Matrix method for solving systems of linear equations Examples. Inverse matrix. The solution of matrix equations. Examples of solving a system of linear equations by the matrix method
Topic 2. Systems of linear algebraic equations.
Basic concepts.
Definition 1.. System m. linear equations from n. Unknowns are called a type system:
where I. 
- Numbers.
Definition 2.. The solution of the system (I) is called such a set of unknown, in which each equation of this system addresses the identity.
Definition 3.. System (I) is called jointif it has at least one solution and non-stopif it does not have solutions. The joint system is called definedif it has the only solution and uncertain otherwise.
Definition 4.. View equation
called zero, and equation of type
called uncommon. It is obvious that the system of equations containing an incontestary equation is incomplete.
Definition 5.. Two systems of linear equations are called equivalentIf each solution of one system serves as a solution to another and, on the contrary, any decision of the second system is the decision of the first.
Matrix record system of linear equations.
Consider the system (I) (see §1).
Denote:
Matrix of coefficients at unknown
,
Matrix - column of free members
Matrix - the column of unknown
.
Definition 1. The matrix is \u200b\u200bcalled the main matrix of the system (I), and the matrix is \u200b\u200ban extended system matrix (I).
By determining the equality of matrices system (I), matrix equality corresponds to:
.
The right-hand part of this equality to determine the work of the matrices ( see Definition 3 § 5 chapters 1) You can decompose on multipliers:
.
Equality (2) called matrix System Recording (I).
Solution of the system of linear equations by the Cramer method.
Let in the system (i) (see §1) m \u003d N. . The number of equations is equal to the number of unknown, and the main matrix of the system is nondegenerate, i.e. . Then the system (I) from §1 has a single decision
where Δ. \u003d Det A. called the main thing determinant system (I), Δ I. It turns out from the determinant Δ replacement i.-to column on a column from free members of the system (I).
Example. The system by the control system:
.
According to formulas (3)
.
Calculate system determinants:
,
,
,
.
To obtain the determinant, we replaced the first column on the column from free members; replacing the 2nd column in the column on the column from free members, we get; Similarly, replacing a 3rd column on a column from free members, we get. Solution Solution:
Solving system of linear equations using a reverse matrix.
Let in the system (i) (see §1) m \u003d N. and the main matrix of the system is non-degenerate. We write the system (I) in matrix form ( see §2.):
because the matrix A. Non-degenerate, then it has a reverse matrix ( see theorem 1 §6 chapter 1). Multiply both parts of equality (2) on the matrix then
. (3)
By definition of the return matrix. From equality (3) have
Solve the system using a return matrix
.
Denote
; ; .
In the example (§ 3) we calculated the determinant, therefore, the matrix A. It has a reverse matrix. Then in power (4) .
. (5)
We find a matrix ( see §6 chapters 1)
, 
, 
,
, 
, 
,
, 
, 
,
,
.
Gauss method.
Let a system of linear equations set:
. (I)
It is required to find all solutions of the system (i) or make sure that the system is incomplete.
Definition 1.Let's call the elementary conversion of the system (I) Any of three actions:
1) defecting the zero equation;
2) add to both parts of the equation of the corresponding parts of another equation multiplied by the number L;
3) the changes in the places of classes in the system equations so that unknown with the same numbers in all equations occupy the same places, i.e. If, for example, in the 1st equation, we changed the 2nd and 3rd terms, then the same must be done in all equations of the system.
The Gauss method is that the system (I) with the help of elementary transformations is given to the equivalent system, the solution of which is directly or determined.
As described in §2, the system (i) is uniquely determined by its extended matrix and any elementary system transformation (I) corresponds to the elementary conversion of the extended matrix:
.
Conversion 1) corresponds to the trimming of the zero string in the matrix, conversion 2) is equivalent to the addition to the corresponding line of the matrix of another line, multiplied by the number L, conversion 3) is equivalent to permuting the columns in the matrix.
It is easy to see that, on the contrary, each elementary conversion of the matrix corresponds to the elementary conversion of the system (I). By virtue of what was said, instead of operations with the system (i) we will work with an extended matrix of this system.
