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Prove that the system of linear equations. Examples of systems of linear equations: solution method

System m linear equations with n unknowns called a system of the form

Where a ij And b i (i=1,…,m; b=1,…,n) are some known numbers, and x 1 ,…,x n– unknown. In the designation of coefficients a ij first index i denotes the equation number, and the second j– the number of the unknown at which this coefficient stands.

We will write the coefficients for the unknowns in the form of a matrix , which we'll call matrix of the system.

The numbers on the right side of the equations are b 1 ,…,b m are called free members.

Totality n numbers c 1 ,…,c n called decision of a given system, if each equation of the system becomes an equality after substituting numbers into it c 1 ,…,c n instead of the corresponding unknowns x 1 ,…,x n.

Our task will be to find solutions to the system. In this case, three situations may arise:

A system of linear equations that has at least one solution is called joint. Otherwise, i.e. if the system has no solutions, then it is called non-joint.

Let's consider ways to find solutions to the system.

MATRIX METHOD FOR SOLVING SYSTEMS OF LINEAR EQUATIONS

Matrices make it possible to briefly write down a system of linear equations. Let a system of 3 equations with three unknowns be given:

Consider the system matrix and matrices columns of unknown and free terms

Let's find the work

those. as a result of the product, we obtain the left-hand sides of the equations of this system. Then, using the definition of matrix equality, this system can be written in the form

or shorter A∙X=B.

Here are the matrices A And B are known, and the matrix X unknown. It is necessary to find it, because... its elements are the solution to this system. This equation is called matrix equation.

Let the matrix determinant be different from zero | A| ≠ 0. Then the matrix equation is solved as follows. Multiply both sides of the equation on the left by the matrix A-1, inverse of the matrix A: . Because the A -1 A = E And E∙X = X, then we obtain a solution to the matrix equation in the form X = A -1 B .

Note that since the inverse matrix can only be found for square matrices, the matrix method can only solve those systems in which the number of equations coincides with the number of unknowns. However, matrix recording of the system is also possible in the case when the number of equations is not equal to the number of unknowns, then the matrix A will not be square and therefore it is impossible to find a solution to the system in the form X = A -1 B.

Examples. Solve systems of equations.

CRAMER'S RULE

Consider a system of 3 linear equations with three unknowns:

Third-order determinant corresponding to the system matrix, i.e. composed of coefficients for unknowns,

called determinant of the system.

Let's compose three more determinants as follows: replace sequentially 1, 2 and 3 columns in the determinant D with a column of free terms

Then we can prove the following result.

Theorem (Cramer's rule). If the determinant of the system Δ ≠ 0, then the system under consideration has one and only one solution, and

Proof. So, let's consider a system of 3 equations with three unknowns. Let's multiply the 1st equation of the system by the algebraic complement A 11 element a 11, 2nd equation – on A 21 and 3rd – on A 31:

Let's add these equations:

Let's look at each of the brackets and the right side of this equation. By the theorem on the expansion of the determinant in elements of the 1st column

Similarly, it can be shown that and .

Finally, it is easy to notice that

Thus, we obtain the equality: .

Hence, .

The equalities and are derived similarly, from which the statement of the theorem follows.

Thus, we note that if the determinant of the system Δ ≠ 0, then the system has a unique solution and vice versa. If the determinant of the system is equal to zero, then the system either has an infinite number of solutions or has no solutions, i.e. incompatible.

Examples. Solve system of equations

GAUSS METHOD

The previously discussed methods can be used to solve only those systems in which the number of equations coincides with the number of unknowns, and the determinant of the system must be different from zero. The Gauss method is more universal and suitable for systems with any number of equations. It consists in the consistent elimination of unknowns from the equations of the system.

Consider again a system of three equations with three unknowns:

We will leave the first equation unchanged, and from the 2nd and 3rd we will exclude the terms containing x 1. To do this, divide the second equation by A 21 and multiply by – A 11, and then add it to the 1st equation. Similarly, we divide the third equation by A 31 and multiply by – A 11, and then add it with the first one. As a result, the original system will take the form:

Now from the last equation we eliminate the term containing x 2. To do this, divide the third equation by, multiply by and add with the second. Then we will have a system of equations:

From here, from the last equation it is easy to find x 3, then from the 2nd equation x 2 and finally, from 1st - x 1.

When using the Gaussian method, the equations can be swapped if necessary.

Often instead of writing new system equations, are limited to writing out the extended matrix of the system:

and then bring it to a triangular or diagonal form using elementary transformations.

TO elementary transformations matrices include the following transformations:

rearranging rows or columns;
multiplying a string by a number other than zero;
adding other lines to one line.

Examples: Solve systems of equations using the Gauss method.

Thus, the system has an infinite number of solutions.

