Homogeneous systems of linear equations. Fundamental set of solutions to a homogeneous system of linear equations

Helpful Hints 14.10.2019

Helpful Hints

Even at school, each of us studied equations and, for sure, 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, which 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 usual form appeared after the emergence 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. Indeed, there is no better example of equality.

The founder of modern letter designations of unknowns and signs of degrees is a French mathematician. However, his designations differed significantly from today's. For example, he denoted the square of an unknown number with the letter Q (lat. "quadratus"), and the cube with the letter C (lat. "cubus"). These notations seem awkward now, but back then it was the most understandable way to write systems of linear algebraic equations.

However, a drawback in the then methods of solution was that mathematicians considered only positive roots. Perhaps this is due to the fact that negative values did not have any practical application. One way or another, it was the Italian mathematicians Niccolo Tartaglia, Gerolamo Cardano and Rafael Bombelli who were the first to consider negative roots in the 16th century. BUT modern look, the main solution method (through the discriminant) was created only in the 17th century thanks to the work of Descartes and Newton.

In the middle of the 18th century, the Swiss mathematician Gabriel Cramer found new way in order to make a decision of systems linear equations easier. This method was subsequently named after him and to this day we use it. But we will talk about Cramer's method a little later, but for now we will discuss linear equations and methods for solving them separately from the system.

Linear equations

Linear equations are the simplest equalities with variable(s). They are classified as algebraic. write down in general view so: a 1 * x 1 + a 2 * x 2 + ... a n * x n \u003d b. We will need their representation in this form when compiling systems and matrices further.

Systems of linear algebraic equations

The definition of this term is as follows: it is a set of equations that have common unknowns and common decision. As a rule, at school, everything was solved by 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 is convenient to solve them later. First, systems of linear algebraic equations will look better if all variables are written as x with the appropriate index: 1,2,3, and so on. Secondly, all equations should be brought to the canonical form: a 1 * x 1 + a 2 * x 2 + ... a n * x n =b.

After all these actions, we can begin to talk about how to find a solution to systems of linear equations. Matrices are 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 either be specific values or variables. Most often, to designate elements, subscripts are placed under them (for example, a 11 or a 23). The first index means the row number and the second one the column number. On matrices, as well as on any other mathematical element, you can perform various operations. Thus, you can:

2) Multiply a matrix by some 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.

We will discuss all these techniques in more detail, as they will be useful to us in the future. Subtracting and adding matrices is very easy. Since we take matrices of the same size, each element of one table corresponds to each element of another. Thus, we add (subtract) these two elements (it is important that they are in the same places in their matrices). When multiplying a matrix by a number or vector, you simply need to multiply each element of the matrix by that number (or vector). Transposition is a very interesting process. It's very interesting sometimes to see him in real life, for example, when you change the orientation of your tablet or phone. The icons on the desktop are a matrix, and when you change the position, it transposes and becomes wider, but decreases in height.

Let's analyze such a process as Although it will not be useful to us, 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. We 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 further by a similar method.

Now we can begin to consider how the system of linear equations is solved.

Gauss method

This topic starts at school. We know well the concept of "system of two linear equations" 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 can not transform it and solve it in its pure form.

So, how is the system of linear Gaussian equations solved by this method? 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 eventually reduce the entire set to a stepped form. That is, it is necessary that from top to bottom (if placed 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 - three unknowns, in the second - two, in the third - 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 master the skills of addition, subtraction of matrices, and you also need to be able to find determinants. Therefore, if you do all this poorly or do not 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 have to construct a matrix from 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 put them in the table in the order they are written in the system. If the number is preceded by a "-" sign, then we write down a negative coefficient. So, we have compiled the first matrix of the coefficients of the unknowns, not including the numbers after the equal signs (naturally, the equation should be reduced to the 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, in the first matrix, in turn, we replace each column with coefficients 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, the matter is small. We have an initial matrix, and there are several resulting matrices that correspond to different variables. To get the solutions of 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 more methods for obtaining a solution 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 related to the use of matrices. There is also a Jacobi method for solving a system of linear algebraic equations. It is the easiest to adapt to a computer and is used in computer technology.

Difficult cases

Complexity usually arises when the number of equations less than number variables. Then we can say for sure that either the system is inconsistent (that is, it 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 have analyzed what a system and a matrix are, learned how to find a general solution to a system of linear equations. In addition, other options were considered. We found out how a system of linear equations is solved: the Gauss method and We talked about difficult cases and other ways to find solutions.

In fact, this topic is much more extensive, and if you want to better understand it, then we advise you to read more specialized literature.

Homogeneous system of linear equations over a field

DEFINITION. The fundamental system of solutions of system of equations (1) is a non-empty linearly independent system of its solutions, the linear span of which coincides with the set of all solutions of system (1).

Note that a homogeneous system of linear equations that has only a zero solution does not have a fundamental system of solutions.

PROPOSITION 3.11. Any two fundamental decision systems homogeneous system linear equations consist of the same number solutions.

Proof. Indeed, any two fundamental systems of solutions of the homogeneous system of equations (1) are equivalent and linearly independent. Therefore, by Proposition 1.12, their ranks are equal. Therefore, the number of solutions included in one fundamental system is equal to the number of solutions included in any other fundamental system of solutions.

If the main matrix A of the homogeneous system of equations (1) is zero, then any vector from is a solution to system (1); in this case, any collection of linearly independent vectors from is a fundamental system of solutions. If the column rank of matrix A is , then system (1) has only one solution - zero; therefore, in this case, the system of equations (1) does not have a fundamental system of solutions.

THEOREM 3.12. If the rank of the main matrix of a homogeneous system of linear equations (1) is less than the number of variables , then system (1) has a fundamental system of solutions consisting of solutions.

Proof. If the rank of the main matrix A of the homogeneous system (1) is equal to zero or , then it was shown above that the theorem is true. Therefore, it is assumed below that Assuming , we will assume that the first columns of the matrix A are linearly independent. In this case, the matrix A is rowwise equivalent to the reduced step matrix, and system (1) is equivalent to the following reduced step system of equations:

It is easy to check that any system of values of free variables of system (2) corresponds to one and only one solution of system (2) and, therefore, of system (1). In particular, only the zero solution of system (2) and system (1) corresponds to the system of zero values.

In system (2), we will assign a value equal to 1 to one of the free variables, and zero values to the other variables. As a result, we obtain solutions to the system of equations (2), which we write as rows of the following matrix C:

The row system of this matrix is linearly independent. Indeed, for any scalars from the equality

equality follows

and hence equality

Let us prove that the linear span of the system of rows of matrix C coincides with the set of all solutions of system (1).

Arbitrary solution of system (1). Then the vector

is also a solution to system (1), and

Example 1 . Find a general solution and some fundamental system of solutions for the system

Solution find with a calculator. The solution algorithm is the same as for systems of linear inhomogeneous equations.
Operating only with rows, we find the rank of the matrix, the basic minor; we declare dependent and free unknowns and find the general solution.

The first and second lines are proportional, one of them will be deleted:

.
Dependent variables - x 2, x 3, x 5, free - x 1, x 4. From the first equation 10x 5 = 0 we find x 5 = 0, then

; .
The general solution looks like:

We find the fundamental system of solutions, which consists of (n-r) solutions. In our case, n=5, r=3, therefore, fundamental system of solutions consists of two 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, that is, 2. It suffices to give the free unknowns x 1 and x 4 values from the rows of the determinant of the second order, which is different from zero, and calculate x 2 , x 3 , x 5 . The simplest non-zero determinant is .
So the first solution is:

, the second -

.
These two decisions constitute the fundamental decision system. Note that the fundamental system is not unique (determinants other than zero can be composed as many as you like).

Example 2 . Find the general solution and the fundamental system of solutions of the system
Solution.

,
it follows that the rank of the matrix is 3 and is equal to the number of unknowns. This means that the system has no free unknowns, and therefore has a unique solution - a trivial one.

Exercise . Explore and solve a system of linear equations.
Example 4

Exercise . Find general and particular solutions for each system.
Solution. We write the main matrix of the system:

5	-2	9	-4	-1
1	4	2	2	-5
6	2	11	-2	-6
x 1	x2	x 3	x4	x5

We bring the matrix to a triangular form. We will work only with rows, since multiplying a matrix row by a number other than zero and adding to another row for the system means multiplying the equation by the same number and adding it to another equation, which does not change the solution of the system.
Multiply the 2nd row by (-5). Let's add the 2nd line to the 1st:

0	-22	-1	-14	24
1	4	2	2	-5
6	2	11	-2	-6

Multiply the 2nd row by (6). Multiply the 3rd row by (-1). Let's add the 3rd line to the 2nd:
Find the rank of the matrix.

0	22	1	14	-24
6	2	11	-2	-6
x 1	x2	x 3	x4	x5

The selected minor has the highest order (of all possible minors) and is non-zero (it is equal to the product of the elements on the reciprocal diagonal), hence rang(A) = 2.
This minor is basic. It includes coefficients for unknown x 1, x 2, which means that the unknown x 1, x 2 are dependent (basic), and x 3, x 4, x 5 are free.
We transform the matrix, leaving only the basic minor on the left.

0	22	14	-1	-24
6	2	-2	-11	-6
x 1	x2	x4	x 3	x5

The system with the coefficients of this matrix is equivalent to the original system and has the form:
22x2 = 14x4 - x3 - 24x5
6x1 + 2x2 = - 2x4 - 11x3 - 6x5
By the method of elimination of unknowns, we find non-trivial solution:
We obtained relations expressing dependent variables x 1 ,x 2 through free x 3 ,x 4 ,x 5 , that is, we found common decision:
x2 = 0.64x4 - 0.0455x3 - 1.09x5
x 1 = - 0.55x 4 - 1.82x 3 - 0.64x 5
We find the fundamental system of solutions, 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.

1	0	0
0	1	0
0	0	1

A task . Find a fundamental set of solutions to a homogeneous system of linear equations.

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. Sample Sample 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.

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:

Example 1

(1) The first row was added to the second row, multiplied by -2. The first line was added to the third line, multiplied by -3.

(2) The second line was added to the third line, multiplied by -1.

Dividing the third row by 3 doesn't make much sense.

As a result of elementary transformations, an equivalent homogeneous system is obtained , and, applying the reverse move of the Gaussian method, it is easy to verify that the solution is unique.

Answer:

Let us formulate an obvious criterion: a homogeneous system of linear equations has only trivial solution, if system matrix rank(in this case, 3) is equal to the number of variables (in this case, 3 pcs.).

We warm up and tune our radio to a wave of elementary transformations:

Example 2

Solve a homogeneous system of linear equations

To finally fix the algorithm, let's analyze the final task:

Example 7

Solve a homogeneous system, write the answer in vector form.

Solution: we write the matrix of the system and, using elementary transformations, we bring it to a stepped form:

(1) The sign of the first line has been changed. Once again, I draw attention to the repeatedly met technique, which allows you to significantly simplify the following action.

(1) The first line was added to the 2nd and 3rd lines. The first line multiplied by 2 was added to the 4th line.

(3) The last three lines are proportional, two of them have been removed.

As a result, a standard step matrix is obtained, and the solution continues along the knurled track:

– basic variables;
are free variables.

We express the basic variables in terms of free variables. From the 2nd equation:

- substitute in the 1st equation:

So the general solution is:

Since there are three free variables in the example under consideration, the fundamental system contains three vectors.

Let's substitute a triple of values into the general solution and obtain a vector whose coordinates satisfy each equation of the homogeneous system. And again, I repeat that it is highly desirable to check each received vector - it will not take so much time, but it will save one hundred percent from errors.

For a triple of values find the vector

And finally for the triple we get the third vector:

Answer: , where

Those wishing to avoid fractional values may consider triplets and get the answer in the equivalent form:

Speaking of fractions. Let's look at the matrix obtained in the problem and ask the question - is it possible to simplify the further solution? After all, here we first expressed the basic variable in terms of fractions, then the basic variable in terms of fractions, and, I must say, this process was not the easiest and not the most pleasant.

The second solution:

The idea is to try choose other basic variables. Let's look at the matrix and notice two ones in the third column. So why not get zero at the top? Let's make one more elementary transformation: