An example of solving a coupled system using the Gaussian method. Gaussian method for solving matrices. Solving a system of linear equations using the Gauss method

Definition and description of the Gaussian method

The Gaussian transformation 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 systems of algebraic equations (SLAE). This classical method is also used to solve problems such as obtaining inverse matrices and determining the rank of a matrix.

Transformation using the Gaussian method consists of making small (elementary) sequential changes to a 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 that is equivalent to the original one.

Definition 1

This part of the solution is called the forward Gaussian solution, since the entire process is carried out from top to bottom.

After reducing the original system of equations to a triangular one, all variables of the system are found from bottom to top (that is, the first variables found are located precisely on the last lines of the system or matrix). This part of the solution is also known as the inverse of the Gaussian solution. His algorithm is as follows: first, the variables closest to the bottom of the system of equations or matrix are calculated, then the resulting values ​​are substituted higher and thus another variable is found, and so on.

Description of the Gaussian method algorithm

The sequence of actions for the general solution of a system of equations using the Gaussian method consists in alternately applying the forward and backward strokes to the matrix based on the SLAE. Let the initial 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 SLAEs using the Gaussian method, it is necessary to write the original 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 a bar with a column of free terms, is called an extended matrix:

$A = \begin(array)(ccc|c) a_(11) & … & a_(1n) & b_1 \\ \vdots & … & \vdots & ...\\ a_(m1) & … & a_( mn) & b_m \end(array)$

Now it is necessary, using elementary transformations on the system of equations (or on the matrix, since this is more convenient), 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 what 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:

1. All its zero lines come after non-zero lines
2. If some row of a matrix with number $k$ is non-zero, then the previous row of the same matrix has fewer zeros than this one with number $k$.

After obtaining the step matrix, it is necessary to substitute the resulting 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 system of equations using this method, you need to use only elementary transformations.

Such transformations are considered to be operations that can be applied to a matrix or system of equations without changing its meaning:

• rearrangement of several lines,
• adding or subtracting from one row of a matrix another row from it,
• multiplying or dividing a string by a constant 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 Gaussian method to solve systems:

1. When a system is inconsistent, that is, it does not have any solutions
2. The system of equations has a solution, and a unique one, and the number of non-zero rows and columns in the matrix is ​​equal to each other.
3. 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.

Outcome of a solution with an inconsistent system

For this option, when solving a matrix equation using the Gaussian 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 do not have solutions, regardless of the other equations they contain. An example of an inconsistent matrix:

$\begin(array)(ccc|c) 2 & -1 & 3 & 0 \\ 1 & 0 & 2 & 0\\ 0 & 0 & 0 & 1 \end(array)$

In the last line an impossible equality arose: $0 \cdot x_(31) + 0 \cdot x_(32) + 0 \cdot x_(33) = 1$.

A system of equations that has only one solution

These systems, after being reduced to a step matrix and removing rows with zeros, have the same number of rows and columns in the main matrix. Here is the simplest example of 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 lower equation yields the following value for $x$: $x_2 = 3 \frac(1)(3)$. Substitute 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 up to the “=” sign, and the remaining variables must be moved to the right side of the equality.

Such a system has only a certain general solution.

Let us 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 basis variables will be $y_1$ and $y_3$ (for $y_1$ - since it comes first, and in the case of $y_3$ - it is located after the zeros).

As basis variables, we choose exactly those that are the first in the row and are not equal to zero.

The remaining variables are called free; we need to express the basic ones through them.

Using the so-called reverse stroke, we analyze the system from bottom to top; to do 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 slough using the Gaussian method. Examples. An example of solving a system of linear equations given by a 3 by 3 matrix using the Gaussian method

$\begin(cases) 4x_1 + 2x_2 – x_3 = 1 \\ 5x_1 + 3x_2 - 2x^3 = 2\\ 3x_1 + 2x_2 – 3x_3 = 0 \end(cases)$

Let's write our system in the form of an extended matrix:

$\begin(array)(ccc|c) 4 & 2 & -1 & 1 \\ 5 & 3 & -2 & 2 \\ 3 & 2 & -3 & 0\\ \end(array)$

Now, for convenience and practicality, you need to transform the matrix so that $1$ is in the upper corner of the outermost column.

To do this, to the 1st line we need to add the line from the middle, multiplied by $-1$, 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 lines by $-1$, and also swap the last and middle lines:

$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & -2 & 3 & -3\\ \end(array)$

$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 3 & 3\\ \end(array)$

And divide 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 multiply the top line by $-2$ and add to the 2nd and 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 $-1$.

$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & -1 & 0 & -1 & -2 \\ 0 & 0 & 5 & 1 & 10\\ 0 & 0 & 3 & 0 & 6 \\ \end(array)$

We multiply line 2 by $-1$, and divide line 4 by $3$ and replace line 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:

$\begin(cases) m = 0 \\ g = 2\\ y + m = 2\ \ x + 3y + 2g + m = 11\end(cases)$

Let a system of linear algebraic equations be given that needs to be solved (find such values ​​of the unknowns xi that turn each equation of the system into an equality).

We know that a system of linear algebraic equations can:

1) Have no solutions (be non-joint).
2) Have infinitely many solutions.
3) Have a single solution.

As we remember, Cramer's rule and the matrix method are not suitable in cases where the system has infinitely many solutions or is inconsistent. Gauss method – the most powerful and versatile tool for finding solutions to any system of linear equations, which in every case will lead us to the answer! The method algorithm itself works the same in all three cases. If the Cramer and matrix methods require knowledge of determinants, then to apply the Gauss method you only need knowledge of arithmetic operations, which makes it accessible even to primary school students.

Augmented matrix transformations ( this is the matrix of the system - a matrix composed only of the coefficients of the unknowns, plus a column of free terms) systems of linear algebraic equations in the Gauss method:

1) With troki matrices Can rearrange in some places.

2) if proportional (as a special case – identical) rows appear (or exist) in the matrix, then you should delete All these rows are from the matrix except one.

3) if a zero row appears in the matrix during transformations, then it should also be delete.

4) a row of the matrix can be multiply (divide) to any number other than zero.

5) to a row of the matrix you can add another string multiplied by a number, different from zero.

In the Gauss method, elementary transformations do not change the solution of the system of equations.

The Gauss method consists of two stages:

1. “Direct move” - using elementary transformations, bring the extended matrix of a system of linear algebraic equations to a “triangular” step form: the elements of the extended matrix located below the main diagonal are equal to zero (top-down move). For example, to this type:

To do this, perform the following steps:

1) Let us consider the first equation of a system of linear algebraic equations and the coefficient for x 1 is equal to K. The second, third, etc. we transform the equations as follows: we divide each equation (coefficients of the unknowns, including free terms) by the coefficient of the unknown x 1 in each equation, and multiply by K. After this, we subtract the first from the second equation (coefficients of unknowns and free terms). For x 1 in the second equation we obtain the coefficient 0. From the third transformed equation we subtract the first equation until all equations except the first, for unknown x 1, have a coefficient 0.

2) Let's move on to the next equation. Let this be the second equation and the coefficient for x 2 equal to M. We proceed with all “lower” equations as described above. Thus, “under” the unknown x 2 there will be zeros in all equations.

3) Move on to the next equation and so on until one last unknown and the transformed free term remain.

1. The “reverse move” of the Gauss method is to obtain a solution to a system of linear algebraic equations (the “bottom-up” move).

From the last “lower” equation we obtain one first solution - the unknown x n. To do this, we solve the elementary equation A * x n = B. In the example given above, x 3 = 4. We substitute the found value into the “upper” next equation and solve it with respect to the next unknown. For example, x 2 – 4 = 1, i.e. x 2 = 5. And so on until we find all the unknowns.

Let's solve the system of linear equations using the Gauss method, as some authors advise:

Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

We look at the upper left “step”. We should have one there. The problem is that there are no units in the first column at all, so rearranging the rows will not solve anything. In such cases, the unit must be organized using an elementary transformation. This can usually be done in several ways. Let's do this:
1 step . To the first line we add the second line, multiplied by –1. That is, we mentally multiplied the second line by –1 and added the first and second lines, while the second line did not change.

Now at the top left there is “minus one”, which suits us quite well. Anyone who wants to get +1 can perform an additional action: multiply the first line by –1 (change its sign).

Step 2 . The first line, multiplied by 5, was added to the second line. The first line, multiplied by 3, was added to the third line.

Step 3 . The first line was multiplied by –1, in principle, this is for beauty. The sign of the third line was also changed and it was moved to second place, so that on the second “step” we had the required unit.

Step 4 . The third line was added to the second line, multiplied by 2.

Step 5 . The third line was divided by 3.

A sign that indicates an error in calculations (more rarely, a typo) is a “bad” bottom line. That is, if we got something like (0 0 11 |23) below, and, accordingly, 11x 3 = 23, x 3 = 23/11, then with a high degree of probability we can say that an error was made during elementary transformations.

Let’s do the reverse; in the design of examples, the system itself is often not rewritten, but the equations are “taken directly from the given matrix.” The reverse move, I remind you, works from the bottom up. In this example, the result was a gift:

x 3 = 1
x 2 = 3
x 1 + x 2 – x 3 = 1, therefore x 1 + 3 – 1 = 1, x 1 = –1

Answer:x 1 = –1, x 2 = 3, x 3 = 1.

Let's solve the same system using the proposed algorithm. We get

4 2 –1 1
5 3 –2 2
3 2 –3 0

Divide the second equation by 5, and the third by 3. We get:

4 2 –1 1
1 0.6 –0.4 0.4
1 0.66 –1 0

Multiplying the second and third equations by 4, we get:

4 2 –1 1
4 2,4 –1.6 1.6
4 2.64 –4 0

Subtract the first equation from the second and third equations, we have:

4 2 –1 1
0 0.4 –0.6 0.6
0 0.64 –3 –1

Divide the third equation by 0.64:

4 2 –1 1
0 0.4 –0.6 0.6
0 1 –4.6875 –1.5625

Multiply the third equation by 0.4

4 2 –1 1
0 0.4 –0.6 0.6
0 0.4 –1.875 –0.625

Subtracting the second from the third equation, we obtain a “stepped” extended matrix:

4 2 –1 1
0 0.4 –0.6 0.6
0 0 –1.275 –1.225

Thus, since the error accumulated during the calculations, we obtain x 3 = 0.96 or approximately 1.

x 2 = 3 and x 1 = –1.

By solving in this way, you will never get confused in the calculations and, despite the calculation errors, you will get the result.

This method of solving a system of linear algebraic equations is easily programmable and does not take into account the specific features of coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.

I wish you success! See you in class! Tutor.

blog.site, when copying material in full or in part, a link to the original source is required.

Let a system of linear algebraic equations be given that needs to be solved (find such values ​​of the unknowns xi that turn each equation of the system into an equality).

We know that a system of linear algebraic equations can:

1) Have no solutions (be non-joint).
2) Have infinitely many solutions.
3) Have a single solution.

As we remember, Cramer's rule and the matrix method are not suitable in cases where the system has infinitely many solutions or is inconsistent. Gauss method – the most powerful and versatile tool for finding solutions to any system of linear equations, which in every case will lead us to the answer! The method algorithm itself works the same in all three cases. If the Cramer and matrix methods require knowledge of determinants, then to apply the Gauss method you only need knowledge of arithmetic operations, which makes it accessible even to primary school students.

Augmented matrix transformations ( this is the matrix of the system - a matrix composed only of the coefficients of the unknowns, plus a column of free terms) systems of linear algebraic equations in the Gauss method:

1) With troki matrices Can rearrange in some places.

2) if proportional (as a special case – identical) rows appear (or exist) in the matrix, then you should delete All these rows are from the matrix except one.

3) if a zero row appears in the matrix during transformations, then it should also be delete.

4) a row of the matrix can be multiply (divide) to any number other than zero.

5) to a row of the matrix you can add another string multiplied by a number, different from zero.

In the Gauss method, elementary transformations do not change the solution of the system of equations.

The Gauss method consists of two stages:

1. “Direct move” - using elementary transformations, bring the extended matrix of a system of linear algebraic equations to a “triangular” step form: the elements of the extended matrix located below the main diagonal are equal to zero (top-down move). For example, to this type:

To do this, perform the following steps:

1) Let us consider the first equation of a system of linear algebraic equations and the coefficient for x 1 is equal to K. The second, third, etc. we transform the equations as follows: we divide each equation (coefficients of the unknowns, including free terms) by the coefficient of the unknown x 1 in each equation, and multiply by K. After this, we subtract the first from the second equation (coefficients of unknowns and free terms). For x 1 in the second equation we obtain the coefficient 0. From the third transformed equation we subtract the first equation until all equations except the first, for unknown x 1, have a coefficient 0.

