The material of this article is intended for the first acquaintance with systems of equations. Here we introduce the definition of a system of equations and its solutions, and also consider the most common types of systems of equations. As usual, we will give explanatory examples.

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What is a system of equations?

We will gradually approach the definition of the system of equations. First, let's just say that it is convenient to give it, pointing out two points: firstly, the type of record, and, secondly, the meaning embedded in this record. Let us dwell on them in turn, and then generalize the reasoning into the definition of systems of equations.

Let us have some of them in front of us. For example, let's take two equations 2 x+y=−3 and x=5 . We write them one under the other and unite them with a curly bracket on the left:

Records of this kind, which are several equations arranged in a column and united on the left with a curly bracket, are records of systems of equations.

What do such records mean? They define the set of all such solutions of the equations of the system, which are the solution of each equation.

It does not hurt to describe it in other words. Suppose some solutions of the first equation are solutions of all other equations of the system. And so record of system just also designates them.

Now we are ready to adequately accept the definition of a system of equations.

Definition.

Systems of equations are called records, which are equations located one below the other, united on the left by a curly bracket, which denote the set of all solutions of equations that are simultaneously solutions to each equation of the system.

A similar definition is given in the textbook, but there it is given not for the general case, but for two rational equations in two variables.

Main types

It is clear that there are infinitely many different equations. Naturally, there are also infinitely many systems of equations compiled using them. Therefore, for the convenience of studying and working with systems of equations, it makes sense to divide them into groups according to similar characteristics, and then proceed to consider systems of equations of individual types.

The first subdivision suggests itself by the number of equations included in the system. If there are two equations, then we can say that we have a system of two equations, if there are three, then a system of three equations, etc. It is clear that it makes no sense to talk about a system of one equation, since in this case, in fact, we are dealing with the equation itself, and not with the system.

The next division is based on the number of variables involved in writing the equations of the system. If there is one variable, then we are dealing with a system of equations with one variable (they also say with one unknown), if there are two, then with a system of equations with two variables (with two unknowns), etc. For example, is a system of equations with two variables x and y .

This refers to the number of all different variables involved in the record. They do not have to be all at once included in the record of each equation, it is enough to have them in at least one equation. For example, is a system of equations with three variables x, y, and z. In the first equation, the variable x is present explicitly, while y and z are implicit (we can assume that these variables have zero), and in the second equation, x and z are present, and the variable y is not explicitly represented. In other words, the first equation can be viewed as , and the second as x+0 y−3 z=0 .

The third point in which systems of equations differ is the form of the equations themselves.

At school, the study of systems of equations begins with systems of two linear equations with two variables. That is, such systems constitute two linear equations. Here are a couple of examples: and . On them, the basics of working with systems of equations are learned.

When solving more challenging tasks one can also encounter systems of three linear equations with three unknowns.

Further in the 9th grade, non-linear equations are added to the systems of two equations with two variables, for the most part entire equations of the second degree, less often - of higher degrees. These systems are called systems of nonlinear equations; if necessary, the number of equations and unknowns is specified. Let us show examples of such systems of nonlinear equations: and .

And then in the systems there are also, for example,. They are usually called simply systems of equations, without specifying which equations. Here it is worth noting that most often they simply say “system of equations” about a system of equations, and refinements are added only if necessary.

In high school, as the material is studied, irrational, trigonometric, logarithmic and exponential equations : , , .

If you look even further into the program of the first courses of universities, then the main emphasis is on the study and solution of systems of linear algebraic equations (SLAE), that is, equations, in the left parts of which are polynomials of the first degree, and in the right - some numbers. But there, unlike the school, not two linear equations with two variables are already taken, but an arbitrary number of equations with an arbitrary number of variables, often not coinciding with the number of equations.

What is the solution of a system of equations?

The term “solution of a system of equations” directly refers to systems of equations. The school gives a definition of solving a system of equations with two variables :

Definition.

Solving a system of equations with two variables a pair of values ​​of these variables is called, which turns each equation of the system into the correct one, in other words, which is the solution to each equation of the system.

For example, a pair of variable values ​​x=5 , y=2 (it can be written as (5, 2) ) is a solution to a system of equations by definition, since the equations of the system, when x=5 , y=2 are substituted into them, turn into true numerical equalities 5+2=7 and 5−2=3 respectively. But the pair of values ​​x=3 , y=0 is not a solution to this system, since when these values ​​are substituted into the equations, the first of them will turn into an incorrect equality 3+0=7 .

