General solution of a homogeneous system of linear equations examples. Systems of linear algebraic equations

To understand what is fundamental system decisions you can watch the video tutorial for the same example by clicking . Now let's move on to the description of all the necessary work. This will help you understand the essence of this issue in more detail.

How to find the fundamental system of solutions of a linear equation?

Let's take this system as an example. linear equations:

Let's find a solution to this linear system of equations. To begin with, we write down the coefficient matrix of the system.

Let's transform this matrix to a triangular one. We rewrite the first line without changes. And all the elements that are under $a_(11)$ must be made zero. To make a zero in the place of the element $a_(21)$, you need to subtract the first from the second line, and write the difference in the second line. To make a zero in the place of the element $a_(31)$, you need to subtract the first from the third row and write the difference in the third row. To make a zero in place of the element $a_(41)$, you need to subtract the first multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(31)$, subtract the first multiplied by 2 from the fifth line and write the difference in the fifth line.

We rewrite the first and second lines without changes. And all the elements that are under $a_(22)$ must be made zero. To make a zero in the place of the element $a_(32)$, it is necessary to subtract the second multiplied by 2 from the third row and write the difference in the third row. To make a zero in the place of the element $a_(42)$, it is necessary to subtract the second multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(52)$, subtract the second multiplied by 3 from the fifth line and write the difference in the fifth line.

We see that the last three lines are the same, so if you subtract the third from the fourth and fifth, then they will become zero.

For this matrix write down a new system of equations.

We see that we have only three linearly independent equations, and five unknowns, so the fundamental system of solutions will consist of two vectors. So we move the last two unknowns to the right.

Now, we begin to express those unknowns that are on the left side through those that are on the right side. We start with the last equation, first we express $x_3$, then we substitute the result obtained into the second equation and express $x_2$, and then into the first equation and here we express $x_1$. Thus, we expressed all the unknowns that are on the left side through the unknowns that are on the right side.

After that, instead of $x_4$ and $x_5$, you can substitute any numbers and find $x_1$, $x_2$ and $x_3$. Each such five numbers will be the roots of our original system of equations. To find the vectors that are included in FSR we need to substitute 1 instead of $x_4$, and substitute 0 instead of $x_5$, find $x_1$, $x_2$ and $x_3$, and then vice versa $x_4=0$ and $x_5=1$.

System m linear equations c n unknown is called system of linear homogeneous equations if all free terms are equal to zero. Such a system looks like:

where and ij (i = 1, 2, …, m; j = 1, 2, …, n) - given numbers; x i- unknown.

Linear system homogeneous equations always compatible, because r(A) = r(). It always has at least zero ( trivial) solution (0; 0; ...; 0).

Let us consider under what conditions homogeneous systems have nonzero solutions.

Theorem 1. A system of linear homogeneous equations has nonzero solutions if and only if the rank of its main matrix r less than number unknown n, i.e. r < n.

□ one). Let the system of linear homogeneous equations have a nonzero solution. Since the rank cannot exceed the size of the matrix, it is obvious that r ≤ n. Let r = n. Then one of the minors of size n n different from zero. Therefore, the corresponding system of linear equations has a unique solution: , , . Hence, there are no solutions other than trivial ones. So if there is non-trivial solution, then r < n.

2). Let r < n. Then a homogeneous system, being consistent, is indefinite. Hence, it has an infinite number of solutions, i.e. also has non-zero solutions. ■

Consider a homogeneous system n linear equations c n unknown:

(2)

Theorem 2. homogeneous system n linear equations c n unknowns (2) has nonzero solutions if and only if its determinant is equal to zero: = 0.

□ If system (2) has a non-zero solution, then = 0. For at , the system has only a unique zero solution. If = 0, then the rank r the main matrix of the system is less than the number of unknowns, i.e. r < n. And, therefore, the system has an infinite number of solutions, i.e. also has non-zero solutions. ■

Denote the solution of system (1) X 1 = k 1 , X 2 = k 2 , …, x n = k n as a string .

Solutions to a system of linear homogeneous equations have the following properties:

1. If the string is a solution to system (1), then the string is also a solution to system (1).

2. If the lines and - solutions of system (1), then for any values With 1 and With 2 their linear combination is also a solution to system (1).

You can check the validity of these properties by directly substituting them into the equations of the system.

It follows from the formulated properties that any linear combination of solutions to a system of linear homogeneous equations is also a solution to this system.

System of linearly independent solutions e 1 , e 2 , …, e r called fundamental, if each solution of system (1) is a linear combination of these solutions e 1 , e 2 , …, e r.

Theorem 3. If rank r the matrix of coefficients for the variables of the system of linear homogeneous equations (1) is less than the number of variables n, then any fundamental system of solutions to system (1) consists of n–r solutions.

That's why common decision system of linear homogeneous equations (1) has the form:

where e 1 , e 2 , …, e r is any fundamental system of solutions to system (9), With 1 , With 2 , …, with p- arbitrary numbers, R = n–r.

Theorem 4. General system solution m linear equations c n unknowns is equal to the sum of the general solution of the corresponding system of linear homogeneous equations (1) and an arbitrary particular solution of this system (1).

Example. Solve the system

Solution. For this system m = n= 3. Determinant

by Theorem 2, the system has only a trivial solution: x = y = z = 0.

Example. 1) Find general and particular solutions of the system

2) Find a fundamental system of solutions.

Solution. 1) For this system m = n= 3. Determinant

by Theorem 2, the system has nonzero solutions.

Since there is only one independent equation in the system

x + y – 4z = 0,

then from it we express x =4z- y. From where we get an infinite set of solutions: (4 z- y, y, z) is the general solution of the system.

At z= 1, y= -1, we get one particular solution: (5, -1, 1). Putting z= 3, y= 2, we get the second particular solution: (10, 2, 3), etc.

2) In the general solution (4 z- y, y, z) variables y and z are free, and the variable X- dependent on them. In order to find the fundamental system of solutions, we assign values ​​to the free variables: first y = 1, z= 0, then y = 0, z= 1. We obtain particular solutions (-1, 1, 0), (4, 0, 1), which form the fundamental system of solutions.

Illustrations:

Rice. 1 Classification of systems of linear equations

Rice. 2 Study of systems of linear equations

Presentations:

Solving SLAE_matrix method

Solution SLAU_Cramer's method

Solution SLAE_Gauss method

・Solution packages math problems Mathematica: search for analytical and numerical solution of systems of linear equations

test questions:

1. Define a linear equation

2. What kind of system does m linear equations with n unknown?

3. What is called the solution of systems of linear equations?

4. What systems are called equivalent?

5. What system is called incompatible?

6. What system is called joint?

7. What system is called defined?

8. What system is called indefinite

9. List the elementary transformations of systems of linear equations

10. List the elementary transformations of matrices

11. Formulate a theorem on the application of elementary transformations to a system of linear equations

12. What systems can be solved by the matrix method?

13. What systems can be solved by Cramer's method?

14. What systems can be solved by the Gauss method?

15. List 3 possible cases that arise when solving systems of linear equations using the Gauss method

16. Describe the matrix method for solving systems of linear equations

17. Describe Cramer's method for solving systems of linear equations

18. Describe the Gauss method for solving systems of linear equations

19. What systems can be solved using inverse matrix?

20. List 3 possible cases that arise when solving systems of linear equations using the Cramer method

Literature:

1. Higher mathematics for economists: Textbook for universities / N.Sh. Kremer, B.A. Putko, I.M. Trishin, M.N. Fridman. Ed. N.Sh. Kremer. - M.: UNITI, 2005. - 471 p.

2. General course of higher mathematics for economists: Textbook. / Ed. IN AND. Ermakov. -M.: INFRA-M, 2006. - 655 p.

3. Collection of problems in higher mathematics for economists: Tutorial/ Under the editorship of V.I. Ermakov. M.: INFRA-M, 2006. - 574 p.

4. V. E. Gmurman, Guide to Problem Solving in Probability Theory and Magmatic Statistics. - M.: graduate School, 2005. - 400 p.

5. Gmurman. V.E. Probability Theory and math statistics. - M.: Higher school, 2005.

6. Danko P.E., Popov A.G., Kozhevnikova T.Ya. Higher mathematics in exercises and tasks. Part 1, 2. - M .: Onyx 21st century: World and education, 2005. - 304 p. Part 1; – 416 p. Part 2

7. Mathematics in Economics: Textbook: In 2 hours / A.S. Solodovnikov, V.A. Babaitsev, A.V. Brailov, I.G. Shandara. - M.: Finance and statistics, 2006.

8. Shipachev V.S. Higher Mathematics: Textbook for students. universities - M .: Higher School, 2007. - 479 p.

Similar information.

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:

Solution. 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, because 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:

Solution. 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 basic variables (the rest, the so-called free variables, we transfer 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. Consequently, 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.

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 the general solution of the system 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 distinguished minor has highest order(of the possible minors) and is different from 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, the fundamental system of 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

Homogeneous systems of linear algebraic equations

Within the lessons Gauss method and Incompatible systems/systems with a common solution we considered inhomogeneous systems of linear equations, where free member(which is usually on the right) at least one of the equations was different from zero.
And now, after a good warm-up with matrix rank, 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

From the article How to find the rank of a matrix? we recall the rational method of incidentally reducing the numbers of the matrix. Otherwise, you will have to butcher large, and often biting fish. An example of an assignment at the end of the lesson.

Zeros are good and convenient, but in practice the case is much more common when the rows of the matrix of the system linearly dependent. And then the appearance of a general solution is inevitable:

Example 3

Solve a homogeneous system of linear equations

Solution: we write the matrix of the system and, using elementary transformations, we bring it to a step form. The first action is aimed not only at obtaining a single value, but also at reducing the numbers in the first column:

(1) The third row was added to the first row, multiplied by -1. The third line was added to the second line, multiplied by -2. At the top left, I got a unit with a "minus", which is often much more convenient for further transformations.

(2) The first two lines are the same, one of them has been removed. Honestly, I didn’t adjust the decision - it happened. If you perform transformations in a template, then linear dependence lines would show up a little later.

(3) To the third line, add the second line, multiplied by 3.

(4) The sign of the first line has been changed.

As a result of elementary transformations, an equivalent system is obtained:

The algorithm works exactly the same as for heterogeneous systems. Variables "sitting on the steps" are the main ones, the variable that did not get the "steps" is free.

We express the basic variables in terms of the free variable:

Answer: common decision:

The trivial solution is included in the general formula, and it is unnecessary to write it separately.

The verification is also carried out according to the usual scheme: the resulting general solution must be substituted into the left side of each equation of the system and a legitimate zero is obtained for all substitutions.

This could be quietly ended, but the solution of a homogeneous system of equations often needs to be represented in vector form by using fundamental decision system. Please temporarily forget about analytical geometry, since now we will talk about vectors in the general algebraic sense, which I slightly opened in an article about matrix rank. Terminology is not necessary to shade, everything is quite simple.