Find nonzero solutions of the homogeneous slough. Find the general solution of the system and fsr
We will continue to polish our technology elementary transformations on homogeneous system linear equations
.
Based on the first paragraphs, the material may seem boring and mediocre, but this impression is deceptive. In addition to further development of technical techniques, there will be many 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 equation of the system is zero. For example:
It is absolutely clear that a homogeneous system is always consistent, that is, it always has a solution. And, first of all, what catches your eye is the so-called trivial solution 
. Trivial, for those who do not understand the meaning of the adjective at all, means without a show-off. 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. Please note that here there is no need to write down the vertical bar and the zero column of free terms - after all, no matter what you do with zeros, they will remain zeros: 
(1) The first line was added to the second line, 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 line by 3 doesn't make much sense.
As a result of elementary transformations, an equivalent homogeneous system is obtained 
, and, using the inverse 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 just a trivial solution, If system matrix rank(in this case 3) is equal to the number of variables (in this case – 3 pieces).
Let's warm up and tune our radio to the wave of elementary transformations:
Example 2
Solve a homogeneous system of linear equations 
To finally consolidate the algorithm, let’s analyze the final task:
Example 7
Solve a homogeneous system, write the answer in vector form. 
Solution: let’s write down the matrix of the system and, using elementary transformations, bring it to a stepwise form: 
(1) The sign of the first line has been changed. Once again I draw attention to a technique that has been encountered many times, which allows you to significantly simplify the next 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.
The result is a standard step matrix, and the solution continues along the knurled track:
– basic variables;
– free variables.
Let us express the basic variables in terms of free variables. From the 2nd equation:
– substitute into the 1st equation:
So the general solution is:
Since in the example under consideration there are three free variables, 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 advisable to check each received vector - it will not take much time, but it will completely protect you from errors.
For a triple of values 
find the vector
And finally for the three 
we get the third vector:
Answer: , Where
Those wishing to avoid fractional values may consider triplets 
and get an answer in equivalent form:
Speaking of fractions. Let's look at the matrix obtained in the problem 
and let us ask ourselves: is it possible to simplify the further solution? After all, here we first expressed the basic variable through fractions, then through fractions the basic variable, and, I must say, this process was not the simplest and not the most pleasant.
Second solution:
The idea is to try choose other basis variables. Let's look at the matrix and notice two ones in the third column. So why not have a zero at the top? Let's carry out one more elementary transformation:
Given matrices
Find: 1) aA - bB,
Solution: 1) We find it sequentially, using the rules of multiplying a matrix by a number and adding matrices..
2. Find A*B if
Solution: We use the matrix multiplication rule
Answer: 
3. For a 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 crossing out line 3 and column 1. We find
1*10*3+4*4*4+1*1*2-2*4*10-1*1*4-1*4*3 = 0.
Let's transform matrix A without changing its determinant (let's make zeros in row 1)
| -3*, -, -4* | |||
| -10 | -15 | ||
| -20 | -25 | ||
| -4 | -5 |
Now we calculate the determinant of matrix A by decomposing 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
2x 1 + x 2 + 2x 3 = 5
Solution: Let's check
You can use Cramer's method
Solution of the system: x 1 = D 1 / D = 2, x 2 = D 2 / D = -5, x 3 = D 3 / D = 3
Let's apply the Gaussian method.
Let us reduce the extended matrix of the system to triangular form.
For ease of calculation, let's swap the lines:
Multiply the 2nd line by (k = -1 / 2 = -1 / 2 ) and add to the 3rd:
| 1 / 2 | 7 / 2 |
Multiply the 1st line by (k = -2 / 2 = -1 ) and add to the 2nd:
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 the 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: Let's apply the Gaussian method. Let us reduce the extended matrix of the system to triangular form.
| -4 | -1 | -4 | -6 | |
| -2 | -2 | -3 | ||
| x 1 | x 2 | x 3 | x 4 | x 5 |
Multiply the 1st line by (-11). Let's multiply the 2nd line by (13). Let's add the 2nd line to the 1st:
| -2 | -2 | -3 | ||
Multiply the 2nd line by (-5). Let's multiply the 3rd line by (11). Let's add the 3rd line to the 2nd:
Multiply the 3rd line by (-7). Let's multiply the 4th line by (5). Let's add the 4th line to the 3rd:
The second equation is a linear combination of the others
Let's find the rank of the matrix.
| -18 | -24 | -18 | -27 | |
| x 1 | x 2 | x 3 | x 4 | x 5 |
The selected minor has the highest order (of possible minors) and is non-zero (it is equal to the product of the elements on the reverse diagonal), therefore rang(A) = 2.
This minor is basic. It includes coefficients for the unknowns x 1 , x 2 , which means that the unknowns 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:
18x 2 = 24x 3 + 18x 4 + 27x 5
7x 1 + 2x 2 = - 5x 3 - 2x 4 - 3x 5
Using the method of eliminating 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 fundamental system 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 row elements be equal to the number of rows, that is, 3.
It is enough to give the free unknowns x 3 , x 4 , x 5 values from the lines of the 3rd order determinant, non-zero, and calculate x 1 , x 2 .
The simplest non-zero determinant is the identity matrix.
But it’s more convenient to take here
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 decision of the FSR: (-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 solution: (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 decision of the FSR: (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 = -4 + 5i, z 2 = 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
To understand what it is fundamental decision system you can watch a video tutorial for the same example by clicking. Now let's move on to the actual 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 to a linear equation?
Let's take for example the following system of linear equations:
Let's find a solution to this linear system equations To begin with, we it is necessary to write out the matrix of system coefficients.
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 zeros. To make a zero in 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 place of the element $a_(31)$, you need to subtract the first from the third line and write the difference in the third line. 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 $a_(31)$, you need to 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 place of the element $a_(32)$, you need to subtract the second one multiplied by 2 from the third line and write the difference in the third line. To make a zero in place of the element $a_(42)$, you need 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 $a_(52)$, you need to 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, they will become zero. 
According to this matrix write down new system 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 we need to 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 resulting result 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. 
Then instead of $x_4$ and $x_5$, we can substitute any numbers and find $x_1$, $x_2$ and $x_3$. Each five of these 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 $x_5=1$.
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; Gauss's method is not suitable for systems with letter coefficients.
Let's consider other methods for solving systems of linear equations. These methods use the concept of matrix rank and reduce the solution of any consistent system to the solution of a system to which Cramer's rule applies.
Example 1. Find a general solution next system linear equations using a fundamental system of solutions to the reduced homogeneous system and a particular solution to the inhomogeneous system.
1. Making a matrix A and extended system matrix (1)
2. Explore the system (1) for togetherness. 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 compatibility study is based on the Kronecker-Capelli theorem).
a. We find rA.
To find rA, we will consider sequentially non-zero minors of the first, second, etc. orders of the matrix A and the minors surrounding them.
M1=1≠0 (we take 1 from the upper left corner of the matrix A).
We border 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 M2′ second order.
We have: 
(since the first two columns are the same)
(since the second and third lines are proportional).
We see that rA=2, a is the basis minor of the matrix A.
b. We find.
Quite basic minor M2′ matrices A border with a column of free terms and all rows (we have only the last row).
. It follows that M3′′ remains the basic minor of the matrix https://pandia.ru/text/78/176/images/image019_33.gif" width="168 height=75" height="75"> (2)
Because M2′- 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 M2′ is in the first two rows of matrix A).
(3)
Since the basic minor https://pandia.ru/text/78/176/images/image021_29.gif" width="153" height="51"> (4)
In this system there are two free unknowns ( x2 And x4 ). That's why FSR systems (4) consists of two solutions. To find them, we assign free unknowns in (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 using Cramer's rule or any other method). Subtracting the first from the second equation, we get:
Her solution will be x1= -1 , x3=0 . Given the values x2 And x4 , which we added, we obtain the first fundamental solution of the system (2) : .
Now we believe 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 – arbitrary constants.
4. Let's find one private solution heterogeneous system(1) . As in paragraph 3 , instead of the system (1) Let's consider an equivalent system (5) , consisting of the first two equations of the system (1) .
(5)
Let us move the free unknowns to the right sides x2 And x4.
(6)
Let's give free unknowns x2 And x4 arbitrary values, for example, x2=2 , x4=1 and put them in (6) . Let's get the system
This system has a unique solution (since its determinant M2′0). Solving it (using Cramer’s theorem or Gauss’s 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 all that remains is to write it down general solution α of an inhomogeneous system(1) : it is equal to the sum private solution this system and general solution of its reduced homogeneous system (2) :
α=α1+γ=(3, 2, 3, 1)+(‑С1+5С2, С1, 4С2, С2).
This means: 
(7)
6. Examination. To check if you solved the system correctly (1) , we need a general solution (7) substitute in (1) . If each equation turns into the identity ( C1 And C2 must be destroyed), then the solution is found correctly.
We'll substitute (7) for example, only 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 the other equations of the system (1) .
Comment. The check is usually quite cumbersome. The following “partial check” can be recommended: in the general solution of the system (1) assign some values to arbitrary constants and substitute the resulting partial solution only into the discarded equations (i.e., into those equations from (1) , which were not included in (5) ). If you get identities, then more likely, system solution (1) found correctly (but such a check does not provide a complete 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 basic unknowns in terms of free ones.
Solution. 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 a system consisting of them, 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-hand sides as free terms.
https://pandia.ru/text/78/176/images/image035_21.gif" width="202 height=106" height="106">
Option 2.
https://pandia.ru/text/78/176/images/image039_16.gif" width="192" height="106 src=">
Option 4.
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Option 5.
https://pandia.ru/text/78/176/images/image044_12.gif" width="179 height=106" height="106">
Option 6.
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System m linear equations c n called unknowns 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.
A system of linear homogeneous equations is always consistent, since r(A) = r(). It always has at least zero ( trivial) solution (0; 0; …; 0).
Let us consider under what conditions homogeneous systems have non-zero solutions.
Theorem 1. A system of linear homogeneous equations has nonzero solutions if and only if the rank of its main matrix is r less number unknown n, i.e. r < n.
□
1). Let a system of linear homogeneous equations have a nonzero solution. Since the rank cannot exceed the size of the matrix, then, obviously, r ≤ n. Let r = n. Then one of the minor sizes n n different from zero. Therefore, the corresponding system of linear equations has a unique solution: 
. . . This means that there are no other solutions other than trivial ones. So, if there is a non-trivial solution, then r < n.
2). Let r < n. Then the homogeneous system, being consistent, is uncertain. This means that it has an infinite number of solutions, i.e. 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 non-zero solutions if and only if its determinant is equal to zero: = 0.
□ If system (2) has a non-zero solution, then = 0. Because when the system has only a single 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. has non-zero solutions. ■
Let us denote the solution of system (1) X 1 = k 1 , X 2 = k 2 , …, x n = k n as a string 
.
Solutions of a system of linear homogeneous equations have the following properties:
1.
If the line is a solution to system (1), then the line is 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).
The validity of these properties can be verified by directly substituting them into the equations of the system.
From the formulated properties it follows 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 matrices of coefficients for variables of the system of linear homogeneous equations (1) are less than the number of variables n, then any fundamental system of solutions to system (1) consists of n–r decisions.
That's why common decision system of linear homogeneous equations (1) has the form:
Where e 1 , e 2 , …, e r– any fundamental system of solutions to system (9), With 1 , With 2 , …, with p– arbitrary numbers, R = n–r.
Theorem 4. General solution of the system 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 the 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 will express x =4z- y. Where do we get an infinite number of solutions: (4 z- y, y, z) – this 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, let’s assign values to the free variables: first y = 1, z= 0, then y = 0, z= 1. We obtain partial 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:
· Solution SLAE_matrix method
· Solution SLAE_Cramer method
· Solution SLAE_Gauss method
· Solution packages mathematical problems Mathematica, MathCad: search for analytical and numerical solutions to systems of linear equations
1. Define a linear equation
2. What type of system does it look like? m linear equations with n unknown?
3. What is called solving systems of linear equations?
4. What systems are called equivalent?
5. Which system is called incompatible?
6. What system is called joint?
7. Which system is called definite?
8. Which 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 using 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 Gauss’s 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. Friedman. Ed. N.Sh. Kremer. – M.: UNITY, 2005. – 471 p.
2. General course of higher mathematics for economists: Textbook. / Ed. IN AND. Ermakova. –M.: INFRA-M, 2006. – 655 p.
3. Collection of problems in higher mathematics for economists: Tutorial/ Edited by V.I. Ermakova. M.: INFRA-M, 2006. – 574 p.
4. Gmurman V. E. Guide to solving problems in probability theory and magmatic statistics. - M.: graduate School, 2005. – 400 p.
5. Gmurman. V.E Probability theory and mathematical statistics. - M.: Higher School, 2005.
6. Danko P.E., Popov A.G., Kozhevnikova T.Ya. Higher mathematics in exercises and problems. Part 1, 2. – M.: Onyx 21st century: Peace and Education, 2005. – 304 p. Part 1; – 416 p. Part 2.
7. Mathematics in economics: Textbook: In 2 parts / A.S. Solodovnikov, V.A. Babaytsev, 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.
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