In the matrix, the 1st column consists of coefficients at x 1, 2nd column - from coefficients at x 2etc. In the case of rearrangement, columns should be borne in mind that this condition is broken. For example, if we change the 1st and 2nd columns in some places, now in the 1st column there will be coefficients when x 2, and in the 2nd column - coefficients at x 1.
We will solve the system (i) by the Gauss method.
1. I cross out in the matrix all zero lines, if any (i.e., with all zero equations in the system (i)).
2. Check whether there is a string among the matrix rows, in which all the elements besides the latter are zero (let's call the incomplete line). Obviously, such a line corresponds to the incomplete equation in the system (i), therefore, the system (i) of solutions does not have and the process ends on this.
3. Let the matrix do not contain incomplete lines (system (I) does not contain unconditional equations). If a a 11 \u003d 0, I find in the 1st line some element (except for the latter) other than zero and rearrange the columns so that in the 1st line on 1st place there was no zero. Now we will assume that (i.e., we change the corresponding terms in the system equations (I)).
4. I multiply the 1st line on and fold the result with the 2nd line, then multiply the 1st line on and press the result with the 3rd string, etc. Obviously, this process is equivalent to the exception of the unknown x 1 Of all the equations of the system (I), except for the 1st. In the new matrix we get zeros in the 1st column under the element a 11.:
.
5. I cross out all zero strings in the matrix if they are, check whether there is no incomplete line (if it is available, the system is incomplete and the solution ends). Check whether a 22 / \u003d 0If yes, we find the element other than zero in the 2nd line and rearrange the columns so that. Next, multiply the elements of the 2nd string on 
and fold with the corresponding elements of the 3rd string, then - elements of the 2nd row on 
and we fold with the corresponding elements of the 4th line, etc., until we get zeros under a 22 /
.
Actions are equivalent to the exclusion of an unknown x 2 Of all the equations of the system (I), except for the 1st and 2nd. Since the number of rows of course, therefore, through the final number of steps, we will get that either the system is incomprehensible, or we will come to a stepped matrix ( see Definition 2 §7 chapter 1) :
,
We repel the system of equations corresponding to the matrix. This system is equivalent to the system (I)
.
From the last equation, express; We substitute in the previous equation, we find it, etc., until we get.
Note 1. Thus, when solving the system (i), we arrive at the Gauss method to one of the following cases.
1. The system (I) is incomplete.
2. System (I) has a single solution if the matrix is \u200b\u200bthe number of rows equal to the number of unknown ().
3. The system (I) has countless solutions if the number of rows in the matrix is \u200b\u200bless than the number of unknown ().
From here there is the following theorem.
Theorem. The system of linear equations is either inconsistent, or has a single solution or - infinite set solutions.
Examples. Solve the system of equations by the Gauss method or prove its incompleteness:
but) 
;
b) 
;
in) 
.
a) rewrite for this system as:
.
We changed places 1st and 2nd equation of the source system to simplify the calculations (instead of fractions, using such a permutation, we will only operate only by integers).
We compile an extended matrix:
.
No zero lines; no inclusive lines; We exclude the 1st unknown of all the equations of the system, except for the 1st. To do this, multiply the elements of the 1st line of the matrix to "-2" and lay with the corresponding elements of the 2nd line, which is equivalent to the multiplication of the 1st equation on "-2" and addition with the 2nd equation. Then I will multiply the elements of the 1st line on "-3" and lay with the corresponding elements of the third line, i.e. Multiply 2nd equation of a given system to "-3" and fold with the 3rd equation. Receive
.
The matrix corresponds to the system of equations
Appointment of service. With this online calculator, unknown (x 1, x 2, ..., x n) are calculated in the equation system. The solution is carried out by the method of inverse matrix. Wherein:- the determinant of the matrix A is calculated;
- through algebraic additions is the inverse matrix A -1;
- creating a solution template in Excel;
Instruction. To obtain a solution by the method of inverse matrix, it is necessary to set the dimension of the matrix. Next, in the new dialog box, fill out the matrix A and the vector of the results B.
Recall that by the solution of the system of linear equations, there is any combination of numbers (x 1, x 2, ..., x n), the substitution of which in this system instead of the corresponding unknowns draws each equation of the system into identity.
Linear system algebraic equations Usually recorded as (for 3 variables): see also the solution of matrix equations.
Algorithm Solutions
- The determinant of the matrix A is calculated. If the determinant is zero, the end of the solution. The system has infinite set solutions.
- When the determined is different from zero, the algebraic additions is the reverse matrix A -1.
- The vector of solutions x \u003d (x 1, x 2, ..., x n) is obtained by multiplying the inverse matrix to the result vector b.
Example number 1. Find the solution solving by the matrix method. We write the matrix in the form:
Algebraic additions.
| A 1.1 \u003d (-1) 1 + 1 |
| ∆ 1,1 = (1 (-2)-0 2) = -2 |
| A 1.2 \u003d (-1) 1 + 2 |
| ∆ 1,2 = -(3 (-2)-1 2) = 8 |
| A 1,3 \u003d (-1) 1 + 3 |
| ∆ 1,3 = (3 0-1 1) = -1 |
| A 2.1 \u003d (-1) 2 + 1 |
| ∆ 2,1 = -(-2 (-2)-0 1) = -4 |
| A 2.2 \u003d (-1) 2 + 2 |
| ∆ 2,2 = (2 (-2)-1 1) = -5 |
| A 2.3 \u003d (-1) 2 + 3 |
| ∆ 2,3 = -(2 0-1 (-2)) = -2 |
| A 3.1 \u003d (-1) 3 + 1 |
| ∆ 3,1 = (-2 2-1 1) = -5 |
| A 3.2 \u003d (-1) 3 + 2 |
| ∆ 3,2 = -(2 2-3 1) = -1 |
| 3 |
| -2 |
| -1 |
X T \u003d (1,0,1)
x 1 \u003d -21 / -21 \u003d 1
x 2 \u003d 0 / -21 \u003d 0
x 3 \u003d -21 / -21 \u003d 1
Check:
2 1+3 0+1 1 = 3
-2 1+1 0+0 1 = -2
1 1+2 0+-2 1 = -1
Example number 2. Solve the slam by the method of inverse matrix.
2 x 1 + 3x 2 + 3x 3 + x 4 \u003d 1
3 x 1 + 5x 2 + 3x 3 + 2x 4 \u003d 2
5 x 1 + 7x 2 + 6x 3 + 2x 4 \u003d 3
4 x 1 + 4x 2 + 3x 3 + x 4 \u003d 4
We write the matrix in the form:
Vector B:
B T \u003d (1,2,3,4)
Chief determinant
Minor for (1,1):
= 5 (6 1-3 2)-7 (3 1-3 2)+4 (3 2-6 2) = -3
Minor for (2,1):
= 3 (6 1-3 2)-7 (3 1-3 1)+4 (3 2-6 1) = 0
Minor for (3,1):
= 3 (3 1-3 2)-5 (3 1-3 1)+4 (3 2-3 1) = 3
Minor for (4,1):
= 3 (3 2-6 2)-5 (3 2-6 1)+7 (3 2-3 1) = 3
Determined Minor
∆ = 2 (-3)-3 0+5 3-4 3 = -3
Example number 4. Write a system of equations in matrix form and solve with the help of a reverse matrix.
Solution: XLS.
Example number 5. A system of three linear equations with three unknowns is given. Required: 1) to find its solution using the Cramer formula; 2) Write the system in matrix form and solve it with matrix calculus.
Guidelines. After solving the cramer, find the button "Solution by the method of the return matrix for the source data". You will receive an appropriate solution. Thus, the data is not yet completed.
Decision. Denote by a - matrix of coefficients at unknown; X is an unknown column matrix; B - Matrix-column of free members:
|
B T \u003d (4, -3, -3)
Taking into account these designations, this system of equations takes the following matrix form: A * x \u003d B.
If the matrix A is non-degenerate (its determinant is different from zero, it has a reverse matrix A -1. Multiplying both parts of the equation on A -1, we get: a -1 * a * x \u003d a -1 * b, a -1 * A \u003d E.
This equality is called matrix recording of the solution of the linear equation system. To find a solution of the system of equations, it is necessary to calculate the reverse matrix A -1.
The system will have a solution if the determinant of the matrix A is differ from zero.
We find the main determinant.
∆=-1 (-2 (-1)-1 1)-3 (3 (-1)-1 0)+2 (3 1-(-2 0))=14
So, the determinant is 14 ≠ 0, so we continue the decision. To do this, we find the return matrix through algebraic additions.
Let them have a nondegenerate matrix A:
| A 1.1 \u003d (- 1) 1 + 1 |
|
| A 1.2 \u003d (- 1) 1 + 2 |
|
| A 1,3 \u003d (- 1) 1 + 3 |
|
| A 2.1 \u003d (- 1) 2 + 1 |
|
| A 2.2 \u003d (- 1) 2 + 2 |
|
| A 2.3 \u003d (- 1) 2 + 3 |
|
| A 3.1 \u003d (- 1) 3 + 1 |
|
| 4 |
| -3 |
| -3 |
| X \u003d 1/14 |
|
∆=4 (0 1-3 (-2))-2 (1 1-3 (-1))+0 (1 (-2)-0 (-1))=16
Transposed matrix
| A 1.2 \u003d (- 1) 1 + 2 |
|
| A 1,3 \u003d (- 1) 1 + 3 |
|
| A 2.1 \u003d (- 1) 2 + 1 |
|
| A 2.2 \u003d (- 1) 2 + 2 |
|
| A 2.3 \u003d (- 1) 2 + 3 |
|
| A 3.1 \u003d (- 1) 3 + 1 |
|
| A 3.2 \u003d (- 1) 3 + 2 |
|
| A 3.3 \u003d (- 1) 3 + 3 | 1/16 |
|
| (4 6)+(1 (-2))+(-1 6) | (4 (-4))+(1 4)+(-1 (-12)) | (4 (-2))+(1 6)+(-1 (-2)) |
| (2 6)+(0 (-2))+(-2 6) | (2 (-4))+(0 4)+(-2 (-12)) | (2 (-2))+(0 6)+(-2 (-2)) |
| (0 6)+(3 (-2))+(1 6) | (0 (-4))+(3 4)+(1 (-12)) | (0 (-2))+(3 6)+(1 (-2)) |
| =1/16 |
|
| A * A -1 \u003d |
|
Example number 7. Decision matrix equations.
Denote:
| A \u003d. |
|
| A 1.1 \u003d (-1) 1 + 1 |
|
| A 1.2 \u003d (-1) 1 + 2 |
|
| A 1,3 \u003d (-1) 1 + 3 |
|
| A 2.1 \u003d (-1) 2 + 1 |
|
| A 2.2 \u003d (-1) 2 + 2 |
|
| A 2.3 \u003d (-1) 2 + 3 |
|
| A 3.1 \u003d (-1) 3 + 1 |
|
| -12 | 15 | -5 |
| -4 | 5 | -2 |
| 7 | -9 | 3 |
B T \u003d (31,13.10)
X T \u003d (4.05,6.13,7.54)
x 1 \u003d 158/39 \u003d 4.05
x 2 \u003d 239/39 \u003d 6.13
x 3 \u003d 294/39 \u003d 7.54
Check.
-2 4.05+-1 6.13+6 7.54=31
1 4.05+-1 6.13+2 7.54=13
2 4.05+4 6.13+-3 7.54=10
Example number 9. Denote by a - matrix of coefficients at unknown; X is an unknown column matrix; B - Matrix-column of free members:
|
B T \u003d (31,13.10)
X T \u003d (5.21,4.51,6.15)
x 1 \u003d 276/53 \u003d 5.21
x 2 \u003d 239/53 \u003d 4.51
x 3 \u003d 326/53 \u003d 6.15
Check.
-2 5.21+1 4.51+6 6.15=31
1 5.21+-1 4.51+2 6.15=13
2 5.21+4 4.51+-3 6.15=10
Example number 10. The solution of matrix equations.
Denote:
A 11 \u003d (-1) 1 + 1 · -3 \u003d -3; A 12 \u003d (-1) 1 + 2 · 3 \u003d -3; A 21 \u003d (-1) 2 + 1 · 1 \u003d -1; A 22 \u003d (-1) 2 + 2 · 2 \u003d 2;
inverse matrix A -1.
| -3 | -3 |
| -1 | 2 |
| 1 | -2 |
| 1 | 1 |
| X \u003d. |
|
Reverse matrix method does not represent anything difficult if you know general principles Work with matrix equations and, of course, be able to produce elementary algebraic actions.
Solving the system of equations by the method of inverse matrix. Example.
It is convenient to comprehend the method of the return matrix on visual example. Take the system of equations:
The first step that needs to be done to solve this system of equations is to find the determinant. Therefore, we transform our system of equations in the following matrix:
And find the desired determinant:
The formula used to solve matrix equations is as follows:
Thus, for calculating x, we need to determine the value of the matrix A-1 and multiply it to b. This formula will help us in this:
At in this case will be transposed matrix - That is, the same, initial, but recorded not lines, but by columns.
We should not forget that reverse matrix method, like the Cramer method, is suitable only for systems in which the determinant is greater or less than zero. If the determinant is zero, you need to use the Gauss method.
The next step is the compilation of the Mind Matrix, which is such a scheme:
As a result, we obtained three matrices - minorors, algebraic additions and a transposed matrix of algebraic additions. Now you can proceed to the drawing matrix actually. We already know the formula. For our example, it will look like this.
Matrix method of solving systems of linear equations
Consider the system of linear equations of the following form:
$ \\ left \\ (\\ begin (array) (C) (A_ (11) x_ (1) + a_ (12) x_ (2) + ... + a_ (1n) x_ (n) \u003d b_ (1)) \\\\ (a_ (21) x_ (1) + a_ (22) x_ (2) + ... + a_ (2n) x_ (n) \u003d b_ (2)) \\\\ (...) \\\\ (A_ (N1) x_ (1) + a_ (n2) x_ (2) + ... + a_ (nn) x_ (n) \u003d b_ (n)) \\ End (Array) \\ Right.. $
Numbers $ a_ (ij) (i \u003d 1..n, j \u003d 1..n) $ - system coefficients, numbers $ b_ (i) (i \u003d 1..n) $ - free members.
Definition 1.
In the case when all free members are zero, the system is called homogeneous, otherwise - inhomogeneous.
Each slope can be put into line with several matrices and write the system in the so-called matrix form.
Definition 2.
The system coefficients matrix is \u200b\u200bcalled the system matrix and is indicated, as a rule, the letter $ a $.
A column of free members forms a vector-column, which, as a rule, is denoted by the letter $ b $ and is called the matrix of free members.
Unknown variables form a vector-column, which, as a rule, is indicated by the letter $ x $ and is called an unknown matrix.
The matrix described above have the form:
$ A \u003d \\ left (\\ begin (array) (CCCC) (A_ (11)) & (A_ (12)) & (...) & (A_ (1N)) \\\\ (A_ (21)) & ( A_ (22)) & (...) & (A_ (2N)) \\\\ (...) & (...) & (...) & (...) \\\\ (A_ (N1) ) & (A_ (N2)) & (...) & (A_ (NN)) \\ END (Array) \\ Right), B \u003d \\ Left (\\ Begin (Array) (C) (B_ (1)) \\ \\\\ (x_ (2)) \\\\ (...) \\\\ (x_ (n)) \\ END (Array) \\ Right). $
Using matrices, the slope can be rewritten in the form $ a \\ cdot x \u003d b $. This entry is often called the matrix equation.
Generally speaking, in the matrix form you can record any slope.
Examples of system solutions using reverse matrix
Example 1.
Dana Slaya: $ \\ left \\ (\\ begin (array) (C) (3x_ (1) -2x_ (2) + x_ (3) -x_ (4) \u003d 3) \\\\ (x_ (1) -12x_ (2 ) -X_ (3) -X_ (4) \u003d 7) \\\\ (2x_ (1) -3x_ (2) + x_ (3) -3x_ (4) \u003d 5) \\ END (Array) \\ Right. $. Record The system in matrix form.
Decision:
$ A \u003d \\ left (\\ begin (array) (CCCC) (3) & (-2) & (1) & (-1) \\ (1) & (-12) & (-1) & (-1 -1 ) \\\\ (2) & (-3) & (1) & (-3) \\ END (Array) \\ Right), B \u003d \\ Left (\\ Begin (Array) (C) (3) \\\\ (7) \\\\ (5) \\ END (Array) \\ Right), x \u003d \\ left (\\ begin (array) (c) (x_ (1)) \\\\ (x_ (2)) \\\\ (x_ (3)) \\ $ \\ left (\\ begin (array) (CCCC) (3) & (-2) & (1) & (-1) \\ (1) & (-12) & (-1) & (-1) \\ 2)) \\\\ (x_ (3)) \\ END (Array) \\ Right) \u003d \\ left (\\ begin (array) (C) (3) \\\\ (7) \\\\ (5) \\ END (Array) \\ In the case when the matrix of the system is a square, the slope can be solved by the equation to the matrix method.
{!LANG-f1e4f7676667ba8a024addf0972ed390!}
{!LANG-aa58705924feadb3ba9ea4919ad34fda!}
Having a matrix equation $ a \\ cdot x \u003d b $ can be expressed from it $ X $ in the following way:
$ A ^ (- 1) \\ cdot a \\ cdot x \u003d a ^ (- 1) \\ Cdot b $
$ A ^ (- 1) \\ Cdot A \u003d E $ (the property of the work of matrices)
$ E \\ Cdot X \u003d A ^ (- 1) \\ Cdot B $
$ E \\ Cdot X \u003d X $ (Property of the Matrix Production)
$ X \u003d a ^ (- 1) \\ Cdot b $
The algorithm for solving the system of algebraic equations using a reverse matrix:
- record the system in matrix;
- calculate the determinant of the system matrix;
- if the determinant of the system matrix is \u200b\u200bdifferent from zero, then we find a reverse matrix;
- solution Solution Calculate according to the formula $ x \u003d a ^ (- 1) \\ Cdot b $.
If the system matrix has a determinant that is not equal to zero, then this system has a single solution that can be found in a matrix method.
If the system matrix has a determinant equal to zero, then this system cannot be solved by a matrix method.
Example 2.
Dana Slaya: $ \\ left \\ (\\ begin (array) (C) (x_ (1) + 3x_ (3) \u003d 26) \\\\ (-x_ (1) + 2x_ (2) + x_ (3) \u003d 52) \\\\ (3x_ (1) + 2x_ (2) \u003d 52) \\ END (Array) \\ Right. $. Solve the slam by the method of the return matrix, if possible.
Decision:
$ A \u003d \\ left (\\ begin (array) (CCC) (1) & (0) & (3) \\\\ (-1) & (2) & (1) \\\\ (3) & (2) & ( 0) \\ END (Array) \\ Right), B \u003d \\ Left (\\ Begin (Array) (C) (26) \\\\ (52) \\\\ (52) \\ END (Array) \\ Right), x \u003d \\ left (\\ begin (array) (c) (x_ (1)) \\\\ (x_ (2)) \\\\ (x_ (3)) \\ END (Array) \\ RIGHT). $
Finding the determinant of the system matrix:
$ \\ begin (array) (L) (\\ Det a \u003d \\ left | \\ begin (array) (CCC) (1) & (0) & (3) \\\\ (-1) & (2) & (1) \\\\ (3) & (2) & (0) \\ END (Array) \\ Right | \u003d 1 \\ CDOT 2 \\ CDOT 0 + 0 \\ CDOT 1 \\ CDOT 3 + 2 \\ CDOT (-1) \\ CDOT 3-3 \\ CDOT 2 \\ CDOT 3-2 \\ CDOT 1 \\ CDOT 1-0 \\ CDOT (-1) \\ CDot 0 \u003d 0 + 0-6-18-2-0 \u003d -26 \\ Ne 0) \\ End (Array) $ Since the determinant is not equal to zero, then the matrix of the system has a reverse matrix and, therefore, the system of equations can be solved by the inverse matrix method. The solution obtained will be the only one.
Let the system of equations using the reverse matrix:
$ A_ (11) \u003d (- 1) ^ (1 + 1) \\ Cdot \\ Left | \\ Begin (Array) (CC) (2) & (1) \\\\ (2) & (0) \\ END (Array) \\ Right | \u003d 0-2 \u003d -2; A_ (12) \u003d (- 1) ^ (1 + 2) \\ Cdot \\ Left | \\ Begin (Array) (CC) (-1) & (1) \\\\ (3) & (0) \\ END (Array) \\ Right | \u003d - (0-3) \u003d 3; $
$ A_ (13) \u003d (- 1) ^ (1 + 3) \\ CDOT \\ LEFT | \\ Begin (Array) (CC) (-1) & (2) \\\\ (3) & (2) \\ END (Array ) \\ RIGHT | \u003d -2-6 \u003d -8; A_ (21) \u003d (- 1) ^ (2 + 1) \\ cdot \\ left | \\ begin (array) (CC) (0) & (3) \\\\ (2) & (0) \\ END (Array) \\ $
$ A_ (22) \u003d (- 1) ^ (2 + 2) \\ Cdot \\ Left | \\ Begin (Array) (CC) (1) & (3) \\\\ (3) & (0) \\ END (Array) \\ Right | \u003d 0-9 \u003d -9; A_ (23) \u003d (- 1) ^ (2 + 3) \\ CDOT \\ LEFT | \\ Begin (Array) (CC) (1) & (0) \\\\ (3) & (2) \\ End (Array) \\
$ A_ (31) \u003d (- 1) ^ (3 + 1) \\ CDOT \\ LEFT | \\ Begin (Array) (CC) (0) & (3) \\\\ (2) & (1) \\ END (Array) \\ RIGHT | \u003d 0-6 \u003d -6; A_ (32) \u003d (- 1) ^ (3 + 2) \\ Cdot \\ Left | \\ Begin (Array) (CC) (1) & (3) \\\\ (-1) & (1) \\ END (Array) \\ Right | \u003d - (1 + 3) \u003d - 4; $
$ A_ (33) \u003d (- 1) ^ (3 + 3) \\ Cdot \\ Left | \\ Begin (Array) (CC) (1) & (0) \\\\ (-1) & (2) \\ END (Array ) \\ Right | \u003d 2-0 \u003d $ 2
The desired reverse matrix:
$ A ^ (- 1) \u003d \\ FRAC (1) (- 26) \\ CDOT \\ LEFT (\\ Begin (Array) (CCC) (-2) & (6) & (-6) \\\\ (3) & ( -9) & (-4) \\\\ (-8) & (-2) & (2) \\ END (Array) \\ Right) \u003d \\ FRAC (1) (26) \\ CDOT \\ LEFT (\\ Begin (Array) (CCC) (2) & (-6) & (6) \\\\ (-3) & (9) & (4) \\\\ (8) & (2) & (-2) \\ END (Array) \\ Right ) \u003d \\ left (\\ begin (array) (CCC) (\\ FRAC (2) (26)) & (\\ FRAC (-6) (26)) & (\\ FRAC (6) (26)) \\\\ (\\ (26)) & (\\ FRAC (-2) (26)) \\ END (Array) \\ Right) \u003d \\ Left (\\ Begin (Array) (CCC) (\\ FRAC (1) (13)) & (- \\ (13)) \\\\ (\\ FRAC (4) (13)) & (\\ FRAC (1) (13)) & (- \\ FRAC (1) (13)) \\ END (Array) \\ Right). $
We find the solution of the system:
$ X \u003d \\ left (\\ begin (array) (CCC) (\\ FRAC (1) (13)) & (- \\ FRAC (3) (13)) & (\\ FRAC (3) (13)) \\\\ ( - \\ FRAC (3) (26)) & (\\ FRAC (9) (26)) & (\\ FRAC (2) (13)) \\\\ (\\ FRAC (4) (13)) & (\\ FRAC (1 ) (13)) & (- \\ FRAC (1) (13)) \\ END (Array) \\ Right) \\ Cdot \\ Left (\\ Begin (Array) (C) (26) \\\\ (52) \\\\ (52 ) \\ END (Array) \\ Right) \u003d \\ left (\\ begin (array) (C) (\\ FRAC (1) (13) \\ CDOT 26- \\ FRAC (3) (13) \\ CDOT 52+ \\ FRAC (3 ) (13) \\ CDOT 52) \\\\ (- \\ FRAC (3) (26) \\ CDOT 26+ \\ FRAC (9) (26) \\ CDOT 52+ \\ FRAC (2) (13) \\ CDOT 52) \\\\ Begin (Array) (C) (2-12 + 12) \\\\ (-3 + 18 + 8) \\\\ (8 + 4-4) \\ END (Array) \\ Right) \u003d \\ Left (\\ Begin (Array) (c) (2) \\\\ (23) \\\\ (8) \\ END (Array) \\ Right) $
$ X \u003d \\ left (\\ begin (array) (c) (2) \\\\ (23) \\\\ (8) \\ END (Array) \\ Right) $ is the desired solution of the equation system.
Equations Generally, linear algebraic equations and their systems, as well as methods for solving them, occupy in mathematics, both theoretical and applied, special place.
This is due to the fact that the overwhelming majority of physical, economic, technical, and even pedagogical problems can be described and solved using a variety of equations and their systems. Recently, special popularity among researchers, scientists and practitioners has acquired math modeling In almost all subject areas, which is explained by its advantages over other well-known and tested methods for researching objects of various nature, in particular, so-called, complex systems. There is a great manifold of various definitions of a mathematical model, data by scientists at different times, but in our opinion, the most successful thing is the following statement. Mathematical model - This is an idea expressed by the equation. Thus, the ability to draw up and solve equations and their systems is an integral characteristic of a modern specialist.
To solve systems of linear algebraic equations, methods are most common: Cramer, Jordan-Gauss and the matrix method.
Matrix method Solutions - a solution method using a reverse matrix of systems of linear algebraic equations with a non-zero determinant.
If you write the coefficients at unknown xi values \u200b\u200bin the matrix A, unknown values \u200b\u200bto assemble in the vector column x, and free members in the vector column B, then the system of linear algebraic equations can be written in the form of the next matrix equation A · x \u003d B, which has the only solution only Then, when the determinant of the matrix A will not be zero. In this case, the solution of the system of equations can be found in the following way. X. = A. -one · B.where A. -1 - reverse matrix.
The matrix solution method is as follows.
Let the system of linear equations with n.unknown:
It can be rewritten in matrix form: AX. = B.where A. - the main matrix of the system, B. and X. - columns of free members and solutions of the system, respectively:
Multiply this matrix equation on the left to A. -1 - matrix, back to the matrix A.: A. -1 (AX.) = A. -1 B.
As A. -1 A. = E.Receive X. \u003d A. -1 B.. The right side of this equation will give a column solutions of the source system. Condition of applicability this method (as in general, the existence of the solution of a non-uniform system of linear equations with the number of equations, equal number unknown) is non-degenerate matrix A.. Necessary and sufficient condition of this is the inequality zero of the matrix determinant A.: Det. A.≠ 0.
For a homogeneous system of linear equations, that is, when vector B. = 0 , really reverse rule: system AX. = 0 has a nontrivial (that is, no zero) solution only if Det A. \u003d 0. Such a link between solutions of homogeneous and inhomogeneous systems of linear equations is called the alternative to Fredholma.
Example solutions of a non-uniform system of linear algebraic equations.
It is convinced that the determinant of the matrix made of coefficients at unknown system of linear algebraic equations is not zero.
The next step will be the calculation of algebraic supplements for the elements of the matrix consisting of coefficients at unknown. They will be needed to find the reverse matrix.
