Establish that the system of equations has a unique solution. Solving systems of linear algebraic equations, solution methods, examples
Solving systems of linear algebraic equations (SLAE) is undoubtedly the most important topic of the linear algebra course. A huge number of problems from all branches of mathematics are reduced to solving systems of linear equations. These factors explain the reason for creating this article. The material of the article is selected and structured so that with its help you can
- choose the optimal method for solving your system of linear algebraic equations,
- study the theory of the chosen method,
- solve your system of linear equations, having considered in detail the solutions of typical examples and problems.
Brief description of the material of the article.
First, we give all the necessary definitions, concepts, and introduce some notation.
Next, we consider methods for solving systems of linear algebraic equations in which the number of equations is equal to the number of unknown variables and which have a unique solution. First, let's focus on the Cramer method, secondly, we will show the matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of successive elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in various ways.
After that, we turn to solving systems of linear algebraic equations of a general form, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is degenerate. We formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (in the case of their compatibility) using the concept of the basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions of the examples.
Be sure to dwell on the structure of the general solution of homogeneous and inhomogeneous systems of linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how the general solution of the SLAE is written using the vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.
In conclusion, we consider systems of equations that are reduced to linear ones, as well as various problems, in the solution of which SLAEs arise.
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Definitions, concepts, designations.
We will consider systems of p linear algebraic equations with n unknown variables (p may be equal to n ) of the form
Unknown variables, - coefficients (some real or complex numbers), - free members (also real or complex numbers).
This form of SLAE is called coordinate.
AT matrix form this system of equations has the form ,
where 
- the main matrix of the system, - the matrix-column of unknown variables, - the matrix-column of free members.
If we add to the matrix A as the (n + 1)-th column the matrix-column of free terms, then we get the so-called expanded matrix systems of linear equations. Usually, the augmented matrix is denoted by the letter T, and the column of free members is separated by a vertical line from the rest of the columns, that is, 
By solving a system of linear algebraic equations called a set of values of unknown variables , which turns all the equations of the system into identities. The matrix equation for the given values of the unknown variables also turns into an identity.
If a system of equations has at least one solution, then it is called joint.
If the system of equations has no solutions, then it is called incompatible.
If a SLAE has a unique solution, then it is called certain; if there is more than one solution, then - uncertain.
If the free terms of all equations of the system are equal to zero 
, then the system is called homogeneous, otherwise - heterogeneous.
Solution of elementary systems of linear algebraic equations.
If the number of system equations is equal to the number of unknown variables and the determinant of its main matrix is not equal to zero, then we will call such SLAEs elementary. Such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.
We started studying such SLAE in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the addition method, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are essentially modifications of the Gauss method.
The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's sort them out.
Solving systems of linear equations by Cramer's method.
Let us need to solve a system of linear algebraic equations 
in which the number of equations is equal to the number of unknown variables and the determinant of the main matrix of the system is different from zero, that is, .
Let be the determinant of the main matrix of the system, and 
are determinants of matrices that are obtained from A by replacing 1st, 2nd, …, nth column respectively to the column of free members: 
With such notation, the unknown variables are calculated by the formulas of Cramer's method as 
. This is how the solution of a system of linear algebraic equations is found by the Cramer method.
Example.
Cramer method 
.
Decision.
The main matrix of the system has the form 
. Calculate its determinant (if necessary, see the article): 
Since the determinant of the main matrix of the system is nonzero, the system has a unique solution that can be found by Cramer's method.
Compose and calculate the necessary determinants 
(the determinant is obtained by replacing the first column in matrix A with a column of free members, the determinant - by replacing the second column with a column of free members, - by replacing the third column of matrix A with a column of free members): 
Finding unknown variables using formulas 
:
Answer:
The main disadvantage of Cramer's method (if it can be called a disadvantage) is the complexity of calculating the determinants when the number of system equations is more than three.
Solving systems of linear algebraic equations by the matrix method (using the inverse matrix).
Let the system of linear algebraic equations be given in matrix form , where the matrix A has dimension n by n and its determinant is nonzero.
Since , then the matrix A is invertible, that is, there is an inverse matrix . If we multiply both parts of the equality by on the left, then we get a formula for finding the column matrix of unknown variables. So we got the solution of the system of linear algebraic equations by the matrix method.
Example.
Solve System of Linear Equations matrix method.
Decision.
Let's rewrite the system of equations in matrix form: 
As
then the SLAE can be solved by the matrix method. Using the inverse matrix, the solution to this system can be found as 
.
Let's build an inverse matrix using a matrix of algebraic complements of the elements of matrix A (if necessary, see the article): 
It remains to calculate - the matrix of unknown variables by multiplying the inverse matrix 
on the matrix-column of free members (if necessary, see the article): 
Answer:
or in another notation x 1 = 4, x 2 = 0, x 3 = -1.
The main problem in finding solutions to systems of linear algebraic equations by the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than the third.
Solving systems of linear equations by the Gauss method.
Suppose we need to find a solution to a system of n linear equations with n unknown variables 
the determinant of the main matrix of which is different from zero.
The essence of the Gauss method consists in the successive exclusion of unknown variables: first, x 1 is excluded from all equations of the system, starting from the second, then x 2 is excluded from all equations, starting from the third, and so on, until only the unknown variable x n remains in the last equation. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called direct Gauss method. After the completion of the forward run of the Gaussian method, x n is found from the last equation, x n-1 is calculated from the penultimate equation using this value, and so on, x 1 is found from the first equation. The process of calculating unknown variables when moving from the last equation of the system to the first is called reverse Gauss method.
Let us briefly describe the algorithm for eliminating unknown variables.
We will assume that , since we can always achieve this by rearranging the equations of the system. We exclude the unknown variable x 1 from all equations of the system, starting from the second one. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, add the first multiplied by to the nth equation. The system of equations after such transformations will take the form 
where , a 
.
We would come to the same result if we expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.
