The joint system ax in is called indefinite if. Solving systems of linear algebraic equations, solution methods, examples

The system is called joint, or solvable if it has at least one solution. The system is called incompatible, or insoluble if it has no solutions.

Definite, indefinite SLAE.

If a SLAE has a solution and is unique, then it is called certain and if the solution is not unique, then uncertain.

MATRIX 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 matrix of the system and matrix columns of unknown and free members

Let's find the product

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 as

or shorter A∙X=B.

Here matrices A And B are known, and the matrix X unknown. She needs to be found, because. its elements are the solution of 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, the inverse of the matrix A: . Insofar as A -1 A = E And E∙X=X, then we obtain the solution of 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 is the same as the number of unknowns.

Cramer's formulas

Cramer's method is that we successively find master system identifier, i.e. determinant of matrix A: D = det (a i j) and n auxiliary determinants D i (i= ), which are obtained from the determinant D by replacing the i-th column with a column of free members.

Cramer's formulas look like: D × x i = D i (i = ).

This implies Cramer's rule, which gives an exhaustive answer to the question of system compatibility: if the main determinant of the system is different from zero, then the system has a unique solution, determined by the formulas: x i = D i / D.

If the main determinant of the system D and all auxiliary determinants D i = 0 (i= ), then the system has an infinite number of solutions. If the main determinant of the system D = 0, and at least one auxiliary determinant is different from zero, then the system is inconsistent.

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

Proof: So, consider a system of 3 equations with three unknowns. Multiply the 1st equation of the system by the algebraic complement A 11 element a 11, 2nd equation - on A21 and 3rd - on A 31:

Let's add these equations:

Consider each of the brackets and the right side of this equation. According to the theorem on the expansion of the determinant in terms of the elements of the 1st column.

Similarly, it can be shown that and .

Finally, it is easy to see that

Thus, we get the equality: . Consequently, .

The equalities and are derived similarly, whence the assertion of the theorem follows.

Kronecker-Capelli theorem.

A system of linear equations is consistent if and only if the rank of the system's matrix is ​​equal to the rank of the augmented matrix.

Proof: It breaks down into two stages.

1. Let the system have a solution. Let's show that .

Let the set of numbers is the solution to the system. Denote by the -th column of the matrix , . Then , that is, the column of free terms is a linear combination of the columns of the matrix . Let be . Let's pretend that . Then by . We choose in the basic minor . He has order. The column of free members must pass through this minor, otherwise it will be the basis minor of the matrix. The column of free terms in minor is a linear combination of the columns of the matrix. By virtue of the properties of the determinant , where is the determinant that is obtained from the minor by replacing the column of free terms with the column . If the column passed through the minor M, then in , there will be two identical columns and, therefore, . If the column did not pass through the minor, then it will differ from the minor of order r + 1 of the matrix only by the order of the columns. Since , then . Thus, which contradicts the definition of a basis minor. Hence, the assumption that , is false.

2. Let . Let us show that the system has a solution. Since , then the basis minor of the matrix is ​​the basis minor of the matrix . Let the columns pass through the minor . Then, by the basis minor theorem in a matrix, the column of free terms is a linear combination of the indicated columns:

(1)

We set , , , , and take the remaining unknowns equal to zero. Then for these values ​​we get

By virtue of equality (1) . The last equality means that the set of numbers is the solution to the system. The existence of a solution is proved.

In the system discussed above , and the system is consistent. In the system , , and the system is inconsistent.

Note: Although the Kronecker-Capelli theorem makes it possible to determine whether a system is consistent, it is used quite rarely, mainly in theoretical studies. The reason is that the calculations performed when finding the rank of a matrix are basically the same as the calculations when finding a solution to the system. Therefore, usually instead of finding and , one looks for a solution to the system. If it can be found, then we learn that the system is consistent and simultaneously obtain its solution. If a solution cannot be found, then we conclude that the system is inconsistent.

Algorithm for finding solutions to an arbitrary system of linear equations (Gauss method)

Let a system of linear equations with unknowns be given. It is required to find its general solution if it is consistent, or establish its inconsistency. The method that will be presented in this section is close to the method of calculating the determinant and to the method of finding the rank of a matrix. The proposed algorithm is called Gauss method or method of successive elimination of unknowns.

Let us write the augmented matrix of the system

We call the following operations with matrices elementary operations:

1. permutation of lines;

2. multiplying a string by a non-zero number;

3. addition of a string with another string multiplied by a number.

Note that when solving a system of equations, in contrast to calculating the determinant and finding the rank, one cannot operate with columns. If the system of equations is restored from the matrix obtained from the elementary operation, then new system will be equal to the original.

The purpose of the algorithm is, by applying a sequence of elementary operations to the matrix, to ensure that each row, except perhaps the first one, starts with zeros, and the number of zeros up to the first non-zero element in each next row is greater than in the previous one.

The step of the algorithm is as follows. Find the first non-zero column in the matrix. Let it be a column with number . We find a non-zero element in it and swap the line with this element with the first line. In order not to pile up additional notation, we will assume that such a change of rows in the matrix has already been made, that is, . Then to the second line we add the first one multiplied by the number , to the third line we add the first one multiplied by the number , etc. As a result, we get the matrix

(The first null columns are usually missing.)

If there is a row with number k in the matrix, in which all elements are equal to zero, and , then we stop the algorithm execution and conclude that the system is inconsistent. Indeed, restoring the system of equations from the extended matrix, we get that the -th equation will have the form

This equation does not satisfy any set of numbers .

The matrix can be written as

With respect to the matrix, we perform the described step of the algorithm. Get the matrix

where , . This matrix can again be written as

and the above step of the algorithm is again applied to the matrix.

The process stops if after the execution of the next step the new reduced matrix consists of only zeros or if all rows are exhausted. Note that the conclusion about the incompatibility of the system could stop the process even earlier.

If we did not reduce the matrix, then in the end we would come to a matrix of the form

Next, the so-called reverse pass of the Gaussian method is performed. Based on the matrix, we compose a system of equations. On the left side, we leave the unknowns with numbers corresponding to the first non-zero elements in each line, that is, . Notice, that . The remaining unknowns are transferred to the right side. Considering the unknowns on the right side to be some fixed quantities, it is easy to express the unknowns on the left side in terms of them.

Now, giving arbitrary values ​​to the unknowns on the right side and calculating the values ​​of the variables on the left side, we will find various solutions original system Ax=b. To write down the general solution, it is necessary to denote the unknowns on the right side in any order by letters , including those unknowns that are not explicitly written on the right side due to zero coefficients, and then the column of unknowns can be written as a column, where each element is a linear combination of arbitrary values (in particular, just an arbitrary value ). This entry will be the general solution of the system.

If the system was homogeneous, then we obtain the general solution of the homogeneous system. The coefficients at taken in each element of the column of the general solution will make up the first solution from the fundamental system of solutions, the coefficients at - the second solution, and so on.

Method 2: The fundamental system of solutions of a homogeneous system can be obtained in another way. To do this, one variable, transferred to the right side, must be assigned the value 1, and the rest - zeros. Calculating the values ​​of the variables on the left side, we obtain one solution from the fundamental system. By assigning the value 1 to the other variable on the right side, and zeros to the others, we obtain the second solution from the fundamental system, and so on.

Definition: the system is called jointly th, if it has at least one solution, and inconsistent - otherwise, that is, in the case when the system has no solutions. The question of whether a system has a solution or not is connected not only with the ratio of the number of equations and the number of unknowns. For example, a system of three equations with two unknowns

has a solution , and even has infinitely many solutions, but a system of two equations with three unknowns.

