Calculation of matrices by Cramer's method. Solve a system of equations using the Cramer, Gauss and inverse matrix methods. Solving systems of linear equations using the ordinary Jordan elimination method

Let the system of linear equations contain as many equations as the number of independent variables, i.e. looks like

Such systems linear equations are called square. The determinant, composed of coefficients for the independent variables of the system (1.5), is called the main determinant of the system. We will denote it Greek letter D. So

. (1.6)

If the main determinant contains an arbitrary ( j th) column, replace with a column of free terms of system (1.5), then you can get n auxiliary qualifiers:

(j = 1, 2, …, n). (1.7)

Cramer's Rule solving quadratic systems of linear equations is as follows. If the main determinant D of system (1.5) is different from zero, then the system has a unique solution, which can be found using the formulas:

(1.8)

Example 1.5. Solve the system of equations using Cramer's method

.

Let us calculate the main determinant of the system:

Since D¹0, the system has a unique solution, which can be found using formulas (1.8):

Thus,

Actions on matrices

1. Multiplying a matrix by a number. The operation of multiplying a matrix by a number is defined as follows.

2. In order to multiply a matrix by a number, you need to multiply all its elements by this number. That is

. (1.9)

Example 1.6. .

Matrix addition.

This operation is introduced only for matrices of the same order.

In order to add two matrices, it is necessary to add the corresponding elements of another matrix to the elements of one matrix:

(1.10)
The operation of matrix addition has the properties of associativity and commutativity.

Example 1.7. .

Matrix multiplication.

If the number of matrix columns A coincides with the number of matrix rows IN, then for such matrices the multiplication operation is introduced:

2

Thus, when multiplying a matrix A dimensions m´ n to the matrix IN dimensions n´ k we get a matrix WITH dimensions m´ k. In this case, the matrix elements WITH are calculated using the following formulas:

Problem 1.8. Find, if possible, the product of matrices AB And B.A.:

Solution. 1) In order to find a work AB, you need matrix rows A multiply by matrix columns B:

2) Work B.A. does not exist, because the number of matrix columns B does not match the number of matrix rows A.

Inverse matrix. Solving systems of linear equations using the matrix method

Matrix A- 1 is called the inverse of a square matrix A, if the equality is satisfied:

where through I denotes the identity matrix of the same order as the matrix A:

.

In order for a square matrix to have an inverse, it is necessary and sufficient that its determinant be different from zero. The inverse matrix is ​​found using the formula:

, (1.13)

Where A ij- algebraic additions to elements a ij matrices A(note that algebraic additions to matrix rows A are located in the inverse matrix in the form of corresponding columns).

Example 1.9. Find the inverse matrix A- 1 to matrix

.

We find the inverse matrix using formula (1.13), which for the case n= 3 has the form:

.

Let's find det A = | A| = 1 × 3 × 8 + 2 × 5 × 3 + 2 × 4 × 3 - 3 × 3 × 3 - 1 × 5 × 4 - 2 × 2 × 8 = 24 + 30 + 24 - 27 - 20 - 32 = - 1. Since the determinant of the original matrix is ​​nonzero, the inverse matrix exists.

1) Find algebraic complements A ij:

For the convenience of finding the inverse matrix, we have placed the algebraic additions to the rows of the original matrix in the corresponding columns.

From the obtained algebraic additions we compose a new matrix and divide it by the determinant det A. Thus, we get the inverse matrix:

Quadratic systems of linear equations with a nonzero principal determinant can be solved using the inverse matrix. To do this, system (1.5) is written in matrix form:

Where

Multiplying both sides of equality (1.14) from the left by A- 1, we get the solution to the system:

, where

Thus, in order to find a solution to a square system, you need to find the inverse matrix of the main matrix of the system and multiply it on the right by the column matrix of free terms.

Problem 1.10. Solve a system of linear equations

using the inverse matrix.

Solution. Let us write the system in matrix form: ,

Where - the main matrix of the system, - the column of unknowns and - the column of free terms. Since the main determinant of the system , then the main matrix of the system A has an inverse matrix A-1 . To find the inverse matrix A-1 , we calculate the algebraic complements to all elements of the matrix A:

From the obtained numbers we will compose a matrix (and algebraic additions to the rows of the matrix A write it in the appropriate columns) and divide it by the determinant D. Thus, we have found the inverse matrix:

We find the solution to the system using formula (1.15):

Thus,

Solving systems of linear equations using the ordinary Jordan elimination method

Let an arbitrary (not necessarily quadratic) system of linear equations be given:

(1.16)

It is required to find a solution to the system, i.e. such a set of variables that satisfies all the equalities of system (1.16). IN general case system (1.16) can have not only one solution, but also countless solutions. It may also have no solutions at all.

When solving such problems, the well-known school course method of eliminating unknowns is used, which is also called the ordinary Jordan elimination method. The essence of this method is that in one of the equations of system (1.16) one of the variables is expressed in terms of other variables. This variable is then substituted into other equations in the system. The result is a system containing one equation and one variable less than the original system. The equation from which the variable was expressed is remembered.

This process is repeated until one last equation remains in the system. Through the process of eliminating unknowns, some equations may become true identities, e.g. Such equations are excluded from the system, since they are satisfied for any values ​​of the variables and, therefore, do not affect the solution of the system. If, in the process of eliminating unknowns, at least one equation becomes an equality that cannot be satisfied for any values ​​of the variables (for example), then we conclude that the system has no solution.

If no contradictory equations arise during the solution, then one of the remaining variables in it is found from the last equation. If there is only one variable left in the last equation, then it is expressed as a number. If other variables remain in the last equation, then they are considered parameters, and the variable expressed through them will be a function of these parameters. Then the so-called “reverse move” takes place. The found variable is substituted into the last remembered equation and the second variable is found. Then the two found variables are substituted into the penultimate memorized equation and the third variable is found, and so on, up to the first memorized equation.

As a result, we obtain a solution to the system. This decision will be unique if the found variables are numbers. If the first variable found, and then all the others, depend on the parameters, then the system will have an infinite number of solutions (each set of parameters corresponds to a new solution). Formulas that allow you to find a solution to a system depending on a particular set of parameters are called the general solution of the system.

Example 1.11.

x

After memorizing the first equation and bringing similar terms in the second and third equations we arrive at the system:

Let's express y from the second equation and substitute it into the first equation:

Let us remember the second equation, and from the first we find z:

Working backwards, we consistently find y And z. To do this, we first substitute into the last remembered equation, from where we find y:

.

Then we’ll substitute it into the first memorized equation where we can find it x:

Problem 1.12. Solve a system of linear equations by eliminating unknowns:

. (1.17)

Solution. Let us express the variable from the first equation x and substitute it into the second and third equations:

.

Let's remember the first equation

In this system, the first and second equations contradict each other. Indeed, expressing y , we get that 14 = 17. This equality does not hold for any values ​​of the variables x, y, And z. Consequently, system (1.17) is inconsistent, i.e. has no solution.

We invite readers to check for themselves that the main determinant of the original system (1.17) is equal to zero.

Let us consider a system that differs from system (1.17) by only one free term.

Problem 1.13. Solve a system of linear equations by eliminating unknowns:

. (1.18)

Solution. As before, we express the variable from the first equation x and substitute it into the second and third equations:

.

Let's remember the first equation and present similar terms in the second and third equations. We arrive at the system:

Expressing y from the first equation and substituting it into the second equation , we get the identity 14 = 14, which does not affect the solution of the system, and, therefore, it can be excluded from the system.

In the last remembered equality, the variable z we will consider it a parameter. We believe. Then

Let's substitute y And z into the first remembered equality and find x:

.

Thus, system (1.18) has an infinite number of solutions, and any solution can be found using formulas (1.19), choosing an arbitrary value of the parameter t:

(1.19)
So the solutions of the system, for example, are the following sets of variables (1; 2; 0), (2; 26; 14), etc. Formulas (1.19) express the general (any) solution of the system (1.18).

In the case when the original system (1.16) has sufficient a large number of equations and unknowns, the indicated method of ordinary Jordan elimination seems cumbersome. However, it is not. It is enough to derive an algorithm for recalculating the system coefficients at one step in general view and formulate the solution to the problem in the form of special Jordan tables.

Let a system of linear forms (equations) be given:

, (1.20)
Where x j- independent (sought) variables, a ij- constant odds
(i = 1, 2,…, m; j = 1, 2,…, n). Right parts of the system y i (i = 1, 2,…, m) can be either variables (dependent) or constants. It is required to find solutions to this system by eliminating the unknowns.

