Solution of a square matrix by the Gauss method. Gauss method or why children do not understand mathematics

In this article, the method is considered as a way to solve systems linear equations(SLAU). The method is analytical, that is, it allows you to write a solution algorithm in general view, and then substitute values ​​from specific examples there. Unlike the matrix method or Cramer's formulas, when solving a system of linear equations using the Gauss method, you can also work with those that have infinitely many solutions. Or they don't have it at all.

What does Gauss mean?

First you need to write down our system of equations in It looks like this. The system is taken:

The coefficients are written in the form of a table, and on the right in a separate column - free members. The column with free members is separated for convenience. The matrix that includes this column is called extended.

Further, the main matrix with coefficients must be reduced to the upper triangular shape. This is the main point of solving the system by the Gauss method. Simply put, after certain manipulations, the matrix should look like this, so that there are only zeros in its lower left part:

Then, if you write the new matrix again as a system of equations, you will notice that the last row already contains the value of one of the roots, which is then substituted into the equation above, another root is found, and so on.

This description of the solution by the Gauss method in the most in general terms. And what happens if suddenly the system does not have a solution? Or are there an infinite number of them? To answer these and many more questions, it is necessary to consider separately all the elements used in the solution by the Gauss method.

Matrices, their properties

There is no hidden meaning in the matrix. It's simple convenient way recording data for subsequent operations with them. Even schoolchildren should not be afraid of them.

The matrix is ​​always rectangular, because it is more convenient. Even in the Gauss method, where everything boils down to building a triangular matrix, a rectangle appears in the entry, only with zeros in the place where there are no numbers. Zeros can be omitted, but they are implied.

The matrix has a size. Its "width" is the number of rows (m), its "length" is the number of columns (n). Then the size of the matrix A (capital Latin letters are usually used for their designation) will be denoted as A m×n . If m=n, then this matrix is ​​square, and m=n is its order. Accordingly, any element of the matrix A can be denoted by the number of its row and column: a xy ; x - row number, changes , y - column number, changes .

B is not the main point of the solution. In principle, all operations can be performed directly with the equations themselves, but the notation will turn out to be much more cumbersome, and it will be much easier to get confused in it.

Determinant

The matrix also has a determinant. This is a very important feature. Finding out its meaning now is not worth it, you can simply show how it is calculated, and then tell what properties of the matrix it determines. The easiest way to find the determinant is through diagonals. Imaginary diagonals are drawn in the matrix; the elements located on each of them are multiplied, and then the resulting products are added: diagonals with a slope to the right - with a "plus" sign, with a slope to the left - with a "minus" sign.

It is extremely important to note that the determinant can only be calculated for a square matrix. For a rectangular matrix, you can do the following: choose the smallest of the number of rows and the number of columns (let it be k), and then randomly mark k columns and k rows in the matrix. The elements located at the intersection of the selected columns and rows will form a new square matrix. If the determinant of such a matrix is ​​a number other than zero, then it is called the basis minor of the original rectangular matrix.

Before proceeding with the solution of the system of equations by the Gauss method, it does not hurt to calculate the determinant. If it turns out to be zero, then we can immediately say that the matrix has either an infinite number of solutions, or there are none at all. In such a sad case, you need to go further and find out about the rank of the matrix.

System classification

There is such a thing as the rank of a matrix. This is the maximum order of its non-zero determinant (remembering the basis minor, we can say that the rank of a matrix is ​​the order of the basis minor).

According to how things are with the rank, SLAE can be divided into:

• Joint. At of joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended one (with a column of free terms). Such systems have a solution, but not necessarily one, therefore, joint systems are additionally divided into:
• - certain- having only decision. In certain systems, the rank of the matrix and the number of unknowns (or the number of columns, which is the same thing) are equal;
• - indefinite - with an infinite number of solutions. The rank of matrices for such systems is less than the number of unknowns.
• Incompatible. At such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.

The Gauss method is good in that it allows one to obtain either an unambiguous proof of the inconsistency of the system (without calculating the determinants of large matrices) or a general solution for a system with an infinite number of solutions during the solution.