……. … ……

A m 1 x 1 + … + a mn x n = 0

This system is always consistent since it has a trivial solution x 1 =…=x n =0

For nontrivial solutions to exist, it is necessary and sufficient that

conditions r = r(A)< n , что равносильно условию det(A)=0, когда матрица А – квадратная.

Th The set of SLAE solutions forms a linear space of dimension (n-r). This means that the product of its solution by a number, as well as the sum and linear combination of a finite number of its solutions, are solutions of this system. The linear solution space of any SLAE is a subspace of the space R n .

Any set of (n-r) linearly independent solutions of a SLAE (which is a basis in the solution space) is called fundamental set of solutions (FSR).

Let х 1 ,…,х r be basic unknowns, х r +1 ,…,х n be free unknowns. We give the free variables in turn the following values:

……. … ……

A m 1 x 1 + … + a mn x n = 0

Forms a linear space S (space of solutions), which is a subspace in R n (n is the number of unknowns), and dims=k=n-r, where r is the rank of the system. The basis in the solution space(x (1) ,…, x (k) ) is called the fundamental system of solutions, and the general solution has the form:

X=c 1 x (1) + … + c k x (k) , c (1) ,…, c (k) ? R

Higher Mathematics » Systems of Linear algebraic equations» Basic terms. Matrix notation.

System of linear algebraic equations. Basic terms. Matrix notation.

1. Definition of a system of linear algebraic equations. System solution. Classification of systems.
2. Matrix form of writing systems of linear algebraic equations.

Definition of a system of linear algebraic equations. System solution. Classification of systems.

Under system of linear algebraic equations(SLAE) imply a system

\begin(equation) \left \( \begin(aligned) & a_(11)x_1+a_(12)x_2+a_(13)x_3+\ldots+a_(1n)x_n=b_1;\\ & a_(21) x_1+a_(22)x_2+a_(23)x_3+\ldots+a_(2n)x_n=b_2;\\ & \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ ldots \\ & a_(m1)x_1+a_(m2)x_2+a_(m3)x_3+\ldots+a_(mn)x_n=b_m.\end(aligned) \right.\end(equation)

The parameters $a_(ij)$ ($i=\overline(1,m)$, $j=\overline(1,n)$) are called coefficients, and $b_i$ ($i=\overline(1,m)$) - free members SLAU. Sometimes, to emphasize the number of equations and unknowns, they say "$m\times n$ system of linear equations" - thereby indicating that the SLAE contains $m$ equations and $n$ unknowns.

If all free terms $b_i=0$ ($i=\overline(1,m)$), then the SLAE is called homogeneous. If among the free members there is at least one other than zero, the SLAE is called heterogeneous.

SLAU decision(1) any ordered collection of numbers ($\alpha_1, \alpha_2,\ldots,\alpha_n$) is called if the elements of this collection, substituted in a given order for the unknowns $x_1,x_2,\ldots,x_n$, invert each SLAE equation into identity.

Any homogeneous SLAE has at least one solution: zero(in a different terminology - trivial), i.e. $x_1=x_2=\ldots=x_n=0$.

If SLAE (1) has at least one solution, it is called joint if there are no solutions, incompatible. If a joint SLAE has exactly one solution, it is called certain, if an infinite number of solutions - uncertain.

Example #1

Consider SLAE

\begin(equation) \left \( \begin(aligned) & 3x_1-4x_2+x_3+7x_4-x_5=11;\\ & 2x_1+10x_4-3x_5=-65;\\ & 3x_2+19x_3+8x_4-6x_5= 0.\\ \end(aligned)\right.\end(equation)

We have a system of linear algebraic equations containing $3$ equations and $5$ unknowns: $x_1,x_2,x_3,x_4,x_5$. One can say that a system of $3\times 5$ linear equations is given.

The coefficients of system (2) are the numbers in front of the unknowns. For example, in the first equation these numbers are: $3,-4,1,7,-1$. The free members of the system are represented by the numbers $11,-65.0$. Since there is at least one among the free members, it is not zero, then SLAE (2) is inhomogeneous.

The ordered collection $(4;-11;5;-7;1)$ is the solution to this SLAE. This is easy to verify if you substitute $x_1=4; x_2=-11; x_3=5; x_4=-7; x_5=1$ into the equations of the given system:

\begin(aligned) & 3x_1-4x_2+x_3+7x_4-x_5=3\cdot4-4\cdot(-11)+5+7\cdot(-7)-1=11;\\ & 2x_1+10x_4-3x_5 =2\cdot 4+10\cdot (-7)-3\cdot 1=-65;\\ & 3x_2+19x_3+8x_4-6x_5=3\cdot (-11)+19\cdot 5+8\cdot ( -7)-6\cdot 1=0. \\ \end(aligned)

Naturally, the question arises whether the verified solution is the only one. The issue of the number of SLAE solutions will be discussed in the relevant topic.

Example #2

Consider SLAE

\begin(equation) \left \( \begin(aligned) & 4x_1+2x_2-x_3=0;\\ & 10x_1-x_2=0;\\ & 5x_2+4x_3=0; \\ & 3x_1-x_3=0; \\ & 14x_1+25x_2+5x_3=0.\end(aligned) \right.\end(equation)

System (3) is a SLAE containing $5$ equations and $3$ unknowns: $x_1,x_2,x_3$. Since all free terms of this system are equal to zero, then SLAE (3) is homogeneous. It is easy to check that the collection $(0;0;0)$ is a solution to the given SLAE. Substituting $x_1=0, x_2=0,x_3=0$, for example, into the first equation of system (3), we get the correct equality: $4x_1+2x_2-x_3=4\cdot 0+2\cdot 0-0=0$ . Substitution into other equations is done in a similar way.

Matrix form of writing systems of linear algebraic equations.

Several matrices can be associated with each SLAE; moreover, the SLAE itself can be written as a matrix equation. For SLAE (1), consider the following matrices:

The matrix $A$ is called system matrix. The elements of this matrix are the coefficients of the given SLAE.

The matrix $\widetilde(A)$ is called expanded matrix system. It is obtained by adding to the system matrix a column containing free members $b_1,b_2,…,b_m$. Usually this column is separated by a vertical line - for clarity.

The column matrix $B$ is called matrix of free members, and the column matrix $X$ - matrix of unknowns.

Using the notation introduced above, SLAE (1) can be written in the form of a matrix equation: $A\cdot X=B$.

Note

The matrices associated with the system can be written different ways: it all depends on the order of the variables and equations of the considered SLAE. But in any case, the order of the unknowns in each equation of a given SLAE must be the same (see example No. 4).

Example #3

Write SLAE $ \left \( \begin(aligned) & 2x_1+3x_2-5x_3+x_4=-5;\\ & 4x_1-x_3=0;\\ & 14x_2+8x_3+x_4=-11. \end(aligned) \right.$ in matrix form and specify the augmented matrix of the system.

We have four unknowns, which in each equation follow in this order: $x_1,x_2,x_3,x_4$. The matrix of unknowns will be: $\left(\begin(array) (c) x_1 \\ x_2 \\ x_3 \\ x_4 \end(array) \right)$.

The free members of this system are expressed by the numbers $-5,0,-11$, therefore the matrix of free members has the form: $B=\left(\begin(array) (c) -5 \\ 0 \\ -11 \end(array )\right)$.

Let's move on to compiling the matrix of the system. The first row of this matrix will contain the coefficients of the first equation: $2.3,-5.1$.

In the second line we write the coefficients of the second equation: $4.0,-1.0$. In this case, it should be taken into account that the coefficients of the system with the variables $x_2$ and $x_4$ in the second equation are equal to zero (because these variables are absent in the second equation).

In the third row of the matrix of the system, we write the coefficients of the third equation: $0.14.8.1$. We take into account the equality to zero of the coefficient at the variable $x_1$ (this variable is absent in the third equation). The system matrix will look like:

$$ A=\left(\begin(array) (cccc) 2 & 3 & -5 & 1\\ 4 & 0 & -1 & 0 \\ 0 & 14 & 8 & 1 \end(array) \right) $$

To make the relationship between the system matrix and the system itself clearer, I will write down the given SLAE and its system matrix side by side:

In matrix form, the given SLAE will look like $A\cdot X=B$. In the expanded entry:

$$ \left(\begin(array) (cccc) 2 & 3 & -5 & 1\\ 4 & 0 & -1 & 0 \\ 0 & 14 & 8 & 1 \end(array) \right) \cdot \left(\begin(array) (c) x_1 \\ x_2 \\ x_3 \\ x_4 \end(array) \right) = \left(\begin(array) (c) -5 \\ 0 \\ -11 \end(array) \right) $$

Let us write the augmented matrix of the system. To do this, to the system matrix $ A=\left(\begin(array) (cccc) 2 & 3 & -5 & 1\\ 4 & 0 & -1 & 0 \\ 0 & 14 & 8 & 1 \end(array ) \right) $ add a column of free terms (i.e. $-5,0,-11$). We get: $\widetilde(A)=\left(\begin(array) (cccc|c) 2 & 3 & -5 & 1 & -5 \\ 4 & 0 & -1 & 0 & 0\\ 0 & 14 & 8 & 1 & -11 \end(array) \right) $.

Example #4

Write SLAE $ \left \(\begin(aligned) & 3y+4a=17;\\ & 2a+4y+7c=10;\\ & 8c+5y-9a=25; \\ & 5a-c=-4 .\end(aligned)\right.$ in matrix form and specify the augmented matrix of the system.

As you can see, the order of the unknowns in the equations of this SLAE is different. For example, in the second equation the order is: $a,y,c$, but in the third equation: $c,y,a$. Before writing the SLAE in matrix form, the order of the variables in all equations must be made the same.

You can order the variables in the equations of a given SLAE different ways(the number of ways to arrange three variables is $3!=6$). I will consider two ways of ordering unknowns.

Method number 1

Let's introduce the following order: $c,y,a$. Let us rewrite the system, placing the unknowns in necessary order: $\left \(\begin(aligned) & 3y+4a=17;\\ & 7c+4y+2a=10;\\ & 8c+5y-9a=25; \\ & -c+5a=-4 .\end(aligned)\right.$

For clarity, I will write the SLAE as follows: $\left \(\begin(aligned) & 0\cdot c+3\cdot y+4\cdot a=17;\\ & 7\cdot c+4\cdot y+ 2\cdot a=10;\\ & 8\cdot c+5\cdot y-9\cdot a=25; \\ & -1\cdot c+0\cdot y+5\cdot a=-4. \ end(aligned)\right.$

The system matrix is: $ A=\left(\begin(array) (ccc) 0 & 3 & 4 \\ 7 & 4 & 2\\ 8 & 5 & -9 \\ -1 & 0 & 5 \end( array) \right) $. Free member matrix: $B=\left(\begin(array) (c) 17 \\ 10 \\ 25 \\ -4 \end(array) \right)$. When writing the matrix of unknowns, remember the order of the unknowns: $X=\left(\begin(array) (c) c \\ y \\ a \end(array) \right)$. So, the matrix form of the given SLAE is as follows: $A\cdot X=B$. Expanded:

$$ \left(\begin(array) (ccc) 0 & 3 & 4 \\ 7 & 4 & 2\\ 8 & 5 & -9 \\ -1 & 0 & 5 \end(array) \right) \ cdot \left(\begin(array) (c) c \\ y \\ a \end(array) \right) = \left(\begin(array) (c) 17 \\ 10 \\ 25 \\ -4 \end(array) \right) $$

The extended system matrix is: $\left(\begin(array) (ccc|c) 0 & 3 & 4 & 17 \\ 7 & 4 & 2 & 10\\ 8 & 5 & -9 & 25 \\ -1 & 0 & 5 & -4 \end(array) \right) $.

Method number 2

Let's introduce the following order: $a,c,y$. Let's rewrite the system, putting the unknowns in the required order: $\left \( \begin(aligned) & 4a+3y=17;\\ & 2a+7c+4y=10;\\ & -9a+8c+5y=25; \ \ & 5a-c=-4.\end(aligned)\right.$

For clarity, I will write the SLAE as follows: $\left \( \begin(aligned) & 4\cdot a+0\cdot c+3\cdot y=17;\\ & 2\cdot a+7\cdot c+ 4\cdot y=10;\\ & -9\cdot a+8\cdot c+5\cdot y=25; \\ & 5\cdot c-1\cdot c+0\cdot y=-4. \ end(aligned)\right.$

The system matrix is: $ A=\left(\begin(array) (ccc) 4 & 0 & 3 \\ 2 & 7 & 4\\ -9 & 8 & 5 \\ 5 & -1 & 0 \end( array)\right)$. Free member matrix: $B=\left(\begin(array) (c) 17 \\ 10 \\ 25 \\ -4 \end(array) \right)$. When writing the matrix of unknowns, remember the order of the unknowns: $X=\left(\begin(array) (c) a \\ c \\ y \end(array) \right)$. So, the matrix form of the given SLAE is as follows: $A\cdot X=B$. Expanded:

$$ \left(\begin(array) (ccc) 4 & 0 & 3 \\ 2 & 7 & 4\\ -9 & 8 & 5 \\ 5 & -1 & 0 \end(array) \right) \ cdot \left(\begin(array) (c) a \\ c \\ y \end(array) \right) = \left(\begin(array) (c) 17 \\ 10 \\ 25 \\ -4 \end(array) \right) $$

The extended system matrix is: $\left(\begin(array) (ccc|c) 4 & 0 & 3 & 17 \\ 2 & 7 & 4 & 10\\ -9 & 8 & 5 & 25 \\ 5 & - 1 & 0 & -4 \end(array) \right) $.

As you can see, changing the order of the unknowns is equivalent to rearranging the columns of the system matrix. But whatever this arrangement of unknowns may be, it must match in all equations of a given SLAE.

Linear equations

Linear equations- a relatively simple mathematical topic, quite often found in assignments in algebra.

Systems of linear algebraic equations: basic concepts, types

Let's figure out what it is and how linear equations are solved.

Usually, linear equation is an equation of the form ax + c = 0, where a and c are arbitrary numbers, or coefficients, and x is an unknown number.

For example, a linear equation would be:

Solution of linear equations.

How to solve linear equations?

Solving linear equations is quite easy. For this, a mathematical technique is used, such as identity transformation. Let's figure out what it is.

An example of a linear equation and its solution.

Let ax + c = 10, where a = 4, c = 2.

Thus, we get the equation 4x + 2 = 10.

In order to solve it was easier and faster, we will use the first method identity transformation- that is, we transfer all the numbers to the right side of the equation, and leave the unknown 4x on the left side.

Get:

Thus, the equation is reduced to a very simple problem for beginners. It remains only to use the second method of identical transformation - leaving x on the left side of the equation, transfer the numbers to the right side. We get:

Examination:

4x + 2 = 10, where x = 2.

The answer is correct.

Linear equation graph.

When solving linear equations with two variables, the plotting method is also often used. The fact is that an equation of the form ax + wy + c \u003d 0, as a rule, has many solutions, because many numbers fit in the place of variables, and in all cases the equation remains true.

Therefore, to facilitate the task, a graph of a linear equation is built.

To build it, it is enough to take one pair of variable values ​​- and, marking them with points on the coordinate plane, draw a straight line through them. All points on this line will be variants of the variables in our equation.

Expressions, expression conversion

The order of actions, rules, examples.

Numeric, literal and expressions with variables in their record may contain characters of various arithmetic operations. When converting expressions and calculating the values ​​of expressions, actions are performed in a certain order, in other words, you must observe order of actions.

In this article, we will figure out which actions should be performed first, and which ones after them. Let's start with the most simple cases when the expression contains only numbers or variables connected by plus, minus signs, multiply and divide. Next, we will explain what order of execution of actions should be followed in expressions with brackets. Finally, consider the sequence in which actions are performed in expressions containing powers, roots, and other functions.

First multiplication and division, then addition and subtraction

The school provides the following a rule that determines the order in which actions are performed in expressions without parentheses:

• actions are performed in order from left to right,
• where multiplication and division are performed first, and then addition and subtraction.

The stated rule is perceived quite naturally. Performing actions in order from left to right is explained by the fact that it is customary for us to keep records from left to right. And the fact that multiplication and division is performed before addition and subtraction is explained by the meaning that these actions carry in themselves.

Let's look at a few examples of the application of this rule. For examples, we will take the simplest numerical expressions so as not to be distracted by calculations, but to focus on the order in which actions are performed.

Follow steps 7−3+6.

The original expression does not contain parentheses, nor does it contain multiplication and division. Therefore, we should perform all the actions in order from left to right, that is, first we subtract 3 from 7, we get 4, after which we add 6 to the difference 4 obtained, we get 10.

Briefly, the solution can be written as follows: 7−3+6=4+6=10.

Indicate the order in which actions are performed in the expression 6:2·8:3.

To answer the question of the problem, let's turn to the rule that indicates the order in which actions are performed in expressions without brackets. The original expression contains only the operations of multiplication and division, and according to the rule, they must be performed in order from left to right.

First, divide 6 by 2, multiply this quotient by 8, and finally, divide the result by 3.

Basic concepts. Systems of linear equations

Calculate the value of the expression 17−5 6:3−2+4:2.

First, let's determine in what order the actions in the original expression should be performed. It includes both multiplication and division and addition and subtraction.

First, from left to right, you need to perform multiplication and division. So we multiply 5 by 6, we get 30, we divide this number by 3, we get 10. Now we divide 4 by 2, we get 2. We substitute the found value 10 instead of 5 6: 3 in the original expression, and the value 2 instead of 4: 2, we have 17−5 6:3−2+4:2=17−10−2+2.

In the resulting expression, there is no longer multiplication and division, so it remains to perform the remaining actions in order from left to right: 17−10−2+2=7−2+2=5+2=7.

17−5 6:3−2+4:2=7.

At first, in order not to confuse the order of performing actions when calculating the value of an expression, it is convenient to place numbers above the signs of actions corresponding to the order in which they are performed. For the previous example, it would look like this: .

The same order of operations - first multiplication and division, then addition and subtraction - should be followed when working with literal expressions.

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Steps 1 and 2

In some textbooks on mathematics, there is a division of arithmetic operations into operations of the first and second steps. Let's deal with this.

In these terms, the rule from the previous paragraph, which determines the order in which actions are performed, will be written as follows: if the expression does not contain brackets, then in order from left to right, the actions of the second stage (multiplication and division) are performed first, then the actions of the first stage (addition and subtraction).

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Order of execution of arithmetic operations in expressions with brackets

Expressions often contain parentheses to indicate the order in which the actions are to be performed. In this case a rule that specifies the order in which actions are performed in expressions with brackets, is formulated as follows: first, the actions in brackets are performed, while multiplication and division are also performed in order from left to right, then addition and subtraction.

So, expressions in brackets are considered as components of the original expression, and the order of actions already known to us is preserved in them. Consider the solutions of examples for greater clarity.

Do the indicated steps 5+(7−2 3) (6−4):2.

The expression contains brackets, so let's first perform the operations in the expressions enclosed in these brackets. Let's start with the expression 7−2 3. In it, you must first perform the multiplication, and only then the subtraction, we have 7−2 3=7−6=1. We pass to the second expression in brackets 6−4. There is only one action here - subtraction, we perform it 6−4=2.

We substitute the obtained values ​​into the original expression: 5+(7−2 3) (6−4):2=5+1 2:2. In the resulting expression, first we perform multiplication and division from left to right, then subtraction, we get 5+1 2:2=5+2:2=5+1=6. On this, all the actions are completed, we adhered to the following order of their execution: 5+(7−2 3) (6−4):2.

Let's write down short solution: 5+(7−2 3)(6−4):2=5+1 2:2=5+1=6.

5+(7−2 3)(6−4):2=6.

It happens that an expression contains brackets within brackets. You should not be afraid of this, you just need to consistently apply the voiced rule for performing actions in expressions with brackets. Let's show an example solution.

Perform actions in the expression 4+(3+1+4 (2+3)).

This is an expression with brackets, which means that the execution of actions must begin with an expression in brackets, that is, with 3 + 1 + 4 (2 + 3).

This expression also contains parentheses, so you must first perform actions in them. Let's do this: 2+3=5. Substituting the found value, we get 3+1+4 5. In this expression, we first perform multiplication, then addition, we have 3+1+4 5=3+1+20=24. The initial value, after substituting this value, takes the form 4+24, and it remains only to complete the actions: 4+24=28.

4+(3+1+4 (2+3))=28.

In general, when parentheses within parentheses are present in an expression, it is often convenient to start with the inner parentheses and work your way to the outer ones.

For example, let's say we need to perform operations in the expression (4+(4+(4−6:2))−1)−1. First, we perform actions in internal brackets, since 4−6:2=4−3=1, then after that the original expression will take the form (4+(4+1)−1)−1. Again, we perform the action in the inner brackets, since 4+1=5, we arrive at the following expression (4+5−1)−1. Again, we perform the actions in brackets: 4+5−1=8, while we arrive at the difference 8−1, which is equal to 7.

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The order in which operations are performed in expressions with roots, powers, logarithms, and other functions

If the expression includes powers, roots, logarithms, sine, cosine, tangent and cotangent, as well as other functions, then their values ​​are calculated before the other actions are performed, while the rules from the previous paragraphs that specify the order in which the actions are performed are also taken into account. In other words, the listed things, roughly speaking, can be considered enclosed in brackets, and we know that the actions in brackets are performed first.

Let's consider examples.

Perform the operations in the expression (3+1) 2+6 2:3−7.

This expression contains a power of 6 2 , its value must be calculated before performing the rest of the steps. So, we perform exponentiation: 6 2 \u003d 36. We substitute this value into the original expression, it will take the form (3+1) 2+36:3−7.

Then everything is clear: we perform actions in brackets, after which an expression without brackets remains, in which, in order from left to right, we first perform multiplication and division, and then addition and subtraction. We have (3+1) 2+36:3−7=4 2+36:3−7=8+12−7=13.

(3+1) 2+6 2:3−7=13.

Others, including more complex examples performing actions in expressions with roots, degrees, etc., you can see the calculation of expression values ​​in the article.

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First step actions are called addition and subtraction, and multiplication and division are called second step actions.

• Maths: studies. for 5 cells. general education institutions / N. Ya. Vilenkin, V. I. Zhokhov, A. S. Chesnokov, S. I. Shvartsburd. - 21st ed., erased. — M.: Mnemozina, 2007. — 280 p.: ill. ISBN 5-346-00699-0.

Write down the system of linear algebraic equations in general form

What is a SLAE solution?

The solution of a system of equations is a set of n numbers,

When which is substituted into the system, each equation becomes an identity.

What system is called joint (non-joint)?

A system of equations is called consistent if it has at least one solution.

A system is called inconsistent if it has no solutions.

What system is called definite (indefinite)?

A joint system is called definite if it has a unique solution.

A joint system is called indeterminate if it has more than one solution.

Matrix form of writing a system of equations

Rank of the vector system

The rank of a system of vectors is the maximum number of linearly independent vectors.

Matrix rank and ways to find it

Matrix rank- the highest of the orders of the minors of this matrix, the determinant of which is different from zero.

The first method, the edging method, is as follows:

If all minors are of the 1st order, i.e. matrix elements are equal to zero, then r=0 .

If at least one of the minors of the 1st order is not equal to zero, and all the minors of the 2nd order are equal to zero, then r=1.

If the 2nd order minor is nonzero, then we investigate the 3rd order minors. In this way, the k-th order minor is found and it is checked whether the k+1-th order minors are not equal to zero.

If all k+1 order minors are equal to zero, then the rank of the matrix is ​​equal to the number k. Such k+1 order minors are usually found by "edging" the k-th order minor.

The second method for determining the rank of a matrix is ​​to apply elementary transformations of the matrix when it is raised to a diagonal form. The rank of such a matrix is ​​equal to the number of non-zero diagonal elements.

General solution of an inhomogeneous system of linear equations, its properties.

Property 1. The sum of any solution to a system of linear equations and any solution to the corresponding homogeneous system is a solution to the system of linear equations.

Property 2.

Systems of linear equations: basic concepts

The difference of any two solutions of an inhomogeneous system of linear equations is a solution of the corresponding homogeneous system.

Gauss method for solving SLAE

Subsequence:

1) an expanded matrix of the equation system is compiled

2) with the help of elementary transformations, the matrix is ​​reduced to a step form

3) the rank of the extended matrix of the system and the rank of the matrix of the system are determined and the pact of compatibility or incompatibility of the system is established

4) in case of compatibility, the equivalent system of equations is written

5) the solution of the system is found. The main variables are expressed in terms of free

Kronecker-Capelli theorem

Kronecker - Capelli theorem- criterion of compatibility of the system of linear algebraic equations:

A system of linear algebraic equations is consistent if and only if the rank of its main matrix is ​​equal to the rank of its extended matrix, and the system has a unique solution if the rank is equal to the number of unknowns, and an infinite set of solutions if the rank less than number unknown.

For a linear system to be consistent, it is necessary and sufficient that the rank of the extended matrix of this system be equal to the rank of its main matrix.

When does the system have no solution, when does it have a single solution, does it have many solutions?

If the number of system equations is equal to the number of unknown variables and the determinant of its main matrix is ​​not equal to zero, then such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.

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

linear equations is called consistent if it has at least one solution, and inconsistent if there are no solutions. In example 14 the system is compatible, the column is its solution:

This solution can also be written without matrices: x = 2, y = 1.

A system of equations will be called indefinite if it has more than one solution, and definite if the solution is unique.

Example 15. The system is indeterminate. For example, ... are its solutions. The reader can find many other solutions to this system.

Formulas relating the coordinates of vectors in the old and new bases

Let's learn how to solve systems of linear equations first in a particular case. A system of equations AX = B will be called Cramer's if its main matrix А is square and nondegenerate. In other words, the number of unknowns in the Cramerian system coincides with the number of equations and |A| = 0.

Theorem 6 (Cramer's rule). The Cramer system of linear equations has a unique solution given by the formulas:

where Δ = |A| is the determinant of the main matrix, Δi is the determinant obtained from A by replacing the i-th column with a column of free terms.

We will carry out the proof for n = 3, since in the general case the arguments are similar.

So, there is a Cramer system:

Let us first assume that a solution to the system exists, i.e., there are

Let's multiply the first one. equality on the algebraic complement to the element aii, the second equality - on A2i, the third - on A3i and add the resulting equalities:

System of linear equations ~ Solution of the system ~ Consistent and inconsistent systems ~ Homogeneous system ~ Compatibility of a homogeneous system ~ Rank of the system matrix ~ Condition of non-trivial compatibility ~ Fundamental system of solutions. General solution ~ Study of a homogeneous system

Consider the system m linear algebraic equations with respect to n unknown
x 1 , x 2 , …, x n :

Decision system is called the totality n unknown values

x 1 \u003d x’ 1, x 2 \u003d x’ 2, ..., x n \u003d x’ n,

upon substitution of which all equations of the system turn into identities.

The system of linear equations can be written in matrix form:

where A- system matrix, b- right part, x- desired solution Ap - expanded matrix systems:

.

A system that has at least one solution is called joint; system that has no solution incompatible.

A homogeneous system of linear equations is a system whose right side is equal to zero:

Matrix view of a homogeneous system: ax=0.

A homogeneous system is always consistent, since any homogeneous linear system has at least one solution:

x 1 \u003d 0, x 2 \u003d 0, ..., x n \u003d 0.

If a homogeneous system has a unique solution, then this unique solution is zero, and the system is called trivially joint. If a homogeneous system has more than one solution, then there are non-zero solutions among them, and in this case the system is called non-trivially joint.

It has been proven that at m=n for non-trivial system compatibility necessary and sufficient so that the determinant of the matrix of the system is equal to zero.

EXAMPLE 1. Non-trivial compatibility of a homogeneous system of linear equations with a square matrix.

Applying the Gaussian elimination algorithm to the system matrix, we reduce the system matrix to the step form

.

Number r non-zero rows in the step form of a matrix is ​​called matrix rank, denote
r=rg(A) or r=Rg(A).

The following assertion is true.

System of linear algebraic equations

For a homogeneous system to be nontrivially consistent, it is necessary and sufficient that the rank r system matrix was less than the number of unknowns n.

EXAMPLE 2. Non-trivial compatibility of a homogeneous system of three linear equations with four unknowns.

If a homogeneous system is non-trivially consistent, then it has an infinite number of solutions, and a linear combination of any solutions of the system is also its solution.
It is proved that among the infinite set of solutions of a homogeneous system, exactly n-r linearly independent solutions.
Aggregate n-r linearly independent solutions of a homogeneous system is called fundamental decision system. Any solution of the system is linearly expressed in terms of the fundamental system. Thus, if the rank r matrices A homogeneous linear system ax=0 fewer unknowns n and vectors
e 1 , e 2 , …, e n-r form its fundamental system of solutions ( Ae i =0, i=1,2, …, n-r), then any solution x systems ax=0 can be written in the form

x=c 1 e 1 + c 2 e 2 + … + c n-r e n-r ,

where c 1 , c 2 , …, c n-r are arbitrary constants. The written expression is called common solution homogeneous system .

Research

homogeneous system means to establish whether it is non-trivially consistent, and if it is, then find a fundamental system of solutions and write down an expression for the general solution of the system.

We study a homogeneous system by the Gauss method.

matrix of the homogeneous system under study, the rank of which is r< n .

Such a matrix is ​​reduced by the Gaussian elimination to the stepped form

.

The corresponding equivalent system has the form

From here it is easy to obtain expressions for variables x 1 , x 2 , …, x r across x r+1 , x r+2 , …, x n. Variables
x 1 , x 2 , …, x r called basic variables and variables x r+1 , x r+2 , …, x n - free variables.

Transferring the free variables to the right side, we obtain the formulas

which determine the overall solution of the system.

Let us successively set the values ​​of the free variables equal to

and calculate the corresponding values ​​of the basic variables. Received n-r solutions are linearly independent and, therefore, form a fundamental system of solutions of the homogeneous system under study:

Investigation of a homogeneous system for compatibility by the Gauss method.

However, two more cases are widespread in practice:

– 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 tasks, it is required to preliminarily examine the system for compatibility, how to do this - see the article on matrix rank.