Let us consider the following operation, henceforth called “one step of ordinary Jordan eliminations”. From arbitrary ( r th) equality we express an arbitrary variable ( xs) and substitute into all other equalities. Of course, this is only possible if a rs¹ 0. Coefficient a rs called the resolving (sometimes guiding or main) element.

We will get the following system:

. (1.21)

From s- equality of system (1.21), we subsequently find the variable xs(after the remaining variables have been found). S The -th line is remembered and subsequently excluded from the system. The remaining system will contain one equation and one less independent variable than the original system.

Let us calculate the coefficients of the resulting system (1.21) through the coefficients of the original system (1.20). Let's start with r th equation, which after expressing the variable xs through the remaining variables it will look like this:

Thus, the new coefficients r th equations are calculated using the following formulas:

(1.23)
Let us now calculate the new coefficients b ij(i¹ r) of an arbitrary equation. To do this, let us substitute the variable expressed in (1.22) xs V i th equation of system (1.20):

After bringing similar terms, we get:

(1.24)
From equality (1.24) we obtain formulas by which the remaining coefficients of system (1.21) are calculated (with the exception r th equation):

(1.25)
The transformation of systems of linear equations by the method of ordinary Jordan elimination is presented in the form of tables (matrices). These tables are called “Jordan tables”.

Thus, problem (1.20) is associated with the following Jordan table:

Table 1.1

	x 1	x 2	…	x j	…	xs	…	x n
y 1 =	a 11	a 12		a 1j		a 1s		a 1n
…………………………………………………………………..
y i=	a i 1	a i 2		a ij		a is		a in
…………………………………………………………………..
y r=	a r 1	a r 2		a rj		a rs		arn
………………………………………………………………….
y n=	a m 1	a m 2		a mj		a ms		a mn

Jordan table 1.1 contains a left header column in which the right parts of the system (1.20) are written and an upper header row in which independent variables are written.

The remaining elements of the table form the main matrix of coefficients of system (1.20). If you multiply the matrix A to the matrix consisting of the elements of the top title row, you get a matrix consisting of the elements of the left title column. That is, essentially, the Jordan table is a matrix form of writing a system of linear equations: . System (1.21) corresponds to the following Jordan table:

Table 1.2

	x 1	x 2	…	x j	…	y r	…	x n
y 1 =	b 11	b 12		b 1 j		b 1 s		b 1 n
…………………………………………………………………..
y i =	b i 1	b i 2		b ij		b is		b in
…………………………………………………………………..
x s =	b r 1	b r 2		b rj		b rs		brn
………………………………………………………………….
y n =	b m 1	b m 2		b mj		bms		b mn

Permissive element a rs We will highlight them in bold. Recall that to implement one step of Jordan elimination, the resolving element must be non-zero. The table row containing the enabling element is called the enabling row. The column containing the enable element is called the enable column. When moving from a given table to the next table, one variable ( xs) from the top header row of the table is moved to the left header column and, conversely, one of the free members of the system ( y r) moves from the left head column of the table to the top head row.

Let us describe the algorithm for recalculating the coefficients when moving from the Jordan table (1.1) to the table (1.2), which follows from formulas (1.23) and (1.25).

1. The resolving element is replaced by the inverse number:

2. The remaining elements of the resolving string are divided into the resolving element and change the sign to the opposite:

3. The remaining elements of the resolution column are divided into the resolution element:

4. Elements that are not included in the allowing row and allowing column are recalculated using the formulas:

The last formula is easy to remember if you notice that the elements that make up the fraction , are at the intersection i-oh and r th lines and j th and s th columns (resolving row, resolving column, and the row and column at the intersection of which the recalculated element is located). More precisely, when memorizing the formula you can use the following diagram:

-21 -26 -13 -37

When performing the first step of Jordan exceptions, you can select any element of Table 1.3 located in the columns as a resolving element x 1 ,…, x 5 (all specified elements are not zero). Just don't select the enabling element in the last column, because you need to find independent variables x 1 ,…, x 5 . For example, we choose the coefficient 1 with variable x 3 in the third line of Table 1.3 (the enabling element is shown in bold). When moving to table 1.4, the variable x The 3 from the top header row is swapped with the constant 0 of the left header column (third row). In this case, the variable x 3 is expressed through the remaining variables.

String x 3 (Table 1.4) can, after remembering in advance, be excluded from Table 1.4. The third column with a zero in the top title line is also excluded from Table 1.4. The point is that regardless of the coefficients of a given column b i 3 all corresponding terms of each equation 0 b i 3 systems will be equal to zero. Therefore, these coefficients need not be calculated. Eliminating one variable x 3 and remembering one of the equations, we arrive at a system corresponding to Table 1.4 (with the line crossed out x 3). Selecting in table 1.4 as a resolving element b 14 = -5, go to table 1.5. In Table 1.5, remember the first row and exclude it from the table along with the fourth column (with a zero at the top).

Table 1.5 Table 1.6

From the last table 1.7 we find: x 1 = - 3 + 2x 5 .

Consistently substituting the already found variables into the remembered lines, we find the remaining variables:

Thus, the system has infinitely many solutions. Variable x 5, arbitrary values ​​can be assigned. This variable acts as a parameter x 5 = t. We proved the compatibility of the system and found its general solution:

x 1 = - 3 + 2t

x 2 = - 1 - 3t

x 3 = - 2 + 4t . (1.27)
x 4 = 4 + 5t

x 5 = t

Giving parameter t different meanings, we will obtain an infinite number of solutions to the original system. So, for example, the solution to the system is the following set of variables (- 3; - 1; - 2; 4; 0).

Gabriel Kramer is a Swiss mathematician, student and friend of Johann Bernoulli, one of the creators of linear algebra. Cramer considered a system of an arbitrary number of linear equations with a square matrix. He presented the solution to the system as a column of fractions with a common denominator - the determinant of the matrix. Cramer's method is based on the use of determinants in solving systems of linear equations, which significantly speeds up the solution process. This method can be used to solve a system of as many linear equations as there are unknowns in each equation. The main thing is that the determinant of the system is not equal to “0”, then Cramer’s method can be used in the solution, if “0” - this method cannot be used. This method can also be used to solve systems of linear equations with a unique solution.

Cramer's theorem. If the determinant of the system is nonzero, then the system of linear equations has one unique solution, and the unknown is equal to the ratio of the determinants. The denominator contains the determinant of the system, and the numerator contains the determinant obtained from the determinant of the system by replacing the coefficients of this unknown with free terms. This theorem holds for a system of linear equations of any order.

Suppose we are given a SLAE of this type:

\[\left\(\begin(matrix) 3x_1 + 2x_2 =1\\ x_1 + 4x_2 = -3 \end(matrix)\right.\]

According to Cramer's theorem we get:

Answer: \

Where can I solve an equation using Cramer's method using an online solver?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.

With the same number of equations as the number of unknowns with the main determinant of the matrix, which is not equal to zero, the coefficients of the system (for such equations there is a solution and there is only one).

Cramer's theorem.

When the determinant of the matrix of a square system is non-zero, it means that the system is consistent and it has one solution and it can be found by Cramer's formulas:

where Δ - determinant of the system matrix,

Δ i is the determinant of the system matrix, in which instead of i The th column contains the column of right sides.

When the determinant of a system is zero, it means that the system can become cooperative or incompatible.

This method is usually used for small systems with extensive calculations and if it is necessary to determine one of the unknowns. The complexity of the method is that many determinants need to be calculated.

Description of the Cramer method.

There is a system of equations:

A system of 3 equations can be solved using the Cramer method, which was discussed above for a system of 2 equations.

We compose a determinant from the coefficients of the unknowns:

It will be system determinant. When D≠0, which means the system is consistent. Now let's create 3 additional determinants:

,,

We solve the system by Cramer's formulas:

Examples of solving systems of equations using Cramer's method.

Example 1.

Given system:

Let's solve it using Cramer's method.

First you need to calculate the determinant of the system matrix:

Because Δ≠0, which means that from Cramer’s theorem the system is consistent and it has one solution. We calculate additional determinants. The determinant Δ 1 is obtained from the determinant Δ by replacing its first column with a column of free coefficients. We get:

In the same way, we obtain the determinant of Δ 2 from the determinant of the system matrix by replacing the second column with a column of free coefficients:

In the first part, we looked at some theoretical material, the substitution method, as well as the method of term-by-term addition of system equations. I recommend everyone who accessed the site through this page to read the first part. Perhaps some visitors will find the material too simple, but in the process of solving systems of linear equations I made a number of very important comments and conclusions concerning the solution of mathematical problems in general.

Now we will analyze Cramer’s rule, as well as solving a system of linear equations using an inverse matrix (matrix method). All materials are presented simply, in detail and clearly; almost all readers will be able to learn how to solve systems using the above methods.

First, we will take a closer look at Cramer's rule for a system of two linear equations in two unknowns. For what? – After all, the simplest system can be solved using the school method, the method of term-by-term addition!

The fact is that, albeit sometimes, such a task occurs - to solve a system of two linear equations with two unknowns using Cramer's formulas. Secondly, a simpler example will help you understand how to use Cramer's rule for a more complex case - a system of three equations with three unknowns.

In addition, there are systems of linear equations with two variables, which are advisable to solve using Cramer’s rule!

Consider the system of equations

At the first step, we calculate the determinant, it is called main determinant of the system.

Gauss method.

If , then the system has a unique solution, and to find the roots we must calculate two more determinants:
And

In practice, the above qualifiers can also be denoted Latin letter.

We find the roots of the equation using the formulas:
,

Example 7

Solve a system of linear equations

Solution: We see that the coefficients of the equation are quite large; on the right side there are decimal fractions with a comma. The comma is a rather rare guest in practical tasks in mathematics, I took this system from an econometric problem.

How to solve such a system? You can try to express one variable in terms of another, but in this case you will probably end up with terrible fancy fractions that are extremely inconvenient to work with, and the design of the solution will look simply terrible. You can multiply the second equation by 6 and subtract term by term, but the same fractions will arise here too.

What to do? In such cases, Cramer's formulas come to the rescue.

;

;

Answer: ,

Both roots have infinite tails and are found approximately, which is quite acceptable (and even commonplace) for econometrics problems.

No comments are needed here, since the task is solved according to ready-made formulas However, there is one caveat. When using this method, compulsory A fragment of the task design is the following fragment: “This means that the system has a unique solution”. Otherwise, the reviewer may punish you for disrespect for Cramer's theorem.

It would not be superfluous to check, which can be conveniently carried out on a calculator: we substitute approximate values ​​into the left side of each equation of the system. As a result, with a small error, you should get numbers that are on the right sides.

Example 8

Present the answer in ordinary improper fractions. Do a check.

This is an example for independent decision(example of finishing and answer at the end of the lesson).

Let's move on to consider Cramer's rule for a system of three equations with three unknowns:

We find the main determinant of the system:

If , then the system has infinitely many solutions or is inconsistent (has no solutions). In this case, Cramer's rule will not help; you need to use the Gauss method.

If , then the system has a unique solution and to find the roots we must calculate three more determinants:
, ,

And finally, the answer is calculated using the formulas:

As you can see, the “three by three” case is fundamentally no different from the “two by two” case; the column of free terms sequentially “walks” from left to right along the columns of the main determinant.

Example 9

Solve the system using Cramer's formulas.

Solution: Let's solve the system using Cramer's formulas.

, which means the system has a unique solution.

Answer: .

Actually, here again there is nothing special to comment on, due to the fact that the solution follows ready-made formulas. But there are a couple of comments.

It happens that as a result of calculations, “bad” irreducible fractions are obtained, for example: .
I recommend the following “treatment” algorithm. If you don’t have a computer at hand, do this:

1) There may be an error in the calculations. As soon as you encounter a “bad” fraction, you immediately need to check Is the condition rewritten correctly?. If the condition is rewritten without errors, then you need to recalculate the determinants using expansion in another row (column).

2) If no errors are identified as a result of checking, then most likely there was a typo in the task conditions. In this case, calmly and CAREFULLY work through the task to the end, and then be sure to check and we draw it up on a clean sheet after the decision. Of course, checking a fractional answer is an unpleasant task, but it will be a disarming argument for the teacher, who really likes to give a minus for any bullshit like . How to handle fractions is described in detail in the answer to Example 8.