Elementary transformations

Before proceeding directly to the solution of the system, it is possible to make it less cumbersome and more convenient for calculations. This is achieved through elementary transformations - such that their implementation does not change the final answer in any way. It should be noted that some of the above elementary transformations are valid only for matrices, the source of which was precisely the SLAE. Here is a list of these transformations:

1. String permutation. It is obvious that if we change the order of the equations in the system record, then this will not affect the solution in any way. Consequently, it is also possible to interchange rows in the matrix of this system, not forgetting, of course, about the column of free members.
2. Multiplying all elements of a string by some factor. Very useful! With it, you can reduce large numbers in the matrix or remove zeros. The set of solutions, as usual, will not change, and it will become more convenient to perform further operations. The main thing is that the coefficient should not be zero.
3. Delete rows with proportional coefficients. This partly follows from the previous paragraph. If two or more rows in the matrix have proportional coefficients, then when multiplying / dividing one of the rows by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and you can remove the extra ones, leaving only one.
4. Removing the null line. If in the course of transformations a string is obtained somewhere in which all elements, including the free member, are zero, then such a string can be called zero and thrown out of the matrix.
5. Adding to the elements of one row the elements of another (in the corresponding columns), multiplied by a certain coefficient. The most obscure and most important transformation of all. It is worth dwelling on it in more detail.

Adding a string multiplied by a factor

For ease of understanding, it is worth disassembling this process step by step. Two rows are taken from the matrix:

a 11 a 12 ... a 1n | b1

a 21 a 22 ... a 2n | b 2

Suppose you need to add the first to the second, multiplied by the coefficient "-2".

a" 21 \u003d a 21 + -2 × a 11

a" 22 \u003d a 22 + -2 × a 12

a" 2n \u003d a 2n + -2 × a 1n

Then in the matrix the second row is replaced with a new one, and the first one remains unchanged.

a 11 a 12 ... a 1n | b1

a" 21 a" 22 ... a" 2n | b 2

It should be noted that the multiplication factor can be chosen in such a way that, as a result of the addition of two strings, one of the elements of the new string is equal to zero. Therefore, it is possible to obtain an equation in the system, where there will be one less unknown. And if you get two such equations, then the operation can be done again and get an equation that will already contain two less unknowns. And if each time we turn to zero one coefficient for all rows that are lower than the original one, then we can, like steps, go down to the very bottom of the matrix and get an equation with one unknown. This is called solving the system using the Gaussian method.

In general

Let there be a system. It has m equations and n unknown roots. You can write it down like this:

The main matrix is ​​compiled from the coefficients of the system. A column of free members is added to the extended matrix and separated by a bar for convenience.

• the first row of the matrix is ​​multiplied by the coefficient k = (-a 21 / a 11);
• the first modified row and the second row of the matrix are added;
• instead of the second row, the result of the addition from the previous paragraph is inserted into the matrix;
• now the first coefficient in the new second row is a 11 × (-a 21 /a 11) + a 21 = -a 21 + a 21 = 0.

Now the same series of transformations is performed, only the first and third rows are involved. Accordingly, in each step of the algorithm, the element a 21 is replaced by a 31 . Then everything is repeated for a 41 , ... a m1 . The result is a matrix where the first element in the rows is equal to zero. Now we need to forget about line number one and execute the same algorithm starting from the second line:

• coefficient k \u003d (-a 32 / a 22);
• the second modified line is added to the "current" line;
• the result of the addition is substituted in the third, fourth, and so on lines, while the first and second remain unchanged;
• in the rows of the matrix, the first two elements are already equal to zero.

The algorithm must be repeated until the coefficient k = (-a m,m-1 /a mm) appears. This means that the algorithm was last run only for the lower equation. Now the matrix looks like a triangle, or has a stepped shape. The bottom line contains the equality a mn × x n = b m . The coefficient and free term are known, and the root is expressed through them: x n = b m /a mn. The resulting root is substituted into the top row to find x n-1 = (b m-1 - a m-1,n ×(b m /a mn))÷a m-1,n-1 . And so on by analogy: each next line contains new root, and, having reached the "top" of the system, one can find many solutions . It will be the only one.

When there are no solutions

If in one of the matrix rows all elements, except for the free term, are equal to zero, then the equation corresponding to this row looks like 0 = b. It has no solution. And since such an equation is included in the system, then the set of solutions of the entire system is empty, that is, it is degenerate.