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

The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. First, consider a couple of examples where the system has no solutions (inconsistent).

Example 1

What immediately catches your eye in 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 it remains only to find out.

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

(1) On the upper left step, we need to get +1 or -1. There are no such numbers in the first column, so rearranging the rows will not work. The unit will have to be organized independently, and this can be done in several ways. I did this: To the first line, 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 is done, 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 -1 on the second step. Divide the third line by -3.

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

Probably, everyone paid attention to the bad line, which turned out as a result of elementary transformations: . It is clear that this cannot be so. Indeed, we 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 non-zero number, then the system is inconsistent (has no solutions) .

How to record the end of a task? Let's draw with white chalk: "as a result of elementary transformations, a line of the form is obtained, where" and give the answer: the system has no solutions (inconsistent).

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

Please note that there is no reverse motion 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 a do-it-yourself example. Complete Solution and the answer at the end of the lesson. Again, I remind you that your solution path may differ from my solution path, the Gaussian algorithm does not have a strong “rigidity”.

One more technical feature solutions: elementary transformations can be stopped immediately, as soon as a line like , where . Consider conditional example: suppose that after the first transformation we get a matrix . The matrix has not yet been reduced to a stepped form, but there is no need for further elementary transformations, since a line of the form has appeared, where . It should be immediately answered that the system is incompatible.

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

But everything in this world is balanced, and the 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. Whatever it was, but the Gauss method in any case will lead us to the answer. Therein lies its versatility.

The beginning is again standard. We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

That's all, and you were afraid.

(1) Note that all the numbers in the first column are divisible by 2, so a 2 is fine on the top left rung. 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 from 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 up: 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 it is necessary to show increased attention, but are the lines really proportional? For reinsurance (especially for a teapot), it would not be superfluous to multiply the second row by -1, and divide the fourth row by 2, resulting in three identical rows. And only after that remove two of them.

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

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

We rewrite the corresponding system of equations:

The “usual” only solution of the system does not smell here. There is no bad line either. This means that this is the third remaining case - the system has infinitely many solutions. Sometimes, by condition, it is necessary to investigate the compatibility of the system (i.e., to 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 break down the basics:

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

We will find the general solution of the system using the reverse motion of the Gauss method.

First we need to determine what variables we have basic, and which variables free. It is not necessary to bother with the terms of linear algebra, it is enough to remember that there are such basis 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 the remaining variables that did not get a step. In our case, there are two of them: – free variables.

Now you need all basis variables express only through free variables.

The reverse move 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:

The result is what you need - all the basis variables ( and ) are expressed only through free variables :

Actually, the general solution is ready:

How to write down the general solution?
Free variables are written into the general solution "on their own" 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, there are infinitely many private decisions. The most popular values ​​are zeros, since the particular solution is the easiest to obtain. Substitute in the general solution:

is a private decision.

Ones are another sweet couple, let's substitute into the general solution:

is another particular solution.

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

Each a particular solution must satisfy to each system equation. 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 in the original system:

Everything has to come together. And with any particular solution you get, everything should also converge.

But, strictly speaking, the verification of a particular solution sometimes deceives; some particular solution can satisfy each equation of the system, and the general solution itself is actually found incorrectly.

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

It's easy, 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 Gauss method. Find a general solution and two private ones. Check the overall solution.

This is a do-it-yourself example. Here, by the way, again the number of equations is less than the number of unknowns, which means that 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 again attention. 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 the augmented matrix of the system and with the help of elementary transformations we bring it to the step 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, 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.

Here is such a beauty:

Basis variables sit on steps, so they are base variables.
There is only one free variable, which did not get a step:

Reverse move:
We express the basic variables in terms of the free variable:
From the third equation:

Consider the second equation and substitute the found expression into it:

Consider the first equation and substitute the found expressions and into it:

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

So the general solution is:

Once again, how did it happen? 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. Work for blacks, but I have already done it, so catch =)

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

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

Now from the found general solution we get two particular solutions. The chef here is the only free variable . You don't need to break your head.

Let then is a private decision.
Let , then be another particular solution.

Answer: Common decision: , particular solutions: , .

I shouldn't have remembered about blacks here ... ... because all sorts of sadistic motives came into my head and I remembered the well-known fotozhaba, in which Ku Klux Klansmen in white overalls run across the field after a black football player. I sit and smile quietly. You know how distracting….

A lot of math is harmful, so a similar final example for an independent solution.

Example 6

Find the general solution of 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 match.

Probably, many people noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to fiddle with ordinary fractions. In practice, this is true, 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 can sometimes include a constant (or constants), for example: . Here one of the basic variables is equal to a constant number: . There is nothing exotic in 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 number of equations more quantity variables. The Gaussian method works in the most severe conditions; one should calmly bring the extended matrix of the system to a stepped form according to the standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a unique solution.

Service assignment. The online calculator is designed to study a system of linear equations. Usually in the condition of the problem it is required to find general and particular solution of the system. When studying systems of linear equations, the following problems are solved:

1. whether the system is collaborative;
2. if the system is consistent, then it is definite or indefinite (the criterion of system compatibility is determined by the theorem);
3. if the system is defined, then how to find its unique solution (the Cramer method, the inverse matrix method or the Jordan-Gauss method are used);
4. if the system is indefinite, then how to describe the set of its solutions.

Classification of systems of linear equations

An arbitrary system of linear equations has the form:
a 1 1 x 1 + a 1 2 x 2 + ... + a 1 n x n = b 1
a 2 1 x 1 + a 2 2 x 2 + ... + a 2 n x n = b 2
...................................................
a m 1 x 1 + a m 2 x 2 + ... + a m n x n = b m

1. Systems of linear inhomogeneous equations (the number of variables is equal to the number of equations, m = n).
2. Arbitrary systems of linear inhomogeneous equations (m > n or m< n).

Definition. A solution of a system is any set of numbers c 1 ,c 2 ,...,c n , whose substitution into the system instead of the corresponding unknowns turns each equation of the system into an identity.

Definition. Two systems are said to be equivalent if the solution to the first is the solution to the second and vice versa.

Definition. A system that has at least one solution is called joint. A system that does not have any solution is called inconsistent.

Definition. A system with a unique solution is called certain, and having more than one solution is indefinite.

Algorithm for solving systems of linear equations

1. Find the ranks of the main and extended matrices. If they are not equal, then, by the Kronecker-Capelli theorem, the system is inconsistent, and this is where the study ends.
2. Let rank(A) = rank(B) . We select the basic minor. In this case, all unknown systems of linear equations are divided into two classes. The unknowns, the coefficients of which are included in the basic minor, are called dependent, and the unknowns, the coefficients of which are not included in the basic minor, are called free. Note that the choice of dependent and free unknowns is not always unique.
3. We cross out those equations of the system whose coefficients were not included in the basic minor, since they are consequences of the rest (according to the basic minor theorem).
4. The terms of the equations containing free unknowns will be transferred to the 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 different from zero.
5. The resulting system is solved in one of the following ways: the Cramer method, the inverse matrix method, or the Jordan-Gauss method. Relations are found that express the dependent variables in terms of the free ones.

Solving systems of linear algebraic equations (SLAE) is undoubtedly the most important topic of the linear algebra course. A huge number of problems from all branches of mathematics are reduced to solving systems of linear equations. These factors explain the reason for creating this article. The material of the article is selected and structured so that with its help you can

• choose the optimal method for solving your system of linear algebraic equations,
• study the theory of the chosen method,
• solve your system of linear equations, having considered in detail the solutions of typical examples and problems.

Brief description of the material of the article.

Let's give it all first necessary definitions, concepts, and introduce notation.

Next, we consider methods for solving systems of linear algebraic equations in which the number of equations is equal to the number of unknown variables and which have a unique solution. First, let's focus on the Cramer method, secondly, we will show the matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of successive elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in various ways.

After that, we turn to solving systems of linear algebraic equations general view, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is degenerate. We formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (in the case of their compatibility) using the concept of the basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions of the examples.

Be sure to dwell on the structure of the general solution of homogeneous and inhomogeneous systems of linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how the general solution of the SLAE is written using the vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.

In conclusion, we consider systems of equations that are reduced to linear ones, as well as various problems, in the solution of which SLAEs arise.

Page navigation.

Definitions, concepts, designations.

We will consider systems of p linear algebraic equations with n unknown variables (p may be equal to n ) of the form

Unknown variables, - coefficients (some real or complex numbers), - free members (also real or complex numbers).

This form of SLAE is called coordinate.

IN matrix form this system of equations has the form ,
where - the main matrix of the system, - the matrix-column of unknown variables, - the matrix-column of free members.

If we add to the matrix A as the (n + 1)-th column the matrix-column of free terms, then we get the so-called expanded matrix systems of linear equations. Usually, the augmented matrix is ​​denoted by the letter T, and the column of free members is separated by a vertical line from the rest of the columns, that is,

By solving a system of linear algebraic equations called a set of values ​​of unknown variables , which turns all the equations of the system into identities. The matrix equation for the given values ​​of the unknown variables also turns into an identity.

If a system of equations has at least one solution, then it is called joint.

If the system of equations has no solutions, then it is called incompatible.

If a SLAE has a unique solution, then it is called certain; if there is more than one solution, then - uncertain.

If the free terms of all equations of the system are equal to zero , then the system is called homogeneous, otherwise - heterogeneous.

Solution of elementary systems of linear algebraic equations.

If the number of system equations is equal to the number of unknown variables and the determinant of its main matrix is ​​not equal to zero, then we will call such SLAEs elementary. Such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.

We began to study such SLAEs in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the addition method, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are essentially modifications of the Gauss method.

The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's sort them out.

Solving systems of linear equations by Cramer's method.

Let us need to solve a system of linear algebraic equations

in which the number of equations is equal to the number of unknown variables and the determinant of the main matrix of the system is different from zero, that is, .

Let be the determinant of the main matrix of the system, and are determinants of matrices that are obtained from A by replacing 1st, 2nd, …, nth column respectively to the column of free members:

With such notation, the unknown variables are calculated by the formulas of Cramer's method as . This is how the solution of a system of linear algebraic equations is found by the Cramer method.

Example.

Cramer method .

Solution.

The main matrix of the system has the form . Calculate its determinant (if necessary, see the article):

Since the determinant of the main matrix of the system is nonzero, the system has a unique solution that can be found by Cramer's method.

Compose and calculate the necessary determinants (the determinant is obtained by replacing the first column in matrix A with a column of free members, the determinant - by replacing the second column with a column of free members, - by replacing the third column of matrix A with a column of free members):

Finding unknown variables using formulas :

Answer:

The main disadvantage of Cramer's method (if it can be called a disadvantage) is the complexity of calculating the determinants when the number of system equations is more than three.

Solving systems of linear algebraic equations by the matrix method (using the inverse matrix).

Let the system of linear algebraic equations be given in matrix form , where the matrix A has dimension n by n and its determinant is nonzero.

Since , then the matrix A is invertible, that is, there is an inverse matrix . If we multiply both parts of the equality by on the left, then we get a formula for finding the column matrix of unknown variables. So we got the solution of the system of linear algebraic equations by the matrix method.

Example.

Solve System of Linear Equations matrix method.

Solution.

Let's rewrite the system of equations in matrix form:

Because

then the SLAE can be solved by the matrix method. Using the inverse matrix, the solution to this system can be found as .

Let's build an inverse matrix using a matrix of algebraic complements of the elements of matrix A (if necessary, see the article):

It remains to calculate - the matrix of unknown variables by multiplying the inverse matrix on the matrix-column of free members (if necessary, see the article):

Answer:

or in another notation x 1 = 4, x 2 = 0, x 3 = -1.

The main problem in finding solutions to systems of linear algebraic equations by the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than the third.

Solving systems of linear equations by the Gauss method.

Suppose we need to find a solution to a system of n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.

The essence of the Gauss method consists in the successive exclusion of unknown variables: first, x 1 is excluded from all equations of the system, starting from the second, then x 2 is excluded from all equations, starting from the third, and so on, until only the unknown variable x n remains in the last equation. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called direct Gauss method. After the forward run of the Gauss method is completed, x n is found from the last equation, x n-1 is calculated from the penultimate equation using this value, and so on, x 1 is found from the first equation. The process of calculating unknown variables when moving from the last equation of the system to the first is called reverse Gauss method.

Let us briefly describe the algorithm for eliminating unknown variables.

We will assume that , since we can always achieve this by rearranging the equations of the system. We exclude the unknown variable x 1 from all equations of the system, starting from the second one. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, add the first multiplied by to the nth equation. The system of equations after such transformations will take the form

where , a .

We would come to the same result if we expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.

Next, we act similarly, but only with a part of the resulting system, which is marked in the figure

To do this, add the second equation multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, add the second multiplied by to the nth equation. The system of equations after such transformations will take the form

where , a . Thus, the variable x 2 is excluded from all equations, starting from the third.

Next, we proceed to the elimination of the unknown x 3, while acting similarly with the part of the system marked in the figure

So we continue the direct course of the Gauss method until the system takes the form

From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as , using the obtained value x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Example.

Solve System of Linear Equations Gaussian method.

Solution.

Let's exclude the unknown variable x 1 from the second and third equations of the system. To do this, to both parts of the second and third equations, we add the corresponding parts of the first equation, multiplied by and by, respectively:

Now we exclude x 2 from the third equation by adding to its left and right parts the left and right parts of the second equation, multiplied by:

On this, the forward course of the Gauss method is completed, we begin the reverse course.

From the last equation of the resulting system of equations, we find x 3:

From the second equation we get .

From the first equation we find the remaining unknown variable and this completes the reverse course of the Gauss method.

Answer:

X 1 \u003d 4, x 2 \u003d 0, x 3 \u003d -1.

Solving systems of linear algebraic equations of general form.

In the general case, the number of equations of the system p does not coincide with the number of unknown variables n:

Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations whose main matrix is ​​square and degenerate.

Kronecker-Capelli theorem.

Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when SLAE is compatible, and when it is incompatible, gives Kronecker–Capelli theorem:
for a system of p equations with n unknowns (p can be equal to n ) to be consistent it is necessary and sufficient that the rank of the main matrix of the system is equal to the rank of the extended matrix, that is, Rank(A)=Rank(T) .

Let us consider the application of the Kronecker-Cappelli theorem for determining the compatibility of a system of linear equations as an example.

Example.

Find out if the system of linear equations has solutions.

Solution.

. Let us use the method of bordering minors. Minor of the second order different from zero. Let's go over the third-order minors surrounding it:

Since all bordering third-order minors are equal to zero, the rank of the main matrix is ​​two.

In turn, the rank of the augmented matrix is equal to three, since the minor of the third order

different from zero.

In this way, Rang(A) , therefore, according to the Kronecker-Capelli theorem, we can conclude that the original system of linear equations is inconsistent.

Answer:

There is no solution system.

So, we have learned to establish the inconsistency of the system using the Kronecker-Capelli theorem.

But how to find the solution of the SLAE if its compatibility is established?

To do this, we need the concept of the basis minor of a matrix and the theorem on the rank of a matrix.

The highest order minor of the matrix A, other than zero, is called basic.

It follows from the definition of the basis minor that its order is equal to the rank of the matrix. For a non-zero matrix A, there can be several basic minors; there is always one basic minor.

For example, consider the matrix .

All third-order minors of this matrix are equal to zero, since the elements of the third row of this matrix are the sum of the corresponding elements of the first and second rows.

The following minors of the second order are basic, since they are nonzero

Minors are not basic, since they are equal to zero.

Matrix rank theorem.

If the rank of a matrix of order p by n is r, then all elements of the rows (and columns) of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding elements of the rows (and columns) that form the basis minor.

What does the matrix rank theorem give us?

If, by the Kronecker-Capelli theorem, we have established the compatibility of the system, then we choose any basic minor of the main matrix of the system (its order is equal to r), and exclude from the system all equations that do not form the chosen basic minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still redundant (according to the matrix rank theorem, they are a linear combination of the remaining equations).

As a result, after discarding the excessive equations of the system, two cases are possible.

If the number of equations r in the resulting system is equal to the number of unknown variables, then it will be definite and the only solution can be found by the Cramer method, the matrix method or the Gauss method.

Example.

.

Solution.

Rank of the main matrix of the system is equal to two, since the minor of the second order different from zero. Extended matrix rank is also equal to two, since the only minor of the third order is equal to zero

and the minor of the second order considered above is different from zero. Based on the Kronecker-Capelli theorem, one can assert the compatibility of the original system of linear equations, since Rank(A)=Rank(T)=2 .

As a basis minor, we take . It is formed by the coefficients of the first and second equations:


The third equation of the system does not participate in the formation of the basic minor, so we exclude it from the system based on the matrix rank theorem:


So we got elementary system linear algebraic equations. Let's solve it by Cramer's method:


Answer:

x 1 \u003d 1, x 2 \u003d 2.

If the number of equations r in the resulting SLAE is less than the number of unknown variables n, then we leave the terms forming the basic minor in the left parts of the equations, and transfer the remaining terms to the right parts of the equations of the system with the opposite sign.

The unknown variables (there are r of them) remaining on the left-hand sides of the equations are called main.

Unknown variables (there are n - r of them) that ended up on the right side are called free.

Now we assume that the free unknown variables can take arbitrary values, while the r main unknown variables will be expressed in terms of the free unknown variables in a unique way. Their expression can be found by solving the resulting SLAE by the Cramer method, the matrix method, or the Gauss method.

Let's take an example.

Example.

Solve System of Linear Algebraic Equations .

Solution.

Find the rank of the main matrix of the system by the bordering minors method. Let us take a 1 1 = 1 as a non-zero first-order minor. Let's start searching for a non-zero second-order minor surrounding this minor:


So we found a non-zero minor of the second order. Let's start searching for a non-zero bordering minor of the third order:


Thus, the rank of the main matrix is ​​three. The rank of the augmented matrix is ​​also equal to three, that is, the system is consistent.

The found non-zero minor of the third order will be taken as the basic one.

For clarity, we show the elements that form the basis minor:


We leave the terms participating in the basic minor on the left side of the equations of the system, and transfer the rest with opposite signs to the right sides:


We give free unknown variables x 2 and x 5 arbitrary values, that is, we take , where are arbitrary numbers. In this case, the SLAE takes the form


We solve the obtained elementary system of linear algebraic equations by the Cramer method:


Consequently, .

In the answer, do not forget to indicate free unknown variables.

Answer:

Where are arbitrary numbers.

Summarize.

To solve a system of linear algebraic equations of a general form, we first find out its compatibility using the Kronecker-Capelli theorem. If the rank of the main matrix is ​​not equal to the rank of the extended matrix, then we conclude that the system is inconsistent.

If the rank of the main matrix is ​​equal to the rank of the extended matrix, then we choose the basic minor and discard the equations of the system that do not participate in the formation of the chosen basic minor.

If the order of the basis minor is equal to the number of unknown variables, then the SLAE has a unique solution, which can be found by any method known to us.

If the order of the basis minor is less than the number of unknown variables, then on the left side of the equations of the system we leave the terms with the main unknown variables, transfer the remaining terms to the right sides and assign arbitrary values ​​to the free unknown variables. From the resulting system of linear equations, we find the main unknown variables by the Cramer method, the matrix method or the Gauss method.

Gauss method for solving systems of linear algebraic equations of general form.

Using the Gauss method, one can solve systems of linear algebraic equations of any kind without their preliminary investigation for compatibility. The process of successive elimination of unknown variables makes it possible to draw a conclusion about both the compatibility and inconsistency of the SLAE, and if a solution exists, it makes it possible to find it.

From the point of view of computational work, the Gaussian method is preferable.

Watch it detailed description and analyzed examples in the article Gauss method for solving systems of linear algebraic equations of general form.

Recording the general solution of homogeneous and inhomogeneous linear algebraic systems using the vectors of the fundamental system of solutions.

In this section, we will focus on joint homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.

Let's deal with homogeneous systems first.

Fundamental decision system of a homogeneous system of p linear algebraic equations with n unknown variables is a set of (n – r) linearly independent solutions of this system, where r is the order of the basis minor of the main matrix of the system.

If we denote linearly independent solutions of a homogeneous SLAE as X (1) , X (2) , …, X (nr) (X (1) , X (2) , …, X (nr) are n by 1 column matrices) , then the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary constant coefficients С 1 , С 2 , …, С (nr) , that is, .

What does the term general solution of a homogeneous system of linear algebraic equations (oroslau) mean?

The meaning is simple: the formula sets everything possible solutions the original SLAE, in other words, taking any set of values ​​of arbitrary constants С 1 , С 2 , …, С (n-r) , according to the formula we get one of the solutions of the original homogeneous SLAE.

Thus, if we find a fundamental system of solutions, then we can set all solutions of this homogeneous SLAE as .

Let us show the process of constructing a fundamental system of solutions for a homogeneous SLAE.

We choose the basic minor of the original system of linear equations, exclude all other equations from the system, and transfer to the right-hand side of the equations of the system with opposite signs all terms containing free unknown variables. Let's give the free unknown variables the values ​​1,0,0,…,0 and calculate the main unknowns by solving the resulting elementary system of linear equations in any way, for example, by the Cramer method. Thus, X (1) will be obtained - the first solution of the fundamental system. If we give the free unknowns the values ​​0,1,0,0,…,0 and calculate the main unknowns, then we get X (2) . Etc. If we give the free unknown variables the values ​​0,0,…,0,1 and calculate the main unknowns, then we get X (n-r) . This is how the fundamental system of solutions of the homogeneous SLAE will be constructed and its general solution can be written in the form .

For inhomogeneous systems of linear algebraic equations, the general solution is represented as

Let's look at examples.

Example.

Find the fundamental system of solutions and the general solution of a homogeneous system of linear algebraic equations .

Solution.

The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let us find the rank of the main matrix by the method of fringing minors. As a nonzero minor of the first order, we take the element a 1 1 = 9 of the main matrix of the system. Find the bordering non-zero minor of the second order:

A minor of the second order, different from zero, is found. Let's go through the third-order minors bordering it in search of a non-zero one:

All bordering minors of the third order are equal to zero, therefore, the rank of the main and extended matrix is ​​two. Let's take the basic minor. For clarity, we note the elements of the system that form it:

The third equation of the original SLAE does not participate in the formation of the basic minor, therefore, it can be excluded:

We leave the terms containing the main unknowns on the right-hand sides of the equations, and transfer the terms with free unknowns to the right-hand sides:

Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basic minor is two. To find X (1), we give the free unknown variables the values ​​x 2 \u003d 1, x 4 \u003d 0, then we find the main unknowns from the system of equations
.