Next, we act similarly, but only with a part of the resulting system, which is marked in the figure 
To do this, add the second multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, add the second multiplied by to the nth equation. The system of equations after such transformations will take the form 
where , a 
. Thus, the variable x 2 is excluded from all equations, starting from the third.
Next, we proceed to the elimination of the unknown x 3, while acting similarly with the part of the system marked in the figure 
So we continue the direct course of the Gauss method until the system takes the form 
From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as , using the obtained value x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Example.
Solve System of Linear Equations Gaussian method.
Decision.
Let's exclude the unknown variable x 1 from the second and third equations of the system. To do this, to both parts of the second and third equations, we add the corresponding parts of the first equation, multiplied by and by, respectively: 
Now we exclude x 2 from the third equation by adding to its left and right parts the left and right parts of the second equation, multiplied by: 
On this, the forward course of the Gauss method is completed, we begin the reverse course.
From the last equation of the resulting system of equations, we find x 3: 
From the second equation we get .
From the first equation we find the remaining unknown variable and this completes the reverse course of the Gauss method.
Answer:
X 1 \u003d 4, x 2 \u003d 0, x 3 \u003d -1.
Solving systems of linear algebraic equations of general form.
In the general case, the number of equations of the system p does not coincide with the number of unknown variables n:
Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations whose main matrix is square and degenerate.
Kronecker-Capelli theorem.
Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when SLAE is compatible, and when it is incompatible, gives Kronecker–Capelli theorem:
for a system of p equations with n unknowns (p can be equal to n ) to be consistent it is necessary and sufficient that the rank of the main matrix of the system is equal to the rank of the extended matrix, that is, Rank(A)=Rank(T) .
Let us consider the application of the Kronecker-Cappelli theorem for determining the compatibility of a system of linear equations as an example.
Example.
Find out if the system of linear equations has 
solutions.
Decision.
. Let us use the method of bordering minors. Minor of the second order 
different from zero. Let's go over the third-order minors surrounding it: 
Since all bordering third-order minors are equal to zero, the rank of the main matrix is two.
In turn, the rank of the augmented matrix 
is equal to three, since the minor of the third order 
different from zero.
In this way, Rang(A) , therefore, according to the Kronecker-Capelli theorem, we can conclude that the original system of linear equations is inconsistent.
Answer:
There is no solution system.
So, we have learned to establish the inconsistency of the system using the Kronecker-Capelli theorem.
But how to find the solution of the SLAE if its compatibility is established?
To do this, we need the concept of the basis minor of a matrix and the theorem on the rank of a matrix.
The highest order minor of the matrix A, other than zero, is called basic.
It follows from the definition of the basis minor that its order is equal to the rank of the matrix. For a non-zero matrix A, there can be several basic minors; there is always one basic minor.
For example, consider the matrix 
.
All third-order minors of this matrix are equal to zero, since the elements of the third row of this matrix are the sum of the corresponding elements of the first and second rows.
The following minors of the second order are basic, since they are nonzero 
Minors 
are not basic, since they are equal to zero.
Matrix rank theorem.
If the rank of a matrix of order p by n is r, then all elements of the rows (and columns) of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding elements of the rows (and columns) that form the basis minor.
What does the matrix rank theorem give us?
If, by the Kronecker-Capelli theorem, we have established the compatibility of the system, then we choose any basic minor of the main matrix of the system (its order is equal to r), and exclude from the system all equations that do not form the chosen basic minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still redundant (according to the matrix rank theorem, they are a linear combination of the remaining equations).
As a result, after discarding the excessive equations of the system, two cases are possible.
If the number of equations r in the resulting system is equal to the number of unknown variables, then it will be definite and the only solution can be found by the Cramer method, the matrix method or the Gauss method.
Example.
.
Decision.
Rank of the main matrix of the system 
is equal to two, since the minor of the second order 
different from zero. Extended matrix rank 
is also equal to two, since the only minor of the third order is equal to zero 
and the minor of the second order considered above is different from zero. Based on the Kronecker-Capelli theorem, one can assert the compatibility of the original system of linear equations, since Rank(A)=Rank(T)=2 .
As a basis minor, we take . It is formed by the coefficients of the first and second equations: 
The third equation of the system does not participate in the formation of the basic minor, so we exclude it from the system based on the matrix rank theorem: 
Thus we have obtained an elementary system of linear algebraic equations. Let's solve it by Cramer's method: 
Answer:
x 1 \u003d 1, x 2 \u003d 2.
If the number of equations r in the resulting SLAE is less than the number of unknown variables n , then we leave the terms that form the basic minor in the left parts of the equations, and transfer the remaining terms to the right parts of the equations of the system with the opposite sign.
The unknown variables (there are r of them) remaining on the left-hand sides of the equations are called main.
Unknown variables (there are n - r of them) that ended up on the right side are called free.
Now we assume that the free unknown variables can take arbitrary values, while the r main unknown variables will be expressed in terms of the free unknown variables in a unique way. Their expression can be found by solving the resulting SLAE by the Cramer method, the matrix method, or the Gauss method.
Let's take an example.
Example.
Solve System of Linear Algebraic Equations 
.
Decision.
Find the rank of the main matrix of the system 
by the bordering minors method. Let us take a 1 1 = 1 as a non-zero first-order minor. Let's start searching for a non-zero second-order minor surrounding this minor: 
So we found a non-zero minor of the second order. Let's start searching for a non-zero bordering minor of the third order: 
Thus, the rank of the main matrix is three. The rank of the augmented matrix is also equal to three, that is, the system is consistent.
The found non-zero minor of the third order will be taken as the basic one.
For clarity, we show the elements that form the basis minor: 
We leave the terms participating in the basic minor on the left side of the equations of the system, and transfer the rest with opposite signs to the right sides: 
We give free unknown variables x 2 and x 5 arbitrary values, that is, we take 
, where are arbitrary numbers. In this case, the SLAE takes the form 
We solve the obtained elementary system of linear algebraic equations by the Cramer method: 
Consequently, .