When there are an infinite number of solutions

It may turn out that in the reduced triangular matrix there are no rows with one element-the coefficient of the equation, and one - a free member. There are only strings that, when rewritten, would look like an equation with two or more variables. This means that the system has an infinite number of solutions. In this case, the answer can be given in the form of a general solution. How to do it?

All variables in the matrix are divided into basic and free. Basic - these are those that stand "on the edge" of the rows in the stepped matrix. The rest are free. In the general solution, the basic variables are written in terms of the free ones.

For convenience, the matrix is ​​first rewritten back into a system of equations. Then in the last of them, where exactly only one basic variable remained, it remains on one side, and everything else is transferred to the other. This is done for each equation with one basic variable. Then, in the rest of the equations, where possible, instead of the basic variable, the expression obtained for it is substituted. If, as a result, an expression again appears containing only one basic variable, it is again expressed from there, and so on, until each basic variable is written as an expression with free variables. That's what it is common decision SLAU.

You can also find the basic solution of the system - give the free variables any values, and then for this particular case calculate the values ​​of the basic variables. There are infinitely many particular solutions.

Solution with specific examples

Here is the system of equations.

For convenience, it is better to immediately create its matrix

It is known that when solving by the Gauss method, the equation corresponding to the first row will remain unchanged at the end of the transformations. Therefore, it will be more profitable if the left top element matrix will be the smallest - then the first elements of the remaining rows after the operations will turn to zero. This means that in the compiled matrix it will be advantageous to put the second in place of the first row.

second line: k = (-a 21 / a 11) = (-3/1) = -3

a" 21 \u003d a 21 + k × a 11 \u003d 3 + (-3) × 1 \u003d 0

a" 22 \u003d a 22 + k × a 12 \u003d -1 + (-3) × 2 \u003d -7

a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11

b "2 \u003d b 2 + k × b 1 \u003d 12 + (-3) × 12 \u003d -24

third line: k = (-a 3 1 /a 11) = (-5/1) = -5

a" 3 1 = a 3 1 + k×a 11 = 5 + (-5)×1 = 0

a" 3 2 = a 3 2 + k×a 12 = 1 + (-5)×2 = -9

a" 3 3 = a 33 + k×a 13 = 2 + (-5)×4 = -18

b "3 \u003d b 3 + k × b 1 \u003d 3 + (-5) × 12 \u003d -57

Now, in order not to get confused, it is necessary to write down the matrix with the intermediate results of the transformations.

It is obvious that such a matrix can be made more convenient for perception with the help of some operations. For example, you can remove all "minuses" from the second line by multiplying each element by "-1".

It is also worth noting that in the third row all elements are multiples of three. Then you can reduce the string by this number, multiplying each element by "-1/3" (minus - at the same time to remove negative values).

Looks much nicer. Now we need to leave alone the first line and work with the second and third. The task is to add the second row to the third row, multiplied by such a factor that the element a 32 becomes equal to zero.

k = (-a 32 / a 22) = (-3/7) = -3/7 fractions, and only then, when the answers are received, decide whether to round up and translate into another form of notation)

a" 32 = a 32 + k × a 22 = 3 + (-3/7) × 7 = 3 + (-3) = 0

a" 33 \u003d a 33 + k × a 23 \u003d 6 + (-3/7) × 11 \u003d -9/7

b "3 \u003d b 3 + k × b 2 \u003d 19 + (-3/7) × 24 \u003d -61/7

The matrix is ​​written again with new values.

1	2	4	12
0	7	11	24
0	0	-9/7	-61/7

As you can see, the resulting matrix already has a stepped form. Therefore, further transformations of the system by the Gauss method are not required. What can be done here is to remove from the third line overall coefficient "-1/7".

Now everything is beautiful. The point is small - write the matrix again in the form of a system of equations and calculate the roots

x + 2y + 4z = 12(1)

7y + 11z = 24 (2)

The algorithm by which the roots will now be found is called the reverse move in the Gauss method. Equation (3) contains the value of z:

y = (24 - 11×(61/9))/7 = -65/9

And the first equation allows you to find x:

x = (12 - 4z - 2y)/1 = 12 - 4x(61/9) - 2x(-65/9) = -6/9 = -2/3

We have the right to call such a system joint, and even definite, that is, having a unique solution. The response is written in the following form:

x 1 \u003d -2/3, y \u003d -65/9, z \u003d 61/9.

