Systems of equations with four unknowns. Examples of systems of linear equations: solution method

X 1 =a 1, r+1 x r+1 +...+a 1 n X n+b 1 ,

............................................

X r=a r, r+1 x r+1 +...+a r n X n+b r.

which is essentially general decision systems (1).

The unknowns x r+1, ..., x n are called free. From system (5) it will be possible to find the values ​​x1,..., x r.

Matrix reduction Ā to the form (3) is possible only in the case when the original system of equations (1) is consistent. If system (1) is inconsistent, then such a reduction is impossible. This circumstance is expressed in the fact that in the process of matrix transformations Ā a line appears in it in which all elements are equal to zero, except the last one. This line corresponds to an equation of the form:

0*x 1 +0*x 2 +...+0*x n=b,

which is not satisfied by any values ​​of the unknowns, since b≠0. In this case the system is inconsistent.

In the process of bringing system (1) to a stepwise form, equations of the form 0=0 can be obtained. They can be discarded, since this leads to a system of equations equivalent to the previous one.

When solving a system of linear equations using the Gaussian method, it is more convenient to reduce not the system of equations itself, but the extended matrix of this system to a stepwise form, performing all transformations on its rows. Sequential matrices obtained during transformations are usually connected by an equivalence sign.

Let us solve the following system of equations with 4 unknowns:

2x 1 +5x 2 +4x 3 +x 4 =20,

x 1 +3x 2 +2x 3 +x 4 =11,

2x 1 +10x 2 +9x 3 +7x 4 =40,

3x 1 +8x 2 +9x 3 +2x 4 =37.

Let's write out the extended matrix of coefficients for unknowns with the addition of a column of free terms.

Let's analyze the rows of the extended matrix:

To the elements of the 2nd line we add the elements of the 1st, divided by (-2);

From the 3rd line, subtract the 1st line;

To the 4th line we add the 1st, multiplied by (-3/2).

As a computational tool, we will use the program tools Excel-97.

1. Turn on your computer.

2. Wait until the operating system boots Windows, after which open a Microsoft Excel window.

3. Fill in the cells tables with values ​​of the extended matrix (Fig. 11.1)

Rice. 11.1 Fig. 11.2

4. To perform the selected verbal algorithm, perform the following actions.

· Activate cell A5 and from the keyboard enter into it a formula of the form =A2+A1/(-2), after which autocomplete enter the numerical results in cells B5¸E5;

· In cell A6 we will place the result of subtracting the 1st line from the 3rd, and again, using autocomplete, fill in cells B6¸E6;

· in cell A7 we write a formula of the form =A4+A1*(-3/2) and autocomplete Let's enter the numerical results in cells B7¸E7.

5. Let us again analyze the rows resulting from elementary transformations of the matrix in order to bring it to a triangular form.

·To the 6th line add the 5th, multiplied by the number (-10);

· subtract the 5th from the 7th line.

We implement the recorded algorithm in cells A8, A9, after which let's hide 6 and 7 – lines (see Fig. 11.3).

Rice. 11.3 Fig. 11.4

6. And the last thing you need to do to bring the matrix to triangular form is to add the 8th to the 9th row, multiplied by (-3/5), after which hide 9th line (Fig. 11.4).

As you can see, the elements of the resulting matrix are in rows 1, 5, 8 and 10, and the rank of the resulting matrix is r = 4, therefore, this system of equations has a unique solution. Let us write out the resulting system:

2x 1 +5x 2 +4x 3 + x 4 =20,

0.5x 2 + 0.5x 4 =1,

5x 3 +x 4 =10,

From the last equation we easily find x 4 =0; from the 3rd equation we find x 3 =2; from the 2nd – x 2 =2 and from the 1st – x 1 =1, respectively.

Assignments for independent work.

Use the Gauss method to solve the systems of equations:

Laboratory work No. 15. Finding the roots of the equation f(x)=0

Methods for solving linear and quadratic equations were known to the ancient Greeks. The solution to equations of the third and fourth degrees was obtained through the efforts of Italian mathematicians S. Ferro, N. Tartaglia, G. Cartano, L. Ferrari during the Renaissance. Then it was time to search for formulas for finding the roots of equations of the fifth and higher degrees. Persistent but fruitless attempts continued for about 300 years and ended in the 20s of the 21st century thanks to the work of the Norwegian mathematician N. Abel. He proved that the general equation of the fifth and higher powers are unsolvable in radicals. Solution of the general equation of the nth degree

a 0 x n +a 1 x n -1 +…+a n -1 x+a n =0, a 0 ¹0 (1)

when n³5 cannot be expressed through coefficients using the operations of addition, subtraction, multiplication, division, exponentiation and root extraction.

For non-algebraic equations like

x–cos(x)=0 (2)

the task becomes even more difficult. In this case, it is rarely possible to find explicit expressions for the roots.

In conditions when formulas “do not work”, when you can count on them only in the simplest cases, universal computational algorithms acquire special importance. There are a number of known algorithms that allow solving the problem under consideration.

Systems of equations are widely used in the economic sector for mathematical modeling of various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.

Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.

A system of linear equations is two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.

Linear equation

Equations of the form ax+by=c are called linear. The designations x, y are the unknowns whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.

Types of systems of linear equations

The simplest examples are considered to be systems of linear equations with two variables X and Y.

F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.

Solve system of equations - this means finding values ​​(x, y) at which the system turns into a true equality or establishing that suitable values ​​of x and y do not exist.

A pair of values ​​(x, y), written as the coordinates of a point, is called a solution to a system of linear equations.

If systems have one common solution or no solution exists, they are called equivalent.

Homogeneous systems of linear equations are systems whose right-hand side is equal to zero. If the right part after the equal sign has a value or is expressed by a function, such a system is heterogeneous.

The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three or more variables.

When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.

Simple and complex methods for solving systems of equations

There is no general analytical method for solving such systems; all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as graphical and matrix methods, solution by the Gaussian method.

The main task when teaching solution methods is to teach how to correctly analyze the system and find the optimal solution algorithm for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of using a particular method

Solving examples of systems of linear equations in the 7th grade general education curriculum is quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.

Solving systems using the substitution method

The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system

Let us give a solution to an example of a system of linear equations of class 7 using the substitution method:

As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Solving this example is easy and allows you to get the Y value. The last step is to check the obtained values.

It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, solving by substitution is also inappropriate.

Solution of an example of a system of linear inhomogeneous equations:

Solution using algebraic addition

When searching for solutions to systems using the addition method, equations are added term by term and multiplied by various numbers. The ultimate goal of mathematical operations is an equation in one variable.

Application of this method requires practice and observation. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.

Solution algorithm:

1. Multiply both sides of the equation by a certain number. As a result of the arithmetic operation, one of the coefficients of the variable should become equal to 1.
2. Add the resulting expression term by term and find one of the unknowns.
3. Substitute the resulting value into the 2nd equation of the system to find the remaining variable.

Method of solution by introducing a new variable

A new variable can be introduced if the system requires finding a solution for no more than two equations; the number of unknowns should also be no more than two.

The method is used to simplify one of the equations by introducing a new variable. The new equation is solved for the introduced unknown, and the resulting value is used to determine the original variable.

The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard quadratic trinomial. You can solve a polynomial by finding the discriminant.

It is necessary to find the value of the discriminant using the well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. In the given example, a=1, b=16, c=39, therefore D=100. If the discriminant is greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant is less than zero, then there is one solution: x = -b / 2*a.

The solution for the resulting systems is found by the addition method.

Visual method for solving systems

Suitable for 3 equation systems. The method consists in constructing graphs of each equation included in the system on the coordinate axis. The coordinates of the intersection points of the curves will be the general solution of the system.

The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.

As can be seen from the example, for each line two points were constructed, the values ​​of the variable x were chosen arbitrarily: 0 and 3. Based on the values ​​of x, the values ​​for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.

The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.

The following example requires finding a graphical solution to a system of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.

As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.

The systems from examples 2 and 3 are similar, but when constructed it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether a system has a solution or not; it is always necessary to construct a graph.

The matrix and its varieties

Matrices are used to concisely write a system of linear equations. A matrix is ​​a special type of table filled with numbers. n*m has n - rows and m - columns.

A matrix is ​​square when the number of columns and rows are equal. A matrix-vector is a matrix of one column with an infinitely possible number of rows. A matrix with ones along one of the diagonals and other zero elements is called identity.

An inverse matrix is ​​a matrix when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.

Rules for converting a system of equations into a matrix

In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.

A matrix row is said to be nonzero if at least one element of the row is not zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.

The matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all elements of the matrix are sequentially multiplied by a number.

Options for finding the inverse matrix

The formula for finding the inverse matrix is ​​quite simple: K -1 = 1 / |K|, where K -1 is the inverse matrix, and |K| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements by each other. For the “three by three” option there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the numbers of columns and rows of elements are not repeated in the work.

Solving examples of systems of linear equations using the matrix method

The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with a large number of variables and equations.

In the example, a nm are the coefficients of the equations, the matrix is ​​a vector x n are variables, and b n are free terms.

Solving systems using the Gaussian method

In higher mathematics, the Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variables of systems with a large number of linear equations.

The Gauss method is very similar to solutions by substitution and algebraic addition, but is more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce the system to the form of an inverted trapezoid. By means of algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, while 3 and 4 are, respectively, with 3 and 4 variables.

After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.

In school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:

As can be seen from the example, at step (3) two equations were obtained: 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. Solving any of the equations will allow you to find out one of the variables x n.

Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.

The Gaussian method is difficult for middle school students to understand, but it is one of the most interesting ways to develop the ingenuity of children enrolled in advanced learning programs in math and physics classes.

For ease of recording, calculations are usually done as follows:

The coefficients of the equations and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right. Roman numerals indicate the numbers of equations in the system.

First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is ​​written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.

The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is ​​reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.

This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.

The free use of any solution method will require care and some experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of ​​human activity, while others exist for educational purposes.

The case when the number of equations m more variables n, by sequentially eliminating unknowns from the equations leads to the case m= n or m n. The first case was discussed earlier.

In the second case, when the number of equations is less than the number of unknowns m n and the equations are independent, stand out m main variables And ( n- m)non-core variables . The main variables are those that satisfy the condition: the determinant, made up of the coefficients of these variables, is not equal to zero. The main ones can be different groups of variables. Total number of such groups N equal to the number of combinations of n elements by m:

If a system has at least one group of basic variables, then this system is uncertain , that is, it has many solutions.

If the system does not have a single group of basic variables, then the system is non-joint , that is, it does not have a single solution.

In the case when a system has many solutions, a basic solution is distinguished among them.

Basic solution is a solution in which the minor variables are equal to zero. The system has no more than basic solutions.

System solutions are divided into acceptable And unacceptable .

Acceptable These are solutions in which the values ​​of all variables are non-negative.

If at least one value of the variable is negative, then the solution is called unacceptable .

Example 4.5

Find basic solutions to the system of equations

Let's find the number of basic solutions

.

So, among the many solutions of the system there are no more than three basic ones. Let us highlight two main variables among the three. Let's assume it's X 1 and X 2. Let's check the determinant from the coefficients of them

.

Since this determinant is not equal to zero, then the variables X 1 ,X 2 are the main ones.

Now let's assume that X 3 =0. Then we obtain a system in the form

Let's solve it using Cramer's formulas:

,
.

So, the first basic solution has the form

X 1 =1,X 2 =0,X 3 =0 .

Let us now check whether the variables belong to the main ones X 1 and X 3 .

.

We get that X 1 and X 3 - second group of main variables. Let's put X 2 =0 and solve the system

,
.

The second basic solution has the form

X 1 =1,X 2 =0,X 3 =0.

Now let’s check whether the variables belong to the main ones X 2 and X 3 .

that is, variables X 2 and X 3 minor. So, this system has two basic solutions in total. Both of these solutions are acceptable.

The compatibility condition for a system of m linear equations with n variables is given using the concept of matrix rank.

Matrix rank – this is a number equal to the highest order of a minor other than zero.

For matrix A

minor k -th order serves as a determinant composed of elements of any k lines and k columns.

For example,

Example 2

Find the rank of a matrix

Let's calculate the determinant of the matrix

To do this, multiply the first line by (-4) and add it with the second line, then multiply the first line by (-7) and add it with the third line, as a result we get the determinant

Because the rows of the resulting determinant are proportional, then
.

From this we can see that the 3rd order minor is equal to 0, and the 2nd order minor is not equal to 0.

Therefore, the rank of the matrix is ​​r=2.

Extended Matrix system has the form

Kronecker-Capelli theorem

In order for a linear system to be consistent, it is necessary and sufficient that the rank of the extended matrix be equal to the rank of the main matrix
.

If
, then the system is inconsistent.

For a simultaneous system of linear equations, three cases are possible:

1)If
, then the LU system has (m-r) linearly dependent equations, they can be excluded from the system;

2) If
, then the LU system has a unique solution;

3) If
, then the LU system has many solutions

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Equations with four unknowns can have many possible solutions. In mathematics, one often encounters equations of this type. To correctly solve such equations, it is necessary to use all the features of the equations in order to simplify and shorten its solution.

Let's look at the solution to the following example:

By adding the first and second equations by parts, you can get a very simple equation:

\ or \

Let's perform similar actions with equations 2 and 3:

\ or \

We solve the resulting equations \ and \

We get \ and \

We substitute the resulting numbers into equations 1 and 3:

\ or \

\ or \

Replacing these numbers with the second and fourth equations will give exactly the same equations.

But that's not all, since there are 2 equations with 2 unknowns left to solve. You can see the solution to this type of equation in the articles here.

Where can I solve an equation with four unknowns online?

You can solve equations with unknowns online at 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.