A homogeneous system is always consistent and has a trivial solution 
. For a nontrivial solution to exist, it is necessary that the rank of the matrix 
was less number unknown:
.
Fundamental system of solutions
homogeneous system 
call a system of solutions in the form of column vectors 
, which correspond to the canonical basis, i.e. basis in which arbitrary constants 
are alternately set equal to one, while the rest are set to zero.
Then the general solution of the homogeneous system has the form:
Where 
- arbitrary constants. In other words, the overall solution is a linear combination of the fundamental system of solutions.
Thus, basic solutions can be obtained from the general solution if the free unknowns are given the value of one in turn, setting all others equal to zero.
Example. Let's find a solution to the system
Let's accept , then we get a solution in the form:
Let us now construct a fundamental system of solutions:
.
The general solution will be written as:
Solutions of a system of homogeneous linear equations have properties:
In other words, any linear combination of solutions to a homogeneous system is again a solution.
Solving systems of linear equations using the Gauss method
Solving systems of linear equations has interested mathematicians for several centuries. The first results were obtained in the 18th century. In 1750, G. Kramer (1704–1752) published his works on the determinants of square matrices and proposed an algorithm for finding the inverse matrix. In 1809, Gauss outlined a new solution method known as the method of elimination.
The Gauss method, or the method of sequential elimination of unknowns, consists in the fact that, using elementary transformations, a system of equations is reduced to an equivalent system of a step (or triangular) form. Such systems make it possible to sequentially find all unknowns in a certain order.
Let us assume that in system (1) 
(which is always possible).
(1)
Multiplying the first equation one by one by the so-called suitable numbers
and adding the result of multiplication with the corresponding equations of the system, we obtain an equivalent system in which in all equations except the first there will be no unknown X 1
(2)
Let us now multiply the second equation of system (2) by suitable numbers, assuming that 
,
and adding it with the lower ones, we eliminate the variable 
from all equations, starting from the third.
Continuing this process, after 
step we get:
(3)
If at least one of the numbers 
is not equal to zero, then the corresponding equality is contradictory and system (1) is inconsistent. Conversely, for any joint number system 
are equal to zero. Number 
is nothing more than the rank of the matrix of system (1).
The transition from system (1) to (3) is called straight ahead Gauss method, and finding the unknowns from (3) – in reverse .
Comment : It is more convenient to carry out transformations not with the equations themselves, but with the extended matrix of the system (1).
Example. Let's find a solution to the system
.
Let's write the extended matrix of the system:
.
Let's add the first one to lines 2,3,4, multiplied by (-2), (-3), (-2) respectively:
.
Let's swap rows 2 and 3, then in the resulting matrix add row 2 to row 4, multiplied by 
:
.
Add to line 4 line 3 multiplied by 
:
.
It's obvious that 
, therefore, the system is consistent. From the resulting system of equations
we find the solution by reverse substitution:
,
,
,
.
Example 2. Find a solution to the system:
.
It is obvious that the system is inconsistent, because 
, A 
.
Advantages of the Gauss method :
Less labor intensive than Cramer's method.
Unambiguously establishes the compatibility of the system and allows you to find a solution.
Makes it possible to determine the rank of any matrices.
Homogeneous systems of linear algebraic equations
As part of the lessons Gaussian 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 from 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 homogeneous system of linear equations.
Based on the first paragraphs, the material may seem boring and mediocre, but this impression is deceptive. In addition to further development 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 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 stepwise 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, applying reverse stroke Gauss's 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(V in this case 3) 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 
From the article How to find the rank of a matrix? Let us recall the rational technique of simultaneously decreasing the matrix numbers. Otherwise, you will have to cut large, and often biting fish. Approximate sample completing the 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 system matrix linearly dependent. And then the emergence of a general solution is inevitable:
Example 3
Solve a homogeneous system of linear equations 
Solution: let's write down the matrix of the system and, using elementary transformations, bring it to a stepwise form. The first action is aimed not only at obtaining a single value, but also at decreasing the numbers in the first column:
(1) A third line was added to the first line, 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 was deleted. Honestly, I didn’t push the solution - it turned out that way. If you perform transformations in a template manner, then linear dependence lines would have been revealed a little later.
(3) The second line was added to the third line, multiplied by 3.
(4) The sign of the first line was changed.
As a result of elementary transformations, an equivalent system was obtained: 
The algorithm works exactly the same as for heterogeneous systems. The variables “sitting on the steps” are the main ones, the variable that did not get a “step” is free.
Let's express the basic variables through a free variable: 
Answer: common decision: 
The trivial solution is included in general formula, and it is unnecessary to write it down separately.
The check 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 legal zero must be obtained for all substitutions.
It would be possible to finish this quietly and peacefully, but the solution to a homogeneous system of equations often needs to be represented in vector form by using fundamental system of solutions. Please forget about it for now analytical geometry, since now we will talk about vectors in the general algebraic sense, which I opened a little in the article about matrix rank. There is no need to gloss over the terminology, everything is quite simple.
Linear systems homogeneous equations - has the form ∑a k i x i = 0. where m > n or m A homogeneous system of linear equations is always consistent, since rangA = rangB. It obviously has a solution consisting of zeros, which is called trivial.Purpose of the service. The online calculator is designed to find a non-trivial and fundamental solution to the SLAE. The resulting solution is saved in a Word file (see example solution).
Instructions. Select matrix dimension:
Properties of systems of linear homogeneous equations
In order for the system to have non-trivial solutions, it is necessary and sufficient that the rank of its matrix be less than the number of unknowns.Theorem. A system in the case m=n has a nontrivial solution if and only if the determinant of this system is equal to zero.
Theorem. Any linear combination of solutions to a system is also a solution to that system.
Definition. The set of solutions to a system of linear homogeneous equations is called fundamental system of solutions, if this set consists of linearly independent solutions and any solution to the system is a linear combination of these solutions.
Theorem. If the rank r of the system matrix is less than the number n of unknowns, then there is fundamental system solutions, consisting of (n-r) solutions.
Algorithm for solving systems of linear homogeneous equations
- Finding the rank of the matrix.
- We select the basic minor. We distinguish dependent (basic) and free unknowns.
- We cross out those equations of the system whose coefficients are not included in the basis minor, since they are consequences of the others (according to the theorem on the basis minor).
- We move the terms of the equations containing free unknowns to the right side. As a result, we obtain a system of r equations with r unknowns, equivalent to the given one, the determinant of which is nonzero.
- We solve the resulting system by eliminating unknowns. We find relationships expressing dependent variables through free ones.
- If the rank of the matrix is not equal to the number of variables, then we find the fundamental solution of the system.
- In the case rang = n we have a trivial solution.
Example. Find the basis of the system of vectors (a 1, a 2,...,a m), rank and express the vectors based on the base. If a 1 =(0,0,1,-1), and 2 =(1,1,2,0), and 3 =(1,1,1,1), and 4 =(3,2,1 ,4), and 5 =(2,1,0,3).
Let's write down the main matrix of the system:
Multiply the 3rd line by (-3). Let's add the 4th line to the 3rd:
| 0 | 0 | 1 | -1 |
| 0 | 0 | -1 | 1 |
| 0 | -1 | -2 | 1 |
| 3 | 2 | 1 | 4 |
| 2 | 1 | 0 | 3 |
Multiply the 4th line by (-2). Let's multiply the 5th line by (3). Let's add the 5th line to the 4th:
Let's add the 2nd line to the 1st:
Let's find the rank of the matrix.
The system with the coefficients of this matrix is equivalent to the original system and has the form:
- x 3 = - x 4
- x 2 - 2x 3 = - x 4
2x 1 + x 2 = - 3x 4
Using the method of eliminating unknowns, we find a nontrivial solution:
We obtained relations expressing the dependent variables x 1 , x 2 , x 3 through the free ones x 4 , that is, we found a general solution:
x 3 = x 4
x 2 = - x 4
x 1 = - x 4
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 expansion along row 1
Answer: M 31 = 0, detA = 0
Solve using the Gauss method and 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). 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 highlighted minor has highest order(of the 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 a fundamental system of solutions (FSD), 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
Back in school, each of us studied equations and, most likely, 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 that 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 familiar 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. And it's true best example equality cannot be invented.
The founder of modern letter designations for unknowns and signs of degrees is a French mathematician. However, his designations were significantly different from those of today. For example, he denoted a square of an unknown number with the letter Q (lat. “quadratus”), and a cube with the letter C (lat. “cubus”). This notation seems awkward now, but at the time it was the most understandable way to write systems of linear algebraic equations.
However, a flaw in the solution methods of that time was that mathematicians only considered positive roots. Perhaps this is due to the fact that negative values didn't have any practical application. One way or another, it was the Italian mathematicians Niccolo Tartaglia, Gerolamo Cardano and Raphael Bombelli who were the first to count negative roots in the 16th century. A modern look, the main solution method (via the discriminant) was created only in the 17th century thanks to the work of Descartes and Newton.
In the mid-18th century, Swiss mathematician Gabriel Cramer found new way in order to make solving systems of linear equations easier. This method was later named after him and we still use it to this day. But we’ll talk about Cramer’s method a little later, but for now let’s discuss linear equations and methods for solving them separately from the system.
Linear equations
Linear equations are the simplest equations with a variable (variables). They are classified as algebraic. write to general view so: a 1 *x 1 +a 2* x 2 +...a n *x n =b. We will need to represent them in this form when compiling systems and matrices later.
Systems of linear algebraic equations
The definition of this term is: it is a set of equations that have common unknown quantities and a common solution. As a rule, at school everyone solved 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 will be convenient to solve in the future. First, systems of linear algebraic equations will look better if all variables are written as x with the appropriate subscript: 1,2,3, and so on. Secondly, all equations should be brought to canonical form: a 1 *x 1 +a 2* x 2 +...a n *x n =b.
After all these steps, we can begin to talk about how to find solutions to systems of linear equations. Matrices will be 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 be either specific values or variables. Most often, to indicate elements, subscripts are placed under them (for example, a 11 or a 23). The first index means the row number, and the second - the column number. Various operations can be performed on matrices, like on any other mathematical element. Thus, you can:
2) Multiply a matrix by any 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.
Let's discuss all these techniques in more detail, as they will be useful to us in the future. Subtracting and adding matrices is very simple. Since we take matrices of the same size, each element of one table correlates with each element of the other. Thus, we add (subtract) these two elements (it is important that they stand in the same places in their matrices). When multiplying a matrix by a number or vector, you simply multiply each element of the matrix by that number (or vector). Transposition is a very interesting process. It's very interesting to see him sometimes real life, for example, when changing the orientation of a tablet or phone. The icons on the desktop represent a matrix, and when the position changes, it transposes and becomes wider, but decreases in height.
Let's look at another process like: Although we won't need it, 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. Let's 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 in further using a similar method.
Now we can begin to consider how a system of linear equations is solved.
Gauss method
This topic begins to be covered in school. We know the concept of “a system of two linear equations” well 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 don’t have to transform it and solve it in its pure form.
So, how does this method solve the system of linear Gaussian equations? 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 ultimately reduce the entire set to a stepwise form. That is, it is necessary that from top to bottom (if arranged 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 there are three unknowns, in the second there are two, in the third there is 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 have the skills of adding and subtracting matrices, and you also need to be able to find determinants. Therefore, if you do all this poorly or don’t 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 must construct a matrix of 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 arrange them in a table in the order in which they are written in the system. If there is a “-” sign in front of the number, then we write down a negative coefficient. So, we have compiled the first matrix of coefficients for unknowns, not including the numbers after the equal signs (naturally, the equation should be reduced to canonical form, when only the number is on the right, and all the unknowns with coefficients are on the left). Then you need to create several more matrices - one for each variable. To do this, we replace each column with coefficients in the first matrix in turn 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, it's a small matter. We have an initial matrix, and there are several resulting matrices that correspond to different variables. To obtain solutions to 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 other methods for obtaining solutions to systems of linear equations. For example, the so-called Gauss-Jordan method, which is used to find solutions to the system quadratic equations and is also associated with the use of matrices. There is also the Jacobi method for solving a system of linear algebraic equations. It is the easiest to adapt to a computer and is used in computing.
Complex 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, 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 figured out what a system and a matrix are, and learned how to find a general solution to a system of linear equations. In addition, we considered other options. We found out how to solve a system of linear equations: the Gauss method and talked about difficult cases and other ways to find solutions.
In fact, this topic is much more extensive, and if you want to understand it better, we recommend reading more specialized literature.