An example of an indefinite system

The variant of solving a certain system by the Gauss method has been analyzed, now it is necessary to consider the case if the system is indefinite, that is, infinitely many solutions can be found for it.

x 1 + x 2 + x 3 + x 4 + x 5 = 7 (1)

3x 1 + 2x 2 + x 3 + x 4 - 3x 5 = -2 (2)

x 2 + 2x 3 + 2x 4 + 6x 5 = 23 (3)

5x 1 + 4x 2 + 3x 3 + 3x 4 - x 5 = 12 (4)

The very form of the system is already alarming, because the number of unknowns is n = 5, and the rank of the matrix of the system is already exactly less than this number, because the number of rows is m = 4, that is, the largest order of the square determinant is 4. This means that there are an infinite number of solutions, and it is necessary to look for its general form. The Gauss method for linear equations makes it possible to do this.

First, as usual, the augmented matrix is ​​compiled.

Second line: coefficient k = (-a 21 / a 11) = -3. In the third line, the first element is before the transformations, so you don't need to touch anything, you need to leave it as it is. Fourth line: k = (-a 4 1 /a 11) = -5

Multiplying the elements of the first row by each of their coefficients in turn and adding them to the desired rows, we obtain a matrix of the following form:

As you can see, the second, third and fourth rows consist of elements that are proportional to each other. The second and fourth are generally the same, so one of them can be removed immediately, and the rest multiplied by the coefficient "-1" and get line number 3. And again, leave one of two identical lines.

It turned out such a matrix. The system has not yet been written down, it is necessary here to determine the basic variables - standing at the coefficients a 11 \u003d 1 and a 22 \u003d 1, and free - all the rest.

The second equation has only one basic variable - x 2 . Hence, it can be expressed from there, writing through the variables x 3 , x 4 , x 5 , which are free.

We substitute the resulting expression into the first equation.

It turned out an equation in which the only basic variable is x 1. Let's do the same with it as with x 2 .

All basic variables, of which there are two, are expressed in terms of three free ones, now you can write the answer in a general form.

You can also specify one of the particular solutions of the system. For such cases, as a rule, zeros are chosen as values ​​for free variables. Then the answer will be:

16, 23, 0, 0, 0.

An example of an incompatible system

The solution of inconsistent systems of equations by the Gauss method is the fastest. It ends as soon as at one of the stages an equation is obtained that has no solution. That is, the stage with the calculation of the roots, which is quite long and dreary, disappears. The following system is considered:

x + y - z = 0 (1)

2x - y - z = -2 (2)

4x + y - 3z = 5 (3)

As usual, the matrix is ​​​​compiled:

1	1	-1	0
2	-1	-1	-2
4	1	-3	5

And it is reduced to a stepped form:

k 1 \u003d -2k 2 \u003d -4

1	1	-1	0
0	-3	1	-2
0	0	0	7

After the first transformation, the third line contains an equation of the form

having no solution. Therefore, the system is inconsistent, and the answer is the empty set.

Advantages and disadvantages of the method

If you choose which method to solve SLAE on paper with a pen, then the method that was considered in this article looks the most attractive. In elementary transformations, it is much more difficult to get confused than it happens if you have to manually look for the determinant or some tricky inverse matrix. However, if you use programs for working with data of this type, for example, spreadsheets, then it turns out that such programs already contain algorithms for calculating the main parameters of matrices - determinant, minors, inverse, and so on. And if you are sure that the machine will calculate these values ​​itself and will not make a mistake, it is more expedient to use matrix method or Cramer's formulas, because their application begins and ends with the calculation of determinants and inverse matrices.

Application

Since the Gaussian solution is an algorithm, and the matrix is, in fact, a two-dimensional array, it can be used in programming. But since the article positions itself as a guide "for dummies", it should be said that the easiest place to put the method in is spreadsheets, for example, Excel. Again, any SLAE entered in a table in the form of a matrix will be considered by Excel as a two-dimensional array. And for operations with them, there are many nice commands: addition (you can only add matrices of the same size!), Multiplication by a number, matrix multiplication (also with certain restrictions), finding the inverse and transposed matrices and, most importantly, calculating the determinant. If this time-consuming task is replaced by a single command, it is much faster to determine the rank of a matrix and, therefore, to establish its compatibility or inconsistency.

