Trivial solution of the system. Gauss method for solving systems of linear algebraic equations of general form

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.

Page navigation.

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 forward run of the Gauss method is completed, 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 equation 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 of 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.

Thus, 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 forming 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:


Hence, .

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 basic minor is less than the number of unknown variables, then on the left side of the equations of the system we leave the terms with the main unknown variables, 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 exclusion 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 of 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 defines all possible solutions of 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) . Etc. 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
.

Even at school, each of us studied equations and, for sure, systems of equations. But not many people know that there are several ways to solve them. Today we will analyze in detail all the methods for solving a system of linear algebraic equations, which consist of more than two equalities.

Story

Today it is known that the art of solving equations and their systems originated in ancient Babylon and Egypt. However, equalities in their usual form appeared after the appearance of the equal sign "=", which was introduced in 1556 by the English mathematician Record. By the way, this sign was chosen for a reason: it means two parallel equal segments. Indeed, there is no better example of equality.

The founder of modern letter designations of unknowns and signs of degrees is a French mathematician. However, his designations differed significantly from today's. For example, he denoted the square of an unknown number with the letter Q (lat. "quadratus"), and the cube with the letter C (lat. "cubus"). These notations seem awkward now, but back then it was the most understandable way to write systems of linear algebraic equations.

However, a drawback in the then methods of solution was that mathematicians considered only positive roots. Perhaps this is due to the fact that negative values ​​had no practical use. One way or another, it was the Italian mathematicians Niccolo Tartaglia, Gerolamo Cardano and Rafael Bombelli who were the first to consider negative roots in the 16th century. And the modern view, the main solution method (through the discriminant) was created only in the 17th century thanks to the work of Descartes and Newton.

In the mid-18th century, the Swiss mathematician Gabriel Cramer found a new way to make solving systems of linear equations easier. This method was subsequently named after him and to this day we use it. But we will talk about Cramer's method a little later, but for now we will discuss linear equations and methods for solving them separately from the system.

Linear equations

Linear equations are the simplest equalities with variable(s). They are classified as algebraic. write in general form as follows: a 1 * x 1 + a 2 * x 2 + ... and n * x n \u003d b. We will need their representation in this form when compiling systems and matrices further.

Systems of linear algebraic equations

The definition of this term is as follows: it is a set of equations that have common unknowns and a common solution. As a rule, at school, everything was solved by systems with two or even three equations. But there are systems with four or more components. Let's first figure out how to write them down so that it is convenient to solve them later. First, systems of linear algebraic equations will look better if all variables are written as x with the appropriate index: 1,2,3, and so on. Secondly, all equations should be brought to the canonical form: a 1 * x 1 + a 2 * x 2 + ... a n * x n =b.

After all these actions, we can begin to talk about how to find a solution to systems of linear equations. Matrices are very useful for this.

matrices

A matrix is ​​a table that consists of rows and columns, and at their intersection are its elements. These can either be specific values ​​or variables. Most often, to designate elements, subscripts are placed under them (for example, a 11 or a 23). The first index means the row number and the second one the column number. On matrices, as well as on any other mathematical element, you can perform various operations. Thus, you can:

2) Multiply a matrix by some number or vector.

3) Transpose: turn matrix rows into columns and columns into rows.

4) Multiply matrices if the number of rows of one of them is equal to the number of columns of the other.

We will discuss all these techniques in more detail, as they will be useful to us in the future. Subtracting and adding matrices is very easy. Since we take matrices of the same size, each element of one table corresponds to each element of another. Thus, we add (subtract) these two elements (it is important that they are in the same places in their matrices). When multiplying a matrix by a number or vector, you simply need to multiply each element of the matrix by that number (or vector). Transposition is a very interesting process. It is very interesting sometimes to see it in real life, for example, when changing the orientation of a tablet or phone. The icons on the desktop are a matrix, and when you change the position, it transposes and becomes wider, but decreases in height.

Let's analyze such a process as Although it will not be useful to us, it will still be useful to know it. You can multiply two matrices only if the number of columns in one table is equal to the number of rows in the other. Now let's take the elements of a row of one matrix and the elements of the corresponding column of another. We multiply them by each other and then add them (that is, for example, the product of the elements a 11 and a 12 by b 12 and b 22 will be equal to: a 11 * b 12 + a 12 * b 22). Thus, one element of the table is obtained, and it is filled further by a similar method.

Now we can begin to consider how the system of linear equations is solved.

Gauss method

This topic starts at school. We know well the concept of "system of two linear equations" and know how to solve them. But what if the number of equations is more than two? This will help us

Of course, this method is convenient to use if you make a matrix out of the system. But you can not transform it and solve it in its pure form.

So, how is the system of linear Gaussian equations solved by this method? By the way, although this method is named after him, it was discovered in ancient times. Gauss proposes the following: to carry out operations with equations in order to eventually reduce the entire set to a stepped form. That is, it is necessary that from top to bottom (if placed correctly) from the first equation to the last, one unknown decreases. In other words, we need to make sure that we get, say, three equations: in the first - three unknowns, in the second - two, in the third - one. Then from the last equation we find the first unknown, substitute its value into the second or first equation, and then find the remaining two variables.

Cramer method

To master this method, it is vital to master the skills of addition, subtraction of matrices, and you also need to be able to find determinants. Therefore, if you do all this poorly or do not know how at all, you will have to learn and practice.

What is the essence of this method, and how to make it so that a system of linear Cramer equations is obtained? Everything is very simple. We have to construct a matrix from numerical (almost always) coefficients of a system of linear algebraic equations. To do this, we simply take the numbers in front of the unknowns and put them in the table in the order they are written in the system. If the number is preceded by a "-" sign, then we write down a negative coefficient. So, we have compiled the first matrix from the coefficients of the unknowns, not including the numbers after the equal signs (naturally, the equation should be reduced to the canonical form, when only the number is on the right, and all the unknowns with coefficients on the left). Then you need to create several more matrices - one for each variable. To do this, in the first matrix, in turn, we replace each column with coefficients with a column of numbers after the equal sign. Thus, we obtain several matrices and then find their determinants.

After we have found the determinants, the matter is small. We have an initial matrix, and there are several resulting matrices that correspond to different variables. To get the solutions of the system, we divide the determinant of the resulting table by the determinant of the initial table. The resulting number is the value of one of the variables. Similarly, we find all the unknowns.

Other Methods

There are several more methods for obtaining a solution to systems of linear equations. For example, the so-called Gauss-Jordan method, which is used to find solutions to a system of quadratic equations and is also associated with the use of matrices. There is also a Jacobi method for solving a system of linear algebraic equations. It is the easiest to adapt to a computer and is used in computer technology.

Difficult cases

Complexity usually arises when the number of equations is less than the number of variables. Then we can say for sure that either the system is inconsistent (that is, it has no roots), or the number of its solutions tends to infinity. If we have the second case, then we need to write down the general solution of the system of linear equations. It will contain at least one variable.

Conclusion

Here we come to the end. Let's summarize: we have analyzed what a system and a matrix are, learned how to find a general solution to a system of linear equations. In addition, other options were considered. We found out how a system of linear equations is solved: the Gauss method and We talked about difficult cases and other ways to find solutions.

In fact, this topic is much more extensive, and if you want to better understand it, then we advise you to read more specialized literature.

The Gaussian method has a number of disadvantages: it is impossible to know whether the system is consistent or not until all the transformations necessary in the Gaussian method have been carried out; the Gaussian method is not suitable for systems with letter coefficients.

Consider other methods for solving systems of linear equations. These methods use the concept of the rank of a matrix and reduce the solution of any joint system to the solution of a system to which Cramer's rule applies.

Example 1 Find the general solution of the following system of linear equations using the fundamental system of solutions of the reduced homogeneous system and a particular solution of the inhomogeneous system.