If you have a computer at hand, then use an automated program to check, which can be downloaded for free at the very beginning of the lesson. By the way, it is most profitable to use the program right away (even before starting the solution); you will immediately see the intermediate step where you made a mistake! The same calculator automatically calculates the solution of the system using the matrix method.

Second remark. From time to time there are systems in the equations of which some variables are missing, for example:

Here in the first equation there is no variable , in the second there is no variable . In such cases, it is very important to correctly and CAREFULLY write down the main determinant:
– zeros are placed in place of missing variables.
By the way, it is rational to open determinants with zeros according to the row (column) in which the zero is located, since there are noticeably fewer calculations.

Example 10

Solve the system using Cramer's formulas.

This is an example for an independent solution (a sample of the final design and the answer at the end of the lesson).

For the case of a system of 4 equations with 4 unknowns, Cramer’s formulas are written according to similar principles. You can see a live example in the lesson Properties of Determinants. Reducing the order of the determinant - five 4th order determinants are quite solvable. Although the task is already very reminiscent of a professor’s shoe on the chest of a lucky student.

Solving the system using an inverse matrix

The inverse matrix method is essentially special case matrix equation(See Example No. 3 of the specified lesson).

To study this section, you must be able to expand determinants, find the inverse of a matrix, and perform matrix multiplication. Relevant links will be provided as the explanations progress.

Example 11

Solve the system using the matrix method

Solution: Let's write the system in matrix form:
, Where

Please look at the system of equations and matrices. I think everyone understands the principle by which we write elements into matrices. The only comment: if some variables were missing from the equations, then zeros would have to be placed in the corresponding places in the matrix.

We find the inverse matrix using the formula:
, where is the transposed matrix of algebraic complements of the corresponding elements of the matrix.

First, let's look at the determinant:

Here the determinant is expanded on the first line.

Attention! If , then the inverse matrix does not exist, and it is impossible to solve the system using the matrix method. In this case, the system is solved by the method of eliminating unknowns (Gauss method).

Now we need to calculate 9 minors and write them into the minors matrix

Reference: It is useful to know the meaning of double subscripts in linear algebra. The first digit is the number of the line in which the element is located. The second digit is the number of the column in which the element is located:

That is, a double subscript indicates that the element is in the first row, third column, and, for example, the element is in 3 row, 2 column

2. Solving systems of equations using the matrix method (using an inverse matrix).
3. Gauss method for solving systems of equations.

Cramer's method.

The Cramer method is used to solve systems of linear algebraic equations ( SLAU).

Formulas using the example of a system of two equations with two variables.
Given: Solve the system using Cramer's method

Regarding variables X And at.
Solution:
Let's find the determinant of the matrix, composed of the coefficients of the system Calculation of determinants. :

Let's apply Cramer's formulas and find the values ​​of the variables:
And .
Example 1:
Solve the system of equations:

regarding variables X And at.
Solution:


Let us replace the first column in this determinant with a column of coefficients from the right side of the system and find its value:

Let's do it similar action, replacing the second column in the first determinant:

Applicable Cramer's formulas and find the values ​​of the variables:
And .
Answer:
Comment: This method can solve systems of higher dimensions.

Comment: If it turns out that , but cannot be divided by zero, then they say that the system does not have a unique solution. In this case, the system either has infinitely many solutions or has no solutions at all.

Example 2(infinite number of solutions):

Solve the system of equations:

regarding variables X And at.
Solution:
Let us find the determinant of the matrix, composed of the coefficients of the system:

Solving systems using the substitution method.

The first of the system's equations is an equality that is true for any values ​​of the variables (because 4 is always equal to 4). This means there is only one equation left. This is an equation for the relationship between variables.
We found that the solution to the system is any pair of values ​​of variables related to each other by the equality .
The general solution will be written as follows:
Particular solutions can be determined by choosing an arbitrary value of y and calculating x from this connection equality.

etc.
There are infinitely many such solutions.
Answer: common decision
Private solutions:

Example 3(no solutions, system is incompatible):

Solve the system of equations:

Solution:
Let us find the determinant of the matrix, composed of the coefficients of the system:

Cramer's formulas cannot be used. Let's solve this system using the substitution method

The second equation of the system is an equality that is not true for any values ​​of the variables (of course, since -15 is not equal to 2). If one of the equations of the system is not true for any values ​​of the variables, then the entire system has no solutions.
Answer: no solutions