1. Linear system algebraic equations

1.1 The concept of a system of linear algebraic equations

A system of equations is a condition consisting in the simultaneous execution of several equations in several variables. A system of linear algebraic equations (hereinafter referred to as SLAE) containing m equations and n unknowns is a system of the form:

where the numbers a ij are called the coefficients of the system, the numbers b i are free members, aij and b i(i=1,…, m; b=1,…, n) are some known numbers, and x 1 ,…, x n- unknown. In the notation of the coefficients aij the first index i denotes the number of the equation, and the second index j is the number of the unknown at which this coefficient stands. Subject to finding the number x n . It is convenient to write such a system in a compact matrix form: AX=B. Here A is the matrix of coefficients of the system, called the main matrix;

is a column vector of unknown xj.
is a column vector of free members bi.

The product of matrices A * X is defined, since there are as many columns in matrix A as there are rows in matrix X (n pieces).

The extended matrix of the system is the matrix A of the system, supplemented by a column of free members

1.2 Solution of a system of linear algebraic equations

The solution of a system of equations is an ordered set of numbers (values ​​of variables), when substituting them instead of variables, each of the equations of the system turns into a true equality.

The solution of the system is n values ​​of the unknowns x1=c1, x2=c2,…, xn=cn, substituting which all equations of the system turn into true equalities. Any solution of the system can be written as a matrix-column

A system of equations is called consistent if it has at least one solution, and inconsistent if it has no solutions.

A joint system is called definite if it has a unique solution, and indefinite if it has more than one solution. In the latter case, each of its solutions is called a particular solution of the system. The set of all particular solutions is called the general solution.

To solve a system means to find out whether it is consistent or inconsistent. If the system is compatible, find its general solution.

Two systems are called equivalent (equivalent) if they have the same general solution. In other words, systems are equivalent if every solution to one of them is a solution to the other, and vice versa.

A transformation, the application of which turns a system into a new system equivalent to the original one, is called an equivalent or equivalent transformation. The following transformations can serve as examples of equivalent transformations: swapping two equations of the system, swapping two unknowns together with the coefficients of all equations, multiplying both parts of any equation of the system by a non-zero number.

A system of linear equations is called homogeneous if all free terms are equal to zero:

A homogeneous system is always consistent, since x1=x2=x3=…=xn=0 is a solution to the system. This solution is called null or trivial.

2. Gaussian elimination method

2.1 The essence of the Gaussian elimination method

The classical method for solving systems of linear algebraic equations is the method of successive elimination of unknowns - Gauss method(It is also called the Gaussian elimination method). This is a method of successive elimination of variables, when, with the help of elementary transformations, a system of equations is reduced to an equivalent system of a stepped (or triangular) form, from which all other variables are found sequentially, starting from the last (by number) variables.

The Gaussian solution process consists of two stages: forward and backward moves.

1. Direct move.

At the first stage, the so-called direct move is carried out, when, by means of elementary transformations over rows, the system is brought to a stepped or triangular form, or it is established that the system is inconsistent. Namely, among the elements of the first column of the matrix, a nonzero one is chosen, it is moved to the uppermost position by permuting the rows, and the first row obtained after the permutation is subtracted from the remaining rows, multiplying it by a value equal to the ratio of the first element of each of these rows to the first element of the first row, zeroing thus the column below it.

After the indicated transformations have been made, the first row and the first column are mentally crossed out and continue until a zero-size matrix remains. If at some of the iterations among the elements of the first column there was not found a non-zero one, then go to the next column and perform a similar operation.

At the first stage (forward run), the system is reduced to a stepped (in particular, triangular) form.

The system below is stepwise:

,

The coefficients aii are called the main (leading) elements of the system.

(if a11=0, rearrange the rows of the matrix so that a 11 was not equal to 0. This is always possible, because otherwise the matrix contains a zero column, its determinant is equal to zero and the system is inconsistent).

We transform the system by eliminating the unknown x1 in all equations except the first one (using elementary transformations of the system). To do this, multiply both sides of the first equation by

and add term by term with the second equation of the system (or from the second equation we subtract term by term the first multiplied by ). Then we multiply both parts of the first equation by and add it to the third equation of the system (or subtract the first one multiplied by the third term by term). Thus, we successively multiply the first row by a number and add to i-th line, for i= 2, 3, …,n.

