Cramer Method Examples with a detailed solution. Method of Cramer Solution of Linear Equations Systems
Cramer method or so-called crawler rule is a way to search for unknown values \u200b\u200bfrom equations systems. It can only be used if the number of desired values \u200b\u200bis equivalent to quantity algebraic equations In the system, that is, the main matrix formed from the system should be square and do not contain zero lines, and also if its determinants should not be zero.
Theorem 1.
Kramera theorem If the main determinant of $ D $ is the main matrix, composed on the basis of the coefficients of equations, not equal to zero., then the system of equations is coordinated, and the solution has the only one. Solution of such a system is calculated through the so-called Cramer formulas to solve systems linear equations: $ x_i \u003d \\ FRAC (D_I) (D) $
What is the Cramer method
The essence of the Cramer method is as follows:
- To find the solution of the system by the Cramer method, the first thing is calculated the main determinant of the $ D $ matrix. When the calculated determinant of the main matrix, when calculating the Cramer method turned out to be zero, then the system does not have a single solution or has an endless amount of solutions. In this case, it is recommended to apply the Gauss method to find a common or basic response for the system.
- Then you need to replace the extreme column of the main matrix on the column of free members and calculate the identifier $ d_1 $.
- Repeat the same for all columns, having received the determinants from $ d_1 $ to $ d_n $, where $ n $ is the number of the extreme right column.
- After all the determinants are found $ d_1 $ ... $ d_n $, you can calculate unknown variables by $ x_i \u003d \\ frac formula (d_i) (D) $.
Receptions for calculating the determinant of the matrix
To calculate the determinant of the matrix with a dimension of more than 2 to 2, you can use several ways:
- The rule of triangles, or the Sarrusus rule, resembling the same rule. The essence of the triangle method is that when calculating the determinant, the product of all numbers connected in the figure of the red line on the right is recorded with a plus sign, and all the numbers connected in the same way in the figure on the left - with a minus sign. B, then, and another rule is suitable for matrices of 3 x 3. 3. In the case of the Sarruska rule, the matrix itself corresponds first, and next to it is rewritten to its first and second column. Through the matrix and these additional columns are diagonally, the matrix members lying on the main diagonal or on the parallel to it are recorded with the plus sign, and the elements lying on the side diagonal or in parallel to it - with a minus sign.
Figure 1. Triangle rule to calculate the determinant for the Cramer method
- Using the method known as the Gauss method, also sometimes this method is called a decrease in the order of the determinant. In this case, the matrix is \u200b\u200bconverted and driven to a triangular form, and then multiplious all the numbers standing on the main diagonal. It should be remembered that with such a search for the determinant, it is impossible to multiply or divide the lines or columns into numbers without making them as a multiplier or divider. In the case of a search for a determinant, it is possible to only deduct and fold the strings and pillars between themselves, after pre-mowing the subtracted line to the nonzero multiplier. Also, with each permutation of the lines or columns of the matrix, the places should be remembered about the need to change the final sign in the matrix.
- When solving the Cramer method, the Slava with 4 unknowns is best to use exactly the Gauss method for searching and finding identifiers or define the determinant through the search for minorors.
Solution of systems of equations by Cramer
Applicable Cramer Method for a system of 2 equations and two desired values:
$ \\ begin (Cases) a_1x_1 + a_2x_2 \u003d b_1 \\\\ a_3x_1 + a_4x_2 \u003d b_2 \\ End (Cases) $
Display it in an extended form for convenience:
$ A \u003d \\ begin (array) (CC | C) A_1 & A_2 & B_1 \\\\ A_3 & A_4 & B_1 \\ END (Array) $
We will find the determinant of the main matrix, also called the main determinant of the system:
$ D \u003d \\ begin (array) (| CC |) A_1 & A_2 \\\\ A_3 & A_4 \\ END (Array) \u003d A_1 \\ CDOT A_4 - A_3 \\ CDOT A_2 $
If the main determinant is not equal to zero, it is necessary to calculate a couple of determinants from two matrices with a replaced column of the main matrix on the line of free members to solve the sample method:
$ D_1 \u003d \\ begin (array) (| cc |) b_1 & a_2 \\\\ b_2 & a_4 \\\\ \\ end (array) \u003d b_1 \\ cdot a_4 - b_2 \\ cdot a_4 $
$ D_2 \u003d \\ begin (array) (| CC |) A_1 & B_1 \\\\ A_3 & B_2 \\\\ \\ END (Array) \u003d A_1 \\ CDOT B_2 - A_3 \\ CDOT B_1 $
Now find unknown $ x_1 $ and $ x_2 $:
$ x_1 \u003d \\ FRAC (D_1) (D) $
$ x_2 \u003d \\ FRAC (D_2) (D) $
Example 1.
Cramer Method for solving a slope with the main matrix 3 of the order (3 x 3) and three is the desired.
Solve the system of equations:
$ \\ begin (Cases) 3x_1 - 2x_2 + 4x_3 \u003d 21 \\\\ 3x_1 + 4x_2 + 2x_3 \u003d 9 \\\\ 2X_1 - X_2 - X_3 \u003d 10 \\ END (Cases) $
Consider the chief determinant of the matrix using the above-mentioned number 1 rule:
$ D \u003d \\ begin (array) (| CCC |) 3 & -2 & 4 \\\\ 3 & 4 & -2 \\ ED (Array) \u003d 3 \\ CDOT 4 \\ CDOT ( -1) + 2 \\ Cdot (-2) \\ CDOT 2 + 4 \\ CDOT 3 \\ CDOT (-1) - 4 \\ CDOT 4 \\ CDOT 2 - 3 \\ CDOT (-2) \\ CDOT (-1) - (- 1) \\ CDOT 2 \\ CDOT 3 \u003d - 12 - 8 -12 -32 - 6 + 6 \u003d - 64 $
And now three other determinants:
$ D_1 \u003d \\ begin (array) (| CCC |) 21 & 2 & 4 \\\\ 9 & 4 & 2 \\\\ 10 & 1 & 1 \\ END (Array) \u003d 21 \\ CDOT 4 \\ CDOT 1 + (- 2) \\ CDOT 2 \\ CDOT 10 + 9 \\ CDOT (-1) \\ CDOT 4 - 4 \\ CDOT 4 \\ CDOT 10 - 9 \\ CDOT (-2) \\ CDOT (-1) - (-1) \\ CDOT 2 \\ $ D_2 \u003d \\ begin (array) (| CCC |) 3 & 21 & 4 \\\\ 3 & 9 & 2 \\ ED (Array) \u003d 3 \\ CDOT 9 \\ CDOT (- 1) + 3 \\ CDOT 10 \\ CDOT 4 + 21 \\ CDOT 2 \\ CDOT 2 - 4 \\ CDOT 9 \\ CDOT 2 - 21 \\ CDOT 3 \\ CDOT (-1) - 2 \\ CDOT 10 \\ CDOT 3 \u003d - 27 + 120 + 84 - 72 + 63 - 60 \u003d 108 $
$ D_3 \u003d \\ begin (array) (| CCC |) 3 & -2 & 21 \\\\ 3 & 4 & 9 \\\\ 2 & 1 & 10 \\ END (Array) \u003d 3 \\ CDOT 4 \\ CDOT 10 + 3 \\ Cdot (-1) \\ CDOT 21 + (-2) \\ CDOT 9 \\ CDOT 2 - 21 \\ CDOT 4 \\ CDOT 2 - (-2) \\ CDOT 3 \\ CDOT 10 - (-1) \\ CDOT 9 \\ CDOT 3 \u003d 120 - 63 - 36 - 168 + 60 + 27 \u003d - $ 60
Find the desired values:
$ X_1 \u003d \\ FRAC (D_1) (D) \u003d \\ FRAC (- 296) (- 64) \u003d 4 \\ FRAC (5) (8) $
$ X_2 \u003d \\ FRAC (D_1) (D) \u003d \\ FRAC (108) (-64) \u003d - 1 \\ FRAC (11) (16) $
$ X_3 \u003d \\ FRAC (D_1) (D) \u003d \\ FRAC (-60) (-64) \u003d \\ FRAC (15) (16) $
2. Solution of systems of equations by the matrix method (using the reverse matrix).
3. Gauss method solving systems of equations.
Cramer method.
The craver method is used to solve linear algebraic equations (
Slough Formulas on the example of a system of two equations with two variables.).
Given:
Solve Cramer Method System Relative to variables
H. and W. decision:.
We find the determinant of the matrix made up of the coefficients of the system calculation of the determinants. :
We use the crawler formulas and find variable values: 
Example 1: 
W. 
.
Solve the system of equations:
relative to variables
and W. decision:.
We find the determinant of the matrix made up of the coefficients of the system calculation of the determinants. :
We replace in this determinant the first column of the column of coefficients from the right side of the system and find its value: 
We will make a similar action, replacing the second column in the first deficient: 
Apply cramer formulas and find the values \u200b\u200bof the variables:
and.
Answer: 
Comment: This method can solve systems and greater dimension.
Comment: If it turns out that, and it is impossible to divide on zero, then they say that the system does not have a single solution. In this case, the system has or infinitely many solutions or has no solutions at all.
Example 2. (infinite number of solutions):
Solve the system of equations:
and W. decision:.
We find the determinant of the matrix made up of the coefficients of the system calculation of the determinants. :
We find the determinant of the matrix made up of system coefficients: 
Solution of systems by substitution. 
The first of the equations of the system is equality, faithful in any values \u200b\u200bof variables (because 4 is always 4). It means that only one equation remains. This is the equation of communication between variables.
Received, solutions of the system are any pairs of variable values \u200b\u200brelated to equality.
The general decision will be recorded like this: 
Private solutions can be determined by choosing an arbitrary value of y and calculating X on this equality of communication. 
etc.
Such solutions are infinitely a lot.
Answer: common decision
Private solutions:
Example 3. (no solutions, the system is incomprehensible):
Solve the system of equations:
We find the determinant of the matrix made up of the coefficients of the system calculation of the determinants. :
We find the determinant of the matrix made up of system coefficients: 
It is impossible to use Cramer formulas. I solve this system by substitution 
The second equation of the system is equality, incorrectly under any values \u200b\u200bof variables (of course, since -15 is not 2). If one of the equations of the system is not true at no matter the values \u200b\u200bof the variables, then the entire system has no solutions.
Answer: No solutions
In the first part we looked at a little theoretical material, substitution method, as well as the method of soil addition of the system equations. Everyone who went to the site through this page recommended to familiarize yourself with the first part. Perhaps some visitors will seem material too simple, but in the course of solving systems of linear equations I made a number of very important comments and conclusions relating to the solution mathematical tasks generally.
And now we will analyze the rule of the crater, as well as the solution of the system of linear equations using reverse matrix (Matrix method). All materials are presented simply, in detail and clearly, almost all readers will be able to learn how to solve the systems in the above methods.
First, we will consider in detail the cramer rule for a system of two linear equations with two unknowns. What for? - After all simpler system You can solve the school method, the method of killing addition!
The fact is that even if it is sometimes, this task is found - to solve the system of two linear equations with two unknowns by crawler formulas. Secondly, a simpler example will help to understand how to use the crawler rule for a more complex case - the systems of three equations with three unknowns.
In addition, there are systems of linear equations with two variables that it is advisable to solve precisely according to the rule of Cramer!
Consider the system of equations
In the first step, we calculate the determinant, it is called the main determinant of the system.
Gauss method.
If, the system has only decisionAnd for finding the roots, we must calculate two more determinants:
W.
In practice, the above determinants may also be denoted by the Latin letter.
The roots of the equations are found by formulas:
,
Example 7.
Solve the system of linear equations 
Decision: We see that the coefficients of the equation are large enough, there are present in the right decimal fractions With comma. Comma - rather rare guest in practical tasks In mathematics, I took this system from an econometric problem.
How to solve such a system? You can try to express one variable across the other, but in this case it will certainly get terrible trousers, with which it is extremely inconvenient to work, and the decoration of the solution will look just awful. You can multiply the second equation on 6 and carry out the soil subtraction, but also the same fractions will arise.
What to do? In such cases, they come to the aid of the formula of the crater.
;
;
Answer: ,
Both roots have endless tails, and are found approximately, which is quite acceptable (and even ordinary) for the problems of econometrics.
Comments are not needed here, since the task is solved on the finished formulas, however, there is one nuance. When you use this method, compulsorythe task design fragment is the following fragment: "So the system has a single decision". Otherwise, the reviewer may punish you for disrespecting the Cramer Theorem.
At all, it will not be superfluous, which is convenient to carry out on the calculator: we substitute approximate values \u200b\u200binto the left part of each equation of the system. As a result, with a small error, the numbers that are in the right parts should be turned out.
Example 8.
Answer to submit to ordinary irregular fractions. Make check.
This is an example for self-decide (An example of a clean design and response at the end of the lesson).
We turn to the consideration of the Cramer rule for a system of three equations with three unknowns: 
We find the main determinant of the system: 
If, the system has infinitely many solutions or inconspicuous (not solutions). In this case, the Rule of Cramer will not help, you need to use the Gauss method.
If, the system has a single solution and for finding the roots, we must calculate three more determinants: 
, 
, 
And finally, the answer is calculated by the formulas: 
As you can see, the case of "three to three" does not differ in principle from the case of "two two", the column of free members consistently "stroll" from left to right through the columns of the main determinant.
Example 9.
Solve the system according to the crawler formulas. 
Decision: Resolving the system according to the crawler formulas. 
So the system has a single solution.
Answer: 
.
Actually, there is nothing more to comment here again, in view of the fact that the decision passes through the finished formulas. But there is a couple of comments.
It happens that as a result of calculations, "bad" non-interpretable fractions are obtained, for example:.
I recommend the next treatment algorithm. If there is no computer at hand, do this:
1) An error in calculations is allowed. As soon as you encountered a "bad" fraction, immediately need to check, conductive conditioner correctly. If the condition is rewritten without errors, then you need to recalculate the determinants using the decomposition on another line (column).
2) If the error check is not detected, it is likely to be a typo in the assignment condition. In this case, calmly and carefully turn the task to the end, and then be sure to check And we make it on the finishing after the decision. Of course, the verification of a fractional response is an unpleasant, but it will be a disarming argument for a teacher who really loves to put minus for any bjaka like. How to manage with fractions, detailed in response for example 8.
If there is a computer at hand, then use the automated program to be downloaded for free at the very beginning of the lesson. By the way, it is most advantageous to immediately use the program (even before the decision), you will immediately see the intermediate step on which the error was allowed! The same calculator automatically calculates the system solution. matrix method.
Remark Second. From time to time there are systems in the equations of which there are no variables, for example: 
Here in the first equation there is no variable, in the second - variable. In such cases, it is very important to correctly and carefully record the main identifier: 
- On the site of the missing variables are zeros.
By the way, the determinants with zeros are rationally disclosed along the line (column), which is zero, since the calculations are noticeably less.
Example 10.
Solve the system according to the crawler formulas. 
This is an example for an independent solution (a sample of clean design and response at the end of the lesson).
For the case of a system of 4 of equations with 4 unknown, the Cramer formula is recorded by similar principles. A living example can be viewed at the lesson properties of the determinant. A decrease in the order of the determinant - the five determinants of the 4th order are completely solid. Although the task is already quite reminded by the professor's boot on the chest at the lucky student.
Solution of the system with a return matrix
The inverse matrix method is essentially a special case matrix equation (See Example number 3 of the specified lesson).
To explore this section, you must be able to disclose the determinants, find a reverse matrix and perform matrix multiplication. Relevant links will be given in the course of the explanation.
Example 11.
Solve the system with a matrix method 
Decision: Write the system in matrix form:
where 
Please look at the system of equations and matrix. According to which principle, write elements in the matrix, I think everyone is understandable. The only comment: if there were no variables in the equations, then at the appropriate places in the matrix it would be necessary to put zeros.
Reverse matrix we find by the formula:
where - a transposed matrix of algebraic additions to the corresponding elements of the matrix.
First we deal with the determinant:
Here the determinant is disclosed on the first line.
Attention! If, then the return matrix does not exist, and it is impossible to solve the system by the matrix method. In this case, the system is solved by the exclusion of unknown (Gauss method).
Now you need to calculate 9 minors and record them in the Mind Matrix 
Reference: It is useful to know the meaning of dual substitution indices in a linear algebra. The first digit is the line number in which this item is located. The second digit is the column number in which this item is: 
That is, a double substitution index indicates that the element is in the first row, the third column, and, for example, the element is in 3 string, 2 columns
Let the system of linear equations contain as many equations, what is the number of independent variables, i.e. Has appearance
Such systems of linear equations are called square. The determinant composed of coefficients with independent system variables (1.5), is called the main determinant of the system. We will denote by its Greek letter D. Thus
. (1.6)
If in the main identifier arbitrary ( j.) column, replace the column of free members of the system (1.5), then you can get more n. Auxiliary identifiers:
(j. = 1, 2, …, n.). (1.7)
Kramer rule Solutions of square systems of linear equations is as follows. If the main determinant D of the system (1.5) is different from zero, the system has and moreover a single solution that can be found by formulas:
(1.8)
Example 1.5. Cramer method solve system equations
.
Calculate the main determinant of the system:
Since D¹0, the system has a single solution that can be found by formulas (1.8):
In this way,
Actions on matrices
1. Multiplying the matrix by number. The multiplication operation of the matrix is \u200b\u200bdetermined as follows.
2. In order to multiply the matrix by the number, all its elements are multiplied by this number. I.e
. (1.9)
Example 1.6. 
.
Addition of matrices.
This operation is entered only for matrices of the same order.
In order to fold two matrices, it is necessary to add the appropriate elements of another matrix to the elements of one matrix:
(1.10)
The operation of the arrangement of matrices has properties of associativity and commutation.
Example 1.7. 
.
Matrix multiplication.
If the number of columns of the matrix BUT coincides with the number of lines of the matrix INFor such matrices, multiplication operation is introduced:
2 
Thus, when multiplying the matrix BUT dimension m.´ n. on the matrix IN dimension n.´ k.we get a matrix FROM dimension m.´ k.. In this case, the elements of the matrix FROM Calculated according to the following formulas:
Task 1.8. Find, if possible, the work of matrices ABand BA.:
Decision. 1) in order to find a work AB, need matrix strings A. multiply on the columns of the matrix B.:
2) Work BA.there is no, since the number of columns of the matrix B. does not coincide with the number of matrix strings A..
Inverse matrix. Solution of systems of linear equations by a matrix method
The matrix A - 1 is called a square matrix BUTIf equality is performed:
where through I. denotes single matrix of the same order as the matrix BUT:
.
In order to square matrix It had a reverse necessary and enough so that its determinant is different from zero. Reverse matrix are found by the formula:
, (1.13)
where A ij. - Algebraic supplements to elements a ij. Matrians BUT(Note that algebraic additions to the matrix rows BUT located in the return matrix in the form of the corresponding columns).
Example 1.9. Find a reverse matrix A - 1 to the matrix
.
Reverse matrix We will find by formula (1.13), which is for the case n. \u003d 3 has the form:
.
We find Det. A. = | A. | \u003d 1 × 3 × 8 + 2 × 5 × 3 + 2 × 4 × 3 - 3 × 3 × 3 - 1 × 5 × 4 - 2 × 2 × 8 \u003d 24 + 30 + 24 - 27 - 20 - 32 \u003d - 1. Since the determinant of the initial matrix is \u200b\u200bdifferent from zero, the reverse matrix exists.
1) We will find algebraic additions A ij.:
For the convenience of finding a reverse matrix, algebraic additions to the rows of the original matrix we are located in the corresponding columns.
From the obtained algebraic additions, we will make a new matrix and divide it to the determinant DET A.. Thus, we will get a reverse matrix:
Square systems of linear equations with non-zero, the main determinant can be solved using a reverse matrix. For this, the system (1.5) is written in a matrix form:
where 
Multiplying both parts of equality (1.14) to the left A - 1, we will get the system solution:
From!
Thus, in order to find the solution of the square system, you need to find a reverse matrix to the main matrix of the system and multiply it to the right to the matrix-column column.
Task 1.10. Solve the system of linear equations
using the reverse matrix.
Decision. We write the system in a matrix form :,
where 
- The main matrix of the system, - the column of unknown and - column of free members. Since the main determinant of the system 
then the main matrix of the system BUT It has a reverse matrix BUT -one . To find a reverse matrix BUT -1, calculate algebraic additions to all elements of the matrix BUT:
Of the numbers obtained, we will make a matrix (and algebraic additions to the rows of the matrix BUT We write to the appropriate columns) and divide it into the determinant D. Thus, we found a reverse matrix:
The solution of the system is found by formula (1.15):
In this way,
Solution of systems of linear equations by the method of ordinary Jordan exceptions
Let an arbitrary (not necessarily square) system of linear equations:
(1.16)
It is required to find a solution system, i.e. Such a set of variables that satisfies all the equalities of the system (1.16). In general, the system (1.16) may not have only one solution, but also countless solutions. It can also not have solutions at all.
When solving such tasks, a method of exclusion of unknown, which is also called the method of ordinary Jordan exceptions is used from the school course. Essence this method It is that in one of the equations of the system (1.16), one of the variables is expressed through other variables. This variable is then substituted into other system equations. The result is a system containing one equation to one equation and one variable less than the source system. The equation from which the variable expressed is remembered.
This process is repeated until one last equation remains in the system. In the process of excluding unknown, some equations can turn into faithful identities, for example. Such equations from the system are excluded, since they are performed in any values \u200b\u200bof variables and, therefore, do not affect the solution of the system. If, in the process of exclusion of unknown, at least one equation becomes equal, which can not be performed under any values \u200b\u200bof variables (for example), then we conclude that the system has no solution.
If during the solution of contradictory equations did not occur, then from the last equation there is one of the variables remaining in it. If only one variable remains in the last equation, it is expressed by the number. If other variables remain in the last equation, they are considered parameters, and the variable expressed through them will be the function of these parameters. Then the so-called " reverse" The variable found is substituted into the last memorized equation and find the second variable. Then the two found variables are substituted into the penultimate stored equation and find the third variable, and so on, up to the first memorized equation.
As a result, we obtain the solution of the system. This solution Will be the only one if the variables found will be numbers. If the first variable found, and then all the others will depend on the parameters, the system will have countless solutions (each set of parameters corresponds to a new solution). Formulas that allow you to find a solution to the system depending on whether or another set of parameters is called the general solution of the system.
Example 1.11.
x.
After memorizing the first equation 
and bringing similar members in the second and third equation we arrive at the system:
Express y. From the second equation and substitute it in the first equation:
We remember the second equation, and from the first we will find z.:
Return a reference, we will consistently find y. W. z.. To do this, we first substitute for the last memorized equation where we will find y.:
.
Then we substitute and in the first memorized equation Where we find x.:
Task 1.12. Solve the system of linear equations by excluding unknowns:
. (1.17)
Decision. Express the variable from the first equation x.and we substitute it in the second and third equation:
.
We remember the first equation 
In this system, the first and second equation contradict each other. Indeed, expressing y. 
, I get that 14 \u003d 17. This equality is not performed, under any values \u200b\u200bof variables x., y., I. z.. Consequently, the system (1.17) is incomputed, i.e. has no solution.
We offer readers to independently verify that the main determinant of the source system (1.17) is zero.
Consider a system that differs from the system (1.17) is just one free member.
Task 1.13. Solve the system of linear equations by excluding unknowns:
. (1.18)
Decision. As before, express from the first equation variable x.and we substitute it in the second and third equation:
.
We remember the first equation and we present similar members in the second and third equation. We come to the system:
Expressing y. from the first equation and substituting it into the second equation We will receive identity 14 \u003d 14, which does not affect the solution of the system, and, therefore, it can be excluded from the system.
In the last memorized equality variable z. We will consider the parameter. We believe. Then
Substitute y. W. z. in the first memorized equality and find x.:
.
Thus, the system (1.18) has countless solutions, and any decision can be found using formulas (1.19), choosing an arbitrary value of the parameter t.:
(1.19)
So the system solutions, for example, are the following sets of variables (1; 2; 0), (2; 26; 14), etc. Formulas (1.19) express a general (any) solution of the system (1.18).
In the case when the initial system (1.16) has a sufficiently large number of equations and unknowns, the specified method of ordinary Jordan exceptions is cumbersome. However, it is not. It is enough to withdraw the algorithm for recalculating the system coefficients at one step in general And make a solution to the problem in the form of special Jordan tables.
Let the system of linear forms (equations) be given:
, (1.20)
Where x J. - independent (sought) variables, a ij.- permanent coefficients
(i \u003d.1, 2,…, m.; j. = 1, 2,…, n.). Right parts of the system y I. (i \u003d.1, 2,…, m.) can be both variables (dependent) and constants. It is required to find solutions to this system by excluding unknown.
Consider the following operation, called in the future "one step of ordinary Jordan exceptions". From arbitrary ( r. -to) equality express an arbitrary variable ( x S.) and substitute for all other equality. Of course, this is possible only when a RS.¹ 0. Coefficient a RS. It is called permissive (sometimes guide or main) element.
We will get the following system:
. (1.21)
Of s.-Ho system equality (1.21) We will later find a variable x S.(After the remaining variables) are found). S.- I remember the string and later from the system is excluded. The remaining system will contain on one equation and one independent variable is less than the source system.
Calculate the coefficients of the obtained system (1.21) through the coefficients of the source system (1.20). Let's start by S. r.-to equation that after the expression of the variable x S.through the remaining variables will look like this:
Thus, new coefficients r.The equations are calculated according to the following formulas:
(1.23)
Calculate now new coefficients b IJ.(i.¹ r.) an arbitrary equation. To do this, we substitute pronounced in (1.22) variable x S. in i.- Equation of the system (1.20):
After bringing such members, we get:
(1.24)
From equality (1.24), we obtain formulas for which the remaining system coefficients (1.21) are calculated (except r.- Equations):
(1.25)
The transformation of system of linear equations by the method of ordinary Jordan exceptions is made in the form of tables (matrices). These tables were called "Jordan".
So, the task (1.20) is put in line with the following Zhordanov Table:
Table 1.1.
| x. 1 | x. 2 | … | x J. | … | x S. | … | x N. | |
| y. 1 = | a. 11 | a. 12 | a. 1j. | a. 1s. | a. 1n. | |||
| ………………………………………………………………….. | ||||||||
| y I.= | a I. 1 | a I. 2 | a ij. | a is. | a IN. | |||
| ………………………………………………………………….. | ||||||||
| y R.= | a R. 1 | a R. 2 | a RJ. | a RS. | a RN. | |||
| …………………………………………………………………. | ||||||||
| y N.= | a M. 1 | a M. 2 | a MJ. | a MS. | a MN. |
Zhortanova Table 1.1 Contains the left capital column, which records the right parts of the system (1.20) and the upper title line into which independent variables are recorded.
The remaining elements of the table form the main matrix of the coefficients of the system (1.20). If you multiply the matrix BUT On the matrix consisting of elements of the upper title line, then the matrix consists of elements of the left capital column. That is, essentially, Zhordanov Table is a matrix form of recording system of linear equations :. The system (1.21) meets the following Zhordanov Table:
Table 1.2.
| x. 1 | x. 2 | … | x J. | … | y R. | … | x N. | |
| y. 1 = | b. 11 | b. 12 | b. 1 J. | b. 1 S. | b. 1 N. | |||
| ………………………………………………………………….. | ||||||||
| y i \u003d. | b I. 1 | b I. 2 | b IJ. | b is. | b in | |||
| ………………………………………………………………….. | ||||||||
| x s \u003d. | b R. 1 | b R. 2 | b RJ. | b RS | b rn. | |||
| …………………………………………………………………. | ||||||||
| y n \u003d | b M. 1 | b M. 2 | b MJ. | b MS. | b Mn. |
Allowing element a RS. We will highlight bold. Recall that in order to implement one step of Jordan exceptions, the allowing element must be different from zero. A string of a table containing the allowing element is called the resolution string. A column containing the allowing element is called the resolution column. When moving from this table to the following table one variable ( x S.) From the rather of the title line, the table moves to the left capital column and, on the contrary, one of the free members of the system ( y R.) From the left capital column of the table moves to the upper title line.
We describe the coefficient recalculation algorithm during the transition from the Jordan table (1.1) to Table (1.2), resulting from formulas (1.23) and (1.25).
1. The resolution element is replaced by the inverse number:
2. The remaining permissive string elements are divided into the allowing element and change the sign to the opposite: 
3. The remaining elements of the resolution column are divided into the allowing item: 
4. Elements that do not fall into the resolution line and the allowing column are recalculated by formulas: 
The latter formula is easily remembered if noted that the elements constituting the fraction 
are at the intersection i.- I. r.Lock I. j.- I. s.-to columns (permissive string allowing column and the row and column, on the intersection of which there is a recalculated element). More precisely, when memorizing the formula 
You can use the following diagram:
Making the first step of Jordan exceptions, you can choose any element of Table 1.3, located in columns as a resolution element. x. 1 ,…, x. 5 (all specified elements are not zero). Do not only choose the resolution element in the last column, because need to find independent variables x. 1 ,…, x. five . We choose, for example, the coefficient 1 With a variable x. 3 In the third line of Table 1.3 (the resolution element is shown in bold). When switching to Table 1.4 variable x. 3 from the upper title string is changing places with a constant of the 0 left capita column (third line). In this case, the variable x. 3 is expressed in the other variables.
Line x. 3 (Table 1.4) can be remembered, eliminating from Table 1.4. From Table 1.4, the third column with zero is also excluded in the upper title line. The fact is that regardless of the coefficients of this column b I. 3 All the corresponding terms of each equation 0 · b I. 3 systems will be zero. Therefore, the specified coefficients can not be calculated. By excluding one variable x. 3 and remembering one of the equations, we arrive at the system, corresponding table 1.4 (with an overclosed string x. 3). Choosing in Table 1.4 as a resolution element b. 14 \u003d -5, go to Table 1.5. Table 1.5 remember the first line and exclude it from the table together with the fourth column (with zero above).
Table 1.5 Table 1.6
From the last table 1.7 we find: x. 1 = - 3 + 2x. 5 .
Sequentially substituting the variables already found in the stored lines, we find the remaining variables:
Thus, the system has countless solutions. Variable x. 5, you can give arbitrary values. This variable acts as a parameter x. 5 \u003d t. We have proven systemage of the system and found its overall solution:
X. 1 = - 3 + 2t.
X. 2 = - 1 - 3t.
X. 3 = - 2 + 4t. . (1.27)
x. 4 = 4 + 5t.
x. 5 = t.
Giving the parameter t. various valuesWe will receive countless solutions of the source system. So, for example, the system solution is the next set of variables (- 3; - 1; - 2; 4; 0).
Consider a system of 3 equations with three unknown
Using the determinants of the 3rd order, the solution of such a system can be written in the same form as for the system of two equations, i.e.
(2.4)
if 0. Here
It is kramer rule solutions of a system of three linear equations with three unknown.
Example 2.3. Solve a system of linear equations using the crawler rule:
Decision . We find the determinant of the main system matrix
Since 0, to find a system solution, you can apply the craver rule, but one more than three more determinants are pre-calculated:
Check:
Consequently, the solution is found correctly.
Cramer rules obtained for linear systems The 2nd and 3rd Order, suggest that the same rules can be formulated for both linear systems of any order. Really takes place
Cramer Theorem. Square system of linear equations with different from zero determinant of the main system matrix (0) it has one and only one solution and this decision is calculated by the formulas.
(2.5)
where – the determinant of the main matrix, i. – the determinant of the matrix, obtained from the main, replacementi.-to column column of free members.
Note that if \u003d 0, then the craver rule is not applicable. This means that the system either does not have any solutions, or has infinitely many solutions.
Formulating the theorem of the Cramer, the question of the calculation of the highest order determinants naturally arises.
2.4. N-order determinants
Additional minor M. iJ. Element a. iJ. called the determinant derived from this by crossing i.- Row I. j.-to column. Algebraic supplement A. iJ. Element a. iJ. called minor of this element taken with a sign (-1) i. + j. . A. iJ. = (–1) i. + j. M. iJ. .
For example, we find minors and algebraic elements supplements a. 23 I. a. 31 determinants
Receive
Using the concept of algebraic supplement can be formulated theorem on the decomposition of the determinantn.-o order row or column.
Theorem 2.1. The determinant of the matrixA. equal to the amount of works of all elements of a certain line (or column) on their algebraic additions:
(2.6)
This theorem underlies one of the main methods for calculating the determinants, the so-called. method lowering order. As a result of the decomposition of the determinant n.-o order for any row or column, it obtains N determinants ( n.-1) -go order. So that such determinants were less, it is advisable to choose the line or column in which the most zeros. In practice, the definition formula of the determinant is usually written in the form:
those. Algebraic additions are recorded explicitly through the minors.
Examples 2.4. Calculate the determinants, pre-laying them on any row or column. Usually in such cases choose such a column or a string in which the most zeros. The selected string or column will be denoted by the arrow.
2.5. The main properties of determinants
Decomposing the determinant for any string or column, we get N determinants ( n.-1) -go order. Then each of these determinants ( n.-1) -go order can also be decomposed in the amount of determinants ( n.-2) -go order. Continuing this process, you can walk to the determinants of the 1st order, i.e. Before the elements of the matrix, the determinant of which is calculated. So, to calculate the determinants of the 2nd order will have to calculate the sum of the two terms, for the determinants of the 3rd order - the amount of the 6 components, for the 4th order determinants - 24 terms. The number of components will sharply increase as the order of the determinant increases. This means that the calculation of identifiers of very high orders becomes a rather laborious task, unbearable even for computer. However, the determinants can be calculated differently using the properties of the determinants.
Property 1. . The determinant will not change if it is swapped in places and columns, i.e. When transposing the matrix:
.
This property indicates equality of the rows and columns of the determinant. In other words, any statement about the columns of the determinant is true for its rows and vice versa.
Property 2. . The determinant changes the sign when rearranging two lines (columns).
Corollary . If the determinant has two identical strings (column), then it is zero.
Property 3. . The total multiplier of all elements in any row (column) can be reached by the identifier sign..
For example,
Corollary . If all the elements of a certain line (column) of the determinant are zero, then the determinant itself is zero.
Property 4. . The determinant will not change if to the elements of one line (column), add the elements of another string (column) multiplied by any number.
For example,
Property 5. . The determinant of the works of matrices is equal to the product of determinants of the matrices:
