Given matrices
Find: 1) aA - bB,
Solution: 1) Find it sequentially, using the rules for multiplying a matrix by a number and adding matrices ..
2. Find A * B if
Solution: Using the matrix multiplication rule
Answer: 
3. For a given matrix, find the minor M 31 and calculate the determinant.
Solution: Minor M 31 is the determinant of the matrix, which is obtained from A
after deleting row 3 and column 1. Find
1*10*3+4*4*4+1*1*2-2*4*10-1*1*4-1*4*3 = 0.
We transform the matrix A without changing its determinant (let's make zeros in row 1)
| -3*, -, -4* | |||
| -10 | -15 | ||
| -20 | -25 | ||
| -4 | -5 |
Now we calculate the determinant of the matrix A by decomposition in row 1
Answer: М 31 = 0, detA = 0
Solve by the Gauss method and the Cramer method.
2x 1 + x 2 + x 3 = 2
x 1 + x 2 + 3x 3 = 6
2x 1 + x 2 + 2x 3 = 5
Solution: Check
Cramer's method can be applied
System solution: x 1 = D 1 / D = 2, x 2 = D 2 / D = -5, x 3 = D 3 / D = 3
Let's apply the Gauss method.
The extended matrix of the system is reduced to a triangular form.
For the convenience of calculations, let's swap the lines:
Multiply the 2nd row by (k = -1 / 2 = -1 / 2 ) and add to the 3rd:
| 1 / 2 | 7 / 2 |
Multiply the 1st row by (k = -2 / 2 = -1 ) and add to the 2nd:
The original system can now be written as:
x 1 = 1 - (1/2 x 2 + 1/2 x 3)
x 2 = 13 - (6x 3)
From the 2nd line we express
From the 1st line we express
The solution is the same.
Answer: (2; -5; 3)
Find a general solution to the system and the SDF
13x 1 - 4x 2 - x 3 - 4x 4 - 6x 5 = 0
11x 1 - 2x 2 + x 3 - 2x 4 - 3x 5 = 0
5x 1 + 4x 2 + 7x 3 + 4x 4 + 6x 5 = 0
7x 1 + 2x 2 + 5x 3 + 2x 4 + 3x 5 = 0
Solution: Let's apply the Gaussian method. The extended matrix of the system is reduced to a triangular form.
| -4 | -1 | -4 | -6 | |
| -2 | -2 | -3 | ||
| x 1 | x 2 | x 3 | x 4 | x 5 |
Multiply the 1st row by (-11). Multiply the 2nd row by (13). Let's add the 2nd line to the 1st:
| -2 | -2 | -3 | ||
Multiply the 2nd row by (-5). Multiply the 3rd row by (11). Let's add the 3rd line to the 2nd:
Multiply the 3rd row by (-7). Multiply the 4th row by (5). Add the 4th line to the 3rd:
The second equation is a linear combination of the rest
Let's find the rank of the matrix.
| -18 | -24 | -18 | -27 | |
| x 1 | x 2 | x 3 | x 4 | x 5 |
The highlighted minor has the highest order (of the possible minors) and is nonzero (it is equal to the product of the elements on the opposite diagonal), therefore, rang (A) = 2.
This minor is basic. It includes the coefficients for the unknowns x 1, x 2, which means that the unknowns x 1, x 2 are dependent (basic), and x 3, x 4, x 5 are free.
The system with the coefficients of this matrix is equivalent to the original system and has the form:
18x 2 = 24x 3 + 18x 4 + 27x 5
7x 1 + 2x 2 = - 5x 3 - 2x 4 - 3x 5
By eliminating unknowns, we find common decision:
x 2 = - 4/3 x 3 - x 4 - 3/2 x 5
x 1 = - 1/3 x 3
We find the fundamental decision system (FDS), which consists of (n-r) solutions. In our case, n = 5, r = 2, therefore, the fundamental system of solutions consists of 3 solutions, and these solutions must be linearly independent.
For the rows to be linearly independent, it is necessary and sufficient that the rank of the matrix composed of the elements of the rows be equal to the number of rows, that is, 3.
It is enough to give the free unknowns x 3, x 4, x 5 values from the rows of the determinant of the 3rd order, other than zero, and calculate x 1, x 2.
The simplest nonzero determinant is the identity matrix.
But here it is more convenient to take
We find using the general solution:
a) x 3 = 6, x 4 = 0, x 5 = 0 Þ x 1 = - 1/3 x 3 = -2, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = - 4 Þ
I solution of the FSR: (-2; -4; 6; 0; 0)
b) x 3 = 0, x 4 = 6, x 5 = 0 Þ x 1 = - 1/3 x 3 = 0, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = - 6 Þ
II solution of the SDF: (0; -6; 0; 6; 0)
c) x 3 = 0, x 4 = 0, x 5 = 6 Þ x 1 = - 1/3 x 3 = 0, x 2 = - 4/3 x 3 - x 4 - 3/2 x 5 = -9 Þ
III solution of the SDF: (0; - 9; 0; 0; 6)
Þ FSR: (-2; -4; 6; 0; 0), (0; -6; 0; 6; 0), (0; - 9; 0; 0; 6)
6. Given: z 1 = -4 + 5i, z 2 = 2 - 4i. Find: a) z 1 - 2z 2 b) z 1 z 2 c) z 1 / z 2
Solution: a) z 1 - 2z 2 = -4 + 5i + 2 (2-4i) = -4 + 5i + 4-8i = -3i
b) z 1 z 2 = (-4 + 5i) (2-4i) = -8 + 10i + 16i-20i 2 = (i 2 = -1) = 12 + 26i
Answer: a) -3i b) 12 + 26i c) -1.4 - 0.3i
Service purpose... The online calculator is designed to find a non-trivial and fundamental solution to a SLAE. The resulting solution is saved in a Word file (see example solution).
Instruction. Select the dimension of the matrix:
Properties of systems of linear homogeneous equations
In order for the system to have non-trivial solutions, it is necessary and sufficient that the rank of its matrix be less than the number of unknowns.Theorem... The system in the case m = n has a nontrivial solution if and only if the determinant of this system is equal to zero.
Theorem... Any linear combination of system solutions is also a solution to this system.
Definition... The set of solutions to a system of linear homogeneous equations is called fundamental decision system if this set consists of linearly independent solutions and any solution to the system is a linear combination of these solutions.
Theorem. If the rank r of the matrix of the system is less than the number n of unknowns, then there is a fundamental system of solutions consisting of (n-r) solutions.
Algorithm for solving systems of linear homogeneous equations
- Find the rank of the matrix.
- Highlight the base minor. We select the dependent (basic) and free unknowns.
- We delete those equations of the system whose coefficients are not included in the basic minor, since they are consequences of the others (by the theorem on the basic minor).
- The terms of the equations containing free unknowns are transferred to the right-hand side. As a result, we obtain a system of r equations with r unknowns, equivalent to the given one, the determinant of which is nonzero.
- We solve the resulting system by eliminating unknowns. We find the relations expressing the dependent variables in terms of free ones.
- If the rank of the matrix is not equal to the number of variables, then we find the fundamental solution to the system.
- In the case of rang = n, we have a trivial solution.
An example. Find the basis of the system of vectors (a 1, a 2, ..., a m), rank and express the vectors in terms of the base. If a 1 = (0,0,1, -1), a 2 = (1,1,2,0), a 3 = (1,1,1,1), and 4 = (3,2,1 , 4), and 5 = (2,1,0,3).
Let's write out the main matrix of the system:
Multiply the 3rd row by (-3). Add the 4th line to the 3rd:
| 0 | 0 | 1 | -1 |
| 0 | 0 | -1 | 1 |
| 0 | -1 | -2 | 1 |
| 3 | 2 | 1 | 4 |
| 2 | 1 | 0 | 3 |
Multiply the 4th row by (-2). Multiply the 5th row by (3). Let's add the 5th line to the 4th:
Let's add the 2nd line to the 1st:
Let's find the rank of the matrix.
The system with the coefficients of this matrix is equivalent to the original system and has the form:
- x 3 = - x 4
- x 2 - 2x 3 = - x 4
2x 1 + x 2 = - 3x 4
Using the method of eliminating unknowns, we find a nontrivial solution:
We got the ratios expressing the dependent variables x 1, x 2, x 3 through free x 4, that is, we found a general solution:
x 3 = x 4
x 2 = - x 4
x 1 = - x 4
Homogeneous systems of linear algebraic equations
Within the lessons Gauss method and Incompatible systems / systems with a common solution we considered inhomogeneous systems of linear equations, where free member(which is usually on the right) at least one of the equations was nonzero.
And now, after a good warm-up with the rank of the matrix, we will continue to polish the technique elementary transformations on the homogeneous system of linear equations.
In the first paragraphs, the material may seem boring and ordinary, but this impression is deceiving. In addition to further developing the techniques, there will be a lot of new information, so please try not to neglect the examples in this article.
What is a homogeneous system of linear equations?
The answer suggests itself. The system of linear equations is homogeneous if the free term of each equations of the system is equal to zero. For example:
It is quite clear that a homogeneous system is always compatible, that is, it always has a solution. And, above all, the so-called trivial solution 
... Trivial, for those who do not understand the meaning of the adjective at all, means bespontov. Not academic, of course, but intelligible =) ... Why beat around the bush, let's find out if this system has any other solutions:
Example 1
Solution: to solve a homogeneous system, it is necessary to write system matrix and with the help of elementary transformations bring it to a stepwise form. Please note that there is no need to write the vertical bar and the zero column of free members here - after all, whatever you do with zeros, they will remain zeros: 
(1) The first line multiplied by –2 was added to the second line. The first line multiplied by –3 was added to the third line.
(2) The second line multiplied by -1 was added to the third line.
Dividing the third row by 3 doesn't make much sense.
As a result of elementary transformations, an equivalent homogeneous system was obtained 
, and, applying the reverse course of the Gauss method, it is easy to verify that the solution is unique.
Answer:
Let us formulate an obvious criterion: the homogeneous system of linear equations has only trivial solution, if system matrix rank(in this case 3) is equal to the number of variables (in this case - 3 pcs.).
We warm up and tune our radio receiver to the wave of elementary transformations:
Example 2
Solve a homogeneous system of linear equations 
From article How do I find the rank of a matrix? remember the rational method of simultaneously decreasing the matrix numbers. Otherwise, you will have to cut up large and often biting fish. An approximate sample of the assignment at the end of the lesson.
Zeros are good and convenient, but in practice the case is much more common when the rows of the matrix of the system linearly dependent... And then the emergence of a common solution is inevitable:
Example 3
Solve a homogeneous system of linear equations 
Solution: we write down the matrix of the system and, using elementary transformations, bring it to a stepwise form. The first action is aimed not only at obtaining a single value, but also at decreasing the numbers in the first column:
(1) The third line multiplied by -1 was added to the first line. The second line was added to the third line multiplied by –2. At the top left, I got a unit with a "minus", which is often much more convenient for further transformations.
(2) The first two lines are the same, one of them has been deleted. Honestly, I did not rush the solution - it just happened. If you perform the transformations in a template, then linear relationship lines would show up a little later.
(3) The second row, multiplied by 3, was added to the third line.
(4) The sign of the first line was changed.
As a result of elementary transformations, an equivalent system was obtained: 
The algorithm works exactly the same as for heterogeneous systems... The variables "sitting on the steps" are the main ones, the variable that did not get the "steps" is free.
Let's express the basic variables in terms of a free variable: 
Answer: common decision: 
A trivial solution is included in the general formula, and it is unnecessary to write it down separately.
The check is also performed according to the usual scheme: the obtained general solution must be substituted into the left side of each equation of the system and a legal zero is obtained for all substitutions.
On this one could quietly and peacefully finish, but the solution of a homogeneous system of equations is often required to be represented in vector form through fundamental decision system... Please temporarily forget about analytic geometry, since now we will talk about vectors in the general algebraic sense, which I opened a little in an article about matrix rank... There is no need to obscure the terminology, everything is quite simple.
The solution of systems of linear algebraic equations (SLAE) is undoubtedly the most important topic of the linear algebra course. A huge number of problems from all branches of mathematics is reduced to solving systems of linear equations. These factors explain the reason for creating this article. The material of the article is selected and structured so that with its help you can
- choose the optimal method for solving your system of linear algebraic equations,
- study the theory of the chosen method,
- solve your system of linear equations by considering in detail the analyzed solutions of typical examples and problems.
Brief description of the article material.
First, we give all the necessary definitions and concepts and introduce the notation.
Next, we will consider methods for solving systems of linear algebraic equations in which the number of equations is equal to the number of unknown variables and which have a unique solution. First, we will dwell on Cramer's method, secondly, we will show a matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of successive elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in different ways.
After that, we turn to solving systems of linear algebraic equations of general form, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is degenerate. Let us formulate the Kronecker - Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (in the case of their compatibility) using the concept of a basic minor of a matrix. We will also consider the Gauss method and describe in detail the solutions of examples.
We will definitely dwell on the structure of the general solution of homogeneous and inhomogeneous systems of linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how the general solution of an SLAE is written using vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.
In conclusion, we consider systems of equations that reduce to linear ones, as well as various problems, in the solution of which SLAEs arise.
Page navigation.
Definitions, concepts, designations.
We will consider systems of p linear algebraic equations with n unknown variables (p can be equal to n) of the form
Unknown variables, - coefficients (some real or complex numbers), - free terms (also real or complex numbers).
This form of SLAE notation is called coordinate.
IN matrix form notation, this system of equations has the form,
where 
- the main matrix of the system, - the matrix-column of unknown variables, - the matrix-column of free members.
If we add to the matrix A as the (n + 1) th column the matrix-column of free terms, then we get the so-called expanded matrix systems of linear equations. Usually, the expanded matrix is denoted by the letter T, and the column of free members is separated by a vertical line from the rest of the columns, that is, 
By solving a system of linear algebraic equations is a set of values of unknown variables that converts all equations of the system into identities. The matrix equation for the given values of the unknown variables also turns into an identity.
If a system of equations has at least one solution, then it is called joint.
If the system of equations has no solutions, then it is called inconsistent.
If the SLAE has a unique solution, then it is called certain; if there is more than one solution, then - undefined.
If the free terms of all equations of the system are equal to zero 
, then the system is called homogeneous, otherwise - heterogeneous.
Solution of elementary systems of linear algebraic equations.
If the number of equations of the system is equal to the number of unknown variables and the determinant of its basic matrix is not equal to zero, then such SLAEs will be called elementary... Such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.
We began to study such SLAEs in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then we took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the method of addition, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are, in fact, modifications of the Gauss method.
The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's analyze them.
Solving systems of linear equations by Cramer's method.
Suppose we need to solve a system of linear algebraic equations 
in which the number of equations is equal to the number of unknown variables and the determinant of the main matrix of the system is nonzero, that is,.
Let be the determinant of the main matrix of the system, and 
- determinants of matrices, which are obtained from A by replacing 1st, 2nd, ..., nth column, respectively, to the column of free members: 
With this notation, the unknown variables are calculated by the formulas of Cramer's method as 
... This is how the solution of a system of linear algebraic equations is found by Cramer's method.
Example.
Cramer's method 
.
Solution.
The main matrix of the system has the form 
... Let's calculate its determinant (if necessary, see the article): 
Since the determinant of the main matrix of the system is nonzero, the system has a unique solution that can be found by Cramer's method.
Let us compose and calculate the necessary determinants 
(the determinant is obtained by replacing the first column in matrix A with a column of free members, the determinant - by replacing the second column with a column of free members, - by replacing the third column of matrix A with a column of free members): 
Find unknown variables by the formulas 
:
Answer:
The main drawback of Cramer's method (if it can be called a drawback) is the complexity of calculating determinants when the number of equations in the system is more than three.
Solving systems of linear algebraic equations by the matrix method (using the inverse matrix).
Let the system of linear algebraic equations be given in matrix form, where matrix A has dimension n by n and its determinant is nonzero.
Since, the matrix A is invertible, that is, there is an inverse matrix. If we multiply both sides of the equality by the left, then we get a formula for finding the column matrix of unknown variables. So we got the solution of a system of linear algebraic equations by the matrix method.
Example.
Solve a system of linear equations matrix method.
Solution.
Let's rewrite the system of equations in matrix form: 
As
then SLAE can be solved by the matrix method. Using the inverse matrix, the solution to this system can be found as 
.
Let's construct an inverse matrix using a matrix of algebraic complements of elements of matrix A (if necessary, see the article): 
It remains to calculate - the matrix of unknown variables by multiplying the inverse matrix 
to a column matrix of free members (see the article if necessary): 
Answer:
or in another notation x 1 = 4, x 2 = 0, x 3 = -1.
The main problem in finding a solution to systems of linear algebraic equations by the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than the third.
Solution of systems of linear equations by the Gauss method.
Suppose we need to find a solution to a system of n linear equations with n unknown variables 
the determinant of the main matrix of which is nonzero.
The essence of the Gauss method consists in the successive elimination of unknown variables: first, x 1 is excluded from all equations of the system, starting with the second, then x 2 is excluded from all equations, starting with the third, and so on, until only the unknown variable x n remains in the last equation. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called by the direct course of the Gauss method... After completing the forward run of the Gauss method, x n is found from the last equation, using this value, x n-1 is calculated from the penultimate equation, and so on, x 1 is found from the first equation. The process of calculating unknown variables when moving from the last equation of the system to the first is called backward Gaussian method.
Let us briefly describe the algorithm for eliminating unknown variables.
We will assume that, since we can always achieve this by rearranging the equations of the system. Eliminate the unknown variable x 1 from all equations of the system, starting with the second. To do this, to the second equation of the system we add the first, multiplied by, to the third equation we add the first, multiplied by, and so on, to the n-th equation we add the first, multiplied by. The system of equations after such transformations takes the form 
where, and 
.
We would come to the same result if we expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression in all other equations. Thus, the variable x 1 is excluded from all equations, starting with the second.
Next, we act in a similar way, but only with a part of the resulting system, which is marked in the figure 
To do this, to the third equation of the system we add the second multiplied by, to the fourth equation we add the second multiplied by, and so on, to the n-th equation we add the second multiplied by. The system of equations after such transformations takes the form 
where, and 
... Thus, the variable x 2 is excluded from all equations, starting with the third.
Next, we proceed to the elimination of the unknown x 3, while we act similarly with the part of the system marked in the figure 
So we continue the direct course of the Gauss method until the system takes the form 
From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as, using the obtained value of x n, we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Example.
Solve a system of linear equations by the Gauss method.
Solution.
Eliminate the unknown variable x 1 from the second and third equations of the system. To do this, add to both parts of the second and third equations the corresponding parts of the first equation, multiplied by and by, respectively: 
Now we exclude x 2 from the third equation by adding to its left and right sides the left and right sides of the second equation, multiplied by: 
This completes the forward move of the Gauss method, we begin the reverse move.
From the last equation of the resulting system of equations, we find x 3: 
From the second equation we obtain.
From the first equation, we find the remaining unknown variable and this completes the reverse course of the Gauss method.
Answer:
X 1 = 4, x 2 = 0, x 3 = -1.
Solution of systems of linear algebraic equations of general form.
In the general case, the number of equations in the system p does not coincide with the number of unknown variables n:
Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations, the basic matrix of which is square and degenerate.
The Kronecker - Capelli theorem.
Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when the SLAE is compatible and when it is incompatible is given by the Kronecker - Capelli theorem:
for a system of p equations with n unknowns (p can be equal to n) to be consistent, it is necessary and sufficient that the rank of the main matrix of the system be equal to the rank of the extended matrix, that is, Rank (A) = Rank (T).
Let us consider by example the application of the Kronecker - Capelli theorem to determine the compatibility of a system of linear equations.
Example.
Find out if the system of linear equations has 
solutions.
Solution.
... Let's use the bordering minors method. Minor of the second order 
nonzero. Let's sort out the third-order minors bordering it: 
Since all bordering minors of the third order are equal to zero, the rank of the main matrix is equal to two.
In turn, the rank of the extended matrix 
is equal to three, since the third-order minor 
nonzero.
Thus, Rang (A), therefore, by the Kronecker - Capelli theorem, we can conclude that the original system of linear equations is inconsistent.
Answer:
The system has no solutions.
So, we have learned to establish the inconsistency of the system using the Kronecker - Capelli theorem.
But how to find a solution to a SLAE if its compatibility has been established?
To do this, we need the concept of a basic minor of a matrix and a theorem on the rank of a matrix.
The highest order minor of the matrix A, other than zero, is called basic.
It follows from the definition of a basic minor that its order is equal to the rank of the matrix. For a nonzero matrix A, there can be several basic minors; there is always one basic minor.
For example, consider the matrix 
.
All third-order minors of this matrix are equal to zero, since the elements of the third row of this matrix are the sum of the corresponding elements of the first and second rows.
The following second-order minors are basic, since they are nonzero 
Minors 
are not basic, since they are equal to zero.
Matrix rank theorem.
If the rank of a matrix of order p by n is equal to r, then all elements of the rows (and columns) of the matrix that do not form the selected basic minor are linearly expressed in terms of the corresponding elements of the rows (and columns) that form the basic minor.
What does the matrix rank theorem give us?
If, by the Kronecker - Capelli theorem, we have established the compatibility of the system, then we choose any basic minor of the basic matrix of the system (its order is r), and we exclude from the system all equations that do not form the chosen basic minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still superfluous (according to the matrix rank theorem, they are a linear combination of the remaining equations).
As a result, after discarding unnecessary equations of the system, two cases are possible.
If the number of equations r in the resulting system is equal to the number of unknown variables, then it will be definite and the only solution can be found by Cramer's method, matrix method or Gauss's method.
Example.
.
Solution.
The rank of the main matrix of the system 
is equal to two, since the second order minor 
nonzero. Extended Matrix Rank 
is also equal to two, since the only minor of the third order is equal to zero 
and the second-order minor considered above is nonzero. Based on the Kronecker - Capelli theorem, we can assert the compatibility of the original system of linear equations, since Rank (A) = Rank (T) = 2.
We take as the basic minor ... It is formed by the coefficients of the first and second equations: 
The third equation of the system does not participate in the formation of the basic minor; therefore, we exclude it from the system based on the theorem on the rank of the matrix: 
This is how we got an elementary system of linear algebraic equations. Let's solve it using Cramer's method: 
Answer:
x 1 = 1, x 2 = 2.
If the number of equations r in the obtained SLAE is less than the number of unknown variables n, then in the left-hand sides of the equations we leave the terms forming the basic minor, the rest of the terms are transferred to the right-hand sides of the equations of the system with the opposite sign.
Unknown variables (there are r of them) remaining in the left-hand sides of the equations are called the main.
Unknown variables (there are n - r pieces) that appear in the right-hand sides are called free.
Now we assume that free unknown variables can take arbitrary values, and r basic unknown variables will be expressed in terms of free unknown variables in a unique way. Their expression can be found by solving the obtained SLAE by the Cramer method, by the matrix method, or by the Gaussian method.
Let's take an example.
Example.
Solve a system of linear algebraic equations 
.
Solution.
Find the rank of the main matrix of the system 
by the method of bordering minors. We take a 1 1 = 1 as a nonzero first-order minor. Let's start looking for a nonzero second-order minor that surrounds this minor: 
This is how we found a nonzero second-order minor. Let's start looking for a third-order nonzero bordering minor: 
Thus, the rank of the main matrix is three. The rank of the extended matrix is also three, that is, the system is consistent.
We take the found nonzero third-order minor as the basic one.
For clarity, we show the elements that form the basic minor: 
We leave on the left side of the equations of the system the terms participating in the basic minor, the rest are transferred with opposite signs to the right sides: 
Let us assign arbitrary values to the free unknown variables x 2 and x 5, that is, we take 
, where are arbitrary numbers. In this case, the SLAE will take the form 
The resulting elementary system of linear algebraic equations is solved by Cramer's method: 
Consequently, .
Do not forget to indicate free unknown variables in your answer.
Answer:
Where are arbitrary numbers.
Summarize.
To solve a system of linear algebraic equations of general form, we first find out its compatibility using the Kronecker - Capelli theorem. If the rank of the main matrix is not equal to the rank of the extended matrix, then we conclude that the system is incompatible.
If the rank of the main matrix is equal to the rank of the extended matrix, then we choose the basic minor and discard the equations of the system that do not participate in the formation of the selected basic minor.
If the order of the basic minor is equal to the number of unknown variables, then the SLAE has a unique solution, which we find by any known method.
If the order of the basic minor is less than the number of unknown variables, then on the left-hand side of the equations of the system we leave the terms with the basic unknown variables, transfer the remaining terms to the right-hand sides and give arbitrary values to the free unknown variables. From the resulting system of linear equations, we find the main unknown variables by the Cramer method, the matrix method, or the Gauss method.
Gauss's method for solving systems of linear algebraic equations of general form.
The Gauss method can be used to solve systems of linear algebraic equations of any kind without first examining them for compatibility. The process of successive elimination of unknown variables makes it possible to conclude both the compatibility and the incompatibility of the SLAE, and if a solution exists, it makes it possible to find it.
From the point of view of computational work, the Gaussian method is preferable.
See its detailed description and disassembled examples in the article. Gauss method for solving systems of linear algebraic equations of general form.
Writing the general solution of homogeneous and inhomogeneous linear algebraic systems using vectors of the fundamental system of solutions.
In this section, we will focus on compatible homogeneous and inhomogeneous systems of linear algebraic equations with an infinite set of solutions.
Let us first deal with homogeneous systems.
Fundamental decision system A homogeneous system of p linear algebraic equations with n unknown variables is the set (n - r) of linearly independent solutions of this system, where r is the order of the basic minor of the basic matrix of the system.
If we denote linearly independent solutions of a homogeneous SLAE as X (1), X (2),…, X (nr) (X (1), X (2),…, X (nr) are n-by-1 column matrices) , then the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary constant coefficients С 1, С 2, ..., С (nr), that is,.
What does the term general solution of a homogeneous system of linear algebraic equations (oroslau) mean?
The meaning is simple: the formula specifies all possible solutions to the original SLAE, in other words, taking any set of values of arbitrary constants С 1, С 2, ..., С (n-r), according to the formula we get one of the solutions of the original homogeneous SLAE.
Thus, if we find a fundamental system of solutions, then we will be able to specify all solutions of this homogeneous SLAE as.
Let us show the process of constructing a fundamental system of solutions to a homogeneous SLAE.
We choose the basic minor of the original system of linear equations, exclude all other equations from the system, and transfer all terms containing free unknown variables to the right-hand sides of the equations of the system with opposite signs. Let us give the free unknown variables the values 1,0,0, ..., 0 and calculate the main unknowns by solving the obtained elementary system of linear equations in any way, for example, by Cramer's method. This will give X (1) - the first solution to the fundamental system. If we give the free unknowns the values 0,1,0,0, ..., 0 and calculate the main unknowns, then we get X (2). Etc. If we give the values 0.0, ..., 0.1 to the free unknown variables and calculate the basic unknowns, we get X (n-r). This is how the fundamental system of solutions of a homogeneous SLAE will be constructed and its general solution can be written in the form.
For inhomogeneous systems of linear algebraic equations, the general solution is represented in the form, where is the general solution of the corresponding homogeneous system, and is the particular solution of the original inhomogeneous SLAE, which we obtain by giving the free unknowns the values 0,0, ..., 0 and calculating the values of the main unknowns.
Let's take a look at examples.
Example.
Find the fundamental system of solutions and the general solution of the homogeneous system of linear algebraic equations 
.
Solution.
The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let us find the rank of the main matrix by the bordering minors method. As a nonzero first-order minor, we take the element a 1 1 = 9 of the main matrix of the system. Find a bordering nonzero second-order minor: 
A nonzero second-order minor has been found. Let's go through the third-order minors bordering it in search of a nonzero one: 
All bordering minors of the third order are equal to zero, therefore, the rank of the main and extended matrices is equal to two. Take as a basic minor. Let's note for clarity the elements of the system that form it: 
The third equation of the original SLAE does not participate in the formation of the basic minor, therefore, it can be excluded: 
We leave on the right-hand sides of the equations the terms containing the main unknowns, and on the right-hand sides we transfer the terms with free unknowns:
Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basic minor is two. To find X (1), we assign the free unknown variables the values x 2 = 1, x 4 = 0, then we find the main unknowns from the system of equations 
.
