To study a system of linear agebraic equations (SLAEs) for consistency means to find out whether this system has solutions or does not have them. Well, if there are solutions, then indicate how many there are.
We will need information from the topic "System of linear algebraic equations. Basic terms. Matrix form of notation". In particular, concepts such as system matrix and extended system matrix are needed, since the formulation of the Kronecker-Capelli theorem is based on them. As usual, we will denote the system matrix by the letter $A$, and the extended matrix of the system by the letter $\widetilde(A)$.
Kronecker-Capelli theorem
Linear system algebraic equations is consistent if and only if the rank of the system matrix is equal to the rank of the extended matrix of the system, i.e. $\rang A=\rang\widetilde(A)$.
Let me remind you that a system is called joint if it has at least one solution. The Kronecker-Capelli theorem says this: if $\rang A=\rang\widetilde(A)$, then there is a solution; if $\rang A\neq\rang\widetilde(A)$, then this SLAE has no solutions (inconsistent). The answer to the question about the number of these solutions is given by a corollary of the Kronecker-Capelli theorem. In the formulation of the corollary, the letter $n$ is used, which is equal to the number of variables of the given SLAE.
Corollary to the Kronecker-Capelli theorem
- If $\rang A\neq\rang\widetilde(A)$, then the SLAE is inconsistent (has no solutions).
- If $\rang A=\rang\widetilde(A)< n$, то СЛАУ является неопределённой (имеет бесконечное количество решений).
- If $\rang A=\rang\widetilde(A) = n$, then the SLAE is definite (has exactly one solution).
Please note that the formulated theorem and its corollary do not indicate how to find a solution to the SLAE. With their help, you can only find out whether these solutions exist or not, and if they exist, then how many.
Example No. 1
Explore SLAE $ \left \(\begin(aligned) & -3x_1+9x_2-7x_3=17;\\ & -x_1+2x_2-4x_3=9;\\ & 4x_1-2x_2+19x_3=-42. \end(aligned )\right.$ for compatibility. If the SLAE is compatible, indicate the number of solutions.
To find out the existence of solutions to a given SLAE, we use the Kronecker-Capelli theorem. We will need the matrix of the system $A$ and the extended matrix of the system $\widetilde(A)$, we will write them:
$$ A=\left(\begin(array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \end(array) \right);\; \widetilde(A)=\left(\begin(array) (ccc|c) -3 & 9 &-7 & 17 \\ -1 & 2 & -4 & 9\\ 4 & -2 & 19 & -42 \end(array) \right). $$
We need to find $\rang A$ and $\rang\widetilde(A)$. There are many ways to do this, some of which are listed in the Matrix Rank section. Typically, two methods are used to study such systems: “Calculating the rank of a matrix by definition” or “Calculating the rank of a matrix by the method of elementary transformations”.
Method number 1. Computing ranks by definition.
According to the definition, rank is the highest order of the minors of a matrix, among which there is at least one that is different from zero. Usually, the study begins with first-order minors, but here it is more convenient to immediately begin calculating the third-order minor of the matrix $A$. The third-order minor elements are located at the intersection of three rows and three columns of the matrix in question. Since the matrix $A$ contains only 3 rows and 3 columns, the third order minor of the matrix $A$ is the determinant of the matrix $A$, i.e. $\Delta A$. To calculate the determinant, we apply formula No. 2 from the topic “Formulas for calculating determinants of the second and third orders”:
$$ \Delta A=\left| \begin(array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \end(array) \right|=-21. $$
So, there is a third order minor of the matrix $A$, which is not equal to zero. It is impossible to construct a fourth-order minor, since it requires 4 rows and 4 columns, and the matrix $A$ has only 3 rows and 3 columns. So, the highest order of the minors of the matrix $A$, among which there is at least one that is not equal to zero, is equal to 3. Therefore, $\rang A=3$.
We also need to find $\rang\widetilde(A)$. Let's look at the structure of the matrix $\widetilde(A)$. Up to the line in the matrix $\widetilde(A)$ there are elements of the matrix $A$, and we found out that $\Delta A\neq 0$. Consequently, the matrix $\widetilde(A)$ has a third-order minor, which is not equal to zero. We cannot construct fourth-order minors of the matrix $\widetilde(A)$, so we conclude: $\rang\widetilde(A)=3$.
Since $\rang A=\rang\widetilde(A)$, then according to the Kronecker-Capelli theorem the system is consistent, i.e. has a solution (at least one). To indicate the number of solutions, we take into account that our SLAE contains 3 unknowns: $x_1$, $x_2$ and $x_3$. Since the number of unknowns is $n=3$, we conclude: $\rang A=\rang\widetilde(A)=n$, therefore, according to the corollary of the Kronecker-Capelli theorem, the system is definite, i.e. has a unique solution.
The problem is solved. What are the disadvantages and advantages of this method? First, let's talk about the advantages. Firstly, we only needed to find one determinant. After this, we immediately made a conclusion about the number of solutions. Typically, standard standard calculations give systems of equations that contain three unknowns and have a unique solution. For such systems this method It’s very convenient, because we know in advance that there is a solution (otherwise there wouldn’t be an example in the standard calculation). Those. all we have to do is show the existence of a solution in the most in a fast way. Secondly, the calculated value of the determinant of the system matrix (i.e. $\Delta A$) will be useful later: when we begin to solve a given system using the Cramer method or using the inverse matrix.
However, the method of calculating the rank is by definition undesirable to use if the matrix of the system $A$ is rectangular. In this case, it is better to use the second method, which will be discussed below. In addition, if $\Delta A=0$, then we cannot say anything about the number of solutions of a given inhomogeneous SLAE. Maybe the SLAE has an infinite number of solutions, or maybe none. If $\Delta A=0$, then additional research is required, which is often cumbersome.
To summarize what has been said, I note that the first method is good for those SLAEs whose system matrix is square. Moreover, the SLAE itself contains three or four unknowns and is taken from standard standard calculations or tests.
Method number 2. Calculation of rank by the method of elementary transformations.
This method is described in detail in the corresponding topic. We will begin to calculate the rank of the matrix $\widetilde(A)$. Why matrices $\widetilde(A)$ and not $A$? The fact is that the matrix $A$ is part of the matrix $\widetilde(A)$, therefore, by calculating the rank of the matrix $\widetilde(A)$ we will simultaneously find the rank of the matrix $A$.
\begin(aligned) &\widetilde(A) =\left(\begin(array) (ccc|c) -3 & 9 &-7 & 17 \\ -1 & 2 & -4 & 9\\ 4 & - 2 & 19 & -42 \end(array) \right) \rightarrow \left|\text(swap the first and second lines)\right| \rightarrow \\ &\rightarrow \left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ -3 & 9 &-7 & 17\\ 4 & -2 & 19 & - 42 \end(array) \right) \begin(array) (l) \phantom(0) \\ II-3\cdot I\\ III+4\cdot I \end(array) \rightarrow \left(\begin (array) (ccc|c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 6 & 3 & -6 \end(array) \right) \begin(array) ( l) \phantom(0) \\ \phantom(0)\\ III-2\cdot II \end(array)\rightarrow\\ &\rightarrow \left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 0 & -7 & 14 \end(array) \right) \end(aligned)
We have reduced the matrix $\widetilde(A)$ to trapezoidal form. On the main diagonal of the resulting matrix $\left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 0 & -7 & 14 \end( array) \right)$ contains three non-zero elements: -1, 3 and -7. Conclusion: the rank of the matrix $\widetilde(A)$ is 3, i.e. $\rang\widetilde(A)=3$. When making transformations with the elements of the matrix $\widetilde(A)$, we simultaneously transformed the elements of the matrix $A$ located up to the line. The matrix $A$ is also reduced to trapezoidal form: $\left(\begin(array) (ccc) -1 & 2 & -4 \\ 0 & 3 &5 \\ 0 & 0 & -7 \end(array) \right )$. Conclusion: the rank of matrix $A$ is also 3, i.e. $\rang A=3$.
Since $\rang A=\rang\widetilde(A)$, then according to the Kronecker-Capelli theorem the system is consistent, i.e. has a solution. To indicate the number of solutions, we take into account that our SLAE contains 3 unknowns: $x_1$, $x_2$ and $x_3$. Since the number of unknowns is $n=3$, we conclude: $\rang A=\rang\widetilde(A)=n$, therefore, according to the corollary of the Kronecker-Capelli theorem, the system is defined, i.e. has a unique solution.
What are the advantages of the second method? The main advantage is its versatility. It doesn't matter to us whether the matrix of the system is square or not. In addition, we actually carried out forward transformations of the Gaussian method. There are only a couple of steps left, and we could obtain a solution to this SLAE. To be honest, I like the second method more than the first, but the choice is a matter of taste.
Answer: The given SLAE is consistent and defined.
Example No. 2
Explore SLAE $ \left\( \begin(aligned) & x_1-x_2+2x_3=-1;\\ & -x_1+2x_2-3x_3=3;\\ & 2x_1-x_2+3x_3=2;\\ & 3x_1- 2x_2+5x_3=1;\\ & 2x_1-3x_2+5x_3=-4.\end(aligned) \right.$ for compatibility.
We will find the ranks of the system matrix and the extended system matrix using the method of elementary transformations. Extended system matrix: $\widetilde(A)=\left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ -1 & 2 & -3 & 3 \\ 2 & -1 & 3 & 2 \\ 3 & -2 & 5 & 1 \\ 2 & -3 & 5 & -4 \end(array) \right)$. Let's find the required ranks by transforming the extended matrix of the system:
The extended matrix of the system is reduced to a stepwise form. If a matrix is reduced to echelon form, then its rank is equal to the number of non-zero rows. Therefore, $\rang A=3$. The matrix $A$ (up to the line) is reduced to trapezoidal form and its rank is 2, $\rang A=2$.
Since $\rang A\neq\rang\widetilde(A)$, then according to the Kronecker-Capelli theorem the system is inconsistent (i.e., has no solutions).
Answer: The system is inconsistent.
Example No. 3
Explore SLAE $ \left\( \begin(aligned) & 2x_1+7x_3-5x_4+11x_5=42;\\ & x_1-2x_2+3x_3+2x_5=17;\\ & -3x_1+9x_2-11x_3-7x_5=-64 ;\\ & -5x_1+17x_2-16x_3-5x_4-4x_5=-90;\\ & 7x_1-17x_2+23x_3+15x_5=132. \end(aligned) \right.$ for compatibility.
The extended matrix of the system has the form: $\widetilde(A)=\left(\begin(array) (ccccc|c) 2 & 0 & 7 & -5 & 11 & 42\\ 1 & -2 & 3 & 0 & 2 & 17 \\ -3 & 9 & -11 & 0 & -7 & -64 \\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \end(array) \right)$. Let's swap the first and second rows of this matrix so that the first element of the first row becomes one: $\left(\begin(array) (ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 2 & 0 & 7 & -5 & 11 & 42 \\ -3 & 9 & -11 & 0 & -7 & -64 \\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \end(array) \right)$.
We have reduced the extended matrix of the system and the matrix of the system itself to a trapezoidal form. The rank of the extended matrix of the system is equal to three, the rank of the matrix of the system is also equal to three. Since the system contains $n=5$ unknowns, i.e. $\rang\widetilde(A)=\rang A< n$, то согласно следствия из теоремы Кронекера-Капелли данная система является неопределённой, т.е. имеет бесконечное количество решений.
Answer: The system is uncertain.
In the second part, we will analyze examples that are often included in standard calculations or tests in higher mathematics: consistency research and solution of SLAE depending on the values of the parameters included in it.
Example 1. Find common decision and some particular solution of the systemSolution We do it using a calculator. Let's write out the extended and main matrices: 
The main matrix A is separated by a dotted line. We write unknown systems at the top, keeping in mind the possible rearrangement of terms in the equations of the system. By determining the rank of the extended matrix, we simultaneously find the rank of the main one. In matrix B, the first and second columns are proportional. Of the two proportional columns, only one can fall into the basic minor, so let’s move, for example, the first column beyond the dotted line with the opposite sign. For the system, this means transferring terms from x 1 to the right side of the equations. 
Let's reduce the matrix to triangular view. We will work only with rows, since multiplying a matrix row by a number other than zero and adding it to another row for the system means multiplying the equation by the same number and adding it with another equation, which does not change the solution of the system. We work with the first row: multiply the first row of the matrix by (-3) and add to the second and third rows in turn. Then multiply the first line by (-2) and add it to the fourth. 
The second and third lines are proportional, therefore, one of them, for example the second, can be crossed out. This is equivalent to crossing out the second equation of the system, since it is a consequence of the third. 
Now we work with the second line: multiply it by (-1) and add it to the third. 
The minor circled with a dotted line has the highest order (of possible minors) and is nonzero (it is equal to the product of the elements on the main diagonal), and this minor belongs to both the main matrix and the extended one, therefore rangA = rangB = 3.
Minor 
is basic. It includes coefficients for the unknowns x 2 , x 3 , x 4 , which means that the unknowns x 2 , x 3 , x 4 are dependent, and x 1 , x 5 are free.
Let's transform the matrix, leaving only the basis minor on the left (which corresponds to point 4 of the above solution algorithm). 
The system with the coefficients of this matrix is equivalent to the original system and has the form
Using the method of eliminating unknowns we find: 
, ,
We obtained relations expressing the dependent variables x 2, x 3, x 4 through the free ones x 1 and x 5, that is, we found a general solution: 
By assigning any values to the free unknowns, we obtain any number of particular solutions. Let's find two particular solutions:
1) let x 1 = x 5 = 0, then x 2 = 1, x 3 = -3, x 4 = 3;
2) put x 1 = 1, x 5 = -1, then x 2 = 4, x 3 = -7, x 4 = 7.
Thus, two solutions were found: (0,1,-3,3,0) – one solution, (1,4,-7,7,-1) – another solution.
Example 2. Explore compatibility, find a general and one particular solution to the system 
Solution. Let's rearrange the first and second equations to have one in the first equation and write the matrix B. 
We get zeros in the fourth column by operating with the first row: 
Now we get the zeros in the third column using the second line: 
The third and fourth lines are proportional, so one of them can be crossed out without changing the rank:
Multiply the third line by (–2) and add it to the fourth: 
We see that the ranks of the main and extended matrices are equal to 4, and the rank coincides with the number of unknowns, therefore, the system has a unique solution:
;
x 4 = 10- 3x 1 – 3x 2 – 2x 3 = 11.
Example 3. Examine the system for compatibility and find a solution if it exists. 
Solution. We compose an extended matrix of the system. 
We rearrange the first two equations so that there is 1 in the upper left corner:
Multiplying the first line by (-1), adding it to the third: 
Multiply the second line by (-2) and add it to the third: 
The system is inconsistent, since in the main matrix we received a row consisting of zeros, which is crossed out when the rank is found, but in the extended matrix the last row remains, that is, r B > r A .
Exercise. Investigate this system of equations for compatibility and solve it using matrix calculus.
Solution
Example. Prove system compatibility linear equations and solve it in two ways: 1) the Gauss method; 2) Cramer's method. (enter the answer in the form: x1,x2,x3)
Solution :doc :doc :xls
Answer: 2,-1,3.
Example. A system of linear equations is given. Prove its compatibility. Find a general solution of the system and one particular solution.
Solution
Answer: x 3 = - 1 + x 4 + x 5 ; x 2 = 1 - x 4 ; x 1 = 2 + x 4 - 3x 5
Exercise. Find the general and particular solutions of each system.
Solution. We study this system using the Kronecker-Capelli theorem.
Let's write out the extended and main matrices:
| 1 | 1 | 14 | 0 | 2 | 0 |
| 3 | 4 | 2 | 3 | 0 | 1 |
| 2 | 3 | -3 | 3 | -2 | 1 |
| x 1 | x 2 | x 3 | x 4 | x 5 |
Here matrix A is highlighted in bold.
Let's reduce the matrix to triangular form. We will work only with rows, since multiplying a matrix row by a number other than zero and adding it to another row for the system means multiplying the equation by the same number and adding it with another equation, which does not change the solution of the system.
Let's multiply the 1st line by (3). Multiply the 2nd line by (-1). Let's add the 2nd line to the 1st:
| 0 | -1 | 40 | -3 | 6 | -1 |
| 3 | 4 | 2 | 3 | 0 | 1 |
| 2 | 3 | -3 | 3 | -2 | 1 |
Let's multiply the 2nd line by (2). Multiply the 3rd line by (-3). Let's add the 3rd line to the 2nd:
| 0 | -1 | 40 | -3 | 6 | -1 |
| 0 | -1 | 13 | -3 | 6 | -1 |
| 2 | 3 | -3 | 3 | -2 | 1 |
Multiply the 2nd line by (-1). Let's add the 2nd line to the 1st:
| 0 | 0 | 27 | 0 | 0 | 0 |
| 0 | -1 | 13 | -3 | 6 | -1 |
| 2 | 3 | -3 | 3 | -2 | 1 |
The selected minor has the highest order (of possible minors) and is non-zero (it is equal to the product of the elements on the reverse diagonal), and this minor belongs to both the main matrix and the extended one, therefore rang(A) = rang(B) = 3 Since the rank of the main matrix is equal to the rank of the extended matrix, then the system is collaborative.
This minor is basic. It includes coefficients for the unknowns x 1 , x 2 , x 3 , which means that the unknowns x 1 , x 2 , x 3 are dependent (basic), and x 4 , x 5 are free.
Let's transform the matrix, leaving only the basis minor on the left.
| 0 | 0 | 27 | 0 | 0 | 0 |
| 0 | -1 | 13 | -1 | 3 | -6 |
| 2 | 3 | -3 | 1 | -3 | 2 |
| x 1 | x 2 | x 3 | x 4 | x 5 |
27x 3 =
- x 2 + 13x 3 = - 1 + 3x 4 - 6x 5
2x 1 + 3x 2 - 3x 3 = 1 - 3x 4 + 2x 5
Using the method of eliminating unknowns we find:
We obtained relations expressing the dependent variables x 1 , x 2 , x 3 through the free ones x 4 , x 5 , that is, we found common decision:
x 3 = 0
x 2 = 1 - 3x 4 + 6x 5
x 1 = - 1 + 3x 4 - 8x 5
uncertain, because has more than one solution.
Exercise. Solve the system of equations.
Answer:x 2 = 2 - 1.67x 3 + 0.67x 4
x 1 = 5 - 3.67x 3 + 0.67x 4
By assigning any values to the free unknowns, we obtain any number of particular solutions. The system is uncertain
SYSTEMS OF LINEAR EQUATIONS
I. Statement of the problem.
II. Compatibility of homogeneous and heterogeneous systems.
III. System T equations with T unknown. Cramer's rule.
IV. Matrix method for solving systems of equations.
V. Gauss method.
I. Statement of the problem.
A system of equations of the form
called a system m linear equations with n unknown 
. The coefficients of the equations of this system are written in the form of a matrix
which is called matrix of the system (1).
The numbers on the right sides of the equations form free members column {B}:
.
If column ( B}={0 ), then the system of equations is called homogeneous. Otherwise, when ( B}≠{0 ) - system heterogeneous.
System of linear equations (1) can be written in matrix form
[A]{x}={B}. (2)
Here 
- column of unknowns.
Solving the system of equations (1) means finding the set n
numbers 
such that when substituting into system (1) instead of the unknowns 
Each equation of the system turns into an identity. Numbers 
are called the solution of a system of equations.
A system of linear equations can have one solution
,
can have countless solutions
or have no solutions at all
.
Systems of equations that have no solutions are called incompatible. If a system of equations has at least one solution, then it is called joint. The system of equations is called certain, if it has a unique solution, and uncertain, if it has infinitely many solutions.
II. Compatibility of homogeneous and heterogeneous systems.
The compatibility condition for the system of linear equations (1) is formulated in Kronecker-Capelli theorem: a system of linear equations has at least one solution if and only if the rank of the system matrix is equal to the rank of the extended matrix: 
.
An extended system matrix is a matrix obtained from the system matrix by adding a column of free terms to it on the right:
.
If Rg A
Homogeneous systems of linear equations, in accordance with the Kronecker-Capelli theorem, are always consistent. Let us consider the case of a homogeneous system in which the number of equations is equal to the number of unknowns, that is t=p. If the determinant of the matrix of such a system is not equal to zero, i.e. 
, a homogeneous system has a unique solution, which is trivial (zero). Homogeneous systems have an infinite number of solutions if among the equations of the system there are linearly dependent ones, i.e. 
.
Example. Consider a homogeneous system of three linear equations with three unknowns:
and examine the question of the number of its solutions. Each of the equations can be considered an equation of a plane passing through the origin of coordinates ( D=0 ). The system of equations has a unique solution when all three planes intersect at one point. Moreover, their normal vectors are non-coplanar, and, therefore, the condition is satisfied
.
The solution of the system in this case x=0, y=0, z=0 .
If at least two of the three planes, for example, the first and second, are parallel, i.e. , then the determinant of the system matrix is equal to zero, and the system has an infinite number of solutions. Moreover, the solutions will be the coordinates x, y, z all points lying on a line
If all three planes coincide, then the system of equations will be reduced to one equation
,
and the solution will be the coordinates of all points lying in this plane.
When studying inhomogeneous systems of linear equations, the question of compatibility is solved using the Kronecker-Capelli theorem. If the number of equations in such a system is equal to the number of unknowns, then the system has a unique solution if its determinant is not equal to zero. Otherwise, the system is either inconsistent or has an infinite number of solutions.
Example. We study an inhomogeneous system of two equations with two unknowns
.
The equations of the system can be considered as equations of two lines on a plane. The system is inconsistent when the lines are parallel, i.e.
,
. In this case, the rank of the system matrix is 1:
Rg A=1
, because 
,
and the rank of the extended matrix 
is equal to two, since for it a second-order minor containing the third column can be chosen as a basis minor.
In the case under consideration, Rg A
If the lines coincide, i.e. , then the system of equations has an infinite number of solutions: coordinates of points on a straight line 
. In this case, Rg A=
Rg A *
=1.
The system has a unique solution when the lines are not parallel, i.e. 
. The solution to this system is the coordinates of the point of intersection of the lines 
III. SystemT equations withT unknown. Cramer's rule.
Let us consider the simplest case when the number of equations of the system is equal to the number of unknowns, i.e. m= n. If the determinant of the system matrix is nonzero, the solution to the system can be found using Cramer's rule:
(3)
Here 
- determinant of the system matrix,
is the determinant of the matrix obtained from [ A] replacement i th column to the column of free members:
.
Example. Solve the system of equations using Cramer's method.
Solution :
1) find the determinant of the system
2) find auxiliary determinants
3) find a solution to the system using Cramer’s rule:
The result of the solution can be checked by substituting into the system of equations
The correct identities are obtained.
IV. Matrix method for solving systems of equations.
Let us write the system of linear equations in matrix form (2)
[A]{x}={B}
and multiply the right and left sides of relation (2) on the left by the matrix [ A -1 ], inverse of the system matrix:
[A -1 ][A]{x}=[A -1 ]{B}. (2)
By definition of the inverse matrix, the product [ A -1 ][A]=[E], and according to the properties of the identity matrix [ E]{x}={x). Then from relation (2") we obtain
{x}=[A -1 ]{B}. (4)
Relation (4) underlies the matrix method for solving systems of linear equations: it is necessary to find the matrix inverse to the matrix of the system and multiply the column vector of the right parts of the system by it on the left.
Example. Let us solve the system of equations considered in the previous example using the matrix method.
System Matrix 
its determinant det A==183
.
Right side column
.
To find the matrix [ A -1 ], find the matrix attached to [ A]:
or 
The formula for calculating the inverse matrix includes 
, Then
Now we can find a solution to the system
Then we finally get 
.
V. Gauss method.
With a large number of unknowns, solving a system of equations using the Cramer method or the matrix method involves calculating high-order determinants or inverting large matrices. These procedures are very labor-intensive even for modern computers. Therefore, to solve systems of a large number of equations, the Gauss method is often used.
The Gaussian method consists of sequentially eliminating unknowns through elementary transformations of the extended matrix of the system. Elementary matrix transformations include permutation of rows, addition of rows, multiplication of rows by numbers other than zero. As a result of the transformations, it is possible to reduce the matrix of the system to an upper triangular one, on the main diagonal of which there are ones, and below the main diagonal there are zeros. This is the direct approach of the Gaussian method. The reverse of the method consists in directly determining the unknowns, starting from the last one.
Let us illustrate the Gauss method using the example of solving a system of equations
At the first step of the forward stroke, it is ensured that the coefficient 
transformed system became equal 1
, and the coefficients 
And 
turned to zero. To do this, multiply the first equation by 1/10
, multiply the second equation by 10
and add it to the first one, multiply the third equation by -10/2
and add it to the first one. After these transformations we get
At the second step, we ensure that after transformations the coefficient 
became equal 1
, and the coefficient 
. To do this, divide the second equation by 42
, and multiply the third equation by -42/27
and add it with the second one. We obtain a system of equations
At the third step we should get the coefficient 
. To do this, divide the third equation by (37 - 84/27)
; we get
This is where the direct progression of the Gauss method ends, because the matrix of the system is reduced to the upper triangular one:
Carrying out the reverse move, we find the unknowns
The Gaussian method has a number of disadvantages: it is impossible to know whether the system is consistent or not until all the transformations necessary in the Gaussian method have been carried out; Gauss's method is not suitable for systems with letter coefficients.
Let's consider other methods for solving systems of linear equations. These methods use the concept of matrix rank and reduce the solution of any consistent system to the solution of a system to which Cramer's rule applies.
Example 1. Find a general solution to the following system of linear equations using the fundamental system of solutions given homogeneous system and a particular solution to a heterogeneous system.
1. Making a matrix A and extended system matrix (1)
2. Explore the system (1) for togetherness. To do this, we find the ranks of the matrices A and https://pandia.ru/text/78/176/images/image006_90.gif" width="17" height="26 src=">). If it turns out that , then the system (1) incompatible. If we get that , then this system is consistent and we will solve it. (The compatibility study is based on the Kronecker-Capelli theorem).
a. We find rA.
To find rA, we will consider sequentially non-zero minors of the first, second, etc. orders of the matrix A and the minors surrounding them.
M1=1≠0 (we take 1 from the upper left corner of the matrix A).
We border M1 the second row and second column of this matrix. 
. We continue to border M1 the second line and the third column..gif" width="37" height="20 src=">. Now we border the non-zero minor M2′ second order.
We have: 
(since the first two columns are the same)
(since the second and third lines are proportional).
We see that rA=2, a is the basis minor of the matrix A.
b. We find.
Fairly basic minor M2′ matrices A border with a column of free terms and all rows (we have only the last row).
. It follows that M3′′ remains the basic minor of the matrix https://pandia.ru/text/78/176/images/image019_33.gif" width="168 height=75" height="75"> (2)
Because M2′- basis minor of the matrix A systems (2) , then this system is equivalent to the system (3) , consisting of the first two equations of the system (2) (for M2′ is in the first two rows of matrix A).
(3)
Since the basic minor https://pandia.ru/text/78/176/images/image021_29.gif" width="153" height="51"> (4)
In this system there are two free unknowns ( x2 And x4 ). That's why FSR systems (4) consists of two solutions. To find them, we assign free unknowns in (4) values first x2=1 , x4=0 , and then - x2=0 , x4=1 .
At x2=1 , x4=0 we get:
.
This system already has the only thing solution (it can be found using Cramer's rule or any other method). Subtracting the first from the second equation, we get:
Her solution will be x1= -1 , x3=0 . Given the values x2 And x4 , which we gave, we get the first fundamental solution systems (2) : .
Now we believe in (4) x2=0 , x4=1 . We get:
.
We solve this system using Cramer’s theorem:
.
We obtain the second fundamental solution of the system (2) : .
Solutions β1 , β2 and make up FSR systems (2) . Then its general solution will be
γ= C1 β1+С2β2=С1(‑1, 1, 0, 0)+С2(5, 0, 4, 1)=(‑С1+5С2, С1, 4С2, С2)
Here C1 , C2 – arbitrary constants.
4. Let's find one private solution heterogeneous system(1) . As in paragraph 3 , instead of the system (1) Let's consider an equivalent system (5) , consisting of the first two equations of the system (1) .
(5)
Let us move the free unknowns to the right sides x2 And x4.
(6)
Let's give free unknowns x2 And x4 arbitrary values, for example, x2=2 , x4=1 and put them in (6) . Let's get the system
This system has a unique solution (since its determinant M2′0). Solving it (using Cramer’s theorem or Gauss’s method), we obtain x1=3 , x3=3 . Given the values of the free unknowns x2 And x4 , we get particular solution of an inhomogeneous system(1)α1=(3,2,3,1).
5. Now all that remains is to write it down general solution α of an inhomogeneous system(1) : it is equal to the sum private solution this system and general solution of its reduced homogeneous system (2) :
α=α1+γ=(3, 2, 3, 1)+(‑С1+5С2, С1, 4С2, С2).
This means: 
(7)
6. Examination. To check if you solved the system correctly (1) , we need a general solution (7) substitute in (1) . If each equation turns into the identity ( C1 And C2 must be destroyed), then the solution is found correctly.
We'll substitute (7) for example, only the last equation of the system (1) (x1 + x2 + x3 ‑9 x4 =‑1) .
We get: (3–С1+5С2)+(2+С1)+(3+4С2)–9(1+С2)=–1
(С1–С1)+(5С2+4С2–9С2)+(3+2+3–9)=–1
Where –1=–1. We got an identity. We do this with all the other equations of the system (1) .
Comment. The check is usually quite cumbersome. The following “partial check” can be recommended: in the general solution of the system (1) assign some values to arbitrary constants and substitute the resulting partial solution only into the discarded equations (i.e., into those equations from (1) , which were not included in (5) ). If you get identities, then more likely, system solution (1) found correctly (but such a check does not provide a complete guarantee of correctness!). For example, if in (7) put C2=- 1 , C1=1, then we get: x1=-3, x2=3, x3=-1, x4=0. Substituting into the last equation of system (1), we have: - 3+3 - 1 - 9∙0= - 1 , i.e. –1=–1. We got an identity.
Example 2. Find a general solution to a system of linear equations (1) , expressing the basic unknowns in terms of free ones.
Solution. As in example 1, compose matrices A and https://pandia.ru/text/78/176/images/image010_57.gif" width="156" height="50"> of these matrices. Now we leave only those equations of the system (1) , the coefficients of which are included in this basic minor (i.e., we have the first two equations) and consider a system consisting of them, equivalent to system (1).
Let us transfer the free unknowns to the right-hand sides of these equations.
system (9) We solve by the Gaussian method, considering the right-hand sides as free terms.
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Option 2.
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Option 4.
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Option 5.
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Option 6.
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Let us first consider the case when the number of equations is equal to the number of variables, i.e. m = n. Then the matrix of the system is square, and its determinant is called the determinant of the system.
Inverse matrix method
Let us consider in general form the system of equations AX = B with non-degenerate square matrix A. In this case there is inverse matrix A -1. Let's multiply both sides by A -1 on the left. We get A -1 AX = A -1 B. Hence EX = A -1 B and
The last equality is a matrix formula for finding solutions to such systems of equations. The use of this formula is called the inverse matrix method
For example, let’s use this method to solve the following system:
;
At the end of solving the system, you can check by substituting the found values into the system equations. In doing so, they must turn into true equalities.
For the example considered, let's check:
Method for solving systems of linear equations with a square matrix using Cramer's formulas
Let n= 2:
If we multiply both sides of the first equation by a 22, and both sides of the second by (-a 12), and then add the resulting equations, then we eliminate the variable x 2 from the system. Similarly, you can eliminate the variable x 1 (by multiplying both sides of the first equation by (-a 21), and both sides of the second by a 11). As a result, we get the system:
The expression in brackets is the determinant of the system
Let's denote
Then the system will take the form:
From the resulting system it follows that if the determinant of the system is 0, then the system will be consistent and definite. Its only solution can be calculated using the formulas:
If = 0, a 1 0 and/or 2 0, then the system equations will take the form 0*x 1 = 2 and/or 0*x 1 = 2. In this case, the system will be inconsistent.
In the case when = 1 = 2 = 0, the system will be consistent and indefinite (will have an infinite number of solutions), since it will take the form:
Cramer's theorem(we will omit the proof). If the determinant of the matrix of a system of equations is not equal to zero, then the system has a unique solution, determined by the formulas:
,
where j is the determinant of the matrix obtained from matrix A by replacing the j-th column with a column of free terms.
The above formulas are called Cramer formulas.
As an example, let’s use this method to solve a system that was previously solved using the inverse matrix method:
Disadvantages of the considered methods:
1) significant labor intensity (calculating determinants and finding the inverse matrix);
2) limited scope (for systems with a square matrix).
Real economic situations are often modeled by systems in which the number of equations and variables is quite significant, and there are more equations than variables. Therefore, in practice, the following method is more common.
Gaussian method (method of sequential elimination of variables)
This method is used to solve a system of m linear equations with n variables in general view. Its essence lies in applying a system of equivalent transformations to the extended matrix, with the help of which the system of equations is transformed to a form where its solutions become easy to find (if any).
This is the kind in which the left top part The matrix of the system will be a stepped matrix. This is achieved using the same techniques that were used to obtain a step matrix to determine the rank. In this case, elementary transformations are applied to the extended matrix, which will allow one to obtain an equivalent system of equations. After this, the expanded matrix will take the form:
Obtaining such a matrix is called straight ahead Gauss method.
Finding the values of variables from the corresponding system of equations is called in reverse Gauss method. Let's consider it.
Note that the last (m – r) equations will take the form:
If at least one of the numbers 
is not equal to zero, then the corresponding equality will be false, and the entire system will be inconsistent.
Therefore, for any joint system 
. In this case, the last (m – r) equations for any values of the variables will be identities 0 = 0, and they can be ignored when solving the system (simply discard the corresponding rows).
After this, the system will look like:
Let us first consider the case when r=n. Then the system will take the form:
From the last equation of the system, x r can be uniquely found.
Knowing x r, we can unambiguously express x r -1 from it. Then from the previous equation, knowing x r and x r -1, we can express x r -2, etc. up to x 1 .
So, in this case the system will be joint and determined.
Now consider the case when r
From this equation we can express the basic variable x r in terms of non-basic ones:
The penultimate equation will look like:
By substituting the resulting expression into it instead of x r, it will be possible to express the basic variable x r -1 in terms of non-basic ones. Etc. to variablex 1 . To obtain a solution to the system, you can equate non-basic variables to arbitrary values and then calculate the basic variables using the resulting formulas. Thus, in this case the system will be consistent and indefinite (have an infinite number of solutions).
For example, let's solve the system of equations:
We will call the set of basic variables basis systems. We will also call the set of columns of coefficients for them basis(base columns), or basic minor system matrices. The solution of the system in which all non-basic variables are equal to zero will be called basic solution.
In the previous example, the basic solution will be (4/5; -17/5; 0; 0) (the variables x 3 and x 4 (c 1 and c 2) are set to zero, and the basic variables x 1 and x 2 are calculated through them) . To give an example of a non-basic solution, we need to equate x 3 and x 4 (c 1 and c 2) to arbitrary numbers that are not simultaneously zero, and calculate the remaining variables through them. For example, with 1 = 1 and 2 = 0, we obtain a non-basic solution - (4/5; -12/5; 1; 0). By substitution it is easy to verify that both solutions are correct.
It is obvious that in an indefinite system there can be an infinite number of non-basic solutions. How many basic solutions can there be? Each row of the transformed matrix must correspond to one basis variable. There are n variables in the problem, and r base lines. Therefore, the number of all possible sets of basic variables cannot exceed the number of combinations of n by 2. It may be less than 
, because it is not always possible to transform the system to such a form that this particular set of variables is the basis.
What kind is this? This is the type when the matrix formed from columns of coefficients for these variables will be stepped, and at the same time will consist of r rows. Those. the rank of the coefficient matrix for these variables must be equal to r. It cannot be greater, since the number of columns is equal. If it turns out to be less than r, then this indicates a linear dependence of the columns on the variables. Such columns cannot form a basis.
Let's consider what other basic solutions can be found in the example discussed above. To do this, consider all possible combinations of four variables, two basic ones each. There will be such combinations 
, and one of them (x 1 and x 2) has already been considered.
Let's take the variables x 1 and x 3. Let us find the rank of the matrix of coefficients for them:
Since it is equal to two, they can be basic. Let us equate the non-basic variables x 2 and x 4 to zero: x 2 = x 4 = 0. Then from the formula x 1 = 4/5 – (1/5)*x 4 it follows that x 1 = 4/5, and from the formula x 2 = -17/5 + x 3 - - (7/5)*x 4 = -17/5 + x 3 it follows that x 3 = x 2 +17/5 = 17/5. Thus, we get the basic solution (4/5; 0; 17/5; 0).
Similarly, you can obtain basic solutions for the basic variables x 1 and x 4 – (9/7; 0; 0; -17/7); x 2 and x 4 – (0; -9; 0; 4); x 3 and x 4 – (0; 0; 9; 4).
The variables x 2 and x 3 in this example cannot be taken as basic ones, since the rank of the corresponding matrix is equal to one, i.e. less than two:
.
Another approach to determining whether or not it is possible to construct a basis from certain variables is also possible. When solving the example, as a result of converting the system matrix to a stepwise form, it took the form:
By selecting pairs of variables, it was possible to calculate the corresponding minors of this matrix. It is easy to verify that for all pairs except x 2 and x 3 they are not equal to zero, i.e. the columns are linearly independent. And only for columns with variables x 2 and x 3 
, which indicates their linear dependence.
Let's look at another example. Let's solve the system of equations
So, the equation corresponding to the third row of the last matrix is contradictory - it resulted in the incorrect equality 0 = -1, therefore, this system is inconsistent.
Jordan-Gauss method 3 is a development of the Gaussian method. Its essence is that the extended matrix of the system is transformed to a form where the coefficients of the variables form an identity matrix up to permutation of rows or columns 4 (where r is the rank of the system matrix).
Let's solve the system using this method:
Let's consider the extended matrix of the system:
In this matrix we select a unit element. For example, the coefficient for x 2 in the third constraint is 5. Let's ensure that the remaining rows in this column contain zeros, i.e. Let's make the column single. During the transformation process we will call this columnpermissive(leading, key). The third limitation (third line) we will also call permissive. Myself element, which stands at the intersection of the resolving row and column (here it is one), is also called permissive.
The first line now contains the coefficient (-1). To get a zero in its place, multiply the third line by (-1) and subtract the result from the first line (i.e. simply add the first line to the third).
The second line contains the coefficient 2. To get zero in its place, multiply the third line by 2 and subtract the result from the first line.
The result of the transformation will look like:
From this matrix it is clearly visible that one of the first two restrictions can be crossed out (the corresponding rows are proportional, i.e. these equations follow from each other). Let's cross out, for example, the second:
So, the new system has two equations. A single column (second) is obtained, and the unit here appears in the second row. Let us remember that the second equation of the new system will correspond to the basic variable x 2.
Let's choose a base variable for the first row. This can be any variable except x 3 (because for x 3 the first constraint has a zero coefficient, i.e. the set of variables x 2 and x 3 cannot be basic here). You can take the first or fourth variable.
Let's choose x 1. Then the resolving element will be 5, and both sides of the resolving equation will have to be divided by five to get one in the first column of the first row.
Let's ensure that the remaining rows (i.e., the second row) have zeros in the first column. Since now the second line contains not zero, but 3, we need to subtract from the second line the elements of the transformed first line, multiplied by 3:
From the resulting matrix, one can directly extract one basic solution by equating non-basic variables to zero, and basic ones to the free terms in the corresponding equations: (0.8; -3.4; 0; 0). You can also derive general formulas expressing basic variables through non-basic ones: x 1 = 0.8 – 1.2 x 4; x 2 = -3.4 + x 3 + 1.6x 4. These formulas describe the entire infinite set of solutions to the system (equating x 3 and x 4 to arbitrary numbers, you can calculate x 1 and x 2).
Note that the essence of the transformations at each stage of the Jordan-Gauss method was as follows:
1) the resolution line was divided by the resolution element to obtain a unit in its place,
2) from all other rows, the transformed resolving element was subtracted, multiplied by the element that was in the given line in the resolving column, to obtain a zero in place of this element.
Let us consider again the transformed extended matrix of the system:
From this record it is clear that the rank of the matrix of system A is equal to r.
In the course of our reasoning, we established that the system will be cooperative if and only if 
. This means that the extended matrix of the system will look like:
By discarding zero rows, we obtain that the rank of the extended matrix of the system is also equal to r.
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 extended matrix of this system.
Recall that the rank of a matrix is equal to the maximum number of its linearly independent rows. It follows from this that if the rank of the extended matrix is less than the number of equations, then the equations of the system are linearly dependent, and one or more of them can be excluded from the system (since they are a linear combination of the others). A system of equations will be linearly independent only if the rank of the extended matrix is equal to the number of equations.
Moreover, for simultaneous systems of linear equations, it can be argued that if the rank of the matrix is equal to the number of variables, then the system has a unique solution, and if it is less than the number of variables, then the system is indefinite and has infinitely many solutions.
1For example, let there be five rows in the matrix (the original row order is 12345). We need to change the second line and the fifth. In order for the second line to take the place of the fifth and “move” down, we successively change the adjacent lines three times: the second and third (13245), the second and fourth (13425) and the second and fifth (13452). Then, in order for the fifth row to take the place of the second in the original matrix, it is necessary to “shift” the fifth row upward by only two consecutive changes: the fifth and fourth rows (13542) and the fifth and third (15342).
2Number of combinations from n to r 
they call the number of all different r-element subsets of an n-element set (those that have different compositions of elements are considered different sets; the order of selection is not important). It is calculated using the formula: 
. Let us recall the meaning of the sign “!” (factorial): 
0!=1.)
3 Since this method is more common than the previously discussed Gaussian method, and is essentially a combination of the forward and backward steps of the Gaussian method, it is also sometimes called the Gaussian method, omitting the first part of the name.
4For example, 
.
5If there were no units in the system matrix, then it would be possible, for example, to divide both sides of the first equation by two, and then the first coefficient would become unity; or the like