addition (you can only add matrices of the same size!), multiplication by a number, multiplication of matrices (also with certain restrictions), finding the inverse and transposed matrices and, most importantly, calculating the determinant. If this time-consuming task is replaced by a single command, it is possible to determine the rank of the matrix much more quickly and, therefore, establish its compatibility or incompatibility.
Educational institution "Belarusian State
Agricultural Academy"
Department of Higher Mathematics
to study the topic “Gauss method for solving systems of linear
equations" by students of the accounting faculty of correspondence education (NISPO)
Gorki, 2013
Gauss method for solving systems of linear equations
Equivalent systems of equations
Two systems of linear equations are said to be equivalent if each solution of one of them is a solution of the other. The process of solving a system of linear equations consists of sequentially transforming it into an equivalent system using the so-called elementary transformations , which are:
1) rearrangement of any two equations of the system;
2) multiplying both sides of any equation of the system by a nonzero number;
3) adding to any equation another equation multiplied by any number;
4) crossing out an equation consisting of zeros, i.e. equations of the form
Gaussian elimination
Consider the system m linear equations with n unknown:
The essence of the Gaussian method or the method of sequential elimination of unknowns is as follows.
First, using elementary transformations, the unknown is eliminated from all equations of the system except the first. Such system transformations are called Gaussian elimination step . The unknown is called enabling variable at the first step of transformation. The coefficient is called resolution factor , the first equation is called resolving equation , and the column of coefficients at permission column .
When performing one step of Gaussian elimination, you need to use the following rules:
1) the coefficients and the free term of the resolving equation remain unchanged;
2) the coefficients of the resolution column located below the resolution coefficient become zero;
3) all other coefficients and free terms when performing the first step are calculated according to the rectangle rule:
, Where i=2,3,…,m; j=2,3,…,n.
We will perform similar transformations on the second equation of the system. This will lead to a system in which the unknown will be eliminated in all equations except the first two. As a result of such transformations over each of the equations of the system (direct progression of the Gaussian method), the original system is reduced to an equivalent step system of one of the following types.
Reverse Gaussian Method
Step system
It has triangular view and that's all 
(i=1,2,…,n). Such a system has a unique solution. The unknowns are determined starting from the last equation (reverse of the Gaussian method).
The step system has the form
where, i.e. the number of equations of the system is less than or equal to the number of unknowns. This system has no solutions, since the last equation will not be satisfied for any values of the variable.
Step type system
has countless solutions. From the last equation, the unknown is expressed through the unknowns 
. Then, in the penultimate equation, instead of the unknown, its expression is substituted through the unknowns . Continuing the reverse of the Gaussian method, the unknowns 
can be expressed in terms of unknowns . In this case, the unknowns are called free
and can take any values, and unknown 
basic.
When solving systems in practice, it is convenient to perform all transformations not with a system of equations, but with an extended matrix of the system, consisting of coefficients for unknowns and a column of free terms.
Example 1. Solve system of equations 
Solution. Let's create an extended matrix of the system and perform elementary transformations:
.
In the extended matrix of the system, the number 3 (it is highlighted) is the resolution coefficient, the first row is the resolution row, and the first column is the resolution column. When moving to the next matrix, the resolution row does not change; all elements of the resolution column below the resolution element are replaced by zeros. And all other elements of the matrix are recalculated according to the quadrilateral rule. Instead of element 4 in the second line we write 
, instead of element -3 in the second line it will be written 
etc. Thus, the second matrix will be obtained. The resolution element of this matrix will be the number 18 in the second row. To form the next (third matrix), we leave the second row unchanged, in the column under the resolving element we write zero and recalculate the remaining two elements: instead of the number 1 we write 
, and instead of the number 16 we write .
As a result, the original system was reduced to an equivalent system
From the third equation we find 
. Let's substitute this value into the second equation: 
y=3. Let’s substitute the found values into the first equation y And z: 
, x=2.
Thus, the solution to this system of equations is x=2, y=3, 
.
Example 2. Solve system of equations 
Solution. Let us perform elementary transformations on the extended matrix of the system:
In the second matrix, each element of the third row is divided by 2.
In the fourth matrix, each element of the third and fourth rows was divided by 11.
. The resulting matrix corresponds to the system of equations 
Solving this system, we find 
, 
, .
Example 3. Solve system of equations
Solution. Let's write the extended matrix of the system and perform elementary transformations:
.
In the second matrix, each element of the second, third and fourth rows was divided by 7.
As a result, a system of equations was obtained
equivalent to the original one.
Since there are two fewer equations than unknowns, then from the second equation 
. Let's substitute the expression for into the first equation: , 
.
Thus, the formulas 
give a general solution to this system of equations. Unknowns are free and can take any value.
Let, for example, 
Then 
And 
. Solution 
is one of the particular solutions of the system, of which there are countless.
Questions for self-control of knowledge
1) What transformations linear systems are called elementary?
2) What transformations of the system are called the Gaussian elimination step?
3) What is a resolving variable, resolving coefficient, resolving column?
4) What rules should be used when performing one step of Gaussian elimination?
Carl Friedrich Gauss, greatest mathematician for a long time hesitated, choosing between philosophy and mathematics. Perhaps it was precisely this mindset that allowed him to make such a noticeable “legacy” in world science. In particular, by creating the "Gauss Method" ...
For almost 4 years, articles on this site concerned school education, mainly from the side of philosophy, the principles of (mis)understanding introduced into the consciousness of children. The time is coming for more specifics, examples and methods... I believe that this is exactly the approach to the familiar, confusing and important areas of life gives better results.
We people are designed in such a way that no matter how much we talk about abstract thinking, But understanding Always happens through examples. If there are no examples, then it is impossible to grasp the principles... Just as it is impossible to get to the top of a mountain except by walking the entire slope from the foot.
Same with school: for now living stories It is not enough that we instinctively continue to regard it as a place where children are taught to understand.
For example, teaching the Gaussian method...
Gauss method in 5th grade school
I’ll make a reservation right away: the Gauss method has a much wider application, for example, when solving systems of linear equations. What we will talk about takes place in 5th grade. This started, having understood which, it is much easier to understand the more “advanced options”. In this article we are talking about Gauss's method (method) for finding the sum of a series
Here is an example that I brought from school younger son, attending 5th grade at a Moscow gymnasium.
School demonstration of the Gauss method
Math teacher using interactive whiteboard (modern methods training) showed the children a presentation of the history of the “creation of the method” by little Gauss.
The school teacher whipped little Karl (an outdated method, not used in schools these days) because he
instead of sequentially adding numbers from 1 to 100, find their sum noticed that pairs of numbers equally spaced from the edges of an arithmetic progression add up to the same number. for example, 100 and 1, 99 and 2. Having counted the number of such pairs, little Gauss almost instantly solved the problem proposed by the teacher. For which he was executed in front of an astonished public. So that others would be discouraged from thinking.
What did little Gauss do? developed number sense? Noticed some feature number series with a constant step (arithmetic progression). AND exactly this later made him a great scientist, those who know how to notice, having feeling, instinct of understanding.
This is why mathematics is valuable, developing ability to see general in particular - abstract thinking . Therefore, most parents and employers instinctively consider mathematics an important discipline ...
“Then you need to learn mathematics, because it puts your mind in order.
M.V.Lomonosov".
However, the followers of those who flogged future geniuses with rods turned the Method into something the opposite. As my supervisor said 35 years ago: “The question has been learned.” Or as my youngest son said yesterday about Gauss’s method: “Maybe it’s not worth making a big science out of this, huh?”
The consequences of the creativity of 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 supporter of the Fibonacci theory
Given an arithmetic progression: 4, 10, 16 ... 244, 250, 256 .
Algorithm for finding the sum of numbers in a series using the Gauss method:
4, 10, 16 ... 244, 250, 256
256, 250, 244 ... 16, 10, 4
At the same time, you need to remember plus one rule : we must add one to the resulting quotient: otherwise we will get a result that is less by one than true number pars: 42 + 1 = 43.
This is the required sum of the arithmetic progression from 4 to 256 with a difference of 6!
Gauss method: explanation in 5th grade at a Moscow gymnasium
Here's how to solve the problem of finding the sum of a series:
20+40+60+ ... +460+480+500
in the 5th grade of a Moscow gymnasium, Vilenkin’s textbook (according to my son).
After showing the presentation, the math teacher showed a couple of examples using the Gaussian method and gave the class a task of finding the sum of the numbers in a series in increments of 20.
This required the following:
As you can see, this is more compact and effective technique: number 3 is also a member of the Fibonacci sequence
My comments on the school version of the Gauss method
The great mathematician would definitely have chosen philosophy if he had foreseen what his “method” would be turned into by his followers German teacher, who flogged Karl with rods. He would have seen the symbolism, the dialectical spiral and the undying stupidity of the “teachers”, trying to measure the harmony of living mathematical thought with the algebra of misunderstanding ....
By the way: did you know. that our education system is rooted in the German school of the 18th and 19th centuries?
But Gauss chose mathematics.
What is the essence of his method?
IN simplification. IN observing and grasping simple patterns of numbers. IN turning dry school arithmetic into interesting and exciting activity , activating in the brain the desire to continue, rather than blocking high-cost mental activity.
Is it possible to use one of the given “modifications of Gauss’s method” to calculate the sum of the numbers of an arithmetic progression almost instantly? According to the “algorithms”, little Karl would be guaranteed to avoid spanking, develop an aversion to mathematics and suppress his creative impulses in the bud.
Why did the tutor so persistently advise fifth-graders “not to be afraid of misunderstanding” of the method, convincing them that they would solve “such” problems as early as 9th grade? Psychologically illiterate action. It was a good move to note: "See? You already in 5th grade you can solve problems that you will complete only in 4 years! What a great fellow you are!”
To use the Gaussian method, a level of class 3 is sufficient, when normal children already know how to add, multiply and divide 2-3 digit numbers. Problems arise due to the inability of adult teachers who are “out of touch” to explain the simplest things in normal human language, not to mention mathematical... They are unable to get people interested in mathematics and completely discourage even those who are “capable.”
Or, as my son commented: “making a big science out of it.”
Gauss method, my explanations
My wife and I explained this “method” to our child, it seems, even before school...
Simplicity instead of complexity or a game of questions and answers
"Look, here are the numbers from 1 to 100. What do you see?"
The point is not what exactly the child sees. The trick is to get him to look.
"How can you put them together?" The son realized that such questions are not asked “just like that” and you need to look at the question “somehow differently, differently than he usually does”
It doesn't matter if the child sees the solution right away, it's unlikely. It is important that he stopped being afraid to look, or as I say: “moved the task”. This is the beginning of the journey to understanding
“Which is easier: adding, for example, 5 and 6 or 5 and 95?” A leading question... But any training comes down to “guiding” a person to the “answer” - in any way acceptable to him.
At this stage, guesses may already arise about how to “save” on calculations.
All we did was hint: the “frontal, linear” method of counting is not the only possible one. If a child understands this, then later he will come up with many more such methods, because it's interesting!!! And he will definitely avoid “misunderstanding” mathematics and will not feel disgusted with it. He got the win!
If child discovered that adding pairs of numbers that add up to a hundred is a piece of cake, then "arithmetic progression with difference 1"- a rather dreary and uninteresting thing for a child - suddenly found life for him . Order emerged from chaos, and this always causes enthusiasm: that's how we are made!
A question to answer: why, after the insight a child has received, should he again be forced into the framework of dry algorithms, which are also functionally useless in this case?!
Why force stupid rewrites? sequence numbers in a notebook: so that even the capable do not have a single chance of understanding? Statistically, of course, but mass education is geared towards “statistics”...
Where did the zero go?
And yet, adding numbers that add up to 100 is much more acceptable to the mind than those that add up to 101...
The "Gauss School Method" requires exactly this: mindlessly fold pairs of numbers equidistant from the center of the progression, Despite everything.
What if you look?
Still zero - greatest invention humanity, which is more than 2,000 years old. And math teachers continue to ignore him.
It is much easier to transform a series of numbers starting with 1 into a series starting with 0. The sum will not change, will it? You need to stop “thinking in textbooks” and start looking... And see that pairs with a sum of 101 can be completely replaced by pairs with a sum of 100!
0 + 100, 1 + 99, 2 + 98 ... 49 + 51
How to abolish the "plus 1 rule"?
To be honest, I first heard about such a rule from that YouTube tutor...
What do I still do when I need to determine the number of members of a series?
I look at the sequence:
1, 2, 3, .. 8, 9, 10
and when you’re completely tired, then move on to a simpler row:
1, 2, 3, 4, 5
and I figure: if you subtract one from 5, you get 4, but I’m absolutely clear I see 5 numbers! Therefore, you need to add one! The number sense developed in elementary 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 world organization intellectual property WIPO has achieved a ban on the use of its good name in school textbooks!..
In 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 becomes even easier, which allows you to carve out 2-3 minutes for remote sensing in Russian, while the rest are “counting”. In addition, it retains the number of steps of the method: 5, which does not allow the approach to be criticized for being unscientific.
Obviously this approach is simpler, faster and more universal, in the style of the Method. But... the teacher not only did not praise her, but also forced her to rewrite " in the right way"(see screenshot). That is, she made a desperate attempt to stifle the creative impulse and the ability to understand mathematics at its root! Apparently, so that she could later get hired as a tutor... She attacked the wrong person...
Everything that I described so long and tediously can be explained to a normal child in a maximum of half an hour. Along with examples.
And in such a way that he will never forget it.
And it will be step towards understanding...not just mathematicians.
Admit it: how many times in your life have you added using the Gaussian method? And I never did!
But instinct of understanding, which develops (or is extinguished) in the process of learning mathematical methods at school... Oh!.. This is truly an irreplaceable thing!
Especially in the age of universal digitalization, which we have quietly entered under the strict leadership of the Party and the Government.
A few words in defense of teachers...
It is unfair and wrong to place all responsibility for this style of teaching solely on school teachers. The system is in effect.
Some teachers understand the absurdity of what is happening, but what to do? Law on Education, Federal State Educational Standards, methods, technological maps lessons... Everything must be done “in accordance with and on the basis of” and everything must be documented. Step aside - stood in line to be fired. Let’s not be hypocrites: the salaries of Moscow teachers are very good... If they fire you, where to go?..
Therefore this site not about education. He's about individual education, only possible way get out of the crowd generation Z ...
Solving systems of linear equations using the Gauss method. Suppose we need to find a solution to the system from n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.
The essence of the Gauss method consists of sequentially eliminating unknown variables: first eliminating x 1 from all equations of the system, starting from the second, is further excluded x 2 from all equations, starting with the third, and so on, until only the unknown variable remains in the last equation x n. This process of transforming system equations to sequentially eliminate unknown variables is called direct Gaussian method. After completing the forward progression of the Gaussian method, from the last equation we find x n, using this value from the penultimate equation we calculate xn-1, and so on, from the first equation we find x 1. The process of calculating unknown variables when moving from the last equation of the system to the first is called inverse of the Gaussian 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, 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 nth to the equation we add the first one, multiplied by . The system of equations after such transformations will take the form 
where and 
.
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 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 nth to the equation we add the second one, multiplied by . The system of equations after such transformations will take the form 
where and 
. So the variable x 2 excluded from all equations starting from the third.
Next we proceed to eliminating the unknown x 3, in this case 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 x n we find xn-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Example.
Solve system of linear equations 
Gauss method.