In the answer, do not forget to indicate free unknown variables.
Answer:
Where are arbitrary numbers.
Summarize.
To solve a system of linear algebraic equations of a general form, we first find out its compatibility using the Kronecker-Capelli theorem. If the rank of the main matrix is not equal to the rank of the extended matrix, then we conclude that the system is inconsistent.
If the rank of the main matrix is equal to the rank of the extended matrix, then we choose the basic minor and discard the equations of the system that do not participate in the formation of the chosen basic minor.
If the order of the basis minor is equal to the number of unknown variables, then the SLAE has a unique solution, which can be found by any method known to us.
If the order of the basis minor is less than the number of unknown variables, then we leave the terms with the main unknown variables on the left side of the equations of the system, transfer the remaining terms to the right sides and assign arbitrary values to the free unknown variables. From the resulting system of linear equations, we find the main unknown variables by the Cramer method, the matrix method or the Gauss method.
Gauss method for solving systems of linear algebraic equations of general form.
Using the Gauss method, one can solve systems of linear algebraic equations of any kind without their preliminary investigation for compatibility. The process of successive elimination of unknown variables makes it possible to draw a conclusion about both the compatibility and inconsistency of the SLAE, and if a solution exists, it makes it possible to find it.
From the point of view of computational work, the Gaussian method is preferable.
See its detailed description and analyzed examples in the article Gauss method for solving systems of linear algebraic equations of general form.
Recording the general solution of homogeneous and inhomogeneous linear algebraic systems using the vectors of the fundamental system of solutions.
In this section, we will focus on joint homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.
Let's deal with homogeneous systems first.
Fundamental decision system A homogeneous system of p linear algebraic equations with n unknown variables is a set of (n – r) linearly independent solutions of this system, where r is the order of the basis minor of the main matrix of the system.
If we designate linearly independent solutions of a homogeneous SLAE as X (1) , X (2) , …, X (n-r) (X (1) , X (2) , …, X (n-r) are matrices columns of dimension n by 1 ) , then the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary constant coefficients С 1 , С 2 , …, С (n-r), that is, .
What does the term general solution of a homogeneous system of linear algebraic equations (oroslau) mean?
The meaning is simple: the formula specifies all possible solutions to the original SLAE, in other words, taking any set of values of arbitrary constants C 1 , C 2 , ..., C (n-r) , according to the formula we will get one of the solutions of the original homogeneous SLAE.
Thus, if we find a fundamental system of solutions, then we can set all solutions of this homogeneous SLAE as .
Let us show the process of constructing a fundamental system of solutions for a homogeneous SLAE.
We choose the basic minor of the original system of linear equations, exclude all other equations from the system, and transfer to the right-hand side of the equations of the system with opposite signs all terms containing free unknown variables. Let's give the free unknown variables the values 1,0,0,…,0 and calculate the main unknowns by solving the resulting elementary system of linear equations in any way, for example, by the Cramer method. Thus, X (1) will be obtained - the first solution of the fundamental system. If we give the free unknowns the values 0,1,0,0,…,0 and calculate the main unknowns, then we get X (2) . And so on. If we give the free unknown variables the values 0,0,…,0,1 and calculate the main unknowns, then we get X (n-r) . This is how the fundamental system of solutions of the homogeneous SLAE will be constructed and its general solution can be written in the form .
For inhomogeneous systems of linear algebraic equations, the general solution is represented as
Let's look at examples.
Example.
Find the fundamental system of solutions and the general solution of a homogeneous system of linear algebraic equations 
.
Decision.
The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let us find the rank of the main matrix by the method of fringing minors. As a nonzero minor of the first order, we take the element a 1 1 = 9 of the main matrix of the system. Find the bordering non-zero minor of the second order: 
A minor of the second order, different from zero, is found. Let's go through the third-order minors bordering it in search of a non-zero one: 
All bordering minors of the third order are equal to zero, therefore, the rank of the main and extended matrix is two. Let's take the basic minor. For clarity, we note the elements of the system that form it: 
The third equation of the original SLAE does not participate in the formation of the basic minor, therefore, it can be excluded: 
We leave the terms containing the main unknowns on the right-hand sides of the equations, and transfer the terms with free unknowns to the right-hand sides:
Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basic minor is two. To find X (1), we give the free unknown variables the values x 2 \u003d 1, x 4 \u003d 0, then we find the main unknowns from the system of equations 
.
Let's solve it by Cramer's method: 
In this way, .
Now let's build X (2) . To do this, we give the free unknown variables the values x 2 \u003d 0, x 4 \u003d 1, then we find the main unknowns from the system of linear equations 
.
Let's use Cramer's method again: 
We get .
So we got two vectors of the fundamental system of solutions and , now we can write down the general solution of a homogeneous system of linear algebraic equations: 
, where C 1 and C 2 are arbitrary numbers., are equal to zero. We also take the minor as the basic one, exclude the third equation from the system, and transfer the terms with free unknowns to the right-hand sides of the system equations: 
To find, we give the free unknown variables the values x 2 \u003d 0 and x 4 \u003d 0, then the system of equations takes the form 
, from which we find the main unknown variables using the Cramer method: 
We have 
, Consequently, 
where C 1 and C 2 are arbitrary numbers.
It should be noted that the solutions of an indefinite homogeneous system of linear algebraic equations generate linear space Decision.
The canonical equation of an ellipsoid in a rectangular Cartesian coordinate system has the form 
. Our task is to determine the parameters a , b and c . Since the ellipsoid passes through points A, B and C, then when substituting their coordinates into the canonical equation of the ellipsoid, it should turn into an identity. So we get a system of three equations: 
Denote 
, then the system becomes a system of linear algebraic equations 
.
Let us calculate the determinant of the main matrix of the system: 
Since it is non-zero, we can find the solution by Cramer's method:
). Obviously, x = 0 and x = 1 are the roots of this polynomial. quotient from division 
on 
is . Thus, we have a decomposition and the original expression will take the form 
.