Systems m linear equations with n unknown.
Solving a system of linear equations- this is such a set of numbers ( x 1 , x 2 , …, x n), when substituted into each of the equations of the system, the correct equality is obtained.
Where a ij , i = 1, …, m; j = 1, …, n— system coefficients;
b i , i = 1, …, m- free members;
x j , j = 1, …, n- unknown.
The above system can be written in matrix form: A X = B,

Where ( A|B) is the main matrix of the system;
A— extended system matrix;
X— column of unknowns;
B— column of free members.
If matrix B is not a null matrix ∅, then this system of linear equations is called inhomogeneous.
If matrix B= ∅, then this system of linear equations is called homogeneous. A homogeneous system always has a zero (trivial) solution: x 1 = x 2 = …, x n = 0.
Joint system of linear equations is a system of linear equations that has a solution.
Inconsistent system of linear equations is an unsolvable system of linear equations.
A certain system of linear equations is a system of linear equations that has a unique solution.
Indefinite system of linear equations is a system of linear equations with an infinite number of solutions.
Systems of n linear equations with n unknowns
If the number of unknowns is equal to the number of equations, then the matrix is square. The determinant of a matrix is called the main determinant of a system of linear equations and is denoted by the symbol Δ.
Cramer method for solving systems n linear equations with n unknown.
Cramer's rule.
If the main determinant of a system of linear equations is not equal to zero, then the system is consistent and defined, and the only solution is calculated using the Cramer formulas:
where Δ i are determinants obtained from the main determinant of the system Δ by replacing i th column to the column of free members. .
Systems of m linear equations with n unknowns
Kronecker–Capelli theorem.

In order for a given system of linear equations to be consistent, it is necessary and sufficient that the rank of the system matrix be equal to the rank of the extended matrix of the system, rang(Α) = rang(Α|B).
If rang(Α) ≠ rang(Α|B), then the system obviously has no solutions.
If rang(Α) = rang(Α|B), then two cases are possible:
1) rank(Α) = n(number of unknowns) - the solution is unique and can be obtained using Cramer’s formulas;
2) rank(Α)< n - there are infinitely many solutions.
Gauss method for solving systems of linear equations

Let's create an extended matrix ( A|B) of a given system from the coefficients of the unknowns and the right-hand sides.
The Gaussian method or the method of eliminating unknowns consists of reducing the extended matrix ( A|B) using elementary transformations over its rows to a diagonal form (to the upper triangular form). Returning to the system of equations, all unknowns are determined.
Elementary transformations over strings include the following:
1) swap two lines;
2) multiplying a string by a number other than 0;
3) adding another string to a string, multiplied by an arbitrary number;
4) throwing out a zero line.
The extended matrix reduced to diagonal form corresponds to linear system, equivalent to this one, the solution of which does not cause difficulties. .
System of homogeneous linear equations.
A homogeneous system has the form:

it corresponds to the matrix equation A X = 0.
1) A homogeneous system is always consistent, since r(A) = r(A|B), there is always a zero solution (0, 0, …, 0).
2) In order for a homogeneous system to have a non-zero solution, it is necessary and sufficient that r = r(A)< n , which is equivalent to Δ = 0.
3) If r< n , then obviously Δ = 0, then free unknowns arise c 1 , c 2 , …, c n-r, the system has non-trivial solutions, and there are infinitely many of them.
4) General solution X at r< n can be written in matrix form as follows:
X = c 1 X 1 + c 2 X 2 + … + c n-r X n-r,
where are the solutions X 1, X 2, …, X n-r form a fundamental system of solutions.
5) The fundamental system of solutions can be obtained from the general solution of a homogeneous system:

,
if we sequentially set the parameter values equal to (1, 0, …, 0), (0, 1, …, 0), …, (0, 0, …, 1).
Expansion of the general solution in terms of the fundamental system of solutions is a recording of the general solution in the form linear combination solutions belonging to the fundamental system.
Theorem. In order for a system of linear homogeneous equations to have a non-zero solution, it is necessary and sufficient that Δ ≠ 0.
So, if the determinant Δ ≠ 0, then the system has a unique solution.
If Δ ≠ 0, then the system of linear homogeneous equations has an infinite number of solutions.
Theorem. In order for a homogeneous system to have a nonzero solution, it is necessary and sufficient that r(A)< n .
Proof:
1) r there can't be more n(the rank of the matrix does not exceed the number of columns or rows);
2) r< n , because If r = n, then the main determinant of the system Δ ≠ 0, and, according to Cramer’s formulas, there is a unique trivial solution x 1 = x 2 = … = x n = 0, which contradicts the condition. Means, r(A)< n .
Consequence. In order for a homogeneous system n linear equations with n unknowns had a non-zero solution, it is necessary and sufficient that Δ = 0.

We continue to deal with systems of linear equations. So far we have considered systems that have a unique solution. Such systems can be solved in any way: by substitution method(“school”), according to Cramer's formulas, matrix method, Gaussian method. However, in practice, two more cases are widespread:

1) the system is inconsistent (has no solutions);

2) the system has infinitely many solutions.

For these systems, the most universal of all solution methods is used - Gaussian method. In fact, the “school” method will also lead to the answer, but in higher mathematics It is customary to use the Gaussian method of sequential elimination of unknowns. Those who are not familiar with the Gaussian method algorithm, please study the lesson first Gaussian method

The elementary matrix transformations themselves are exactly the same, the difference will be in the ending of the solution. First, let's look at a couple of examples when the system has no solutions (inconsistent).

Example 1

What immediately catches your eye about this system? The number of equations is less than the number of variables. There is a theorem that states: “If the number of equations in the system is less than the number of variables, then the system is either inconsistent or has infinitely many solutions.” And all that remains is to find out.

The beginning of the solution is completely ordinary - we write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

(1). On the top left step we need to get (+1) or (–1). There are no such numbers in the first column, so rearranging the rows will not give anything. The unit will have to organize itself, and this can be done in several ways. We did this. To the first line we add the third line, multiplied by (–1).

(2). Now we get two zeros in the first column. To the second line we add the first line, multiplied by 3. To the third line we add the first, multiplied by 5.

(3). After the transformation has been completed, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the desired one (–1) on the second step. Divide the third line by (–3).

(4). Add a second line to the third line. Probably everyone noticed the bad line that resulted from elementary transformations:

. It is clear that this cannot be so.

Indeed, let us rewrite the resulting matrix

back to the system of linear equations:

If, as a result of elementary transformations, a string of the form is obtained , Whereλ is a number other than zero, then the system is inconsistent (has no solutions).

How to write down the ending of a task? You need to write down the phrase:

“As a result of elementary transformations, a string of the form was obtained, where λ ≠ 0 " Answer: “The system has no solutions (inconsistent).”

Please note that in this case there is no reversal of the Gaussian algorithm, there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is an example for independent decision. Complete solution and the answer at the end of the lesson.

We remind you again that your solution may differ from our solution; the Gaussian method does not specify an unambiguous algorithm; the order of actions and the actions themselves must be guessed in each case independently.

Another technical feature of the solution: elementary transformations can be stopped At once, as soon as a line like , where λ ≠ 0 . Let's consider a conditional example: suppose that after the first transformation the matrix is obtained

This matrix has not yet been reduced to echelon form, but there is no need for further elementary transformations, since a line of the form has appeared, where λ ≠ 0 . The answer should be given immediately that the system is incompatible.

When a system of linear equations has no solutions, it is almost a gift to the student, due to the fact that a short solution is obtained, sometimes literally in 2-3 steps. But everything in this world is balanced, and a problem in which the system has infinitely many solutions is just longer.

Example 3:

Solve a system of linear equations

There are 4 equations and 4 unknowns, so the system can either have a single solution, or have no solutions, or have infinitely many solutions. Be that as it may, the Gaussian method will in any case lead us to the answer. This is its versatility.

The beginning is again standard. Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

That's all, and you were afraid.

(1). Please note that all numbers in the first column are divisible by 2, so 2 is fine on the top left step. To the second line we add the first line multiplied by (–4). To the third line we add the first line multiplied by (–2). To the fourth line we add the first line, multiplied by (–1).

Attention! Many may be tempted by the fourth line subtract first line. This can be done, but it is not necessary; experience shows that the probability of an error in calculations increases several times. We just add: to the fourth line we add the first line, multiplied by (–1) – exactly!

(2). The last three lines are proportional, two of them can be deleted. Here again we need to show increased attention, but are the lines really proportional? To be on the safe side, it would be a good idea to multiply the second line by (–1), and divide the fourth line by 2, resulting in three identical lines. And only after that remove two of them. As a result of elementary transformations, the extended matrix of the system is reduced to a stepwise form:

When writing a task in a notebook, it is advisable to make the same notes in pencil for clarity.

Let us rewrite the corresponding system of equations:

There is no smell of a “ordinary” single solution to the system here. Bad line where λ ≠ 0, also no. This means that this is the third remaining case - the system has infinitely many solutions.

An infinite set of solutions to a system is briefly written in the form of the so-called general solution of the system.

We find the general solution of the system using the inverse of the Gaussian method. For systems of equations with an infinite set of solutions, new concepts appear: "basic variables" And "free variables". First let's define what variables we have basic, and which variables - free. It is not necessary to explain in detail the terms of linear algebra; it is enough to remember that there are such basic variables And free variables.

Basic variables always “sit” strictly on the steps of the matrix. In this example, the basic variables are x 1 and x 3 .

Free variables are everything remaining variables that did not receive a step. In our case there are two of them: x 2 and x 4 – free variables.

Now you need Allbasic variables express only throughfree variables. The reverse of the Gaussian algorithm traditionally works from the bottom up. From the second equation of the system we express the basic variable x 3:

Now look at the first equation: . First we substitute the found expression into it:

It remains to express the basic variable x 1 via free variables x 2 and x 4:

In the end we got what we needed - All basic variables ( x 1 and x 3) expressed only through free variables ( x 2 and x 4):

Actually, the general solution is ready:

How to write the general solution correctly? First of all, free variables are written into the general solution “by themselves” and strictly in their places. In this case, free variables x 2 and x 4 should be written in the second and fourth positions:

The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:

From the general solution of the system one can find infinitely many private solutions. It's very simple. Free variables x 2 and x 4 are called so because they can be given any final values. The most popular values are zero values, since this is the easiest partial solution to obtain.

Substituting ( x 2 = 0; x 4 = 0) into the general solution, we obtain one of the particular solutions:

, or is a particular solution corresponding to free variables with values ( x 2 = 0; x 4 = 0).

Another sweet pair is ones, let's substitute ( x 2 = 1 and x 4 = 1) into the general solution:

, i.e. (-1; 1; 1; 1) – another particular solution.

It is easy to see that the system of equations has infinitely many solutions since we can give free variables any meanings.

Each the particular solution must satisfy to each equation of the system. This is the basis for a “quick” check of the correctness of the solution. Take, for example, the particular solution (-1; 1; 1; 1) and substitute it into the left side of each equation of the original system:

Everything must come together. And with any particular solution you receive, everything should also agree.

Strictly speaking, checking a particular solution is sometimes deceiving, i.e. some particular solution may satisfy each equation of the system, but the general solution itself is actually found incorrectly. Therefore, first of all, the verification of the general solution is more thorough and reliable.

How to check the resulting general solution ?

It's not difficult, but it does require some lengthy transformations. We need to take expressions basic variables, in this case and , and substitute them into the left side of each equation of the system.

To the left side of the first equation of the system:

The right side of the initial first equation of the system is obtained.

To the left side of the second equation of the system:

The right side of the initial second equation of the system is obtained.

And then - to the left sides of the third and fourth equation of the system. This check takes longer, but guarantees 100% correctness of the overall solution. In addition, some tasks require checking the general solution.

Example 4:

Solve the system using the Gaussian method. Find the general solution and two particular ones. Check the general solution.

Example 5:

Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution

Solution: Let's write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

(1). Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.

(2). To the third line we add the second line, multiplied by (–5). To the fourth line we add the second line, multiplied by (–7).

(3). The third and fourth lines are the same, we delete one of them. This is such a beauty:

Basic variables sit on the steps, therefore - basic variables.

There is only one free variable that did not get a step here: .

(4). Reverse move. Let's express the basic variables through a free variable:

From the third equation:

Let's consider the second equation and substitute the found expression into it:

, , ,

Let's consider the first equation and substitute the found expressions and into it:

Thus, the general solution with one free variable x 4:

Once again, how did it turn out? Free variable x 4 sits alone in its rightful fourth place. The resulting expressions for the basic variables , , are also in place.

Let us immediately check the general solution.

We substitute the basic variables , , into the left side of each equation of the system:

The corresponding right-hand sides of the equations are obtained, thus the correct general solution is found.

Now from the found general solution we obtain two particular solutions. All variables are expressed here through a single free variable x 4 . No need to rack your brains.

Let x 4 = 0 then – the first particular solution.

Let x 4 = 1 then – another private solution.

Answer: Common decision: . Private solutions:

And .

Example 6:

Find the general solution to the system of linear equations.

We have already checked the general solution, the answer can be trusted. Your solution may differ from our solution. The main thing is that the general decisions coincide. Many people probably noticed an unpleasant moment in the solutions: very often, when reversing the Gauss method, we had to tinker with ordinary fractions. In practice, this is indeed the case; cases where there are no fractions are much less common. Be prepared mentally and, most importantly, technically.

Let us dwell on the features of the solution that were not found in the solved examples. The general solution of the system may sometimes include a constant (or constants).

For example, a general solution: . Here one of the basic variables is equal to a constant number: . There is nothing exotic about this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables. However, the Gaussian method works in the harshest conditions. You should calmly reduce the extended matrix of the system to a stepwise form using a standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a single solution.

Let us repeat our advice - in order to feel comfortable when solving a system using the Gaussian method, you should get good at solving at least a dozen systems.

Solutions and answers:

Example 2:

Solution:Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form.

Elementary transformations performed:

(1) The first and third lines have been swapped.

(2) The first line was added to the second line, multiplied by (–6). The first line was added to the third line, multiplied by (–7).

(3) The second line was added to the third line, multiplied by (–1).

As a result of elementary transformations, a string of the form is obtained, Where λ ≠ 0 .This means the system is inconsistent.Answer: there are no solutions.

Example 4:

Solution:Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

Conversions performed:

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

There is no unit for the second step , and transformation (2) is aimed at obtaining it.

(2). The third line was added to the second line, multiplied by –3.

(3). The second and third lines were swapped (we moved the resulting –1 to the second step)

(4). The third line was added to the second line, multiplied by 3.

(5). The first two lines had their sign changed (multiplied by –1), the third line was divided by 14.

Reverse:

(1). Here are the basic variables (which are on the steps), and – free variables (who did not get a step).

(2). Let's express the basic variables in terms of free variables:

From the third equation: .

(3). Consider the second equation:, private solutions:

Answer: Common decision:

Complex numbers

In this section we will introduce the concept complex number, consider algebraic, trigonometric And exponential form complex number. We will also learn how to perform operations with complex numbers: addition, subtraction, multiplication, division, exponentiation and root extraction.

To master complex numbers, no special knowledge from a higher mathematics course is required, and the material is accessible even to schoolchildren. It is enough to be able to perform algebraic operations with “ordinary” numbers, and remember trigonometry.

First, let's remember the “ordinary” Numbers. In mathematics they are called set of real numbers and are designated by the letter R, or R (thickened). All real numbers sit on the familiar number line:

The company of real numbers is very diverse - here there are integers, fractions, and irrational numbers. In this case, each point on the number axis necessarily corresponds to some real number.

To study a system of linear agebraic equations (SLAEs) for consistency means to find out whether this system has solutions or does not have them. Well, if there are solutions, then indicate how many there are.

We will need information from the topic "System of linear algebraic equations. Basic terms. Matrix form of notation". In particular, concepts such as system matrix and extended system matrix are needed, since the formulation of the Kronecker-Capelli theorem is based on them. As usual, we will denote the system matrix by the letter $A$, and the extended matrix of the system by the letter $\widetilde(A)$.

Kronecker-Capelli theorem

Linear system algebraic equations is consistent if and only if the rank of the system matrix is equal to the rank of the extended matrix of the system, i.e. $\rang A=\rang\widetilde(A)$.

Let me remind you that a system is called joint if it has at least one solution. The Kronecker-Capelli theorem says this: if $\rang A=\rang\widetilde(A)$, then there is a solution; if $\rang A\neq\rang\widetilde(A)$, then this SLAE has no solutions (inconsistent). The answer to the question about the number of these solutions is given by a corollary of the Kronecker-Capelli theorem. In the formulation of the corollary, the letter $n$ is used, which is equal to the number of variables of the given SLAE.