Continuing this process, we get the equivalent system:

– new values ​​of the coefficients for unknowns and free terms in the last m-1 equations of the system, which are determined by the formulas:

Thus, at the first step, all coefficients under the first leading element a 11 are destroyed

0, the second step destroys the elements under the second leading element a 22 (1) (if a 22 (1) 0), and so on. Continuing this process further, we will finally reduce the original system to a triangular system at the (m-1) step.

If, in the process of reducing the system to a stepwise form, zero equations appear, i.e. equalities of the form 0=0, they are discarded. If there is an equation of the form

This indicates the incompatibility of the system.

This completes the direct course of the Gauss method.

2. Reverse move.

At the second stage, the so-called reverse move is carried out, the essence of which is to express all the resulting basic variables in terms of non-basic ones and construct fundamental system solutions, or, if all variables are basic, then express in numerical form the only solution of the system of linear equations.

This procedure begins with the last equation, from which the corresponding basic variable is expressed (it is only one in it) and substituted into the previous equations, and so on, going up the "steps" to the top.

Each line corresponds to exactly one basic variable, so at each step, except for the last (topmost), the situation exactly repeats the case of the last line.

Note: in practice, it is more convenient to work not with the system, but with its extended matrix, performing all elementary transformations on its rows. It is convenient that the coefficient a11 be equal to 1 (rearrange the equations, or divide both sides of the equation by a11).

2.2 Examples of solving SLAE by the Gauss method

In this section, three various examples Let us show how the SLAE can be solved by the Gauss method.

Example 1. Solve SLAE of the 3rd order.

Set the coefficients to zero at

in the second and third lines. To do this, multiply them by 2/3 and 1, respectively, and add them to the first line:

The online calculator finds a solution to the system of linear equations (SLE) by the Gauss method. given detailed solution. To calculate, choose the number of variables and the number of equations. Then enter the data in the cells and click on the "Calculate."

x 1 +x2 +x 3 x 1 +x2 +x 3 x 1 +x2 +x 3 = = =

Number representation:

Integers and (or) Common fractions
Integers and/or Decimals

Number of digits after decimal separator

×

A warning

Clear all cells?

Close Clear

Data entry instruction. Numbers are entered as whole numbers (examples: 487, 5, -7623, etc.), decimal numbers (eg. 67., 102.54, etc.) or fractions. The fraction must be typed in the form a/b, where a and b (b>0) are integers or decimal numbers. Examples 45/5, 6.6/76.4, -7/6.7, etc.

Gauss method

The Gauss method is a method of transition from the original system of linear equations (using equivalent transformations) to a system that is easier to solve than the original system.

The equivalent transformations of the system of linear equations are:

• swapping two equations in the system,
• multiplication of any equation in the system by a non-zero real number,
• adding to one equation another equation multiplied by an arbitrary number.

Consider a system of linear equations:

(1)

We write system (1) in matrix form:

ax=b

(2)

(3)

A is called the coefficient matrix of the system, b − right part restrictions x− vector of variables to be found. Let rank( A)=p.

Equivalent transformations do not change the rank of the coefficient matrix and the rank of the augmented matrix of the system. The set of solutions of the system also does not change under equivalent transformations. The essence of the Gauss method is to bring the matrix of coefficients A to diagonal or stepped.

Let's build the extended matrix of the system:

At the next stage, we reset all elements of column 2, below the element. If the given element is null, then this row is interchanged with the row lying below the given row and having a non-zero element in the second column. Next, we zero out all the elements of column 2 below the leading element a 22. To do this, add rows 3, ... m with row 2 multiplied by − a 32 /a 22 , ..., −a m2 / a 22, respectively. Continuing the procedure, we obtain a matrix of a diagonal or stepped form. Let the resulting augmented matrix look like:

(7)

Because rankA=rank(A|b), then the set of solutions (7) is ( n−p) is a variety. Consequently n−p unknowns can be chosen arbitrarily. The remaining unknowns from system (7) are calculated as follows. From the last equation we express x p through the rest of the variables and insert into the previous expressions. Next, from the penultimate equation, we express x p−1 through the rest of the variables and insert into the previous expressions, etc. Consider the Gauss method on concrete examples.

Examples of solving a system of linear equations using the Gauss method

Example 1. Find the general solution of a system of linear equations using the Gauss method:

Denote by a ij elements i-th line and j-th column.

a eleven . To do this, add rows 2,3 with row 1, multiplied by -2/3, -1/2, respectively:

Matrix record type: ax=b, where

Denote by a ij elements i-th line and j-th column.

Exclude the elements of the 1st column of the matrix below the element a eleven . To do this, add rows 2,3 with row 1, multiplied by -1/5, -6/5, respectively:

We divide each row of the matrix by the corresponding leading element (if the leading element exists):

where x 3 , x

Substituting the upper expressions into the lower ones, we obtain the solution.

Then the vector solution can be represented as follows:

where x 3 , x 4 are arbitrary real numbers.

One of the universal and effective methods for solving linear algebraic systems is Gauss method , consisting in the successive elimination of unknowns.

Recall that the two systems are called equivalent (equivalent) if the sets of their solutions are the same. In other words, systems are equivalent if every solution to one of them is a solution to the other, and vice versa. Equivalent systems are obtained with elementary transformations system equations:

multiplying both sides of the equation by a non-zero number;

adding to some equation the corresponding parts of another equation, multiplied by a number other than zero;

permutation of two equations.

Let the system of equations

The process of solving this system by the Gauss method consists of two stages. At the first stage (forward run), the system is reduced by means of elementary transformations to stepped , or triangular mind, and at the second stage (reverse move) there is a sequential, starting from the last variable, the definition of unknowns from the resulting step system.

Let us assume that the coefficient of this system
, otherwise in the system the first row can be interchanged with any other row so that the coefficient at was different from zero.

Let's transform the system, eliminating the unknown in all equations except the first. To do this, multiply both sides of the first equation by and add term by term with the second equation of the system. Then multiply both sides of the first equation by and add it to the third equation of the system. Continuing this process, we obtain an equivalent system

Here
are the new values ​​of the coefficients and free terms, which are obtained after the first step.

Similarly, considering the main element
, exclude the unknown from all equations of the system, except for the first and second. We continue this process as long as possible, as a result we get a step system

,

where ,
,…,- the main elements of the system
.

If in the process of bringing the system to a step form, equations appear, i.e., equalities of the form
, they are discarded, since any set of numbers satisfies them
. If at
an equation of the form that has no solutions appears, this indicates the inconsistency of the system.

At reverse course the first unknown is expressed from the last equation of the transformed step system through all the other unknowns
who are called free . Then the variable expression from the last equation of the system is substituted into the penultimate equation and the variable is expressed from it
. The variables are defined in a similar way
. Variables
, expressed in terms of free variables, are called basic (dependent). As a result, the general solution of the system of linear equations is obtained.

To find private decision systems, free unknown
in the general solution, arbitrary values ​​are assigned and the values ​​of the variables are calculated
.

It is technically more convenient to subject elementary transformations not to the equations of the system, but to the extended matrix of the system

.

The Gauss method is a universal method that allows you to solve not only square, but also rectangular systems in which the number of unknowns
not equal to the number of equations
.

The advantage of this method also lies in the fact that in the process of solving we simultaneously examine the system for compatibility, since, having reduced the augmented matrix
to the stepped form, it is easy to determine the ranks of the matrix and extended matrix
and apply the Kronecker-Capelli theorem .

Example 2.1 Solve the system using the Gauss method

Decision. Number of Equations
and the number of unknowns
.

Let us compose the extended matrix of the system by assigning to the right of the matrix of coefficients free members column .

Let's bring the matrix to a triangular shape; to do this, we will get "0" below the elements on the main diagonal using elementary transformations.

To get "0" in the second position of the first column, multiply the first row by (-1) and add to the second row.

We write this transformation as a number (-1) against the first line and denote it by an arrow going from the first line to the second line.

To get "0" in the third position of the first column, multiply the first row by (-3) and add to the third row; Let's show this action with an arrow going from the first line to the third.

.

In the resulting matrix, written second in the matrix chain, we get "0" in the second column in the third position. To do this, multiply the second line by (-4) and add to the third. In the resulting matrix, we multiply the second row by (-1), and divide the third row by (-8). All elements of this matrix that lie below the diagonal elements are zeros.