1. We make a matrix A and the augmented matrix of the system (1)

2. Explore the system (1) for compatibility. To do this, we find the ranks of the matrices A and https://pandia.ru/text/78/176/images/image006_90.gif" width="17" height="26 src=">). If it turns out that , then the system (1) incompatible. If we get that , then this system is consistent and we will solve it. (The consistency study is based on the Kronecker-Capelli theorem).

a. We find rA.

To find rA, we will consider successively non-zero minors of the first, second, etc. orders of the matrix A and the minors surrounding them.

M1=1≠0 (1 is taken from the upper left corner of the matrix BUT).

Bordering M1 the second row and second column of this matrix. . We continue to border M1 the second line and the third column..gif" width="37" height="20 src=">. Now we border the non-zero minor М2′ second order.

We have: (because the first two columns are the same)

(because the second and third lines are proportional).

We see that rA=2, and is the basis minor of the matrix A.

b. We find .

Sufficiently basic minor М2′ matrices A border with a column of free members and all lines (we have only the last line).

. It follows from this that М3′′ remains the basis minor of the matrix https://pandia.ru/text/78/176/images/image019_33.gif" width="168 height=75" height="75"> (2)

As М2′- basis minor of the matrix A systems (2) , then this system is equivalent to the system (3) , consisting of the first two equations of the system (2) (for М2′ is in the first two rows of matrix A).

(3)

Since the basic minor is https://pandia.ru/text/78/176/images/image021_29.gif" width="153" height="51"> (4)

In this system, two free unknowns ( x2 and x4 ). So FSR systems (4) consists of two solutions. To find them, we assign free unknowns to (4) values ​​first x2=1 , x4=0 , and then - x2=0 , x4=1 .

At x2=1 , x4=0 we get:

.

This system already has the only thing solution (it can be found by Cramer's rule or by any other method). Subtracting the first equation from the second equation, we get:

Her decision will be x1= -1 , x3=0 . Given the values x2 and x4 , which we have given, we obtain the first fundamental solution of the system (2) : .

Now we put in (4) x2=0 , x4=1 . We get:

.

We solve this system using Cramer's theorem:

.

We obtain the second fundamental solution of the system (2) : .

Solutions β1 , β2 and make up FSR systems (2) . Then its general solution will be

γ= C1 β1+С2β2=С1(-1, 1, 0, 0)+С2(5, 0, 4, 1)=(-С1+5С2, С1, 4С2, С2)

Here C1 , C2 are arbitrary constants.

4. Find one private decision heterogeneous system(1) . As in paragraph 3 , instead of the system (1) consider the equivalent system (5) , consisting of the first two equations of the system (1) .

(5)

We transfer the free unknowns to the right-hand sides x2 and x4.

(6)

Let's give free unknowns x2 and x4 arbitrary values, for example, x2=2 , x4=1 and plug them into (6) . Let's get the system

This system has a unique solution (because its determinant М2′0). Solving it (using the Cramer theorem or the Gauss method), we obtain x1=3 , x3=3 . Given the values ​​of the free unknowns x2 and x4 , we get particular solution of an inhomogeneous system(1)α1=(3,2,3,1).

5. Now it remains to write general solution α of an inhomogeneous system(1) : it is equal to the sum private decision this system and general solution of its reduced homogeneous system (2) :

α=α1+γ=(3, 2, 3, 1)+(‑С1+5С2, С1, 4С2, С2).

It means: (7)

6. Examination. To check if you have solved the system correctly (1) , we need a general solution (7) substitute in (1) . If each equation becomes an identity ( C1 and C2 should be destroyed), then the solution is found correctly.

We will substitute (7) for example, only in the last equation of the system (1) (x1 + x2 + x3 ‑9 x4 =‑1) .

We get: (3–С1+5С2)+(2+С1)+(3+4С2)–9(1+С2)=–1

(С1–С1)+(5С2+4С2–9С2)+(3+2+3–9)=–1

Where -1=-1. We got an identity. We do this with all other equations of the system (1) .

Comment. Verification is usually quite cumbersome. We can recommend the following "partial verification": in the overall solution of the system (1) assign some values ​​to arbitrary constants and substitute the resulting particular solution only into the discarded equations (i.e., into those equations from (1) that are not included in (5) ). If you get identities, then more likely, solution of the system (1) found correctly (but such a check does not give a full guarantee of correctness!). For example, if in (7) put C2=- 1 , C1=1, then we get: x1=-3, x2=3, x3=-1, x4=0. Substituting into the last equation of system (1), we have: - 3+3 - 1 - 9∙0= - 1 , i.e. –1=–1. We got an identity.

Example 2 Find a general solution to a system of linear equations (1) , expressing the main unknowns in terms of free ones.

Decision. As in example 1, compose matrices A and https://pandia.ru/text/78/176/images/image010_57.gif" width="156" height="50"> of these matrices. Now we leave only those equations of the system (1) , the coefficients of which are included in this basic minor (i.e., we have the first two equations) and consider the system consisting of them, which is equivalent to system (1).

Let us transfer the free unknowns to the right-hand sides of these equations.

system (9) we solve by the Gaussian method, considering the right parts as free members.

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Option 2.

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Option 4.

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Option 5.

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Option 6.

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

Any homogeneous system is always consistent, since it always has zero (trivial ) solution. The question arises under what conditions a homogeneous system will have a non-trivial solution.

Theorem 5.2.A homogeneous system has a non-trivial solution if and only if the rank of the underlying matrix is ​​less than the number of its unknowns.

Consequence. A square homogeneous system has a non-trivial solution if and only if the determinant of the main matrix of the system is not equal to zero.

Example 5.6. Determine the values ​​of the parameter l for which the system has nontrivial solutions and find these solutions:

Decision. This system will have a non-trivial solution when the determinant of the main matrix is ​​equal to zero:

Thus, the system is nontrivial when l=3 or l=2. For l=3, the rank of the main matrix of the system is 1. Then, leaving only one equation and assuming that y=a and z=b, we get x=b-a, i.e.

For l=2, the rank of the main matrix of the system is 2. Then, choosing as the basic minor:

we get a simplified system

From here we find that x=z/4, y=z/2. Assuming z=4a, we get

The set of all solutions of a homogeneous system has a very important linear property : if X columns 1 and X 2 - solutions of the homogeneous system AX = 0, then any linear combination of them a X 1+b X 2 will also be the solution of this system. Indeed, since AX 1 = 0 and AX 2 = 0 , then A(a X 1+b X 2) = a AX 1+b AX 2 = a · 0 + b · 0 = 0. Due to this property, if a linear system has more than one solution, then there will be infinitely many of these solutions.

Linearly Independent Columns E 1 , E 2 , E k, which are solutions of a homogeneous system, is called fundamental decision system homogeneous system of linear equations if the general solution of this system can be written as a linear combination of these columns:

If a homogeneous system has n variables, and the rank of the main matrix of the system is equal to r, then k = n-r.

Example 5.7. Find the fundamental system of solutions of the following system of linear equations:

Decision. Find the rank of the main matrix of the system:

Thus, the set of solutions of this system of equations forms a linear subspace of dimension n - r= 5 - 2 = 3. We choose as the basic minor

.

Then, leaving only the basic equations (the rest will be a linear combination of these equations) and the basic variables (we transfer the rest, the so-called free variables to the right), we get a simplified system of equations:

Assuming x 3 = a, x 4 = b, x 5 = c, we find

, .

Assuming a= 1, b=c= 0, we obtain the first basic solution; assuming b= 1, a = c= 0, we obtain the second basic solution; assuming c= 1, a = b= 0, we obtain the third basic solution. As a result, the normal fundamental system of solutions takes the form

Using the fundamental system, the general solution of the homogeneous system can be written as

X = aE 1 + bE 2 + cE 3 . a

Let us note some properties of solutions of the inhomogeneous system of linear equations AX=B and their relationship with the corresponding homogeneous system of equations AX = 0.