Similar definitions can be formulated for systems with one variable, as well as for systems with three, four, etc. variables.

Definition.

Solving a system of equations with one variable there will be a variable value that is the root of all the equations of the system, that is, that turns all equations into true numerical equalities.

Let's take an example. Consider a system of equations with one variable t of the form . The number −2 is its solution, since both (−2) 2 =4 and 5·(−2+2)=0 are true numerical equalities. And t=1 is not a solution to the system, since substitution of this value will give two incorrect equalities 1 2 =4 and 5·(1+2)=0 .

Definition.

The solution of a system with three, four, etc. variables called a triple, a quadruple, etc. values ​​of the variables, respectively, which converts all equations of the system into true equalities.

So, by definition, the triple of values ​​of the variables x=1 , y=2 , z=0 is the solution to the system , since 2 1=2 , 5 2=10 and 1+2+0=3 are correct numerical equalities. And (1, 0, 5) is not a solution to this system, since when these values ​​of variables are substituted into the equations of the system, the second of them turns into an incorrect equality 5 0=10 , and the third one is also 1+0+5=3 .

Note that systems of equations may not have solutions, may have a finite number of solutions, for example, one, two, ..., or may have infinitely many solutions. You will see this as you delve deeper into the topic.

Taking into account the definitions of a system of equations and their solutions, we can conclude that the solution of a system of equations is the intersection of the sets of solutions of all its equations.

To conclude, here are a few related definitions:

Definition.

incompatible if it has no solutions, otherwise the system is called joint.

Definition.

The system of equations is called uncertain if it has infinitely many solutions, and certain, if it has a finite number of solutions, or has none at all.

These terms are introduced, for example, in a textbook, but they are rarely used at school, more often they can be heard in higher educational institutions.

Bibliography.

1. Algebra: textbook for 7 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 17th ed. - M. : Education, 2008. - 240 p. : ill. - ISBN 978-5-09-019315-3.
2. Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
3. Mordkovich A. G. Algebra. 7th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 17th ed., add. - M.: Mnemozina, 2013. - 175 p.: ill. ISBN 978-5-346-02432-3.
4. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., Sr. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
5. Mordkovich A. G. Algebra and beginnings mathematical analysis. Grade 11. At 2 pm Part 1. Textbook for students of educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.
6. Algebra and the beginning of the analysis: Proc. for 10-11 cells. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorova.- 14th ed.- M.: Enlightenment, 2004.- 384 p.: ill.- ISBN 5-09-013651-3.
7. A. G. Kurosh. Course of higher algebra.
8. Ilyin V. A., Poznyak E. G. Analytic geometry: Textbook: For universities. – 5th ed. – M.: Science. Fizmatlit, 1999. - 224 p. – (Course of higher mathematics and mathematical physics). – ISBN 5-02-015234 – X (Issue 3)

With this mathematical program, you can solve a system of two linear equations with two variables using the substitution method and the addition method.

The program not only gives the answer to the problem, but also provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.

This program can be useful for high school students in preparation for control work and exams, when testing knowledge before the exam, parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

Rules for Entering Equations

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.

When entering equations you can use brackets. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of the elements.
For example: 6x+1 = 5(x+y)+2

In equations, you can use not only integers, but also fractional numbers in the form of decimal and ordinary fractions.

Rules for entering decimal fractions.
Integer and fractional part decimal fractions can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &

Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)

Solve a system of equations

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A bit of theory.

Solving systems of linear equations. Substitution method

The sequence of actions when solving a system of linear equations by the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression in another equation of the system instead of this variable;

$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$

Let's express from the first equation y through x: y = 7-3x. Substituting the expression 7-3x instead of y into the second equation, we get the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$

It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$

Substituting the number 1 instead of x into the equation y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$

Pair (1;4) - solution of the system

Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.

Solving systems of linear equations by adding

Consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by the substitution method, we pass from a given system to another system equivalent to it, in which one of the equations contains only one variable.

The sequence of actions when solving a system of linear equations by the addition method:
1) multiply the equations of the system term by term, choosing the factors so that the coefficients for one of the variables become opposite numbers;
2) add term by term the left and right parts of the equations of the system;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.

Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$

In the equations of this system, the coefficients of y are opposite numbers. Adding term by term the left and right parts of the equations, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$

From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38 \) we get an equation with the variable y: \(11-3y=38 \). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)

Thus, we found the solution to the system of equations by adding: \(x=11; y=-9 \) or \((11; -9) \)

Taking advantage of the fact that in the equations of the system the coefficients of y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both parts of each of the equations of the original symmeme), in which one of the equations contains only one variable.

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Solve the system with two unknowns - this means finding all pairs of variable values ​​​​that satisfy each of the given equations. Each such pair is called system solution.

Example:
The pair of values ​​\(x=3\);\(y=-1\) is a solution to the first system, because by substituting these triples and minus ones into the system instead of \(x\) and \(y\), both equations become into valid equalities \(\begin(cases)3-2\cdot (-1)=5 \\3 \cdot 3+2 \cdot (-1)=7 \end(cases)\)

But \(x=1\); \(y=-2\) - is not a solution to the first system, because after substitution the second equation "does not converge" \(\begin(cases)1-2\cdot(-2)=5 \\3\cdot1+2 \cdot(-2)≠7 \end(cases)\)

Note that such pairs are often written shorter: instead of "\(x=3\); \(y=-1\)" they write like this: \((3;-1)\).

How to solve a system of linear equations?

There are three main ways to solve systems of linear equations:

1. Substitution method.

\(\begin(cases)x-2y=5\\3x+2y=7 \end(cases)\)\(\Leftrightarrow\) \(\begin(cases)x=5+2y\\3x+2y= 7\end(cases)\)\(\Leftrightarrow\)

Substitute the resulting expression instead of this variable into another equation of the system.

\(\Leftrightarrow\) \(\begin(cases)x=5+2y\\3(5+2y)+2y=7\end(cases)\)\(\Leftrightarrow\)

1. \(\begin(cases)13x+9y=17\\12x-2y=26\end(cases)\)

   In the second equation, each term is even, so we simplify the equation by dividing it by \(2\).
   

   \(\begin(cases)13x+9y=17\\6x-y=13\end(cases)\)

   This system can be solved in any of the ways, but it seems to me that the substitution method is the most convenient here. Let's express y from the second equation.
   

   \(\begin(cases)13x+9y=17\\y=6x-13\end(cases)\)

   Substitute \(6x-13\) for \(y\) in the first equation.
   

   \(\begin(cases)13x+9(6x-13)=17\\y=6x-13\end(cases)\)

   The first equation has become normal. We solve it.

   Let's open the parentheses first.
   

   \(\begin(cases)13x+54x-117=17\\y=6x-13\end(cases)\)

   Let's move \(117\) to the right and give like terms.

   \(\begin(cases)67x=134\\y=6x-13\end(cases)\)

   Divide both sides of the first equation by \(67\).

   \(\begin(cases)x=2\\y=6x-13\end(cases)\)

   Hooray, we found \(x\)! Substitute its value into the second equation and find \(y\).

   \(\begin(cases)x=2\\y=12-13\end(cases)\)\(\Leftrightarrow\)\(\begin(cases)x=2\\y=-1\end(cases )\)

   Let's write down the answer.

Let us first recall the definition of a solution to a system of equations in two variables.

Definition 1

A pair of numbers is called a solution to a system of equations with two variables if, when they are substituted into the equation, the correct equality is obtained.

In what follows, we will consider systems of two equations with two variables.

Exist four basic ways to solve systems of equations: substitution method, addition method, graphical method, new variable management method. Let's take a look at these methods concrete examples. To describe the principle of using the first three methods, we will consider a system of two linear equations with two unknowns:

Substitution method

The substitution method is as follows: any of these equations is taken and $y$ is expressed in terms of $x$, then $y$ is substituted into the equation of the system, from where the variable $x.$ is found. After that, we can easily calculate the variable $y.$

Example 1

Let us express from the second equation $y$ in terms of $x$:

Substitute in the first equation, find $x$:

\ \ \

Find $y$:

Answer: $(-2,\ 3)$

Addition method.

Consider this method with an example:

Example 2

\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]

Multiply the second equation by 3, we get:

\[\left\( \begin(array)(c) (2x+3y=5) \\ (9x-3y=-27) \end(array) \right.\]

Now let's add both equations together:

\ \ \

Find $y$ from the second equation:

\[-6-y=-9\] \

Answer: $(-2,\ 3)$

Remark 1

Note that in this method it is necessary to multiply one or both equations by such numbers that when adding one of the variables "disappears".

Graphical way

The graphical method is as follows: both equations of the system are displayed on the coordinate plane and the point of their intersection is found.

Example 3

\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]

Let us express $y$ from both equations in terms of $x$:

\[\left\( \begin(array)(c) (y=\frac(5-2x)(3)) \\ (y=3x+9) \end(array) \right.\]

Let's draw both graphs on the same plane:

Picture 1.

Answer: $(-2,\ 3)$

How to introduce new variables

We will consider this method in the following example:

Example 4

\[\left\( \begin(array)(c) (2^(x+1)-3^y=-1) \\ (3^y-2^x=2) \end(array) \right .\]

Decision.

This system is equivalent to the system

\[\left\( \begin(array)(c) ((2\cdot 2)^x-3^y=-1) \\ (3^y-2^x=2) \end(array) \ right.\]

Let $2^x=u\ (u>0)$ and $3^y=v\ (v>0)$, we get:

\[\left\( \begin(array)(c) (2u-v=-1) \\ (v-u=2) \end(array) \right.\]

We solve the resulting system by the addition method. Let's add the equations:

\ \

Then from the second equation, we get that

Returning to the substitution, we get new system exponential equations:

\[\left\( \begin(array)(c) (2^x=1) \\ (3^y=3) \end(array) \right.\]

We get:

\[\left\( \begin(array)(c) (x=0) \\ (y=1) \end(array) \right.\]

Let us first consider the case when the number of equations is equal to the number of variables, i.e. m = n. Then the matrix of the system is square, and its determinant is called the determinant of the system.

Inverse matrix method

Consider in general terms the system of equations AX = B with a nonsingular square matrix A. In this case, there exists inverse matrix A -1 . Let's multiply both sides by A -1 on the left. We get A -1 AX \u003d A -1 B. From here EX \u003d A -1 B and

The last equality is a matrix formula for finding solutions to such systems of equations. The use of this formula is called the inverse matrix method

For example, let's use this method to solve the following system:

;

At the end of the solution of the system, a check can be made by substituting the found values ​​into the equations of the system. In this case, they must turn into true equalities.

For this example, let's check:

Method for solving systems of linear equations with a square matrix using Cramer's formulas

Let n=2:

If both parts of the first equation are multiplied by a 22, and both parts of the second by (-a 12), and then the resulting equations are added, then we will exclude the variable x 2 from the system. Similarly, you can eliminate the variable x 1 (by multiplying both sides of the first equation by (-a 21) and both sides of the second by a 11). As a result, we get the system:

The expression in brackets is the determinant of the system

Denote

Then the system will take the form:

It follows from the resulting system that if the determinant of the system is 0, then the system will be consistent and definite. Its unique solution can be calculated by the formulas:

If = 0, a 1 0 and/or  2 0, then the equations of the system will take the form 0*х 1 = 2 and/or 0*х 1 = 2. In this case, the system will be inconsistent.

In the case when = 1 = 2 = 0, the system will be consistent and indefinite (it will have an infinite number of solutions), as it will take the form:

Cramer's theorem(we omit the proof). If the determinant of the matrix of the system n of equations  is not equal to zero, then the system has a unique solution, determined by the formulas:

,

where  j is the determinant of the matrix obtained from the matrix A by replacing the j-th column with a column of free terms.

The above formulas are called Cramer's formulas.

As an example, we will use this method to solve a system that was previously solved using the inverse matrix method:

Disadvantages of the considered methods:

1) significant complexity (calculation of determinants and finding the inverse matrix);

2) limited scope (for systems with a square matrix).

Real economic situations are often modeled by systems in which the number of equations and variables is quite significant, and there are more equations than variables. Therefore, the following method is more common in practice.

Gauss method (method of successive elimination of variables)

This method is used to solve a system of m linear equations with n variables in general view. Its essence lies in applying a system of equivalent transformations to the expanded matrix, with the help of which the system of equations is transformed to the form when its solutions become easy to find (if any).

This is the view in which the left top part system matrix will be a step matrix. This is achieved using the same techniques that were used to obtain a stepped matrix in order to determine the rank. In this case, elementary transformations are applied to the expanded matrix, which will allow one to obtain an equivalent system of equations. After that, the augmented matrix will take the form:

Obtaining such a matrix is ​​called in a straight line Gauss method.

Finding the values ​​of variables from the corresponding system of equations is called backwards Gauss method. Let's consider it.

Note that the last (m – r) equations will take the form:

If at least one of the numbers
is not equal to zero, then the corresponding equality will be false, and the whole system will be inconsistent.

Therefore, for any joint system
. In this case, the last (m – r) equations for any values ​​of the variables will be identities 0 = 0, and they can be ignored when solving the system (just discard the corresponding rows).

After that, the system will look like:

Consider first the case when r=n. Then the system will take the form:

From the last equation of the system one can uniquely find x r .

Knowing x r , one can uniquely express x r -1 from it. Then from the previous equation, knowing x r and x r -1 , we can express x r -2 and so on. up to x 1 .

So, in this case, the system will be collaborative and definite.

Now consider the case when r basic(basic), and all the rest - non-basic(minor, free). The last equation of the system will look like:

From this equation, we can express the basic variable x r in terms of non-basic ones:

The penultimate equation will look like:

By substituting the resulting expression instead of x r, it will be possible to express the basic variable x r -1 through non-basic ones. Etc. to variable x 1 . To obtain a solution to the system, you can equate non-basic variables to arbitrary values ​​and then calculate the basic variables using the obtained formulas. Thus, in this case, the system will be consistent and indeterminate (having an infinite number of solutions).

For example, let's solve the system of equations:

The set of basic variables will be called basis systems. The set of columns of coefficients for them will also be called basis(basic columns), or basic minor system matrices. That solution of the system, in which all non-basic variables are equal to zero, will be called basic solution.

In the previous example, the basic solution will be (4/5; -17/5; 0; 0) (variables x 3 and x 4 (c 1 and c 2) are set to zero, and the basic variables x 1 and x 2 are calculated through them) . To give an example of a non-basic solution, it is necessary to equate x 3 and x 4 (c 1 and c 2) to arbitrary numbers that are not equal to zero at the same time, and calculate the rest of the variables through them. For example, with c 1 = 1 and c 2 = 0, we get a non-basic solution - (4/5; -12/5; 1; 0). By substitution, it is easy to verify that both solutions are correct.

Obviously, in an indefinite system of non-basic solutions, there can be an infinite number of solutions. How many basic solutions can there be? Each row of the transformed matrix must correspond to one basic variable. In total, there are n variables in the problem, and r basic rows. Therefore, the number of possible sets of basic variables cannot exceed the number of combinations from n to 2 . It may be less than , because it is not always possible to transform the system to such a form that this particular set of variables is the basis.

What kind is this? This is such a form when the matrix formed from the columns of the coefficients for these variables will be stepwise and, in this case, will consist of rrows. Those. the rank of the matrix of coefficients for these variables must be equal to r. It cannot be larger, since the number of columns is equal to r. If it turns out to be less than r, then this indicates a linear dependence of the columns with variables. Such columns cannot form a basis.

Let us consider what other basic solutions can be found in the above example. To do this, consider all possible combinations of four variables with two basic ones. Such combinations will
, and one of them (x 1 and x 2) has already been considered.

Let's take variables x 1 and x 3 . Find the rank of the matrix of coefficients for them:

Since it is equal to two, they can be basic. We equate the non-basic variables x 2 and x 4 to zero: x 2 \u003d x 4 \u003d 0. Then from the formula x 1 \u003d 4/5 - (1/5) * x 4 it follows that x 1 \u003d 4/5, and from the formula x 2 \u003d -17/5 + x 3 - - (7/5) * x 4 \u003d -17/5 + x 3 it follows that x 3 \u003d x 2 + 17/5 \u003d 17/5. Thus, we get the basic solution (4/5; 0; 17/5; 0).

Similarly, you can get basic solutions for the basic variables x 1 and x 4 - (9/7; 0; 0; -17/7); x 2 and x 4 - (0; -9; 0; 4); x 3 and x 4 - (0; 0; 9; 4).

The variables x 2 and x 3 in this example cannot be taken as basic ones, since the rank of the corresponding matrix is ​​equal to one, i.e. less than two:

.

Another approach is possible to determine whether or not it is possible to form a basis from some variables. When solving the example, as a result of transforming the system matrix to a stepped form, it took the form:

By choosing pairs of variables, it was possible to calculate the corresponding minors of this matrix. It is easy to see that for all pairs, except for x 2 and x 3 , they are not equal to zero, i.e. the columns are linearly independent. And only for columns with variables x 2 and x 3
, which indicates their linear dependence.

Let's consider one more example. Let's solve the system of equations

So, the equation corresponding to the third row of the last matrix is ​​inconsistent - it led to the wrong equality 0 = -1, therefore, this system is inconsistent.

Jordan-Gauss method 3 is a development of the Gaussian method. Its essence is that the extended matrix of the system is transformed to the form when the coefficients of the variables form an identity matrix up to a permutation of rows or columns 4 (where is the rank of the system matrix).

