Find the general and fundamental solution of the online system. Systems of linear homogeneous equations

Systems of linear homogeneous equations- has the form ∑a k i x i = 0. where m > n or m Homogeneous system 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:

number of variables: 2 3 4 5 6 7 8 and number of lines 2 3 4 5 6

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. The system in the case m=n has no trivial solution if and only if the determinant of this system 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

1. Finding the rank of the matrix.
2. We select the basic minor. We distinguish dependent (basic) and free unknowns.
3. We cross out those equations of the system whose coefficients are not included basic minor, since they are consequences of the others (by the theorem on the basis minor).
4. 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.
5. We solve the resulting system by eliminating unknowns. We find relationships expressing dependent variables through free ones.
6. If the rank of the matrix is ​​not equal to the number of variables, then we find the fundamental solution of the system.
7. 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 unknowns and signs of degrees is a French mathematician. However, his notation was significantly different from today's. 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. This may be due to the fact that negative values ​​did not have any practical application. One way or another, but be the first to count negative roots It was the Italian mathematicians Niccolo Tartaglia, Gerolamo Cardano and Raphael Bombelli who started it 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. written in general form as follows: 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. It could 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 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.

We will continue to polish our technology elementary transformations on homogeneous system of linear equations.
Based on the first paragraphs, the material may seem boring and mediocre, but this impression deceptively. 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 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, 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(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

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:

The linear equation is called homogeneous, if its free term is equal to zero, and inhomogeneous otherwise. A system consisting of homogeneous equations is called homogeneous and has general form:

It is obvious that every homogeneous system is consistent and has a zero (trivial) solution. Therefore, when applied to homogeneous systems of 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 the number of unknowns .

Proof: Let us assume that a system whose rank is equal has a non-zero solution. Obviously it does not exceed . In case the system has a unique solution. Since a system of homogeneous linear equations always has a zero solution, then the zero solution will be this unique solution. Thus, non-zero 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 non-zero solution.

Proof: If a 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 non-zero solution.

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

Proof: Let us assume that a system of linear homogeneous equations, the matrix of which with the determinant , has a non-zero solution. Then, according to the proven theorem, and this means that the matrix is ​​singular, i.e. .

Kronecker-Capelli theorem: An SLU is consistent if and only if the rank of the system matrix is ​​equal to the rank of the extended matrix of this system. A system ur is called consistent if it has at least one solution.

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 consistent, because it always has at least a zero solution. A system of linear homogeneous equations has a non-zero solution if and only if the rank of its matrix of coefficients for variables is less than the number of variables, i.e. for rank A (n. Any linear combination

Lin system solutions. homogeneous. ur-ii is also a solution to this system.

A system of linear independent solutions e1, e2,...,еk 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 for the variables of a system of linear homogeneous equations is less than the number of variables n, then every fundamental system of solutions to the system consists of n-r solutions. Therefore, the general solution of the linear system. one-day ur-th has the form: c1e1+c2e2+...+skek, 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

general solution the corresponding system 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 linear independent vectors, and any system from more vectors are linearly dependent. The number is called dimension (number of dimensions) linear space and is denoted by . In other words, the dimension of a space is the maximum number of linearly independent vectors of this space. If such a number exists, then the space is called finite-dimensional. If for anyone natural number n in space there is a system consisting of linearly independent vectors, then such a space is called infinite-dimensional (written: ). In what follows, unless otherwise stated, finite-dimensional spaces will be considered.

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

Theorem 8.1 on the expansion of a vector in terms of a basis. If is the 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 the only way, i.e. the coefficients are determined uniquely. In other words, any vector of space can be expanded into a basis and, moreover, in a unique way.

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