2) Let's move on to the next equation. Let this be the second equation and the coefficient for x 2 equal to M. We proceed with all “lower” equations as described above. Thus, “under” the unknown x 2 there will be zeros in all equations.

3) Move on to the next equation and so on until one last unknown and the transformed free term remain.

1. The “reverse move” of the Gauss method is to obtain a solution to a system of linear algebraic equations (the “bottom-up” move).

From the last “lower” equation we obtain one first solution - the unknown x n. To do this, we solve the elementary equation A * x n = B. In the example given above, x 3 = 4. We substitute the found value into the “upper” next equation and solve it with respect to the next unknown. For example, x 2 – 4 = 1, i.e. x 2 = 5. And so on until we find all the unknowns.

Let's solve the system of linear equations using the Gauss method, as some authors advise:

Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

We look at the upper left “step”. We should have one there. The problem is that there are no units in the first column at all, so rearranging the rows will not solve anything. In such cases, the unit must be organized using an elementary transformation. This can usually be done in several ways. Let's do this:
1 step . To the first line we add the second line, multiplied by –1. That is, we mentally multiplied the second line by –1 and added the first and second lines, while the second line did not change.

Now at the top left there is “minus one”, which suits us quite well. Anyone who wants to get +1 can perform an additional action: multiply the first line by –1 (change its sign).

Step 2 . The first line, multiplied by 5, was added to the second line. The first line, multiplied by 3, was added to the third line.

Step 3 . The first line was multiplied by –1, in principle, this is for beauty. The sign of the third line was also changed and it was moved to second place, so that on the second “step” we had the required unit.

Step 4 . The third line was added to the second line, multiplied by 2.

Step 5 . The third line was divided by 3.

A sign that indicates an error in calculations (more rarely, a typo) is a “bad” bottom line. That is, if we got something like (0 0 11 |23) below, and, accordingly, 11x 3 = 23, x 3 = 23/11, then with a high degree of probability we can say that an error was made during elementary transformations.

Let’s do the reverse; in the design of examples, the system itself is often not rewritten, but the equations are “taken directly from the given matrix.” The reverse move, I remind you, works from the bottom up. In this example, the result was a gift:

x 3 = 1
x 2 = 3
x 1 + x 2 – x 3 = 1, therefore x 1 + 3 – 1 = 1, x 1 = –1

Answer:x 1 = –1, x 2 = 3, x 3 = 1.

Let's solve the same system using the proposed algorithm. We get

4 2 –1 1
5 3 –2 2
3 2 –3 0

Divide the second equation by 5, and the third by 3. We get:

4 2 –1 1
1 0.6 –0.4 0.4
1 0.66 –1 0

Multiplying the second and third equations by 4, we get:

4 2 –1 1
4 2,4 –1.6 1.6
4 2.64 –4 0

Subtract the first equation from the second and third equations, we have:

4 2 –1 1
0 0.4 –0.6 0.6
0 0.64 –3 –1

Divide the third equation by 0.64:

4 2 –1 1
0 0.4 –0.6 0.6
0 1 –4.6875 –1.5625

Multiply the third equation by 0.4

4 2 –1 1
0 0.4 –0.6 0.6
0 0.4 –1.875 –0.625

Subtracting the second from the third equation, we obtain a “stepped” extended matrix:

4 2 –1 1
0 0.4 –0.6 0.6
0 0 –1.275 –1.225

Thus, since the error accumulated during the calculations, we obtain x 3 = 0.96 or approximately 1.

x 2 = 3 and x 1 = –1.

By solving in this way, you will never get confused in the calculations and, despite the calculation errors, you will get the result.

This method of solving a system of linear algebraic equations is easily programmable and does not take into account the specific features of coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.

I wish you success! See you in class! Tutor Dmitry Aystrakhanov.

website, when copying material in full or in part, a link to the source is required.

The essence of the Gauss method is to transform (1) to a system with a triangular matrix, from which the values ​​of all unknowns are then obtained sequentially (in reverse). Let's consider one of the computational schemes. This circuit is called a single division circuit. So let's look at this diagram. Let a 11 ≠0 (leading element) divide the first equation by a 11. We get
(2)
Using equation (2), it is easy to eliminate the unknowns x 1 from the remaining equations of the system (to do this, it is enough to subtract equation (2) from each equation, previously multiplied by the corresponding coefficient for x 1), that is, in the first step we obtain
.
In other words, at step 1, each element of subsequent rows, starting from the second, is equal to the difference between the original element and the product of its “projection” onto the first column and the first (transformed) row.
Following this, leaving the first equation alone, we perform a similar transformation over the remaining equations of the system obtained in the first step: we select from among them the equation with the leading element and, with its help, exclude x 2 from the remaining equations (step 2).
After n steps, instead of (1), we obtain an equivalent system
(3)
Thus, at the first stage we obtain a triangular system (3). This stage is called forward stroke.
At the second stage (reverse), we find sequentially from (3) the values ​​x n, x n -1, ..., x 1.
Let us denote the resulting solution as x 0 . Then the difference ε=b-A x 0 called residual.
If ε=0, then the found solution x 0 is correct.

Calculations using the Gaussian method are performed in two stages:

1. The first stage is called the forward method. At the first stage, the original system is converted to a triangular form.
2. The second stage is called the reverse stroke. At the second stage, a triangular system equivalent to the original one is solved.

Purpose of the Gauss method

Types of Gaussian method

1. Classical Gaussian method;
2. Modifications of the Gauss method. One of the modifications of the Gaussian method is a scheme with the choice of the main element. A feature of the Gauss method with the choice of the main element is such a rearrangement of the equations so that at the kth step the leading element turns out to be the largest element in the kth column.
3. Jordano-Gauss method;

Example of a solution using the Gaussian method
Let's solve the system:

For ease of calculation, let's swap the lines:

Let's multiply the 2nd line by (2). Add the 3rd line to the 2nd

Multiply the 2nd line by (-1). Add the 2nd line to the 1st

From the 1st line we express x 3:
From the 2nd line we express x 2:
From the 3rd line we express x 1:

An example of a solution using the Jordano-Gauss method
Let us solve the same SLAE using the Jordano-Gauss method.

We will sequentially select the resolving element RE, which lies on the main diagonal of the matrix.
The resolution element is equal to (1).



NE = SE - (A*B)/RE
RE - resolving element (1), A and B - matrix elements forming a rectangle with elements STE and RE.
Let's present the calculation of each element in the form of a table:

x 1	x 2	x 3	B
1 / 1 = 1	2 / 1 = 2	-2 / 1 = -2	1 / 1 = 1

x 1	x 2	x 3	B
0 / 3 = 0	3 / 3 = 1	1 / 3 = 0.33	4 / 3 = 1.33

x 1	x 2	x 3	B
0 / -4 = 0	0 / -4 = 0	-4 / -4 = 1	-4 / -4 = 1

Implementation of the Gaussian method

Using the Gaussian Method

Application of the Gauss method in game theory

Application of the Gauss method in solving differential equations

Application of the Jordano-Gauss method in linear programming

Carl Friedrich Gauss, the greatest mathematician, hesitated for a long time, choosing between philosophy and mathematics. Perhaps it was precisely this mindset that allowed him to make such a noticeable “legacy” in world science. In particular, by creating the "Gauss Method" ...

For almost 4 years, articles on this site dealt with school education, mainly from the point of view of philosophy, the principles of (mis)understanding introduced into the minds of children. The time is coming for more specifics, examples and methods... I believe that this is exactly the approach to the familiar, confusing and important areas of life gives better results.

We people are designed in such a way that no matter how much we talk about abstract thinking, But understanding Always happens through examples. If there are no examples, then it is impossible to grasp the principles... Just as it is impossible to get to the top of a mountain except by walking the entire slope from the foot.

Same with school: for now living stories It is not enough that we instinctively continue to regard it as a place where children are taught to understand.

For example, teaching the Gaussian method...

Gauss method in 5th grade school

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 my youngest son, who attends 5th grade at a Moscow gymnasium, brought from school.

School demonstration of the Gauss method

A mathematics teacher using an interactive whiteboard (modern teaching methods) showed 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 ...

“Mathematics must then be taught, because it puts the 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 the true number of pairs: 42 + 1 = 43.

This is the required sum of the arithmetic progression from 4 to 256 with a difference of 6!

Gauss method: explanation in 5th grade at a Moscow gymnasium

Here's how to solve the problem of finding the sum of a series:

20+40+60+ ... +460+480+500

in the 5th grade of a Moscow gymnasium, Vilenkin’s textbook (according to my son).

After showing the presentation, the math teacher showed a couple of examples using the Gaussian method and gave the class a task of finding the sum of the numbers in a series in increments of 20.

This required the following:

As you can see, this is a 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 “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 driven into the framework of dry algorithms, which are also functionally useless in this case?!

Why force stupid rewrites? sequence numbers in a notebook: so that even the capable do not have a single chance of understanding? Statistically, of course, but mass education is geared towards “statistics”...

Where did the zero go?

And yet, adding numbers that add up to 100 is much more acceptable to the mind than those that add up to 101...

The "Gauss School Method" requires exactly this: mindlessly fold pairs of numbers equidistant from the center of the progression, Despite everything.

What if you look?

Still, zero is the greatest invention of mankind, which is more than 2,000 years old. And math teachers continue to ignore him.

It is much easier to transform a series of numbers starting with 1 into a series starting with 0. The sum will not change, will it? You need to stop “thinking in textbooks” and start looking... And see that pairs with a sum of 101 can be completely replaced by pairs with a sum of 100!

0 + 100, 1 + 99, 2 + 98 ... 49 + 51

How to abolish the "plus 1 rule"?

To be honest, I first heard about such a rule from that YouTube tutor...

What do I still do when I need to determine the number of members of a series?

I look at the sequence:

1, 2, 3, .. 8, 9, 10

and when you’re completely tired, then move on to a simpler row:

1, 2, 3, 4, 5

and I figure: if you subtract one from 5, you get 4, but I’m absolutely clear I see 5 numbers! Therefore, you need to add one! 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 me a bad mark. And she made me write in my notebook “the course of the solution” in their own way. Why make a big science out of this?..”

The main crime of a math teacher

Hardly after that incident Carl Gauss experienced a high sense of respect for his school mathematics teacher. But if he knew how followers of that teacher will distort the very essence of the method... he would roar with indignation and, through the World Intellectual Property Organization WIPO, achieve a ban on the use of his good name in school textbooks!..

What the main mistake of the 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 more efficient counting methods.

Gauss method according to Vilenkin

In school they teach that Gauss's method is to

What, if the number of elements of the series is odd, as in the problem that was assigned to my son?..

The "catch" is that in this case you should find an “extra” number in the series and add it to the sum of the pairs. In our example this number is 260.

How to detect? Copying all pairs of numbers into a notebook!(This is why the teacher made the kids do this stupid job of trying to teach "creativity" using the Gaussian method... And this is why such a "method" is practically inapplicable to large data series, AND this is why it is not the Gaussian method.)

A little creativity in the school routine...

The son acted differently.

(20 + 500, 40 + 480 ...).

0+500, 20+480, 40+460 ...

Not difficult, right?

But in practice it becomes even easier, which allows you to carve out 2-3 minutes for remote sensing in Russian, while the rest are “counting”. In addition, it retains the number of steps of the method: 5, which does not allow the approach to be criticized for being unscientific.

Obviously this approach is simpler, faster and more universal, in the style of the Method. But... the teacher not only did not praise, but also forced me to rewrite it “in the correct way” (see screenshot). That is, she made a desperate attempt to stifle the creative impulse and the ability to understand mathematics at the root! Apparently, so that she could later be hired as a tutor... She attacked the wrong person...

Everything that I described so long and tediously can be explained to a normal child in a maximum of half an hour. Along with examples.

And in such a way that he will never forget it.

And it will be step towards understanding...not just mathematicians.

Admit it: how many times in your life have you added using the Gaussian method? And I never did!

But instinct of understanding, which develops (or is extinguished) in the process of studying 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? The Law on Education, Federal State Educational Standards, methods, lesson plans... Everything must be done “in accordance and on the basis” 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, the only possible way to get out of the crowd generation Z ...