The case when the number of equations m more variables n, by sequentially eliminating unknowns from the equations leads to the case m= n or m n. The first case was discussed earlier.
In the second case, when the number of equations is less than the number of unknowns m n and the equations are independent, stand out m main variables And ( n- m)non-core variables . The main variables are those that satisfy the condition: the determinant, made up of the coefficients of these variables, is not equal to zero. The main ones can be different groups of variables. Total number of such groups N equal to the number of combinations of n elements by m:
If a system has at least one group of basic variables, then this system is uncertain , that is, it has many solutions.
If the system does not have a single group of basic variables, then the system is non-joint , that is, it does not have a single solution.
In the case when a system has many solutions, a basic solution is distinguished among them.
Basic solution
is a solution in which the minor variables are equal to zero. The system has no more than 
basic solutions.
System solutions are divided into acceptable And unacceptable .
Acceptable These are solutions in which the values of all variables are non-negative.
If at least one value of the variable is negative, then the solution is called unacceptable .
Example 4.5
Find basic solutions to the system of equations
Let's find the number of basic solutions
.
So, among the many solutions of the system there are no more than three basic ones. Let us highlight two main variables among the three. Let's assume it's X 1 and X 2. Let's check the determinant from the coefficients of them
.
Since this determinant is not equal to zero, then the variables X 1 ,X 2 are the main ones.
Now let's assume that X 3 =0. Then we obtain a system in the form
Let's solve it using Cramer's formulas:
,
.
So, the first basic solution has the form
X 1 =1,X 2 =0,X 3 =0 .
Let us now check whether the variables belong to the main ones X 1 and X 3 .
.
We get that X 1 and X 3 - second group of main variables. Let's put X 2 =0 and solve the system
,
.
The second basic solution has the form
X 1 =1,X 2 =0,X 3 =0.
Now let’s check whether the variables belong to the main ones X 2 and X 3 .
that is, variables X 2 and X 3 minor. So, this system has two basic solutions in total. Both of these solutions are acceptable.
The compatibility condition for a system of m linear equations with n variables is given using the concept of matrix rank.
Matrix rank – this is a number equal to the highest order of a minor other than zero.
For matrix A
minor k -th order serves as a determinant composed of elements of any k lines and k columns.
For example,
Example 2
Find the rank of a matrix
Let's calculate the determinant of the matrix
To do this, multiply the first line by (-4) and add it with the second line, then multiply the first line by (-7) and add it with the third line, as a result we get the determinant
Because the rows of the resulting determinant are proportional, then 
.
From this we can see that the 3rd order minor is equal to 0, and the 2nd order minor is not equal to 0.
Therefore, the rank of the matrix is r=2.
Extended Matrix system has the form
Kronecker-Capelli theorem
In order for a linear system to be consistent it is necessary and sufficient that the rank of the extended matrix be equal to the rank of the main matrix 
.
If 
,
then the system is inconsistent.
For a simultaneous system of linear equations, three cases are possible:
1)If 
, then the LU system has (m-r) linearly dependent equations, they can be excluded from the system;
2) If 
, then the LU system has a unique solution;
3) If 
, then the LU system has many solutions
A 21 x 1 + a 22 x 2 +...+ a 2p x p= b 2 ,
........................................
A s 1 x 1 + a s 2 x 2 +...+ a s p x p= b s.
We will perform elementary transformations on it. To do this, we write out a matrix of coefficients for the unknowns of system (1) with the addition of a column of free terms, in other words extended matrix Ā for system (1):
Let us assume that with the help of such transformations it was possible to reduce the matrix Ā to the form:
b 22 x 2 +...+b 2 r x r +...+b 2 n x n =c 2,......................................
b rr x r +...+b rn x n =c r ,
which is obtained from system (1) using a certain number of elementary transformations and, therefore, is equivalent to system (1). If in the system (4) r=n, then from the last equation, which has the form b nn x n =c n(Where b nn≠ 0), we find the only value x n, from the penultimate equation – the value xn-1(since x n already known), etc., finally, from the first equation - the value x 1. So, in case) r=n the system has a unique solution. If r
X 1 =a 1, r+1 x r+1 +...+a 1 n X n+b 1,
|
|
............................................
X r=a r, r+1 x r+1 +...+a r n X n+b r.
which is essentially general decision systems (1).
The unknowns x r+1, ..., x n are called free. From system (5) it will be possible to find the values x1,..., x r.
Matrix reduction Ā to the form (3) is possible only in the case when the original system of equations (1) is consistent. If system (1) is inconsistent, then such a reduction is impossible. This circumstance is expressed in the fact that in the process of matrix transformations Ā a line appears in it in which all elements are equal to zero, except the last one. This line corresponds to an equation of the form:
0*x 1 +0*x 2 +...+0*x n=b,
which is not satisfied by any values of the unknowns, since b≠0. In this case the system is inconsistent.
In the process of bringing system (1) to a stepwise form, equations of the form 0=0 can be obtained. They can be discarded, since this leads to a system of equations equivalent to the previous one.
When solving a system of linear equations by the Gaussian method, it is more convenient to reduce not the system of equations itself, but the extended matrix of this system to a stepwise form, performing all transformations on its rows. Sequential matrices obtained during transformations are usually connected by an equivalence sign.
Let us solve the following system of equations with 4 unknowns:
2x 1 +5x 2 +4x 3 +x 4 =20,
x 1 +3x 2 +2x 3 +x 4 =11,
2x 1 +10x 2 +9x 3 +7x 4 =40,
3x 1 +8x 2 +9x 3 +2x 4 =37.
Let's write out the extended matrix of coefficients for unknowns with the addition of a column of free terms.
Let's analyze the rows of the extended matrix:
To the elements of the 2nd line we add the elements of the 1st, divided by (-2);
From the 3rd line, subtract the 1st line;
To the 4th line we add the 1st, multiplied by (-3/2).
As a computational tool, we will use the program tools Excel-97.
1. Turn on your computer.
2. Wait until the operating system boots Windows, after which open a Microsoft Excel window.
3. Fill in the cells tables with values of the extended matrix (Fig. 11.1)
Rice. 11.1 Fig. 11.2
4. To perform the selected verbal algorithm, perform the following actions.
· Activate cell A5 and from the keyboard enter into it a formula of the form =A2+A1/(-2), after which autocomplete enter the numerical results in cells B5¸E5;
· In cell A6 we will place the result of subtracting the 1st line from the 3rd, and again, using autocomplete, fill in cells B6¸E6;
· in cell A7 we write a formula of the form =A4+A1*(-3/2) and autocomplete Let's enter the numerical results in cells B7¸E7.
5. Let us again analyze the rows resulting from elementary transformations of the matrix in order to bring it to a triangular form.
·To the 6th line add the 5th, multiplied by the number (-10);
· subtract the 5th from the 7th line.
We implement the recorded algorithm in cells A8, A9, after which let's hide 6 and 7 – lines (see Fig. 11.3).
Rice. 11.3 Fig. 11.4
6. And the last thing you need to do to bring the matrix to triangular form is to add the 8th to the 9th row, multiplied by (-3/5), after which hide 9th line (Fig. 11.4).
As you can see, the elements of the resulting matrix are in rows 1, 5, 8 and 10, and the rank of the resulting matrix is r = 4, therefore, this system of equations has a unique solution. Let us write out the resulting system:
2x 1 +5x 2 +4x 3 + x 4 =20,
0.5x 2 + 0.5x 4 =1,
5x 3 +x 4 =10,
From the last equation we easily find x 4 =0; from the 3rd equation we find x 3 =2; from the 2nd – x 2 =2 and from the 1st – x 1 =1, respectively.
Assignments for independent work.
Using the Gauss method, solve the systems of equations:
Laboratory work No. 15. Finding the roots of the equation f(x)=0
Methods for solving linear and quadratic equations were known to the ancient Greeks. The solution to equations of the third and fourth degrees was obtained through the efforts of Italian mathematicians S. Ferro, N. Tartaglia, G. Cartano, L. Ferrari during the Renaissance. Then it was time to search for formulas for finding the roots of equations of the fifth and higher degrees. Persistent but fruitless attempts continued for about 300 years and ended in the 20s of the 21st century thanks to the work of the Norwegian mathematician N. Abel. He proved that the general equation of the fifth and higher powers are unsolvable in radicals. Solution of the general equation of the nth degree
a 0 x n +a 1 x n -1 +…+a n -1 x+a n =0, a 0 ¹0 (1)
when n³5 cannot be expressed through coefficients using the operations of addition, subtraction, multiplication, division, exponentiation and root extraction.
For non-algebraic equations like
x–cos(x)=0 (2)
the task becomes even more difficult. In this case, it is rarely possible to find explicit expressions for the roots.
In conditions when formulas “do not work”, when you can count on them only in the simplest cases, universal computational algorithms acquire special importance. There are a number of known algorithms that allow solving the problem under consideration.
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. Equations with four unknowns can have many possible solutions. In mathematics, one often encounters equations of this type. To correctly solve such equations, it is necessary to use all the features of the equations in order to simplify and shorten its solution.
Let's look at the solution to the following example:
By adding the first and second equations by parts, you can get a very simple equation:
\ or \
Let's perform similar actions with equations 2 and 3:
\ or \
We solve the resulting equations \ and \
We get \ and \
We substitute the resulting numbers into equations 1 and 3:
\ or \
\ or \
Replacing these numbers with the second and fourth equations will give exactly the same equations.
But that's not all, since there are 2 equations left to solve with 2 unknowns. You can see the solution to this type of equation in the articles here.
Where can I solve an equation with four unknowns online?
You can solve equations with unknowns online at https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch video instructions and learn how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.