Because , the system is collaborative and specific.

The system of equations corresponding to the last matrix has a triangular form:

From the last (third) equation
. Substitute in the second equation and get
.

Substitute
and
into the first equation, we find


.

Two systems of linear equations are said to be equivalent if the set of all their solutions is the same.

Elementary transformations systems of equations are:

1. Deletion from the system of trivial equations, i.e. those for which all coefficients are equal to zero;
2. Multiplying any equation by a non-zero number;
3. Addition to any i -th equation of any j -th equation, multiplied by any number.

The variable x i is called free if this variable is not allowed, and the whole system of equations is allowed.

Theorem. Elementary transformations transform the system of equations into an equivalent one.

The meaning of the Gauss method is to transform the original system of equations and obtain an equivalent allowed or equivalent inconsistent system.

So, the Gauss method consists of the following steps:

1. Consider the first equation. We choose the first non-zero coefficient and divide the whole equation by it. We obtain an equation in which some variable x i enters with a coefficient of 1;
2. Let us subtract this equation from all the others, multiplying it by numbers such that the coefficients for the variable x i in the remaining equations are set to zero. We get a system that is resolved with respect to the variable x i and is equivalent to the original one;
3. If trivial equations arise (rarely, but it happens; for example, 0 = 0), we delete them from the system. As a result, the equations become one less;
4. We repeat the previous steps no more than n times, where n is the number of equations in the system. Each time we select a new variable for “processing”. If conflicting equations arise (for example, 0 = 8), the system is inconsistent.

As a result, after a few steps we obtain either an allowed system (possibly with free variables) or an inconsistent one. Allowed systems fall into two cases:

1. The number of variables is equal to the number of equations. So the system is defined;
2. Number of variables more number equations. We collect all free variables on the right - we get formulas for allowed variables. These formulas are written in the answer.

That's all! The system of linear equations is solved! This is a fairly simple algorithm, and to master it, you do not need to contact a tutor in mathematics. Consider an example:

A task. Solve the system of equations:

Description of steps:

1. We subtract the first equation from the second and third - we get the allowed variable x 1;
2. We multiply the second equation by (−1), and divide the third equation by (−3) - we get two equations in which the variable x 2 enters with a coefficient of 1;
3. We add the second equation to the first, and subtract from the third. Let's get the allowed variable x 2 ;
4. Finally, we subtract the third equation from the first - we get the allowed variable x 3 ;
5. We have received an authorized system, we write down the answer.

The general solution of the joint system of linear equations is new system, which is equivalent to the original one, in which all allowed variables are expressed in terms of free ones.

When might a general solution be needed? If you have to take fewer steps than k (k is how many equations in total). However, the reasons why the process ends at some step l< k , может быть две:

1. After the l -th step, we get a system that does not contain an equation with the number (l + 1). In fact, this is good, because. the resolved system is received anyway - even a few steps earlier.
2. After the l -th step, an equation is obtained in which all coefficients of the variables are equal to zero, and the free coefficient is different from zero. This is an inconsistent equation, and, therefore, the system is inconsistent.

It is important to understand that the appearance of an inconsistent equation by the Gauss method is a sufficient reason for inconsistency. At the same time, we note that as a result of the l -th step, trivial equations cannot remain - all of them are deleted directly in the process.

Description of steps:

1. Subtract the first equation times 4 from the second. And also add the first equation to the third - we get the allowed variable x 1;
2. We subtract the third equation, multiplied by 2, from the second - we get the contradictory equation 0 = −5.

So, the system is inconsistent, since an inconsistent equation has been found.

A task. Investigate compatibility and find the general solution of the system:

Description of steps:

1. We subtract the first equation from the second (after multiplying by two) and the third - we get the allowed variable x 1;
2. Subtract the second equation from the third. Since all the coefficients in these equations are the same, the third equation becomes trivial. At the same time, we multiply the second equation by (−1);
3. We subtract the second equation from the first equation - we get the allowed variable x 2. The entire system of equations is now also resolved;
4. Since the variables x 3 and x 4 are free, we move them to the right to express the allowed variables. This is the answer.

So, the system is joint and indefinite, since there are two allowed variables (x 1 and x 2) and two free ones (x 3 and x 4).