Let's solve the system using this method:

Consider the augmented matrix of the system:

In this matrix, we select the identity element. For example, the coefficient at x 2 in the third constraint is 5. Let's make sure that in the remaining rows in this column there are zeros, i.e. make the column single. In the process of transformations, we will call this columnpermissive(leading, key). The third constraint (the third string) will also be called permissive. Myself element, which stands at the intersection of the allowing row and column (here it is a unit), is also called permissive.

The first line now contains the coefficient (-1). To get zero in its place, multiply the third row by (-1) and subtract the result from the first row (i.e. just add the first row to the third).

The second line contains a coefficient of 2. To get zero in its place, multiply the third line by 2 and subtract the result from the first line.

The result of the transformations will look like:

This matrix clearly shows that one of the first two constraints can be deleted (the corresponding rows are proportional, i.e. these equations follow from each other). Let's cross out the second one:

So, there are two equations in the new system. A single column (second) is received, and the unit here is in the second row. Let's remember that the basic variable x 2 will correspond to the second equation of the new system.

Let's choose a basic variable for the first row. It can be any variable except x 3 (because at x 3 the first constraint has a zero coefficient, i.e. the set of variables x 2 and x 3 cannot be basic here). You can take the first or fourth variable.

Let's choose x 1. Then the resolving element will be 5, and both parts of the resolving equation will have to be divided by five in order to get one in the first column of the first row.

Let's make sure that the rest of the rows (i.e., the second row) have zeros in the first column. Since now the second line contains not zero, but 3, it is necessary to subtract from the second line the elements of the converted first line, multiplied by 3:

One basic solution can be directly extracted from the resulting matrix by equating the non-basic variables to zero, and the basic variables to the free terms in the corresponding equations: (0.8; -3.4; 0; 0). You can also derive general formulas expressing basic variables through non-basic ones: x 1 \u003d 0.8 - 1.2 x 4; x 2 \u003d -3.4 + x 3 + 1.6x 4. These formulas describe the entire infinite set of solutions to the system (by equating x 3 and x 4 to arbitrary numbers, you can calculate x 1 and x 2).

Note that the essence of the transformations at each stage of the Jordan-Gauss method was as follows:

1) the permissive string was divided by the permissive element to get a unit in its place,

2) from all other rows, the transformed resolving power multiplied by the element that was in the given line in the resolving column was subtracted to get zero in place of this element.

Consider once again the transformed augmented matrix of the system:

It can be seen from this entry that the rank of the matrix of system A is r.

In the course of the above reasoning, we have established that the system is consistent if and only if
. This means that the augmented matrix of the system will look like:

Discarding zero rows, we get that the rank of the extended matrix of the system is also equal to r.

Kronecker-Capelli theorem. A system of linear equations is consistent if and only if the rank of the matrix of the system is equal to the rank of the extended matrix of this system.

Recall that the rank of a matrix is ​​equal to the maximum number of its linearly independent rows. It follows from this that if the rank of the extended matrix is ​​less than the number of equations, then the equations of the system are linearly dependent, and one or more of them can be excluded from the system (because they are a linear combination of the others). The system of equations will be linearly independent only if the rank of the extended matrix is ​​equal to the number of equations.

Moreover, for consistent systems of linear equations, it can be argued that if the rank of the matrix is ​​equal to the number of variables, then the system has a unique solution, and if it is less than the number of variables, then the system is indefinite and has infinitely many solutions.

1For example, suppose there are five rows in the matrix (the initial row order is 12345). We need to change the second line and the fifth. In order for the second line to fall into the place of the fifth, to “move” down, we sequentially change the adjacent lines three times: the second and third (13245), the second and fourth (13425) and the second and fifth (13452). Then, in order for the fifth row to take the place of the second in the original matrix, it is necessary to “shift” the fifth row up by only two consecutive changes: the fifth and fourth rows (13542) and the fifth and third (15342).

2Number of combinations from n to r the number of all different r-element subsets of an n-element set is called (different sets are those that have a different composition of elements, the selection order is not important). It is calculated by the formula:
. Recall the meaning of the sign “!” (factorial):
0!=1.)

3Since this method is more common than the Gaussian method discussed earlier, and in essence is a combination of the forward and reverse Gaussian method, it is also sometimes called the Gaussian method, omitting the first part of the name.

4For example,
.

5If there were no units in the matrix of the system, then it would be possible, for example, to divide both parts of the first equation by two, and then the first coefficient would become unity; or the like.