System of linear homogeneous equations and its solutions. Find the general solution of the system and fsr

Matrix data

Find: 1) aA - bB,

Solution: 1) We find sequentially, using the rules for multiplying a matrix by a number and adding matrices ..

2. Find A*B if

Solution: Use the Matrix Multiplication Rule

Answer:

3. For given matrix find the minor M 31 and calculate the determinant.

Solution: Minor M 31 is the determinant of the matrix that is obtained from A

after deleting row 3 and column 1. Find

1*10*3+4*4*4+1*1*2-2*4*10-1*1*4-1*4*3 = 0.

Let's transform the matrix A without changing its determinant (let's make zeros in row 1)

Now we calculate the determinant of matrix A by expansion along row 1

Answer: M 31 = 0, detA = 0

Solve using the Gauss method and the Cramer method.

2x 1 + x 2 + x 3 = 2

x 1 + x 2 + 3x 3 = 6

2x1 + x2 + 2x3 = 5

Solution: Let's check

You can use Cramer's method

System solution: x 1 = D 1 / D = 2, x 2 = D 2 / D = -5, x 3 = D 3 / D = 3

We apply the Gauss method.

We reduce the extended matrix of the system to a triangular form.

For the convenience of calculations, we swap the lines:

Multiply the 2nd row by (k = -1 / 2 = -1 / 2 ) and add to the 3rd one:

Multiply the 1st row by (k = -2 / 2 = -1 ) and add to the 2nd one:

Now the original system can be written as:

x 1 = 1 - (1/2 x 2 + 1/2 x 3)

x 2 = 13 - (6x 3)

From the 2nd line we express

From the 1st line we express

The solution is the same.

Answer: (2; -5; 3)

Find common decision systems and FSR

13x 1 - 4x 2 - x 3 - 4x 4 - 6x 5 = 0

11x 1 - 2x 2 + x 3 - 2x 4 - 3x 5 = 0

5x 1 + 4x 2 + 7x 3 + 4x 4 + 6x 5 = 0

7x 1 + 2x 2 + 5x 3 + 2x 4 + 3x 5 = 0

Solution: Apply the Gauss method. We reduce the extended matrix of the system to a triangular form.

Multiply the 1st row by (-11). Multiply the 2nd row by (13). Let's add the 2nd line to the 1st:

Multiply the 2nd row by (-5). Multiply the 3rd row by (11). Let's add the 3rd line to the 2nd:

Multiply the 3rd row by (-7). Multiply the 4th row by (5). Let's add the 4th line to the 3rd:

The second equation is a linear combination of the rest

Find the rank of the matrix.

The selected minor has the highest order (of all possible minors) and is non-zero (it is equal to the product of the elements on the reciprocal diagonal), hence rang(A) = 2.

This minor is basic. It includes coefficients for unknown x 1, x 2, which means that the unknown x 1, x 2 are dependent (basic), and x 3, x 4, x 5 are free.

The system with the coefficients of this matrix is ​​equivalent to the original system and has the form:

18x2 = 24x3 + 18x4 + 27x5

7x1 + 2x2 = - 5x3 - 2x4 - 3x5

By the method of elimination of unknowns, we find common decision:

x 2 = - 4 / 3 x 3 - x 4 - 3 / 2 x 5

x 1 = - 1 / 3 x 3

We find the fundamental system of solutions (FSR), which consists of (n-r) solutions. In our case, n=5, r=2, therefore, fundamental system solutions consists of 3 solutions, and these solutions must be linearly independent.

For the rows to be linearly independent, it is necessary and sufficient that the rank of the matrix composed of the elements of the rows be equal to the number of rows, i.e. 3.

It is enough to give the free unknowns x 3 ,x 4 ,x 5 values ​​from the rows of the determinant of the 3rd order, different from zero, and calculate x 1 ,x 2 .

The simplest non-zero determinant is the identity matrix.

But here it is more convenient to take

We find using the general solution:

a) x 3 = 6, x 4 = 0, x 5 = 0 Þ x 1 = - 1/3 x 3 = -2, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = - 4 Þ

I FSR decision: (-2; -4; 6; 0; 0)

b) x 3 = 0, x 4 = 6, x 5 = 0 Þ x 1 = - 1/3 x 3 = 0, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = - 6 Þ

II FSR decision: (0; -6; 0; 6; 0)

c) x 3 = 0, x 4 = 0, x 5 = 6 Þ x 1 = - 1/3 x 3 = 0, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = -9 Þ

III FSR decision: (0; - 9; 0; 0; 6)

Þ FSR: (-2; -4; 6; 0; 0), (0; -6; 0; 6; 0), (0; - 9; 0; 0; 6)

6. Given: z 1 \u003d -4 + 5i, z 2 \u003d 2 - 4i. Find: a) z 1 - 2z 2 b) z 1 z 2 c) z 1 / z 2

Solution: a) z 1 – 2z 2 = -4+5i+2(2-4i) = -4+5i+4-8i = -3i

b) z 1 z 2 = (-4+5i)(2-4i) = -8+10i+16i-20i 2 = (i 2 = -1) = 12 + 26i

Answer: a) -3i b) 12+26i c) -1.4 - 0.3i

The linear equation is called homogeneous if its intercept is zero, and inhomogeneous otherwise. System consisting of homogeneous equations, is called homogeneous and has general form:

Obviously, any homogeneous system is consistent and has a zero (trivial) solution. Therefore, for homogeneous systems linear equations one often has to look for an answer to the question of the existence of nonzero solutions. The answer to this question can be formulated as the following theorem.

Theorem . A homogeneous system of linear equations has a nonzero solution if and only if its rank is less than number unknown .

Proof: Suppose a system whose rank is equal has a nonzero solution. Obviously, does not exceed . In the case the system has a unique solution. Since the system of homogeneous linear equations always has a zero solution, it is precisely the zero solution that will be this unique solution. Thus, nonzero solutions are possible only for .

Corollary 1 : A homogeneous system of equations, in which the number of equations is less than the number of unknowns, always has a nonzero solution.

Proof: If the system of equations has , then the rank of the system does not exceed the number of equations , i.e. . Thus, the condition is satisfied and, therefore, the system has a nonzero solution.

Consequence 2 : A homogeneous system of equations with unknowns has a nonzero solution if and only if its determinant is zero.

Proof: Suppose a system of linear homogeneous equations whose matrix with determinant has a nonzero solution. Then, according to the proved theorem, , which means that the matrix is ​​degenerate, i.e. .

Homogeneous system of linear algebraic equations .

A system of m linear equations with n variables is called a system of linear homogeneous equations if all free terms are equal to 0. A system of linear homogeneous equations is always compatible, because it always has at least a zero solution. A system of linear homogeneous equations has a nonzero solution if and only if the rank of its matrix of coefficients at variables is less than the number of variables, i.e. for rank A (n. Any linear combination

solutions of the system of lines. homogeneous ur-ii is also a solution to this system.

A system of linearly independent solutions e1, e2,…,ek is called fundamental if each solution of the system is a linear combination of solutions. Theorem: if the rank r of the matrix of coefficients at the variables of the system of linear homogeneous equations is less than the number of variables n, then any fundamental system of solutions of the system consists of n-r solutions. Therefore, the general solution of the system of lines. single ur-th has the form: c1e1+c2e2+…+ckek, where e1, e2,…, ek is any fundamental system of solutions, c1, c2,…,ck are arbitrary numbers and k=n-r. The general solution of a system of m linear equations with n variables is equal to the sum

the general solution of the system corresponding to it is homogeneous. linear equations and an arbitrary particular solution of this system.

7. Linear spaces. Subspaces. Basis, dimension. Linear shell. Linear space is called n-dimensional, if it contains a system of linearly independent vectors, and any system of more vectors is linearly dependent. The number is called dimension (number of measurements) linear space and is denoted by . In other words, the dimension of a space is the maximum number of linearly independent vectors in that space. If such a number exists, then the space is said to be finite-dimensional. If for any natural number n in space there is a system consisting of linearly independent vectors, then such a space is called infinite-dimensional (write: ). In what follows, unless otherwise stated, finite-dimensional spaces will be considered.

The basis of an n-dimensional linear space is an ordered set of linearly independent vectors ( basis vectors).

Theorem 8.1 on the expansion of a vector in terms of a basis. If is a basis of an n-dimensional linear space, then any vector can be represented as a linear combination of basis vectors:

V=v1*e1+v2*e2+…+vn+en
and, moreover, in a unique way, i.e. coefficients are uniquely determined. In other words, any space vector can be expanded in a basis and, moreover, in a unique way.

Indeed, the dimension of the space is . The system of vectors is linearly independent (this is the basis). After joining any vector to the basis, we get a linearly dependent system (since this system consists of vectors in an n-dimensional space). By the property of 7 linearly dependent and linearly independent vectors, we obtain the conclusion of the theorem.

Solving systems of linear algebraic equations (SLAE) is undoubtedly the most important topic of the linear algebra course. Great amount problems from all branches of mathematics is 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, we will focus on the Cramer method, secondly, we will show the matrix method for solving such systems of equations, thirdly, we will analyze the Gauss method (the method sequential exclusion 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 basic minor matrices. 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 reduce to linear ones, as well as various tasks, the solution of which gives rise to SLAEs.

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 began to study such SLAEs 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 .

Solution.

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.

Solution.

Let's rewrite the system of equations in matrix form:

Because

then the SLAE can be solved by the matrix method. By using 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 a solution to systems of linear algebraic equations by the matrix method is the complexity of finding the inverse matrix, especially for square matrices 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.

Solution.

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.

Solution.

. 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.

Minor highest order matrix A that is non-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.

.

Solution.

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 .

Solution.

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.

Watch it 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 coefficientsС 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 sets everything possible solutions the original SLAE, in other words, taking any set of values ​​of arbitrary constants С 1 , С 2 , …, С (n-r) , according to the formula we 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 .

Solution.

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
.

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 solution . 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

Solution: 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 - after all, 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.

Solution: 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:

Let M 0 is the set of solutions of the homogeneous system (4) of linear equations.

Definition 6.12. Vectors With 1 ,With 2 , …, with p, which are solutions of a homogeneous system of linear equations, are called fundamental set of solutions(abbreviated FNR) if

1) vectors With 1 ,With 2 , …, with p linearly independent (that is, none of them can be expressed in terms of the others);

2) any other solution of a homogeneous system of linear equations can be expressed in terms of solutions With 1 ,With 2 , …, with p.

Note that if With 1 ,With 2 , …, with p is some f.n.r., then by the expression k 1× With 1 + k 2× With 2 + … + kp× with p can describe the whole set M 0 solutions to system (4), so it is called general view of the system solution (4).

Theorem 6.6. Any indefinite homogeneous system of linear equations has a fundamental set of solutions.

The way to find the fundamental set of solutions is as follows:

Find the general solution of a homogeneous system of linear equations;

Build ( n – r) partial solutions of this system, while the values ​​of the free unknowns must form an identity matrix;

Write out the general form of the solution included in M 0 .

Example 6.5. Find the fundamental set of solutions of the following system:

Solution. Let us find the general solution of this system.

~ ~ ~ ~ Þ Þ Þ This system has five unknowns ( n= 5), of which there are two principal unknowns ( r= 2), three free unknowns ( n – r), that is, the fundamental set of solutions contains three solution vectors. Let's build them. We have x 1 and x 3 - main unknowns, x 2 , x 4 , x 5 - free unknowns

Values ​​of free unknowns x 2 , x 4 , x 5 form the identity matrix E third order. Got that vectors With 1 ,With 2 , With 3 form f.n.r. this system. Then the set of solutions of this homogeneous system will be M 0 = {k 1× With 1 + k 2× With 2 + k 3× With 3 , k 1 , k 2 , k 3 О R).

Let us now find out the conditions for the existence of nonzero solutions of a homogeneous system of linear equations, in other words, the conditions for the existence of a fundamental set of solutions.

A homogeneous system of linear equations has non-zero solutions, that is, it is indefinite if

1) the rank of the main matrix of the system is less than the number of unknowns;

2) in a homogeneous system of linear equations, the number of equations is less than the number of unknowns;

3) if in a homogeneous system of linear equations the number of equations is equal to the number of unknowns, and the determinant of the main matrix is ​​equal to zero (i.e. | A| = 0).

Example 6.6. At what value of the parameter a homogeneous system of linear equations has non-zero solutions?

Solution. Let's compose the main matrix of this system and find its determinant: = = 1×(–1) 1+1 × = – a– 4. The determinant of this matrix is ​​equal to zero when a = –4.

Answer: –4.

7. Arithmetic n-dimensional vector space

Basic concepts

In the previous sections, we already encountered the concept of a set of real numbers arranged in a certain order. This is a row matrix (or column matrix) and a solution to a system of linear equations with n unknown. This information can be summarized.

Definition 7.1. n-dimensional arithmetic vector is called an ordered set of n real numbers.

Means a= (a 1 , a 2 , …, a n), where a iО R, i = 1, 2, …, n is the general view of the vector. Number n called dimension vector, and the numbers a i called him coordinates.

For example: a= (1, –8, 7, 4, ) is a five-dimensional vector.

All set n-dimensional vectors are usually denoted as R n.

Definition 7.2. Two vectors a= (a 1 , a 2 , …, a n) and b= (b 1 , b 2 , …, b n) of the same dimension equal if and only if their respective coordinates are equal, i.e. a 1 = b 1 , a 2 = b 2 , …, a n= b n.

Definition 7.3.sum two n-dimensional vectors a= (a 1 , a 2 , …, a n) and b= (b 1 , b 2 , …, b n) is called a vector a + b= (a 1 + b 1 , a 2 + b 2 , …, a n+b n).

Definition 7.4. work real number k per vector a= (a 1 , a 2 , …, a n) is called a vector k× a = (k×a 1 , k×a 2 , …, k×a n)

Definition 7.5. Vector about= (0, 0, …, 0) is called zero(or null-vector).

It is easy to check that the actions (operations) of adding vectors and multiplying them by a real number have the following properties: a, b, c Î R n, " k, lОR:

1) a + b = b + a;

2) a + (b+ c) = (a + b) + c;

3) a + about = a;

4) a+ (–a) = about;

5) 1× a = a, 1 О R;

6) k×( l× a) = l×( k× a) = (l× k)× a;

7) (k + l)× a = k× a + l× a;

8) k×( a + b) = k× a + k× b.

Definition 7.6. Lots of R n with the operations of adding vectors and multiplying them by a real number given on it is called arithmetic n-dimensional vector space.