Home Oral cavity Find the general solution and write it in terms of fsr. Find the general solution of the system and fsr

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Find the general solution and write it in terms of fsr. Find the general solution of the system and fsr

Homogeneous system linear equations over the field

DEFINITION. A fundamental system of solutions to a 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 to system (1).

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

PROPOSAL 3.11. Any two fundamental systems of solutions to a homogeneous system of linear equations consist of the same number of solutions.

Proof. In fact, any two fundamental systems of solutions to 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 is linear independent vectors of is a fundamental system of solutions. If the column rank of matrix A is equal to , 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 solution system 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, below it is assumed that Assuming , we will assume that the first columns of matrix A are linearly independent. In this case, matrix A is rowwise equivalent to the reduced stepwise matrix, and system (1) is equivalent to the following reduced stepwise system of equations:

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

In system (2) we will assign one of the free variables value, equal to 1, and the remaining variables have zero values. As a result, we obtain solutions to the system of equations (2), which we write in the form of 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, therefore, equality

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

Arbitrary solution of system (1). Then the vector

is also a solution to system (1), and

You can order a detailed solution to your problem!!!

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 description of the whole 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 the solution to this linear system of equations. To begin with, we you need to write out the coefficient matrix of the system.

Let's transform this matrix to a triangular one. We rewrite the first line without changes. And all the elements that are under $a_(11)$ must be made 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 zeros. 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 $x_5=1$.

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 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, 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 only 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.

As a result, a standard step matrix is obtained, 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:

Thus, common decision:

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 can consider triplets and get the 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:

A system of linear equations in which all free terms are equal to zero is called homogeneous :

Any homogeneous system is always consistent, since it always has zero (trivial ) solution. The question arises under what conditions will a homogeneous system have a nontrivial solution.

Theorem 5.2.A homogeneous system has a nontrivial solution if and only if the rank of the underlying matrix is less than the number of its unknowns.

Consequence. A square homogeneous system has a nontrivial solution if and only if the determinant of the main matrix of the system is not equal to zero.

Example 5.6. Determine the values of the parameter l at which the system has nontrivial solutions, and find these solutions:

Solution. This system will have a non-trivial solution when the determinant of the main matrix is equal to zero:

Thus, the system is non-trivial when l=3 or l=2. For l=3, the rank of the main matrix of the system is 1. Then, leaving only one equation and assuming that y=a And z=b, we get x=b-a, i.e.

For l=2, the rank of the main matrix of the system is 2. Then, choosing the minor as the basis:

we get a simplified system

From here we find that x=z/4, y=z/2. Believing z=4a, we get

The set of all solutions of a homogeneous system has a very important linear property : if columns X 1 and X 2 - solutions to a homogeneous system AX = 0, then any linear combination of them a X 1 + b X 2 will also be a solution to this system. Indeed, since AX 1 = 0 And AX 2 = 0 , That A(a X 1 + b X 2) = a AX 1 + b AX 2 = a · 0 + b · 0 = 0. It is because of this property that if a linear system has more than one solution, then there will be an infinite number of these solutions.

Linearly independent columns E 1 , E 2 , Ek, which are solutions of a homogeneous system, are called fundamental system of solutions homogeneous system of linear equations if the general solution of this system can be written as a linear combination of these columns:

If a homogeneous system has n variables, and the rank of the main matrix of the system is equal to r, That k = n-r.

Example 5.7. Find the fundamental system of solutions next system linear equations:

Solution. Let's find the rank of the main matrix of the system:

Thus, the set of solutions to this system of equations forms a linear subspace of dimension n-r= 5 - 2 = 3. Let’s choose minor as the base

Then, leaving only the basic equations (the rest will be a linear combination of these equations) and the basic variables (we move the rest, the so-called free variables to the right), we obtain a simplified system of equations:

Believing x 3 = a, x 4 = b, x 5 = c, we find

, .

Believing a= 1, b = c= 0, we obtain the first basic solution; believing b= 1, a = c= 0, we obtain the second basic solution; believing c= 1, a = b= 0, we obtain the third basic solution. As a result, the normal fundamental system of solutions will take the form

Using the fundamental system, the general solution of a homogeneous system can be written as

X = aE 1 + bE 2 + cE 3. a

Let us note some properties of solutions to an inhomogeneous system of linear equations AX=B and their relationship with the corresponding homogeneous system of equations AX = 0.

General solution of an inhomogeneous systemis equal to the sum of the general solution of the corresponding homogeneous system AX = 0 and an arbitrary particular solution of the inhomogeneous system. Indeed, let Y 0 is an arbitrary particular solution of an inhomogeneous system, i.e. AY 0 = B, And Y- general solution of a heterogeneous system, i.e. AY=B. Subtracting one equality from the other, we get
A(Y-Y 0) = 0, i.e. Y-Y 0 is the general solution of the corresponding homogeneous system AX=0. Hence, Y-Y 0 = X, or Y=Y 0 + X. Q.E.D.

Let the inhomogeneous system have the form AX = B 1 + B 2 . Then the general solution of such a system can be written as X = X 1 + X 2 , where AX 1 = B 1 and AX 2 = B 2. This property expresses the universal property of any linear systems(algebraic, differential, functional, etc.). In physics this property is called superposition principle, in electrical and radio engineering - principle of superposition. For example, in the theory of linear electrical circuits, the current in any circuit can be obtained as the algebraic sum of the currents caused by each energy source separately.

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 than the number of unknowns:

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 the following 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).