Solving systems of linear algebraic equations, solution methods, examples. Find the general solution of the system and fsr

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 selected minor has the highest order (of 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

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).

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.

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 an identity matrix;

Write down the general form of the solution 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 verify that the actions (operations) of adding vectors and multiplying them by a 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.

The Gaussian method has a number of disadvantages: it is impossible to know whether the system is consistent or not until all the transformations necessary in the Gaussian method have been carried out; Gauss's method is not suitable for systems with letter coefficients.

Let's consider other methods for solving systems of linear equations. These methods use the concept of matrix rank and reduce the solution of any consistent system to the solution of a system to which Cramer's rule applies.

Example 1. Find a general solution to the following system of linear equations using the fundamental system of solutions to the reduced homogeneous system and a particular solution to the inhomogeneous system.

1. Making a matrix A and extended system matrix (1)

2. Explore the system (1) for togetherness. To do this, we find the ranks of the matrices A and https://pandia.ru/text/78/176/images/image006_90.gif" width="17" height="26 src=">). If it turns out that , then the system (1) incompatible. If we get that , then this system is consistent and we will solve it. (The compatibility study is based on the Kronecker-Capelli theorem).

a. We find rA.

To find rA, we will consider sequentially non-zero minors of the first, second, etc. orders of the matrix A and the minors surrounding them.

M1=1≠0 (we take 1 from the upper left corner of the matrix A).

We border M1 the second row and second column of this matrix. . We continue to border M1 the second line and the third column..gif" width="37" height="20 src=">. Now we border the non-zero minor M2′ second order.

We have: (since the first two columns are the same)

(since the second and third lines are proportional).

We see that rA=2, a is the basis minor of the matrix A.

b. We find.

Fairly basic minor M2′ matrices A border with a column of free terms and all rows (we have only the last row).

. It follows that M3′′ remains the basic minor of the matrix https://pandia.ru/text/78/176/images/image019_33.gif" width="168 height=75" height="75"> (2)

Because M2′- basis minor of the matrix A systems (2) , then this system is equivalent to the system (3) , consisting of the first two equations of the system (2) (for M2′ is in the first two rows of matrix A).

(3)

Since the basic minor https://pandia.ru/text/78/176/images/image021_29.gif" width="153" height="51"> (4)

In this system there are two free unknowns ( x2 And x4 ). That's why FSR systems (4) consists of two solutions. To find them, we assign free unknowns in (4) values first x2=1 , x4=0 , and then - x2=0 , x4=1 .

At x2=1 , x4=0 we get:

This system already has the only thing solution (it can be found using Cramer's rule or any other method). Subtracting the first from the second equation, we get:

Her solution will be x1= -1 , x3=0 . Given the values x2 And x4 , which we added, we obtain the first fundamental solution of the system (2) : .

Now we believe in (4) x2=0 , x4=1 . We get:

We solve this system using Cramer’s theorem:

We obtain the second fundamental solution of the system (2) : .

Solutions β1 , β2 and make up FSR systems (2) . Then its general solution will be

γ= C1 β1+С2β2=С1(‑1, 1, 0, 0)+С2(5, 0, 4, 1)=(‑С1+5С2, С1, 4С2, С2)

Here C1 , C2 – arbitrary constants.

4. Let's find one private solution heterogeneous system(1) . As in paragraph 3 , instead of the system (1) Let's consider an equivalent system (5) , consisting of the first two equations of the system (1) .

(5)

Let us move the free unknowns to the right sides x2 And x4.

(6)

Let's give free unknowns x2 And x4 arbitrary values, for example, x2=2 , x4=1 and put them in (6) . Let's get the system

This system has a unique solution (since its determinant M2′0). Solving it (using Cramer’s theorem or Gauss’s method), we obtain x1=3 , x3=3 . Given the values of the free unknowns x2 And x4 , we get particular solution of an inhomogeneous system(1)α1=(3,2,3,1).

5. Now all that remains is to write it down general solution α of an inhomogeneous system(1) : it is equal to the sum private solution this system and general solution of its reduced homogeneous system (2) :

α=α1+γ=(3, 2, 3, 1)+(‑С1+5С2, С1, 4С2, С2).

This means: (7)

6. Examination. To check if you solved the system correctly (1) , we need a general solution (7) substitute in (1) . If each equation turns into the identity ( C1 And C2 must be destroyed), then the solution is found correctly.

We'll substitute (7) for example, only the last equation of the system (1) (x1 + x2 + x3 ‑9 x4 =‑1) .

We get: (3–С1+5С2)+(2+С1)+(3+4С2)–9(1+С2)=–1

(С1–С1)+(5С2+4С2–9С2)+(3+2+3–9)=–1

Where –1=–1. We got an identity. We do this with all the other equations of the system (1) .

Comment. The check is usually quite cumbersome. The following “partial check” can be recommended: in the general solution of the system (1) assign some values to arbitrary constants and substitute the resulting partial solution only into the discarded equations (i.e., into those equations from (1) , which were not included in (5) ). If you get identities, then more likely, system solution (1) found correctly (but such a check does not provide a complete guarantee of correctness!). For example, if in (7) put C2=- 1 , C1=1, then we get: x1=-3, x2=3, x3=-1, x4=0. Substituting into the last equation of system (1), we have: - 3+3 - 1 - 9∙0= - 1 , i.e. –1=–1. We got an identity.

Example 2. Find a general solution to a system of linear equations (1) , expressing the basic unknowns in terms of free ones.

Solution. As in example 1, compose matrices A and https://pandia.ru/text/78/176/images/image010_57.gif" width="156" height="50"> of these matrices. Now we leave only those equations of the system (1) , the coefficients of which are included in this basic minor (i.e., we have the first two equations) and consider a system consisting of them, equivalent to system (1).

Let us transfer the free unknowns to the right-hand sides of these equations.

system (9) We solve by the Gaussian method, considering the right-hand sides as free terms.

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Option 2.

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Option 4.

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Option 5.

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Option 6.

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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 to the following system of 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 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 the algebraic sum of the currents caused by each energy source separately.