Corollary to the Kronecker-Capelli theorem

If $\rang A\neq\rang\widetilde(A)$, then the SLAE is inconsistent (has no solutions).
If $\rang A=\rang\widetilde(A)< n$, то СЛАУ является неопределённой (имеет бесконечное количество решений).
If $\rang A=\rang\widetilde(A) = n$, then the SLAE is definite (has exactly one solution).

Please note that the formulated theorem and its corollary do not indicate how to find a solution to the SLAE. With their help, you can only find out whether these solutions exist or not, and if they exist, then how many.

Example No. 1

Explore SLAE $ \left \(\begin(aligned) & -3x_1+9x_2-7x_3=17;\\ & -x_1+2x_2-4x_3=9;\\ & 4x_1-2x_2+19x_3=-42. \end(aligned )\right.$ for compatibility. If the SLAE is compatible, indicate the number of solutions.

To find out the existence of solutions to a given SLAE, we use the Kronecker-Capelli theorem. We will need the matrix of the system $A$ and the extended matrix of the system $\widetilde(A)$, we will write them:

$$ A=\left(\begin(array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \end(array) \right);\; \widetilde(A)=\left(\begin(array) (ccc|c) -3 & 9 &-7 & 17 \\ -1 & 2 & -4 & 9\\ 4 & -2 & 19 & -42 \end(array) \right). $$

We need to find $\rang A$ and $\rang\widetilde(A)$. There are many ways to do this, some of which are listed in the Matrix Rank section. Typically, two methods are used to study such systems: “Calculating the rank of a matrix by definition” or “Calculating the rank of a matrix by the method of elementary transformations”.

Method number 1. Computing ranks by definition.

According to the definition, rank is the highest order of the minors of a matrix, among which there is at least one that is different from zero. Usually, the study begins with first-order minors, but here it is more convenient to immediately begin calculating the third-order minor of the matrix $A$. The third-order minor elements are located at the intersection of three rows and three columns of the matrix in question. Since the matrix $A$ contains only 3 rows and 3 columns, the third order minor of the matrix $A$ is the determinant of the matrix $A$, i.e. $\Delta A$. To calculate the determinant, we apply formula No. 2 from the topic “Formulas for calculating determinants of the second and third orders”:

$$ \Delta A=\left| \begin(array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \end(array) \right|=-21. $$

So, there is a third order minor of the matrix $A$, which is not equal to zero. It is impossible to construct a fourth-order minor, since it requires 4 rows and 4 columns, and the matrix $A$ has only 3 rows and 3 columns. So, the highest order of the minors of the matrix $A$, among which there is at least one that is not equal to zero, is equal to 3. Therefore, $\rang A=3$.

We also need to find $\rang\widetilde(A)$. Let's look at the structure of the matrix $\widetilde(A)$. Up to the line in the matrix $\widetilde(A)$ there are elements of the matrix $A$, and we found out that $\Delta A\neq 0$. Consequently, the matrix $\widetilde(A)$ has a third-order minor, which is not equal to zero. We cannot construct fourth-order minors of the matrix $\widetilde(A)$, so we conclude: $\rang\widetilde(A)=3$.

Since $\rang A=\rang\widetilde(A)$, then according to the Kronecker-Capelli theorem the system is consistent, i.e. has a solution (at least one). To indicate the number of solutions, we take into account that our SLAE contains 3 unknowns: $x_1$, $x_2$ and $x_3$. Since the number of unknowns is $n=3$, we conclude: $\rang A=\rang\widetilde(A)=n$, therefore, according to the corollary of the Kronecker-Capelli theorem, the system is definite, i.e. has a unique solution.

The problem is solved. What disadvantages and advantages does this method have? First, let's talk about the advantages. Firstly, we only needed to find one determinant. After this, we immediately made a conclusion about the number of solutions. Typically, standard standard calculations give systems of equations that contain three unknowns and have a unique solution. For such systems, this method is very convenient, because we know in advance that there is a solution (otherwise the example would not have been in the standard calculation). Those. All we have to do is show the existence of a solution in the fastest way. Secondly, the calculated value of the determinant of the system matrix (i.e. $\Delta A$) will be useful later: when we begin to solve given system Cramer's method or using the inverse matrix.

However, the method of calculating the rank is by definition undesirable to use if the matrix of the system $A$ is rectangular. In this case, it is better to use the second method, which will be discussed below. In addition, if $\Delta A=0$, then we cannot say anything about the number of solutions of a given inhomogeneous SLAE. Maybe the SLAE has an infinite number of solutions, or maybe none. If $\Delta A=0$, then additional research is required, which is often cumbersome.

To summarize what has been said, I note that the first method is good for those SLAEs whose system matrix is square. Moreover, the SLAE itself contains three or four unknowns and is taken from standard standard calculations or tests.

Method number 2. Calculation of rank by the method of elementary transformations.

This method is described in detail in the corresponding topic. We will begin to calculate the rank of the matrix $\widetilde(A)$. Why matrices $\widetilde(A)$ and not $A$? The fact is that the matrix $A$ is part of the matrix $\widetilde(A)$, therefore, by calculating the rank of the matrix $\widetilde(A)$ we will simultaneously find the rank of the matrix $A$.

\begin(aligned) &\widetilde(A) =\left(\begin(array) (ccc|c) -3 & 9 &-7 & 17 \\ -1 & 2 & -4 & 9\\ 4 & - 2 & 19 & -42 \end(array) \right) \rightarrow \left|\text(swap the first and second lines)\right| \rightarrow \\ &\rightarrow \left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ -3 & 9 &-7 & 17\\ 4 & -2 & 19 & - 42 \end(array) \right) \begin(array) (l) \phantom(0) \\ r_2-3r_1\\ r_3+4r_1 \end(array) \rightarrow \left(\begin(array) (ccc| c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 6 & 3 & -6 \end(array) \right) \begin(array) (l) \phantom(0 ) \\ \phantom(0)\\ r_3-2r_2 \end(array)\rightarrow\\ &\rightarrow \left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 0 & -7 & 14 \end(array) \right) \end(aligned)

We have reduced the matrix $\widetilde(A)$ to echelon form. The resulting echelon matrix has three non-zero rows, so its rank is 3. Consequently, the rank of the matrix $\widetilde(A)$ is equal to 3, i.e. $\rang\widetilde(A)=3$. When making transformations with the elements of the matrix $\widetilde(A)$, we simultaneously transformed the elements of the matrix $A$ located up to the line. The matrix $A$ is also reduced to echelon form: $\left(\begin(array) (ccc) -1 & 2 & -4 \\ 0 & 3 &5 \\ 0 & 0 & -7 \end(array) \right )$. Conclusion: the rank of matrix $A$ is also 3, i.e. $\rang A=3$.

Since $\rang A=\rang\widetilde(A)$, then according to the Kronecker-Capelli theorem the system is consistent, i.e. has a solution. To indicate the number of solutions, we take into account that our SLAE contains 3 unknowns: $x_1$, $x_2$ and $x_3$. Since the number of unknowns is $n=3$, we conclude: $\rang A=\rang\widetilde(A)=n$, therefore, according to the corollary of the Kronecker-Capelli theorem, the system is defined, i.e. has a unique solution.

What are the advantages of the second method? The main advantage is its versatility. It doesn't matter to us whether the matrix of the system is square or not. In addition, we actually carried out forward transformations of the Gaussian method. There are only a couple of steps left, and we could obtain a solution to this SLAE. To be honest, I like the second method more than the first, but the choice is a matter of taste.

Answer: The given SLAE is consistent and defined.

Example No. 2

Explore SLAE $ \left\( \begin(aligned) & x_1-x_2+2x_3=-1;\\ & -x_1+2x_2-3x_3=3;\\ & 2x_1-x_2+3x_3=2;\\ & 3x_1- 2x_2+5x_3=1;\\ & 2x_1-3x_2+5x_3=-4.\end(aligned) \right.$ for compatibility.

We will find the ranks of the system matrix and the extended system matrix using the method of elementary transformations. Extended system matrix: $\widetilde(A)=\left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ -1 & 2 & -3 & 3 \\ 2 & -1 & 3 & 2 \\ 3 & -2 & 5 & 1 \\ 2 & -3 & 5 & -4 \end(array) \right)$. Let's find the required ranks by transforming the extended matrix of the system:

$$ \left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ -1 & 2 & -3 & 3 \\ 2 & -3 & 5 & -4 \\ 3 & -2 & 5 & 1 \\ 2 & -1 & 3 & 2 \end(array) \right) \begin(array) (l) \phantom(0)\\r_2+r_1\\r_3-2r_1\\ r_4 -3r_1\\r_5-2r_1\end(array)\rightarrow \left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ 0 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \\ 0 & 1 & -1 & 4 \\ 0 & 1 & -1 & 4 \end(array) \right) \begin(array) (l) \phantom(0)\\ \phantom(0)\\r_3-r_2\\ r_4-r_2\\r_5+r_2\end(array)\rightarrow\\ $$ $$ \rightarrow\left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \end(array) \ right) \begin(array) (l) \phantom(0)\\\phantom(0)\\\phantom(0)\\ r_4-r_3\\\phantom(0)\end(array)\rightarrow \left (\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end(array) \right) $$

The extended matrix of the system is reduced to a stepwise form. Rank step matrix is equal to the number of its non-zero lines, so $\rang\widetilde(A)=3$. The matrix $A$ (up to the line) is also reduced to echelon form, and its rank is 2, $\rang(A)=2$.

Since $\rang A\neq\rang\widetilde(A)$, then according to the Kronecker-Capelli theorem the system is inconsistent (i.e., has no solutions).

Answer: The system is inconsistent.

Example No. 3

Explore SLAE $ \left\( \begin(aligned) & 2x_1+7x_3-5x_4+11x_5=42;\\ & x_1-2x_2+3x_3+2x_5=17;\\ & -3x_1+9x_2-11x_3-7x_5=-64 ;\\ & -5x_1+17x_2-16x_3-5x_4-4x_5=-90;\\ & 7x_1-17x_2+23x_3+15x_5=132. \end(aligned) \right.$ for compatibility.

We bring the extended matrix of the system to a stepwise form:

$$ \left(\begin(array)(ccccc|c) 2 & 0 & 7 & -5 & 11 & 42\\ 1 & -2 & 3 & 0 & 2 & 17 \\ -3 & 9 & -11 & 0 & -7 & -64 \\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \end(array) \right) \overset (r_1\leftrightarrow(r_3))(\rightarrow) $$ $$ \rightarrow\left(\begin(array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 2 & 0 & 7 & -5 & 11 & 42\\ -3 & 9 & -11 & 0 & -7 & -64\\ -5 & 17 & -16 & -5 & -4 & -90\\ 7 & -17 & 23 & 0 & 15 & 132 \end(array) \right) \begin(array) (l) \phantom(0)\\ r_2-2r_1 \\r_3+3r_1 \\ r_4+5r_1 \\ r_5-7r_1 \end( array) \rightarrow \left(\begin(array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 0 & 4 & 1 & -5 & 7 & 8\\ 0 & 3 & - 2 & 0 & -1 & -13\\ 0 & 7 & -1 & -5 & 6 & -5 \\ 0 & -3 & 2 & 0 & 1 & 13 \end(array) \right) \begin( array) (l) \phantom(0)\\ \phantom(0)\\4r_3+3r_2 \\ 4r_4-7r_2 \\ 4r_5+3r_2 \end(array) \rightarrow $$ $$ \rightarrow\left(\begin (array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 0 & 4 & 1 & -5 & 7 & 8\\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & 11 & -15 & 25 & 76 \end(array) \right) \begin(array) (l) \phantom(0 )\\ \phantom(0)\\\phantom(0) \\ r_4-r_3 \\ r_5+r_2 \end(array) \rightarrow \left(\begin(array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 0 & 4 & 1 & -5 & 7 & 8\\ 0 & 0 & -11 & 15 & -25 & -76\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end(array) \right) $$

We have brought the extended matrix of the system and the matrix of the system itself to a stepwise form. The rank of the extended matrix of the system is equal to three, the rank of the matrix of the system is also equal to three. Since the system contains $n=5$ unknowns, i.e. $\rang\widetilde(A)=\rang(A)\lt(n)$, then according to the corollary of the Kronecker-Capelli theorem, this system is indeterminate, i.e. has an infinite number of solutions.

Answer: The system is uncertain.

In the second part we will look at examples that are often included in standard calculations or test papers in higher mathematics: study of consistency and solution of SLAE depending on the values of the parameters included in it.

However, in practice two more cases are widespread:

– The system is inconsistent (has no solutions);
– The system is consistent and has infinitely many solutions.

Note : The term “consistency” implies that the system has at least some solution. In a number of problems, it is necessary to first examine the system for compatibility; how to do this, see the article on rank of matrices.

For these systems, the most universal of all solution methods is used - Gaussian method. In fact, the “school” method will also lead to the answer, but in higher mathematics it is customary to use the Gaussian method of sequential elimination of unknowns. Those who are not familiar with the Gaussian method algorithm, please study the lesson first Gaussian method for dummies.

Example 1

What immediately catches your eye about this system? The number of equations is less than the number of variables. If the number of equations is less than the number of variables, then we can immediately say that the system is either inconsistent or has infinitely many solutions. And all that remains is to find out.

The beginning of the solution is completely ordinary - we write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

(1) On the top left step we need to get +1 or –1. There are no such numbers in the first column, so rearranging the rows will not give anything. The unit will have to organize itself, and this can be done in several ways. I did this: To the first line we add the third line, multiplied by –1.

(2) Now we get two zeros in the first column. To the second line we add the first line multiplied by 3. To the third line we add the first line multiplied by 5.

(3) After the transformation has been completed, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the required –1 on the second step. Divide the third line by –3.

(4) Add the second line to the third line.

Probably everyone noticed the bad line that resulted from elementary transformations: . It is clear that this cannot be so. Indeed, let us rewrite the resulting matrix back to the system of linear equations:

If, as a result of elementary transformations, a string of the form is obtained, where is a number other than zero, then the system is inconsistent (has no solutions).

How to write down the ending of a task? Let’s draw with white chalk: “as a result of elementary transformations, a string of the form , where ” is obtained and give the answer: the system has no solutions (inconsistent).

If, according to the condition, it is required to RESEARCH the system for compatibility, then it is necessary to formalize the solution in a more solid style using the concept matrix rank and the Kronecker-Capelli theorem.

Please note that there is no reversal of the Gaussian algorithm here - there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is an example for you to solve on your own. Full solution and answer at the end of the lesson. I remind you again that your solution may differ from my solution; the Gaussian algorithm does not have strong “rigidity”.

Another technical feature of the solution: elementary transformations can be stopped At once, as soon as a line like , where . Let's consider a conditional example: suppose that after the first transformation the matrix is obtained . The matrix has not yet been reduced to echelon form, but there is no need for further elementary transformations, since a line of the form has appeared, where . The answer should be given immediately that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift, due to the fact that a short solution is obtained, sometimes literally in 2-3 steps.

But everything in this world is balanced, and a problem in which the system has infinitely many solutions is just longer.

Example 3

Solve a system of linear equations

The beginning is again standard. Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

That's all, and you were afraid.

(1) Please note that all numbers in the first column are divisible by 2, so 2 is fine on the top left step. To the second line we add the first line, multiplied by –4. To the third line we add the first line, multiplied by –2. To the fourth line we add the first line, multiplied by –1.

Attention! Many may be tempted by the fourth line subtract first line. This can be done, but it is not necessary; experience shows that the probability of an error in calculations increases several times. Just add: To the fourth line add the first line multiplied by –1 – exactly!

(2) The last three lines are proportional, two of them can be deleted.

Here again we need to show increased attention, but are the lines really proportional? To be on the safe side (especially for a teapot), it would be a good idea to multiply the second line by –1, and divide the fourth line by 2, resulting in three identical lines. And only after that remove two of them.

As a result of elementary transformations, the extended matrix of the system is reduced to a stepwise form:

When writing a task in a notebook, it is advisable to make the same notes in pencil for clarity.

Let us rewrite the corresponding system of equations:

There is no smell of a “ordinary” single solution to the system here. There is no bad line either. This means that this is the third remaining case - the system has infinitely many solutions. Sometimes, according to the condition, it is necessary to investigate the compatibility of the system (i.e. prove that a solution exists at all), you can read about this in the last paragraph of the article How to find the rank of a matrix? But for now let’s go over the basics:

An infinite set of solutions to a system is briefly written in the form of the so-called general solution of the system .

We find the general solution of the system using the inverse of the Gaussian method.

First we need to define what variables we have basic, and what variables free. You don’t have to bother yourself with the terms of linear algebra, just remember that there are such basic variables And free variables.

Basic variables always “sit” strictly on the steps of the matrix.
In this example, the basic variables are and

Free variables are everything remaining variables that did not receive a step. In our case there are two of them: – free variables.

Now you need All basic variables express only through free variables.

The reverse of the Gaussian algorithm traditionally works from the bottom up.
From the second equation of the system we express the basic variable:

Now look at the first equation: . First we substitute the found expression into it:

It remains to express the basic variable in terms of free variables:

In the end we got what we needed - All basic variables ( and ) are expressed only through free variables:

Actually, the general solution is ready:

How to write the general solution correctly?
Free variables are written into the general solution “by themselves” and strictly in their places. In this case, free variables should be written in the second and fourth positions:
.

The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:

Giving free variables arbitrary values, you can find infinitely many private solutions. The most popular values are zeros, since the particular solution is the easiest to obtain. Let's substitute into the general solution:

– private solution.

Another sweet pair are ones, let’s substitute them into the general solution:

– another private solution.

It is easy to see that the system of equations has infinitely many solutions(since we can give free variables any values)

Each the particular solution must satisfy to each equation of the system. This is the basis for a “quick” check of the correctness of the solution. Take, for example, a particular solution and substitute it into the left side of each equation of the original system:

Everything must come together. And with any particular solution you receive, everything should also agree.

But, strictly speaking, checking a particular solution is sometimes deceiving, i.e. some particular solution may satisfy each equation of the system, but the general solution itself is actually found incorrectly.

Therefore, verification of the general solution is more thorough and reliable. How to check the resulting general solution ?

It's not difficult, but quite tedious. We need to take expressions basic variables, in this case and , and substitute them into the left side of each equation of the system.

To the left side of the first equation of the system:

To the left side of the second equation of the system:

The right side of the original equation is obtained.

Example 4

Solve the system using the Gaussian method. Find the general solution and two particular ones. Check the general solution.

This is an example for you to solve on your own. Here, by the way, again the number of equations is less than the number of unknowns, which means it is immediately clear that the system will either be inconsistent or have an infinite number of solutions. What is important in the decision process itself? Attention, and attention again. Full solution and answer at the end of the lesson.

And a couple more examples to reinforce the material

Example 5

Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution

Solution: Let's write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:

(1) Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.
(2) To the third line we add the second line, multiplied by –5. To the fourth line we add the second line, multiplied by –7.
(3) The third and fourth lines are the same, we delete one of them.

This is such a beauty:

Basic variables sit on the steps, therefore - basic variables.
There is only one free variable that did not get a step:

Reverse:
Let's express the basic variables through a free variable:
From the third equation:

Let's consider the second equation and substitute the found expression into it:

Let's consider the first equation and substitute the found expressions and into it:

Yes, a calculator that calculates ordinary fractions is still convenient.

So the general solution is:

Once again, how did it turn out? The free variable sits alone in its rightful fourth place. The resulting expressions for the basic variables also took their ordinal places.

Let us immediately check the general solution. The job is for blacks, but I have already done it, so catch it =)

We substitute three heroes , , into the left side of each equation of the system:

The corresponding right-hand sides of the equations are obtained, thus the general solution is found correctly.

Now from the found general solution we obtain two particular solutions. The only free variable here is the chef. No need to rack your brains.

Let it be then – private solution.
Let it be then – another private solution.

Answer: Common decision: , private solutions: , .

I shouldn’t have remembered about blacks... ...because all sorts of sadistic motives came into my head and I remembered the famous photoshop in which Ku Klux Klansmen in white robes are running across the field after a black football player. I sit and smile quietly. You know how distracting...

A lot of mathematics is harmful, so a similar final example for solving it yourself.

Example 6

Find the general solution to the system of linear equations.

I have already checked the general solution, the answer can be trusted. Your solution may differ from my solution, the main thing is that the general solutions coincide.

Probably, many people noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gaussian method, we had to tinker with ordinary fractions. In practice, this is indeed the case; cases where there are no fractions are much less common. Be prepared mentally and, most importantly, technically.

I will dwell on some features of the solution that were not found in the solved examples.

The general solution of the system may sometimes include a constant (or constants), for example: . Here one of the basic variables is equal to a constant number: . There is nothing exotic about this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables. The Gaussian method works in the most severe conditions; one should calmly reduce the extended matrix of the system to a stepwise form using a standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a single solution.

Prove that the system of linear equations. Examples of systems of linear equations: solution method

Method number 1. Computing ranks by definition.

Method number 2. Calculation of rank by the method of elementary transformations.

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