General solution of an inhomogeneous systemis equal to the sum of the general solution of the corresponding homogeneous system AX = 0 and an arbitrary particular solution of the inhomogeneous system. Indeed, let Y 0 is an arbitrary particular solution of an inhomogeneous system, i.e. AY 0 = B, and Y is the general solution of an inhomogeneous system, i.e. AY=B. Subtracting one equality from the other, we get
A(Y-Y 0) = 0, i.e. Y-Y 0 is the general solution of the corresponding homogeneous system AX=0. Hence, Y-Y 0 = X, or Y=Y 0 + X. Q.E.D.

Let an inhomogeneous system have the form AX = B 1 + B 2 . Then the general solution of such a system can be written as X = X 1 + X 2 , where AX 1 = B 1 and AX 2 = B 2. This property expresses the universal property of any linear systems in general (algebraic, differential, functional, etc.). In physics, this property is called superposition principle, in electrical and radio engineering - overlay principle. For example, in the theory of linear electrical circuits, the current in any circuit can be obtained as an algebraic sum of the currents caused by each energy source separately.

We will continue to polish the technique elementary transformations on the homogeneous system of linear equations.
According to the first paragraphs, the material may seem boring and ordinary, but this impression is deceptive. In addition to further developing techniques, there will be a lot of new information, so please try not to neglect the examples in this article.

What is a homogeneous system of linear equations?

The answer suggests itself. A system of linear equations is homogeneous if the free term everyone system equation is zero. For example:

It is quite clear that homogeneous system is always consistent, that is, it always has a solution. And, first of all, the so-called trivial decision . Trivial, for those who do not understand the meaning of the adjective at all, means bespontovoe. Not academically, of course, but intelligibly =) ... Why beat around the bush, let's find out if this system has any other solutions:

Example 1

Decision: to solve a homogeneous system it is necessary to write system matrix and with the help of elementary transformations bring it to a stepped form. Note that there is no need to write down the vertical bar and zero column of free members here - because whatever you do with zeros, they will remain zero:

(1) The first row was added to the second row, multiplied by -2. The first line was added to the third line, multiplied by -3.

(2) The second line was added to the third line, multiplied by -1.

Dividing the third row by 3 doesn't make much sense.

As a result of elementary transformations, an equivalent homogeneous system is obtained , and, applying the reverse move of the Gaussian method, it is easy to verify that the solution is unique.

Answer:

Let us formulate an obvious criterion: a homogeneous system of linear equations has only trivial solution, if system matrix rank(in this case, 3) is equal to the number of variables (in this case, 3 pcs.).

We warm up and tune our radio to a wave of elementary transformations:

Example 2

Solve a homogeneous system of linear equations

To finally fix the algorithm, let's analyze the final task:

Example 7

Solve a homogeneous system, write the answer in vector form.

Decision: we write the matrix of the system and, using elementary transformations, we bring it to a stepped form:

(1) The sign of the first line has been changed. Once again, I draw attention to the repeatedly met technique, which allows you to significantly simplify the following action.

(1) The first line was added to the 2nd and 3rd lines. The first line multiplied by 2 was added to the 4th line.

(3) The last three lines are proportional, two of them have been removed.

As a result, a standard step matrix is ​​obtained, and the solution continues along the knurled track:

– basic variables;
are free variables.

We express the basic variables in terms of free variables. From the 2nd equation:

- substitute in the 1st equation:

So the general solution is:

Since there are three free variables in the example under consideration, the fundamental system contains three vectors.

Let's substitute a triple of values into the general solution and obtain a vector whose coordinates satisfy each equation of the homogeneous system. And again, I repeat that it is highly desirable to check each received vector - it will not take so much time, but it will save one hundred percent from errors.

For a triple of values find the vector

And finally for the triple we get the third vector:

Answer: , where

Those wishing to avoid fractional values ​​may consider triplets and get the answer in the equivalent form:

Speaking of fractions. Let's look at the matrix obtained in the problem and ask the question - is it possible to simplify the further solution? After all, here we first expressed the basic variable in terms of fractions, then the basic variable in terms of fractions, and, I must say, this process was not the easiest and not the most pleasant.

The second solution:

The idea is to try choose other basic variables. Let's look at the matrix and notice two ones in the third column. So why not get zero at the top? Let's make one more elementary transformation: