Solve slough find a normal fundamental system of solutions. Solving homogeneous systems of linear equations

Systems linear equations, for which all free terms are equal to zero are 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 main matrix less number her 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 , E k, 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 fundamental system the general solution of a homogeneous system can be written in the form

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 systemequal to the sum general solution 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 a universal property of any linear systems in general (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 algebraic sum currents caused by each energy source separately.

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

Solution find using a calculator. The solution algorithm is the same as for systems of linear not homogeneous equations.
Operating only with rows, we find the rank of the matrix, the basis minor; We declare dependent and free unknowns and find a general solution.

The first and second lines are proportional, let’s cross out one of them:

.
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 is:

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

, second –

.
These two decisions constitute a fundamental decision system. Note that the fundamental system is not unique (you can create as many nonzero determinants as you like).

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

,
it follows that the rank of the matrix is 3 and equal to the number unknown. This means that the system does not have free unknowns, and therefore has a unique solution - a trivial one.

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

Exercise . Find the general and particular solutions of each system.
Solution. Let's write down the main matrix of the system:

5	-2	9	-4	-1
1	4	2	2	-5
6	2	11	-2	-6
x 1	x 2	x 3	x 4	x 5

Let's reduce the matrix to triangular view. We will work only with rows, since multiplying a matrix row by a number other than zero and adding it to another row for the system means multiplying the equation by the same number and adding it with another equation, which does not change the solution of the system.
Multiply the 2nd line 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

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

0	22	1	14	-24
6	2	11	-2	-6
x 1	x 2	x 3	x 4	x 5

The highlighted minor has highest order(of possible minors) and is nonzero (it equal to the product elements on the reverse diagonal), hence rank(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.
Let's transform the matrix, leaving only the basis minor on the left.

0	22	14	-1	-24
6	2	-2	-11	-6
x 1	x 2	x 4	x 3	x 5

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

1	0	0
0	1	0
0	0	1

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

You can order detailed solution your task!!!

To understand what it is fundamental decision system you can watch a video tutorial for the same example by clicking. Now let's move on to the actual description of all the necessary work. This will help you understand the essence of this issue in more detail.

How to find the fundamental system of solutions to a linear equation?

Let's take for example the following system of linear equations:

Let's find a solution to this linear system equations To begin with, we 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 $x_5=1$.

Let M 0 – set of solutions to a homogeneous system (4) of linear equations.

Definition 6.12. Vectors With 1 ,With 2 , …, with p, which are solutions of a homogeneous system of linear equations are called fundamental set of solutions(abbreviated FNR), if

1) vectors With 1 ,With 2 , …, with p linearly independent (i.e., none of them can be expressed in terms of the others);

2) any other solution to a homogeneous system of linear equations can be expressed in terms of solutions With 1 ,With 2 , …, with p.

Note that if With 1 ,With 2 , …, with p– any f.n.r., then the expression k 1× With 1 + k 2× With 2 + … + k p× with p you can describe the whole set M 0 solutions to system (4), so it is called general view of the system solution (4).

Theorem 6.6. Any indeterminate homogeneous system of linear equations has a fundamental set of solutions.

The way to find the fundamental set of solutions is as follows:

Find a general solution to a homogeneous system of linear equations;

Build ( n – r) partial solutions of this system, while the values of the free unknowns must form identity matrix;

Write out general form solutions included in M 0 .

Example 6.5. Find a fundamental set of solutions to the following system:

Solution. Let's find a general solution to this system.

~ ~ ~ ~ Þ Þ Þ There are five unknowns in this system ( n= 5), of which there are two main unknowns ( r= 2), there are three free unknowns ( n – r), that is, the fundamental solution set contains three solution vectors. Let's build them. We have x 1 and x 3 – main unknowns, x 2 , x 4 , x 5 – free unknowns

Values of free unknowns x 2 , x 4 , x 5 form the identity matrix E third order. Got that vectors With 1 ,With 2 , With 3 form f.n.r. of this system. Then the set of solutions of this homogeneous system will be M 0 = {k 1× With 1 + k 2× With 2 + k 3× With 3 , k 1 , k 2 , k 3 О R).

Let us now find out the conditions for the existence of nonzero solutions of a homogeneous system of linear equations, in other words, the conditions for the existence of a fundamental set of solutions.

A homogeneous system of linear equations has non-zero solutions, that is, it is uncertain if

1) the rank of the main matrix of the system is less than the number of unknowns;

2) in a homogeneous system of linear equations, the number of equations is less than the number of unknowns;

3) if in a homogeneous system of linear equations the number of equations is equal to the number of unknowns, and the determinant of the main matrix is equal to zero (i.e. | A| = 0).

Example 6.6. At what parameter value a homogeneous system of linear equations has non-zero solutions?

Solution. Let's compose the main matrix of this system and find its determinant: = = 1×(–1) 1+1 × = – A– 4. The determinant of this matrix is equal to zero at a = –4.

Answer: –4.

7. Arithmetic n-dimensional vector space

Basic Concepts

In previous sections we have already encountered the concept of a set of real numbers arranged in a certain order. This is a row matrix (or column matrix) and a solution to a system of linear equations with n unknown. This information can be summarized.

Definition 7.1. n-dimensional arithmetic vector called an ordered set of n real numbers.

Means A= (a 1 , a 2 , …, a n), where a iО R, i = 1, 2, …, n– general view of the vector. Number n called dimension vectors, and numbers a i are called his coordinates.

For example: A= (1, –8, 7, 4, ) – five-dimensional vector.

All set n-dimensional vectors are usually denoted as Rn.

Definition 7.2. Two vectors A= (a 1 , a 2 , …, a n) And b= (b 1 , b 2 , …, b n) of the same dimension equal if and only if their corresponding coordinates are equal, i.e. a 1 = b 1 , a 2 = b 2 , …, a n= b n.

Definition 7.3.Amount two n-dimensional vectors A= (a 1 , a 2 , …, a n) And b= (b 1 , b 2 , …, b n) is called a vector a + b= (a 1 + b 1, a 2 + b 2, …, a n+b n).

Definition 7.4. The work real number k to vector A= (a 1 , a 2 , …, a n) is called a vector k× A = (k×a 1, k×a 2 , …, k×a n)

Definition 7.5. Vector O= (0, 0, …, 0) is called zero(or null vector).

It is easy to check that the actions (operations) of adding vectors and multiplying them by real number have the following properties: " a, b, c Î Rn, " k, lО R:

1) a + b = b + a;

2) a + (b+ c) = (a + b) + c;

3) a + O = a;

4) a+ (–a) = O;

5) 1× a = a, 1 О R;

6) k×( l× a) = l×( k× a) = (l× k)× a;

7) (k + l)× a = k× a + l× a;

8) k×( a + b) = k× a + k× b.

Definition 7.6. A bunch of Rn with the operations of adding vectors and multiplying them by a real number given on it is called arithmetic n-dimensional vector space.

Systems of linear 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 exists a fundamental system of 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 basic minor, since they are consequences of the others (by the theorem on the basis minor).
We transfer the terms of the equations containing free unknowns to 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 fundamental solution systems.
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