Let's use the method of indefinite coefficients. 
Equating the corresponding coefficients of the numerators, we arrive at a system of linear algebraic equations 
. Its solution will give us the desired indefinite coefficients A, B, C and D.
We solve the system using the Gauss method: 
In the reverse course of the Gauss method, we find D = 0, C = -2, B = 1, A = 1 .
We get 
Answer:
.
- Systems m linear equations with n unknown.
Solving a system of linear equations is such a set of numbers ( x 1 , x 2 , …, x n), substituting which into each of the equations of the system, the correct equality is obtained.
where a ij , i = 1, …, m; j = 1, …, n are the coefficients of the system;
b i , i = 1, …, m- free members;
x j , j = 1, …, n- unknown.
The above system can be written in matrix form: A X = B,
where ( A|B) is the main matrix of the system;
A— extended matrix of the system;
X— column of unknowns;
B is a column of free members.
If the matrix B is not a null matrix ∅, then this system of linear equations is called inhomogeneous.
If the matrix B= ∅, then this system of linear equations is called homogeneous. A homogeneous system always has a zero (trivial) solution: x 1 \u003d x 2 \u003d ..., x n \u003d 0.
Joint system of linear equations is a system of linear equations that has a solution.
Inconsistent system of linear equations is a system of linear equations that has no solution.
Certain system of linear equations is a system of linear equations that has a unique solution.
Indefinite system of linear equations is a system of linear equations that has an infinite number of solutions. - Systems of n linear equations with n unknowns
If the number of unknowns is equal to the number of equations, then the matrix is square. The matrix determinant is called the main determinant of the system of linear equations and is denoted by the symbol Δ.
Cramer method for solving systems n linear equations with n unknown.
Cramer's rule.
If the main determinant of a system of linear equations is not equal to zero, then the system is consistent and defined, and the only solution is calculated using the Cramer formulas:
where Δ i are the determinants obtained from the main determinant of the system Δ by replacing i th column to the column of free members. . - Systems of m linear equations with n unknowns
Kronecker-Cappelli theorem.
In order for this system of linear equations to be consistent, it is necessary and sufficient that the rank of the matrix of the system be equal to the rank of the extended matrix of the system, rank(Α) = rank(Α|B).
If rang(Α) ≠ rang(Α|B), then the system obviously has no solutions.
If rank(Α) = rank(Α|B), then two cases are possible:
1) rang(Α) = n(to the number of unknowns) - the solution is unique and can be obtained by Cramer's formulas;
2) rank(Α)< n − there are infinitely many solutions. - Gauss method for solving systems of linear equations
Let's compose the augmented matrix ( A|B) of the given system of coefficients at the unknown and right-hand sides.
The Gaussian method or the elimination of unknowns method consists in reducing the augmented matrix ( A|B) with the help of elementary transformations over its rows to a diagonal form (to an upper triangular form). Returning to the system of equations, all unknowns are determined.
Elementary transformations on strings include the following:
1) swapping two lines;
2) multiplying a string by a number other than 0;
3) adding to the string another string multiplied by an arbitrary number;
4) discarding a null string.
An extended matrix reduced to a diagonal form corresponds to a linear system equivalent to the given one, the solution of which does not cause difficulties. . - System of homogeneous linear equations.
The homogeneous system has the form:
it corresponds to the matrix equation A X = 0.
1) A homogeneous system is always consistent, since r(A) = r(A|B), there is always a zero solution (0, 0, …, 0).
2) For a homogeneous system to have a nonzero solution, it is necessary and sufficient that r = r(A)< n , which is equivalent to Δ = 0.
3) If r< n , then Δ = 0, then there are free unknowns c 1 , c 2 , …, c n-r, the system has nontrivial solutions, and there are infinitely many of them.
4) General solution X at r< n can be written in matrix form as follows:
X \u003d c 1 X 1 + c 2 X 2 + ... + c n-r X n-r,
where are the solutions X 1 , X 2 , …, X n-r form a fundamental system of solutions.
5) The fundamental system of solutions can be obtained from the general solution of the homogeneous system:
,
if we sequentially assume the values of the parameters to be (1, 0, …, 0), (0, 1, …, 0), …, (0, 0, …, 1).
Decomposition of the general solution in terms of the fundamental system of solutions is a record of the general solution as a linear combination of solutions belonging to the fundamental system.
Theorem. For a system of linear homogeneous equations to have a nonzero solution, it is necessary and sufficient that Δ ≠ 0.
So, if the determinant is Δ ≠ 0, then the system has a unique solution.
If Δ ≠ 0, then the system of linear homogeneous equations has an infinite number of solutions.
Theorem. For a homogeneous system to have a nonzero solution, it is necessary and sufficient that r(A)< n .
Proof:
1) r can't be more n(matrix rank does not exceed the number of columns or rows);
2) r< n , because if r=n, then the main determinant of the system Δ ≠ 0, and, according to Cramer's formulas, there is a unique trivial solution x 1 \u003d x 2 \u003d ... \u003d x n \u003d 0, which contradicts the condition. Means, r(A)< n .
Consequence. In order for a homogeneous system n linear equations with n unknowns has a nonzero solution, it is necessary and sufficient that Δ = 0.
Sections: Mathematics
If a problem has fewer than three variables, it's not a problem; if more than eight, it is undecidable. Enon.
Problems with parameters are found in all variants of the USE, since when solving them, it is most clearly revealed how deep and informal the knowledge of the graduate is. The difficulties that students have in performing such tasks are caused not only by their relative complexity, but also by the fact that insufficient attention is paid to them in textbooks. In variants of KIMs in mathematics, there are two types of assignments with parameters. First: "for each value of the parameter, solve the equation, inequality, or system." Second: "find all values of the parameter, for each of which the solutions of the inequality, equation or system satisfy the given conditions." Accordingly, the answers in these two types of problems differ in essence. In the first case, all possible values of the parameter are listed in the answer, and solutions to the equation are written for each of these values. The second one lists all parameter values under which the conditions of the problem are met. Recording the answer is an essential stage of the solution, it is very important not to forget to reflect all the stages of the decision in the answer. This needs to be brought to the attention of students.
The appendix to the lesson contains additional material on the topic "Solving systems of linear equations with parameters", which will help in preparing students for the final certification.
Lesson Objectives:
- systematization of students' knowledge;
- development of skills to apply graphical representations in solving systems of equations;
- formation of the ability to solve systems of linear equations containing parameters;
- implementation of operational control and self-control of students;
- development of research and cognitive activity of schoolchildren, the ability to evaluate the results obtained.
The lesson is designed for two teaching hours.
During the classes
- Organizing time
Message topics, goals and objectives of the lesson.
- Updating the basic knowledge of students
Checking homework. As homework, students were asked to solve each of the three systems of linear equations
a) b) 
in)
graphically and analytically; draw a conclusion about the number of solutions obtained for each case
The conclusions made by the students are heard and analyzed. The results of the work under the guidance of the teacher are summarized in notebooks.
In general, a system of two linear equations with two unknowns can be represented as: .
To solve a given system of equations graphically means to find the coordinates of the intersection points of the graphs of these equations or to prove that there are none. The graph of each equation of this system on the plane is some straight line.
There are three cases of mutual arrangement of two straight lines on a plane:
<Рисунок1>;
<Рисунок2>;
<Рисунок3>.
For each case, it is useful to draw a picture.
- Learning new material
Today in the lesson we will learn how to solve systems of linear equations containing parameters. We will call a parameter an independent variable, the value of which in the problem is considered to be a given fixed or arbitrary real number, or a number belonging to a predetermined set. To solve a system of equations with a parameter means to establish a correspondence that allows for any value of the parameter to find the corresponding set of solutions to the system.
The solution of a problem with a parameter depends on the question posed in it. If you just need to solve a system of equations for various values of a parameter or investigate it, then you need to give a reasonable answer for any value of the parameter or for the value of a parameter that belongs to the set specified in advance in the problem. If it is necessary to find the values of the parameter that satisfy certain conditions, then a complete study is not required, and the solution of the system is limited to finding these particular values of the parameter.
Example 1 For each parameter value, we solve the system of equations
Decision.
- The system has a unique solution if
In this case we have
- If a = 0, then the system takes the form
The system is inconsistent, i.e. has no solutions.
- If then the system can be written in the form
It is obvious that in this case the system has infinitely many solutions of the form x = t; where t is any real number.
Answer:
Example 2
- has a unique solution;
- has many solutions;
- has no solutions?
Decision.
Answer:
Example 3 Let us find the sum of parameters a and b for which the system
has an infinite number of solutions.
Decision. The system has an infinite number of solutions if
That is, if a = 12, b = 36; a + b = 12 + 36 = 48.
Answer: 48.
- Consolidation of what was learned in the course of solving problems
- No. 15.24(a) . For each parameter value, solve the system of equations
- #15.25(a) For each parameter value, solve the system of equations
- For what values of the parameter a the system of equations
a) has no solutions; b) has infinitely many solutions.
Answer: for a = 2 there are no solutions, for a = -2 there are an infinite number of solutions
- Practical work in groups
The class is divided into groups of 4-5 people. Each group includes students with different levels of mathematical training. Each group receives a card with a task. You can invite all groups to solve one system of equations, and draw up the solution. The group that completed the task correctly first presents its solution; the rest hand over the decision to the teacher.
Card. Solve System of Linear Equations
for all values of the parameter a.
Answer: when 
the system has a unique solution 
; when there are no solutions; for a = -1 there are infinitely many solutions of the form, (t; 1- t) where t R
If the class is strong, groups can be offered different systems of equations, a list of which is in Appendix 1. Then each group presents its solution to the class.
Report of the group that first correctly completed the task
Participants voice and explain their version of the solution and answer questions that have arisen from representatives of other groups.
- Independent work
Option 1
Option 2
- Lesson summary
Solving systems of linear equations with parameters can be compared to a study that includes three main conditions. The teacher asks the students to formulate them.
When deciding, keep in mind:
- in order for the system to have a unique solution, it is necessary that the lines corresponding to the equation of the system intersect, i.e. it is necessary to fulfill the condition;
- to have no solutions, the lines must be parallel, i.e. the condition was met
- and, finally, for the system to have infinitely many solutions, the lines must coincide, i.e. condition was met.
The teacher evaluates the work in the lesson of the class as a whole and sets marks for the lesson for individual students. After checking independent work, each student will receive an assessment for the lesson.
- Homework
For what values of the parameter b the system of equations 
- has infinitely many solutions;
- has no solutions?
The graphs of the functions y = 4x + b and y = kx + 6 are symmetrical about the y-axis.
- Find b and k,
- find the coordinates of the point of intersection of these graphs.
Solve the system of equations for all values of m and n.
Solve a system of linear equations for all values of the parameter a (any choice).
Literature
- Algebra and the beginning of mathematical analysis: textbook. for 11 cells. general education institutions: basic and profile. levels / S. M. Nikolsky, M. K. Potapov, N. N. Reshetnikov, A. V. Shevkin - M .: Education, 2008.
- Mathematics: Grade 9: Preparation for the state final certification / M. N. Korchagina, V. V. Korchagin - M .: Eksmo, 2008.
- Getting ready for university. Mathematics. Part 2 state technol. un-t; Institute of modern technol. and economy; Compiled by: S. N. Gorshkova, L. M. Danovich, N. A. Naumova, A.V. Martynenko, I.A. Palshchikov. – Krasnodar, 2006.
- Collection of problems in mathematics for preparatory courses TUSUR: Study guide / Z. M. Goldstein, G. A. Kornievskaya, G. A. Korotchenko, S. N. Kudinov. – Tomsk: Tomsk. State. University of Control Systems and Radioelectronics, 1998.
- Mathematics: an intensive course of preparation for the exam / O. Yu. Cherkasov, A.G. Yakushev. - M .: Rolf, Iris-press, 1998.
We continue to deal with systems of linear equations. So far, we have considered systems that have a unique solution. Such systems can be solved in any way: substitution method("school") by Cramer's formulas, matrix method, Gauss method. However, two more cases are widespread in practice when:
1) the system is inconsistent (has no solutions);
2) the system has infinitely many solutions.
For these systems, the most universal of all solution methods is used - Gauss method. In fact, the "school" method will also lead to the answer, but in higher mathematics it is customary to use the Gaussian method of successive elimination of unknowns. Those who are not familiar with the Gauss method algorithm, please study the lesson first Gauss method
The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. First, consider a couple of examples where the system has no solutions (inconsistent).
Example 1
What immediately catches your eye in this system? The number of equations is less than the number of variables. There is a theorem that says: “If the number of equations in the system is less than the number of variables, then the system is either inconsistent or has infinitely many solutions. And it remains only to find out.
The beginning of the solution is quite ordinary - we write the extended matrix of the system and, using elementary transformations, we bring it to a stepwise form:
(1). On the upper left step, we need to get (+1) or (-1). There are no such numbers in the first column, so rearranging the rows will not work. The unit will have to be organized independently, and this can be done in several ways. We did so. To the first line we add the third line, multiplied by (-1).
(2). Now we get two zeros in the first column. To the second line, add the first line, multiplied by 3. To the third line, add the first, multiplied by 5.
(3). After the transformation is done, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the desired one (-1) on the second step. Divide the third line by (-3).
(four). Add the second line to the third line. Probably, everyone paid attention to the bad line, which turned out as a result of elementary transformations:
. It is clear that this cannot be so.
Indeed, we rewrite the resulting matrix
back to the system of linear equations:
If as a result of elementary transformations a string of the form , whereλ is a non-zero number, then the system is inconsistent (has no solutions).
How to record the end of a task? You need to write down the phrase:
“As a result of elementary transformations, a string of the form is obtained, where λ ≠ 0 ". Answer: "The system has no solutions (inconsistent)."
Please note that in this case there is no reverse move of the Gaussian algorithm, there are no solutions and there is simply nothing to find.
Example 2
Solve a system of linear equations 
This is a do-it-yourself example. Full solution and answer at the end of the lesson.
Again, we remind you that your solution process may differ from our solution process, the Gauss method does not set an unambiguous algorithm, you need to guess the procedure and the actions themselves in each case yourself.
One more technical feature of the solution: elementary transformations can be stopped At once, as soon as a line like , where λ ≠ 0 . Consider a conditional example: suppose that after the first transformation we get a matrix
.
This matrix has not yet been reduced to a stepped form, but there is no need for further elementary transformations, since a line of the form has appeared, where λ ≠ 0 . It should be immediately answered that the system is incompatible.
When a system of linear equations has no solutions, this is almost a gift to the student, because a short solution is obtained, sometimes literally in 2-3 steps. But everything in this world is balanced, and the problem in which the system has infinitely many solutions is just longer.
Example 3:
Solve a system of linear equations
There are 4 equations and 4 unknowns, so the system can either have a single solution, or have no solutions, or have infinitely many solutions. Whatever it was, but the Gauss method in any case will lead us to the answer. This is its versatility.
The beginning is again standard. We write the extended matrix of the system and, using elementary transformations, bring it to a step form:
That's all, and you were afraid.
(1). Please note that all the numbers in the first column are divisible by 2, so on the upper left step we are also satisfied with a deuce. To the second line we add the first line, multiplied by (-4). To the third line we add the first line, multiplied by (-2). To the fourth line we add the first line, multiplied by (-1).
Attention! Many may be tempted from the fourth line subtract first line. This can be done, but it is not necessary, experience shows that the probability of an error in calculations increases several times. We just add: to the fourth line we add the first line, multiplied by (-1) - exactly!
(2). The last three lines are proportional, two of them can be deleted. Here again it is necessary to show increased attention, but are the lines really proportional? For reinsurance, it will not be superfluous to multiply the second row by (-1), and divide the fourth row by 2, resulting in three identical rows. And only after that remove two of them. As a result of elementary transformations, the extended matrix of the system is reduced to a stepped form:
When completing a task in a notebook, it is advisable to make the same notes in pencil for clarity.
We rewrite the corresponding system of equations: 
The “usual” only solution of the system does not smell here. Bad line where λ ≠ 0, also no. Hence, this is the third remaining case - the system has infinitely many solutions.
The infinite set of solutions of the system is briefly written in the form of the so-called general system solution.
We will find the general solution of the system using the reverse motion of the Gauss method. For systems of equations with an infinite set of solutions, new concepts appear: "basic variables" and "free variables". First, let's define what variables we have basic, and what variables - free. It is not necessary to explain in detail the terms of linear algebra, it is enough to remember that there are such basis variables and free variables.
Basic variables always "sit" strictly on the steps of the matrix. In this example, the base variables are x 1 and x 3 .
Free variables are everything remaining variables that did not get a step. In our case, there are two: x 2 and x 4 - free variables.
Now you need Everybodybasis variables express only throughfree variables. The reverse move of the Gaussian algorithm traditionally works from the bottom up. From the second equation of the system, we express the basic variable x 3:
Now look at the first equation: 
. First, we substitute the found expression into it:
It remains to express the basic variable x 1 through free variables x 2 and x 4:
The result is what you need - Everybody basis variables ( x 1 and x 3) expressed only through free variables ( x 2 and x 4):
Actually, the general solution is ready:
.
How to write down the general solution? First of all, free variables are written into the general solution “on their own” and strictly in their places. In this case, the free variables x 2 and x 4 should be written in the second and fourth positions:
.
The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:
From the general solution of the system, one can find infinitely many private decisions. It's very simple. free variables x 2 and x 4 are called so because they can be given any final values. The most popular values are zero values, since this is the easiest way to obtain a particular solution.
Substituting ( x 2 = 0; x 4 = 0) into the general solution, we get one of the particular solutions:
, or is a particular solution corresponding to free variables with values ( x 2 = 0; x 4 = 0).
Ones are another sweet couple, let's substitute ( x 2 = 1 and x 4 = 1) into the general solution:
, i.e. (-1; 1; 1; 1) is another particular solution.
It is easy to see that the system of equations has infinitely many solutions since we can give free variables any values.
Each a particular solution must satisfy to each system equation. This is the basis for a “quick” check of the correctness of the solution. Take, for example, a particular solution (-1; 1; 1; 1) and substitute it into the left side of each equation in the original system:
Everything has to come together. And with any particular solution you get, everything should also converge.
Strictly speaking, the verification of a particular solution sometimes deceives, i.e. some particular solution can satisfy each equation of the system, and the general solution itself is actually found incorrectly. Therefore, first of all, the verification of the general solution is more thorough and reliable.
How to check the resulting general solution 
?
It's not difficult, but it requires quite a long transformation. We need to take expressions basic variables, in this case 
and , and substitute them into the left side of each equation of the system.
To the left side of the first equation of the system:
The right side of the original first equation of the system is obtained.
To the left side of the second equation of the system:
The right side of the original second equation of the system is obtained.
And further - to the left parts of the third and fourth equations of the system. This check is longer, but it guarantees the 100% correctness of the overall solution. In addition, in some tasks it is required to check the general solution.
Example 4:
Solve the system using the Gauss method. Find a general solution and two private ones. Check the overall solution.
This is a do-it-yourself example. Here, by the way, again the number of equations is less than the number of unknowns, which means that it is immediately clear that the system will be either inconsistent or with an infinite number of solutions.
Example 5:
Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution
Decision: Let us write down the extended matrix of the system and, with the help of elementary transformations, bring it to a stepped form:
(1). Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.
(2). To the third line we add the second line, multiplied by (-5). To the fourth line we add the second line, multiplied by (-7).
(3). The third and fourth lines are the same, we delete one of them. Here is such a beauty:
Basis variables sit on steps, so they are base variables.
There is only one free variable, which did not get a step: .
(four). Reverse move. We express the basic variables in terms of the free variable:
From the third equation:
Consider the second equation and substitute the found expression into it:
, 
, ,
Consider the first equation and substitute the found expressions and into it:
Thus, the general solution with one free variable x 4:
Once again, how did it happen? free variable x 4 sits alone in its rightful fourth place. The resulting expressions for the basic variables , , are also in their places.
Let us immediately check the general solution.
We substitute the basic variables , , into the left side of each equation of the system:
The corresponding right-hand sides of the equations are obtained, thus, the correct general solution is found.
Now from the found general solution 
we get two particular solutions. All variables are expressed here through a single free variable x four . You don't need to break your head.
Let x 4 = 0, then 
is the first particular solution.
Let x 4 = 1, then 
is another particular solution.
Answer: Common decision: . Private Solutions:
and .
Example 6:
Find the general solution of the system of linear equations.
We have already checked the general solution, the answer can be trusted. Your course of action may differ from our course of action. The main thing is that the general solutions coincide. Probably, many people noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to fiddle with ordinary fractions. In practice, this is true, cases where there are no fractions are much less common. Be prepared mentally, and most importantly, technically.
Let us dwell on the features of the solution that were not found in the solved examples. The general solution of the system may sometimes include a constant (or constants).
For example, the general solution: . Here one of the basic variables is equal to a constant number: . There is nothing exotic in this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.
Rarely, but there are systems in which the number of equations is greater than the number of variables. However, the Gauss method works under the most severe conditions. You should calmly bring the extended matrix of the system to a stepped form according to the standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a unique solution.
We repeat in our advice - in order to feel comfortable when solving a system using the Gauss method, you should fill your hand and solve at least a dozen systems.
Solutions and answers:
Example 2:
Decision:Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepped form.
Performed elementary transformations:
(1) The first and third lines have been swapped.
(2) The first line was added to the second line, multiplied by (-6). The first line was added to the third line, multiplied by (-7).
(3) The second line was added to the third line, multiplied by (-1).
As a result of elementary transformations, a string of the form, where λ ≠ 0 .So the system is inconsistent.Answer: there are no solutions.
Example 4:
Decision:We write the extended matrix of the system and, using elementary transformations, bring it to a step form:
Conversions performed:
(1). The first line multiplied by 2 was added to the second line. The first line multiplied by 3 was added to the third line.
There is no unit for the second step , and transformation (2) is aimed at obtaining it.
(2). The second line was added to the third line, multiplied by -3.
(3). The second and third rows were swapped (the resulting -1 was moved to the second step)
(four). The second line was added to the third line, multiplied by 3.
(five). The sign of the first two lines was changed (multiplied by -1), the third line was divided by 14.
Reverse move:
(1). Here are the basic variables (which are on steps), and are free variables (who did not get the step).
(2). We express the basic variables in terms of free variables:
From the third equation: .
(3). Consider the second equation:, particular solutions:
Answer: Common decision: 
Complex numbers
In this section, we will introduce the concept complex number, consider algebraic, trigonometric and show form complex number. We will also learn how to perform operations with complex numbers: addition, subtraction, multiplication, division, exponentiation and root extraction.
To master complex numbers, you do not need any special knowledge from the course of higher mathematics, and the material is available even to a schoolboy. It is enough to be able to perform algebraic operations with "ordinary" numbers, and remember trigonometry.
First, let's remember the "ordinary" Numbers. In mathematics they are called set of real numbers and are marked with the letter R, or R (thick). All real numbers sit on the familiar number line:
The company of real numbers is very colorful - here are integers, and fractions, and irrational numbers. In this case, each point of the numerical axis necessarily corresponds to some real number.
Systems of equations are widely used in the economic industry in the mathematical modeling of various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.
Equation systems are used not only in the field of mathematics, but also in physics, chemistry and biology, when solving problems of finding the population size.
A system of linear equations is a term for 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, the value of which must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving the equation by plotting its graph will look like a straight line, all points of which are the solution of the polynomial.
Types of systems of linear equations
The simplest are examples of 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 a system of equations - it means to find such values (x, y) for which the system becomes a true equality, or to establish that there are no suitable values of x and y.
A pair of values (x, y), written as point coordinates, is called a solution to a system of linear equations.
If the systems have one common solution or there is no solution, they are called equivalent.
Homogeneous systems of linear equations are systems whose right 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 not homogeneous.
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 variables or more.
Faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not so. The number of equations in the system does not depend on the variables, there can be an arbitrarily large number of them.
Simple and complex methods for solving systems of equations
There is no general analytical way to solve such systems, all methods are based on numerical solutions. The school course of mathematics describes in detail such methods as permutation, algebraic addition, substitution, as well as the graphical and matrix method, the solution by the Gauss method.
The main task in teaching methods of solving 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 applying a particular method.
The solution of examples of systems of linear equations of the 7th grade of the general education school program is quite simple and is explained in great detail. In any textbook on mathematics, this section is given enough attention. The solution of examples of systems of linear equations by the method of Gauss and Cramer is studied in more detail in the first courses of higher educational institutions.
Solution of systems by the substitution method
The actions of the substitution method are aimed at expressing the value of one variable through the second. The expression is substituted into the remaining equation, then it is reduced to a single variable form. The action is repeated depending on the number of unknowns in the system
Let's give an example of a system of linear equations of the 7th class by 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. The solution of this example does not cause difficulties 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 the expression of 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, the substitution solution is also impractical.
Solution of an example of a system of linear inhomogeneous equations:
Solution using algebraic addition
When searching for a solution to systems by the addition method, term-by-term addition and multiplication of equations by various numbers are performed. The ultimate goal of mathematical operations is an equation with one variable.
Applications of this method require practice and observation. It is not easy to solve a system of linear equations using the addition method with the number of variables 3 or more. Algebraic addition is useful when the equations contain fractions and decimal numbers.
Solution action algorithm:
- Multiply both sides of the equation by some number. As a result of the arithmetic operation, one of the coefficients of the variable must become equal to 1.
- Add the resulting expression term by term and find one of the unknowns.
- Substitute the resulting value into the 2nd equation of the system to find the remaining variable.
Solution method by introducing a new variable
A new variable can be introduced if the system needs to find 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 with respect to the entered unknown, and the resulting value is used to determine the original variable.
It can be seen from the example that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard square 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 multipliers of the polynomial. In the given example, a=1, b=16, c=39, hence 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 only one solution: x= -b / 2*a.
The solution for the resulting systems is found by the addition method.
A visual method for solving systems
Suitable for systems with 3 equations. The method consists in plotting graphs of each equation included in the system on the coordinate axis. The coordinates of the points of intersection of the curves will be the general solution of the system.
The graphic method has a number of nuances. Consider several examples of solving systems of linear equations in a visual way.
As can be seen from the example, two points were constructed for each line, 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.
In the following example, it is required to find a graphical solution to the 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 the system has a solution or not, it is always necessary to build a graph.
Matrix and its varieties
Matrices are used to briefly write down 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 is equal. A matrix-vector is a single-column matrix with an infinitely possible number of rows. A matrix with units along one of the diagonals and other zero elements is called identity.
An inverse matrix is such a matrix, when multiplied by which the original one turns into a unit one, such a matrix exists only for the original square one.
Rules for transforming a system of equations into a matrix
With regard to systems of equations, the coefficients and free members of the equations are written as numbers of the matrix, one equation is one row of the matrix.
A matrix row is called non-zero if at least one element of the row is not equal to 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 columns of the matrix must strictly correspond to the variables. This means that the coefficients of the variable x can only be written in one column, for example the first, the coefficient of the unknown y - only in the second.
When multiplying a matrix, all matrix elements are successively 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| - matrix determinant. |K| must not be equal to zero, then the system has a solution.
The determinant is easily calculated for a two-by-two matrix, it is only necessary to multiply the elements diagonally 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 column and row numbers of the elements do not repeat in the product.
Solution of examples of systems of linear equations by the matrix method
The matrix method of finding a solution makes it possible 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 the variables, and b n are the free terms.
Solution of systems by the Gauss method
In higher mathematics, the Gauss method is studied together with the Cramer method, and the process of finding a solution to systems is called the Gauss-Cramer method of solving. These methods are used to find the variables of systems with a large number of linear equations.
The Gaussian method is very similar to substitution and algebraic addition solutions, but is more systematic. In the school course, the Gaussian solution is used for systems of 3 and 4 equations. The purpose of the method is to bring the system to the form of an inverted trapezoid. By 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, and 3 and 4 - with 3 and 4 variables, respectively.
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 Gaussian solution 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. The solution of 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 is one of the most interesting ways to develop the ingenuity of children studying in the advanced study program in math and physics classes.
For ease of recording calculations, it is customary to do the following:
Equation coefficients 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 side. Roman numerals denote the numbers of equations in the system.
First, they write down the matrix with which to work, then all the actions carried out with one of the rows. The resulting matrix is written after the "arrow" sign and continue to perform the necessary algebraic operations until the result is achieved.
As a result, a matrix should be obtained in which one of the diagonals is 1, and all other coefficients are equal to zero, that is, the matrix is reduced to a single form. We must not forget to make calculations with the numbers of both sides of the equation.
This notation is less cumbersome and allows you not to be distracted by listing numerous unknowns.
The free application of any method of solution will require care and a certain amount of experience. Not all methods are applied. Some ways of finding solutions are more preferable in a particular area of human activity, while others exist for the purpose of learning.
