System of linear equations, examples of solutions. System of equations
Solving systems of linear algebraic equations (SLAEs) is undoubtedly the most important topic in a linear algebra course. A huge number of problems from all branches of mathematics come down to solving systems of linear equations. These factors explain the reason for 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 detailed solutions to typical examples and problems.
Brief description of the article material.
First, we give all the necessary definitions, concepts and introduce notations.
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. Firstly, we will focus on Cramer’s method, secondly, we will show the matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of sequential elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in different ways.
After this, we will move on 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 singular. Let us formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (if they are compatible) using the concept of a basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions to the 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 a SLAE is written using the vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.
In conclusion, we will consider systems of equations that can be reduced to linear ones, as well as various problems in the solution of which SLAEs arise.
Page navigation.
Definitions, concepts, designations.
We will consider systems of p linear algebraic equations with n unknown variables (p 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 recording SLAE is called coordinate.
IN matrix form This system of equations has the form,
Where 
- the main matrix of the system, - a column matrix of unknown variables, - a column matrix of free terms.
If we add a matrix-column of free terms to matrix A as the (n+1)th column, we get the so-called extended matrix systems of linear equations. Typically, an extended matrix is denoted by the letter T, and the column of free terms is separated by a vertical line from the remaining columns, that is, 
Solving a system of linear algebraic equations called a set of values of unknown variables that turns all equations of the system into identities. The matrix equation for given values of the unknown variables also becomes an identity.
If a system of equations has at least one solution, then it is called joint.
If a system of equations has no solutions, then it is called non-joint.
If a SLAE has a unique solution, then it is called certain; if there is more than one solution, then – uncertain.
If the free terms of all equations of the system are equal to zero 
, then the system is called homogeneous, otherwise - heterogeneous.
Solving elementary systems of linear algebraic equations.
If the number of equations of a system is equal to the number of unknown variables and the determinant of its main 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 started studying such SLAEs in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the addition method, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are essentially modifications of the Gauss method.
The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's sort them out.
Solving systems of linear equations using 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 different from zero, that is, .
Let be the determinant of the main matrix of the system, and 
- determinants of matrices that are obtained from A by replacement 1st, 2nd, …, nth column respectively to the column of free members: 
With this notation, unknown variables are calculated using the formulas of Cramer’s method as 
. This is how the solution to a system of linear algebraic equations is found using 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's compose and calculate the necessary determinants 
(we obtain the determinant by replacing the first column in matrix A with a column of free terms, the determinant by replacing the second column with a column of free terms, and by replacing the third column of matrix A with a column of free terms): 
Finding unknown variables using formulas 
:
Answer:
The main disadvantage of Cramer's method (if it can be called a disadvantage) is the complexity of calculating determinants when the number of equations in the system is more than three.
Solving systems of linear algebraic equations using the matrix method (using an inverse matrix).
Let a system of linear algebraic equations be given in matrix form, where the matrix A has dimension n by n and its determinant is nonzero.
Since , matrix A is invertible, that is, there is an inverse matrix. If we multiply both sides of the equality by the left, we get a formula for finding a matrix-column of unknown variables. This is how we obtained a solution to a system of linear algebraic equations using the matrix method.
Example.
Solve system of linear equations matrix method.
Solution.
Let's rewrite the system of equations in matrix form: 
Because
then the SLAE can be solved using 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 from algebraic additions 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 matrix-column of free members (if necessary, see the article): 
Answer:
or in another notation x 1 = 4, x 2 = 0, x 3 = -1.
The main problem when finding solutions to systems of linear algebraic equations using the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than third.
Solving systems of linear equations using the Gauss method.
Suppose we need to find a solution to a system of n linear equations with n unknown variables 
the determinant of the main matrix of which is different from zero.
The essence of the Gauss method consists of sequential exclusion of unknown variables: first, x 1 is excluded from all equations of the system, starting from the second, then x 2 is excluded from all equations, starting from the third, and so on, until only the unknown variable x n remains in the last equation. This process of transforming system equations to sequentially eliminate unknown variables is called direct Gaussian method. After completing the forward stroke of the Gaussian method, x n is found from the last equation, using this value from the penultimate equation, x n-1 is calculated, 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 inverse of the 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. Let's 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 nth equation we add the first, multiplied by . The system of equations after such transformations will take the form 
where , and 
.
We would have arrived at the same result if we had expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.
Next, we proceed in a similar way, but only with 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 nth equation we add the second, multiplied by . The system of equations after such transformations will take the form 
where , and 
. Thus, the variable x 2 is excluded from all equations, starting from the third.
Next, we proceed to eliminate the unknown x 3, while we act similarly with the part of the system marked in the figure 
So we continue the direct progression of the Gaussian method until the system takes the form 
From this moment we begin the reverse of the Gaussian 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 system of linear equations Gauss method.
Solution.
Let us exclude the unknown variable x 1 from the second and third equations of the system. To do this, to both sides of the second and third equations we add the corresponding parts of the first equation, multiplied by and by, respectively: 
Now we eliminate 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 stroke of the Gauss method; we begin the reverse stroke.
From the last equation of the resulting system of equations we find x 3: 
From the second equation we get .
From the first equation we find the remaining unknown variable and thereby complete the reverse of the Gauss method.
Answer:
X 1 = 4, x 2 = 0, x 3 = -1.
Solving systems of linear algebraic equations of general form.
In general, the number of equations of the system p does not coincide with the number of unknown variables n:
Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations whose main matrix is square and singular.
Kronecker–Capelli theorem.
Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when SLAE is compatible and when it is inconsistent is given by Kronecker–Capelli theorem:
In order 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, as an example, the application of the Kronecker–Capelli theorem to determine the compatibility of a system of linear equations.
Example.
Find out whether the system of linear equations has 
solutions.
Solution.
. Let's use the method of bordering minors. Minor of the second order 
different from zero. Let's look at the third-order minors bordering it: 
Since all the 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 minor is of third order 
different from zero.
Thus, Rang(A), therefore, using 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 a system using the Kronecker–Capelli theorem.
But how to find a solution to an SLAE if its compatibility is established?
To do this, we need the concept of a basis minor of a matrix and a theorem about the rank of a matrix.
The minor of the highest order of the matrix A, different from zero, is called basic.
From the definition of a basis minor it follows that its order is equal to the rank of the matrix. For a non-zero matrix A there can be several basis minors; there is always one basis 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 non-zero 
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 row (and column) elements of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding row (and column) elements forming the basis minor.
What does the matrix rank theorem tell us?
If, according to the Kronecker–Capelli theorem, we have established the compatibility of the system, then we choose any basis minor of the main matrix of the system (its order is equal to r), and exclude from the system all equations that do not form the selected basis minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still redundant (according to the matrix rank theorem, they are a linear combination of the remaining equations).
As a result, after discarding 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 the Cramer method, the matrix method or the Gauss method.
Example.
.
Solution.
Rank of the main matrix of the system 
is equal to two, since the minor is of second order 
different from zero. Rank of the extended matrix 
is also equal to two, since the only third order minor is zero 
and the second-order minor considered above is different from zero. Based on the Kronecker–Capelli theorem, we can assert the compatibility of the original system of linear equations, since Rank(A)=Rank(T)=2.
As a basis minor we take . It is formed by the coefficients of the first and second equations: 
The third equation of the system does not participate in the formation of the basis minor, so we exclude it from the system based on the theorem on the rank of the matrix: 
This is how we obtained 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 resulting SLAE is less than the number of unknown variables n, then on the left sides of the equations we leave the terms that form the basis minor, and we transfer the remaining terms to the right sides of the equations of the system with the opposite sign.
The unknown variables (r of them) remaining on the left sides of the equations are called main.
Unknown variables (there are n - r pieces) that are on the right sides are called free.
Now we believe that free unknown variables can take arbitrary values, while the r main unknown variables will be expressed through free unknown variables in a unique way. Their expression can be found by solving the resulting SLAE using the Cramer method, the matrix method, or the Gauss method.
Let's look at it with an example.
Example.
Solve a system of linear algebraic equations 
.
Solution.
Let's find the rank of the main matrix of the system 
by the method of bordering minors. Let's take a 1 1 = 1 as a non-zero minor of the first order. Let's start searching for a non-zero minor of the second order bordering this minor: 
This is how we found a non-zero minor of the second order. Let's start searching for a non-zero bordering minor of the third order: 
Thus, the rank of the main matrix is three. The rank of the extended matrix is also equal to three, that is, the system is consistent.
We take the found non-zero minor of the third order as the basis one.
For clarity, we show the elements that form the basis minor: 
We leave the terms involved in the basis minor on the left side of the system equations, and transfer the rest with opposite signs to the right sides: 
Let's give the free unknown variables x 2 and x 5 arbitrary values, that is, we accept 
, where are arbitrary numbers. In this case, the SLAE will take the form 
Let us solve the resulting elementary system of linear algebraic equations using Cramer’s method: 
Hence, .
In your answer, do not forget to indicate free unknown variables.
Answer:
Where are arbitrary numbers.
Summarize.
To solve a system of general linear algebraic equations, we first determine 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 select a basis minor and discard the equations of the system that do not participate in the formation of the selected basis minor.
If the order of the basis minor is equal to the number of unknown variables, then the SLAE has a unique solution, which can be found by any method known to us.
If the order of the basis minor is less than the number of unknown variables, then on the left side of the system equations we leave the terms with the main unknown variables, transfer the remaining terms to the right sides and give arbitrary values to the free unknown variables. From the resulting system of linear equations we find the main unknown variables using the Cramer method, the matrix method or the Gauss method.
Gauss 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 testing them for consistency. The process of sequential elimination of unknown variables makes it possible to draw a conclusion about both the compatibility and incompatibility of the SLAE, and if a solution exists, it makes it possible to find it.
From a computational point of view, the Gaussian method is preferable.
See its detailed description and analyzed examples in the article Gauss method for solving systems of general linear algebraic equations.
Writing a general solution to homogeneous and inhomogeneous linear algebraic systems using vectors of the fundamental system of solutions.
In this section we will talk about simultaneous homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.
Let us first deal with homogeneous systems.
Fundamental system of solutions homogeneous system of p linear algebraic equations with n unknown variables is a collection of (n – r) linearly independent solutions of this system, where r is the order of the basis minor of the main matrix of the system.
If we denote linearly independent solutions of a homogeneous SLAE as X (1) , X (2) , …, X (n-r) (X (1) , X (2) , …, X (n-r) are columnar matrices of dimension n by 1) , 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 C 1, C 2, ..., C (n-r), 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 of the original SLAE, in other words, taking any set of values of arbitrary constants C 1, C 2, ..., C (n-r), using the formula we will obtain one of the solutions of the original homogeneous SLAE.
Thus, if we find a fundamental system of solutions, then we can define all solutions of this homogeneous SLAE as .
Let us show the process of constructing a fundamental system of solutions to a homogeneous SLAE.
We select the basis 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's give the free unknown variables the values 1,0,0,...,0 and calculate the main unknowns by solving the resulting elementary system of linear equations in any way, for example, using the Cramer method. This will result in X (1) - the first solution of the fundamental system. If we give the free unknowns the values 0,1,0,0,…,0 and calculate the main unknowns, we get X (2) . And so on. If we assign the values 0.0,...,0.1 to the free unknown variables and calculate the main unknowns, we obtain X (n-r) . In this way, a fundamental system of solutions to 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 look at examples.
Example.
Find the fundamental system of solutions and the general solution of a homogeneous system of linear algebraic equations 
.
Solution.
The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let's find the rank of the main matrix using the method of bordering minors. As a non-zero minor of the first order, we take element a 1 1 = 9 of the main matrix of the system. Let's find the bordering non-zero minor of the second order: 
A minor of the second order, different from zero, has been found. Let's go through the third-order minors bordering it in search of a non-zero one: 
All third-order bordering minors are equal to zero, therefore, the rank of the main and extended matrix is equal to two. Let's take . For clarity, let us note the elements of the system that form it: 
The third equation of the original SLAE does not participate in the formation of the basis minor, therefore, it can be excluded: 
We leave the terms containing the main unknowns on the right sides of the equations, and transfer the terms with free unknowns to the right sides:
Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basis minor is equal to two. To find X (1), we give the free unknown variables the values x 2 = 1, x 4 = 0, then we find the main unknowns from the system of equations 
.
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 to the reduced homogeneous system and a particular solution to the inhomogeneous 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.
Quite 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 added, we obtain the first fundamental solution of the system (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.
https://pandia.ru/text/78/176/images/image035_21.gif" width="202 height=106" height="106">
Option 2.
https://pandia.ru/text/78/176/images/image039_16.gif" width="192" height="106 src=">
Option 4.
https://pandia.ru/text/78/176/images/image042_14.gif" width="172" height="80">
Option 5.
https://pandia.ru/text/78/176/images/image044_12.gif" width="179 height=106" height="106">
Option 6.
https://pandia.ru/text/78/176/images/image046_11.gif" width="195" height="106">
Solution. A= 
. Let's find r(A). Because matrix And has order 3x4, then the highest order of minors is 3. Moreover, all third-order minors are equal to zero (check it yourself). Means, r(A)< 3. Возьмем главный basic minor = -5-4 = -9 ≠
0. Therefore r(A) =2.
Let's consider matrix WITH = 
.
Minor third order ≠ 0. So r(C) = 3.
Since r(A) ≠ r(C) , then the system is inconsistent.
Example 2. Determine the compatibility of a system of equations 
Solve this system if it turns out to be consistent.
Solution.
A = , C = 
. It is obvious that r(A) ≤ 3, r(C) ≤ 4. Since detC = 0, then r(C)< 4. Let's consider minor third order, located in the upper left corner of the matrix A and C: = -23 ≠
0. So r(A) = r(C) = 3.
Number unknown in system n=3. This means that the system has a unique solution. In this case, the fourth equation represents the sum of the first three and can be ignored.
According to Cramer's formulas we get x 1 = -98/23, x 2 = -47/23, x 3 = -123/23.
2.4. Matrix method. Gaussian method
system n linear equations With n unknowns can be solved matrix method according to the formula X = A -1 B (at Δ ≠ 0), which is obtained from (2) by multiplying both parts by A -1.
Example 1. Solve a system of equations 
matrix method (in section 2.2 this system was solved using Cramer’s formulas)
Solution. Δ = 10 ≠ 0 A = - non-degenerate matrix.
= 
(check this yourself by making the necessary calculations).
A -1 = (1/Δ)х= 
.
X = A -1 V = x= .
Answer: .
From a practical point of view matrix method and formulas Kramer are associated with a large amount of calculations, so preference is given Gaussian method, which consists in the sequential elimination of unknowns. To do this, the system of equations is reduced to an equivalent system with a triangular extended matrix (all elements below the main diagonal are equal to zero). These actions are called forward movement. From the resulting triangular system, the variables are found using successive substitutions (reverse).
Example 2. Solve the system using the Gauss method
(Above, this system was solved using Cramer’s formula and the matrix method).
Solution.
Direct move. Let's write down the extended matrix and, using elementary transformations, reduce it to triangular form:
~ 
~ 
~ 
~ 
.
We get system 
Reverse move. From the last equation we find X 3 = -6 and substitute this value into the second equation:
X 2 = - 11/2 - 1/4X 3 = - 11/2 - 1/4(-6) = - 11/2 + 3/2 = -8/2 = -4.
X 1 = 2 -X 2 + X 3 = 2+4-6 = 0.
Answer: .
2.5. General solution of a system of linear equations
Let a system of linear equations be given = b i(i=). Let r(A) = r(C) = r, i.e. the system is collaborative. Any minor of order r other than zero is basic minor. Without loss of generality, we will assume that the basis minor is located in the first r (1 ≤ r ≤ min(m,n)) rows and columns of matrix A. Having discarded the last m-r equations of the system, we write a shortened system:
which is equivalent to the original one. Let's name the unknowns x 1 ,….x r basic, and x r +1 ,…, x r free and move the terms containing free unknowns to the right side of the equations of the truncated system. We obtain a system with respect to the basic unknowns:
which for each set of values of free unknowns x r +1 = С 1 ,…, x n = С n-r has only one solution x 1 (C 1 ,…, C n-r),…, x r (C 1 ,…, C n-r), found by Cramer's rule.
Corresponding Solution the shortened, and therefore the original system has the form:
X(C 1 ,…, C n-r) = 
-
general solution of the system.
If in the general solution we assign some numerical values to the free unknowns, we obtain a solution to the linear system, called a partial solution.
Example. Establish compatibility and find a general solution of the system
Solution. A = 
, C = 
.
So How r(A)= r(C) = 2 (see this for yourself), then the original system is consistent and has an infinite number of solutions (since r< 4).
Example 1. Find a general solution 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 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. 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 non-zero (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 the compatibility of the system of linear equations and solve it in two ways: 1) by 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 one, 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
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