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. A system of three equations with three unknowns does not have a solution in all cases, despite the large number of equations. As a rule, this type of system is solved using the substitution method or using the Cramer method. The second method makes it possible to determine at the first stages whether the system has a solution.
Let's say we are given the following system of three equations with three unknowns:
\[\left\(\begin(matrix) x_1+x_2+2x_3=6\\ 2x_1+3x_2+7x_3=16\\ 5x_1+2x_2+x_3=16& \end(matrix)\right.\]
You can solve this inhomogeneous system of linear algebraic equations Ax = B using the Cramer method:
\[\Delta _A\begin(vmatrix) 1 & 1 & -2\\ 2 & 3 & -7\\ 5 & 2 & 1 \end(vmatrix)=2\]
The determinant of the system \ is not equal to zero. Let's find auxiliary determinants \ if they are not equal to zero, then there are no solutions, if they are equal, then there are an infinite number of solutions
\[\Delta _1\begin(vmatrix) 6 & 1 & -2\\ 16 & 3 & -7\\ 16 & 2 & 1 \end(vmatrix)=6\]
\[\Delta _2\begin(vmatrix) 1 & 6 & -2\\ 2 & 16 & -7\\ 5 & 16 & 1 \end(vmatrix)=2\]
\[\Delta _3\begin(vmatrix) 1 & 1 & 6\\ 2 & 3 & 16\\ 5 & 2 & 16 \end(vmatrix)=-2\]
A system of 3 linear equations with 3 unknowns, the determinant of which is nonzero, is always consistent and has a unique solution, calculated by the formulas:
Answer: got a solution
\[\left\(\begin(matrix) X_1=3\\ X_2=1\\ X_3=-1\\ \end(matrix)\right.\]
Where can I solve a system of equations with three unknowns online?
You can solve the equation on our website 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.
Lesson contentLinear equations in two variables
A schoolchild has 200 rubles to eat lunch at school. A cake costs 25 rubles, and a cup of coffee costs 10 rubles. How many cakes and cups of coffee can you buy for 200 rubles?
Let us denote the number of cakes by x, and the number of cups of coffee through y. Then the cost of the cakes will be denoted by the expression 25 x, and the cost of cups of coffee in 10 y .
25x— price x cakes
10y — price y cups of coffee
The total amount should be 200 rubles. Then we get an equation with two variables x And y
25x+ 10y= 200
How many roots does this equation have?
It all depends on the student’s appetite. If he buys 6 cakes and 5 cups of coffee, then the roots of the equation will be the numbers 6 and 5.
The pair of values 6 and 5 are said to be the roots of equation 25 x+ 10y= 200 . Written as (6; 5), with the first number being the value of the variable x, and the second - the value of the variable y .
6 and 5 are not the only roots that reverse equation 25 x+ 10y= 200 to identity. If desired, for the same 200 rubles a student can buy 4 cakes and 10 cups of coffee:
In this case, the roots of equation 25 x+ 10y= 200 is a pair of values (4; 10).
Moreover, a schoolchild may not buy coffee at all, but buy cakes for the entire 200 rubles. Then the roots of equation 25 x+ 10y= 200 will be the values 8 and 0
Or vice versa, don’t buy cakes, but buy coffee for the entire 200 rubles. Then the roots of equation 25 x+ 10y= 200 the values will be 0 and 20
Let's try to list all possible roots of equation 25 x+ 10y= 200 . Let us agree that the values x And y belong to the set of integers. And let these values be greater than or equal to zero:
x∈Z, y∈ Z;
x ≥ 0, y ≥ 0
This will be convenient for the student himself. It is more convenient to buy whole cakes than, for example, several whole cakes and half a cake. It is also more convenient to take coffee in whole cups than, for example, several whole cups and half a cup.
Note that for odd x it is impossible to achieve equality under any circumstances y. Then the values x the following numbers will be 0, 2, 4, 6, 8. And knowing x can be easily determined y
Thus, we received the following pairs of values (0; 20), (2; 15), (4; 10), (6; 5), (8; 0). These pairs are solutions or roots of Equation 25 x+ 10y= 200. They turn this equation into an identity.
Equation of the form ax + by = c called linear equation with two variables. The solution or roots of this equation are a pair of values ( x; y), which turns it into identity.
Note also that if a linear equation with two variables is written in the form ax + b y = c , then they say that it is written in canonical(normal) form.
Some linear equations in two variables can be reduced to canonical form.
For example, the equation 2(16x+ 3y − 4) = 2(12 + 8x − y) can be brought to mind ax + by = c. Let's open the brackets on both sides of this equation and get 32x + 6y − 8 = 24 + 16x − 2y . We group terms containing unknowns on the left side of the equation, and terms free of unknowns - on the right. Then we get 32x− 16x+ 6y+ 2y = 24 + 8 . We present similar terms in both sides, we get equation 16 x+ 8y= 32. This equation is reduced to the form ax + by = c and is canonical.
Equation 25 discussed earlier x+ 10y= 200 is also a linear equation with two variables in canonical form. In this equation the parameters a , b And c are equal to the values 25, 10 and 200, respectively.
Actually the equation ax + by = c has countless solutions. Solving the equation 25x+ 10y= 200, we looked for its roots only on the set of integers. As a result, we obtained several pairs of values that turned this equation into an identity. But on the set of rational numbers, equation 25 x+ 10y= 200 will have infinitely many solutions.
To obtain new pairs of values, you need to take an arbitrary value for x, then express y. For example, let's take for the variable x value 7. Then we get an equation with one variable 25×7 + 10y= 200 in which one can express y
Let x= 15. Then the equation 25x+ 10y= 200 becomes 25 × 15 + 10y= 200. From here we find that y = −17,5
Let x= −3 . Then the equation 25x+ 10y= 200 becomes 25 × (−3) + 10y= 200. From here we find that y = −27,5
System of two linear equations with two variables
For the equation ax + by = c you can take arbitrary values for as many times as you like x and find values for y. Taken separately, such an equation will have countless solutions.
But it also happens that the variables x And y connected not by one, but by two equations. In this case they form the so-called system of linear equations in two variables. Such a system of equations can have one pair of values (or in other words: “one solution”).
It may also happen that the system has no solutions at all. A system of linear equations can have countless solutions in rare and exceptional cases.
Two linear equations form a system when the values x And y enter into each of these equations.
Let's go back to the very first equation 25 x+ 10y= 200 . One of the pairs of values for this equation was the pair (6; 5) . This is a case when for 200 rubles you could buy 6 cakes and 5 cups of coffee.
Let's formulate the problem so that the pair (6; 5) becomes the only solution for equation 25 x+ 10y= 200 . To do this, let’s create another equation that would connect the same x cakes and y cups of coffee.
Let us state the text of the problem as follows:
“The student bought several cakes and several cups of coffee for 200 rubles. A cake costs 25 rubles, and a cup of coffee costs 10 rubles. How many cakes and cups of coffee did the student buy if it is known that the number of cakes is one unit greater than the number of cups of coffee?
We already have the first equation. This is equation 25 x+ 10y= 200 . Now let's create an equation for the condition “the number of cakes is one unit greater than the number of cups of coffee” .
The number of cakes is x, and the number of cups of coffee is y. You can write this phrase using the equation x−y= 1. This equation will mean that the difference between cakes and coffee is 1.
x = y+ 1 . This equation means that the number of cakes is one more than the number of cups of coffee. Therefore, to obtain equality, one is added to the number of cups of coffee. This can be easily understood if we use the model of scales that we considered when studying the simplest problems:
We got two equations: 25 x+ 10y= 200 and x = y+ 1. Since the values x And y, namely 6 and 5 are included in each of these equations, then together they form a system. Let's write down this system. If the equations form a system, then they are framed by the system sign. The system symbol is a curly brace:
Let's solve this system. This will allow us to see how we arrive at the values 6 and 5. There are many methods for solving such systems. Let's look at the most popular of them.
Substitution method
The name of this method speaks for itself. Its essence is to substitute one equation into another, having previously expressed one of the variables.
In our system, nothing needs to be expressed. In the second equation x = y+ 1 variable x already expressed. This variable is equal to the expression y+ 1 . Then you can substitute this expression into the first equation instead of the variable x
After substituting the expression y+ 1 into the first equation instead x, we get the equation 25(y+ 1) + 10y= 200 . This is a linear equation with one variable. This equation is quite easy to solve:
We found the value of the variable y. Now let's substitute this value into one of the equations and find the value x. For this it is convenient to use the second equation x = y+ 1 . Let’s substitute the value into it y
This means that the pair (6; 5) is a solution to the system of equations, as we intended. We check and make sure that the pair (6; 5) satisfies the system:
Example 2
Let's substitute the first equation x= 2 + y into the second equation 3 x− 2y= 9. In the first equation the variable x equal to the expression 2 + y. Let’s substitute this expression into the second equation instead of x
Now let's find the value x. To do this, let's substitute the value y into the first equation x= 2 + y
This means that the solution to the system is the pair value (5; 3)
Example 3. Solve the following system of equations using the substitution method:
Here, unlike previous examples, one of the variables is not expressed explicitly.
To substitute one equation into another, you first need .
It is advisable to express the variable that has a coefficient of one. The variable has a coefficient of one x, which is contained in the first equation x+ 2y= 11. Let's express this variable.
After variable expression x, our system will take the following form:
Now let's substitute the first equation into the second and find the value y
Let's substitute y x
This means that the solution to the system is a pair of values (3; 4)
Of course, you can also express a variable y. This will not change the roots. But if you express y, The result is not a very simple equation, which will take more time to solve. It will look like this:
We see that in this example we express x much more convenient than expressing y .
Example 4. Solve the following system of equations using the substitution method:
Let us express in the first equation x. Then the system will take the form:
y
Let's substitute y into the first equation and find x. You can use the original equation 7 x+ 9y= 8, or use the equation in which the variable is expressed x. We will use this equation because it is convenient:
This means that the solution to the system is a pair of values (5; −3)
Addition method
The addition method consists of adding the equations included in the system term by term. This addition results in a new equation with one variable. And solving such an equation is quite simple.
Let's solve the following system of equations:
Let's add the left side of the first equation with the left side of the second equation. And the right side of the first equation with the right side of the second equation. We get the following equality:
Let's look at similar terms:
As a result, we got the simplest equation 3 x= 27 whose root is 9. Knowing the value x you can find the value y. Let's substitute the value x into the second equation x−y= 3 . We get 9 − y= 3 . From here y= 6 .
This means that the solution to the system is a pair of values (9; 6)
Example 2
Let's add the left side of the first equation with the left side of the second equation. And the right side of the first equation with the right side of the second equation. In the resulting equality we present similar terms:
As a result, we got the simplest equation 5 x= 20, whose root is 4. Knowing the value x you can find the value y. Let's substitute the value x into the first equation 2 x+y= 11. Let's get 8+ y= 11. From here y= 3 .
This means that the solution to the system is a pair of values (4;3)
The addition process is not described in detail. It must be done mentally. When adding, both equations must be reduced to canonical form. That is, by the way ac + by = c .
From the examples considered, it is clear that the main purpose of adding equations is to get rid of one of the variables. But it is not always possible to immediately solve a system of equations using the addition method. Most often, the system is first brought to a form in which the equations included in this system can be added.
For example, the system 
can be solved immediately by addition. When adding both equations, the terms y And −y will disappear because their sum is zero. As a result, the simplest equation 11 is formed x= 22, whose root is 2. It will then be possible to determine y equal to 5.
And the system of equations 
The addition method cannot be solved immediately, since this will not lead to the disappearance of one of the variables. Addition will result in equation 8 x+ y= 28, which has an infinite number of solutions.
If both sides of the equation are multiplied or divided by the same number, not equal to zero, you get an equation equivalent to the given one. This rule is also true for a system of linear equations with two variables. One of the equations (or both equations) can be multiplied by any number. The result will be an equivalent system, the roots of which will coincide with the previous one.
Let's return to the very first system, which described how many cakes and cups of coffee a schoolchild bought. The solution to this system was a pair of values (6; 5).
Let's multiply both equations included in this system by some numbers. Let's say we multiply the first equation by 2, and the second by 3
As a result, we got a system 
The solution to this system is still the pair of values (6; 5)
This means that the equations included in the system can be reduced to a form suitable for applying the addition method.
Let's return to the system , which we could not solve using the addition method.
Multiply the first equation by 6, and the second by −2
Then we get the following system:
Let's add up the equations included in this system. Adding components 12 x and −12 x will result in 0, addition 18 y and 4 y will give 22 y, and adding 108 and −20 gives 88. Then we get equation 22 y= 88, from here y = 4 .
If at first it’s hard to add equations in your head, then you can write down how the left side of the first equation adds up with the left side of the second equation, and the right side of the first equation with the right side of the second equation:
Knowing that the value of the variable y equals 4, you can find the value x. Let's substitute y into one of the equations, for example into the first equation 2 x+ 3y= 18. Then we get an equation with one variable 2 x+ 12 = 18. Let's move 12 to the right side, changing the sign, we get 2 x= 6, from here x = 3 .
Example 4. Solve the following system of equations using the addition method:
Let's multiply the second equation by −1. Then the system will take the following form:
Let's add both equations. Adding components x And −x will result in 0, addition 5 y and 3 y will give 8 y, and adding 7 and 1 gives 8. The result is equation 8 y= 8 whose root is 1. Knowing that the value y equals 1, you can find the value x .
Let's substitute y into the first equation, we get x+ 5 = 7, hence x= 2
Example 5. Solve the following system of equations using the addition method:
It is desirable that terms containing the same variables be located one below the other. Therefore, in the second equation the terms 5 y and −2 x Let's swap places. As a result, the system will take the form:
Let's multiply the second equation by 3. Then the system will take the form:
Now let's add both equations. As a result of addition we obtain equation 8 y= 16, whose root is 2.
Let's substitute y into the first equation, we get 6 x− 14 = 40. Let's move the term −14 to the right side, changing the sign, and get 6 x= 54 . From here x= 9.
Example 6. Solve the following system of equations using the addition method:
Let's get rid of fractions. Multiply the first equation by 36, and the second by 12
In the resulting system 
the first equation can be multiplied by −5, and the second by 8
Let's add up the equations in the resulting system. Then we get the simplest equation −13 y= −156 . From here y= 12. Let's substitute y into the first equation and find x
Example 7. Solve the following system of equations using the addition method:
Let us bring both equations to normal form. Here it is convenient to apply the rule of proportion in both equations. If in the first equation the right side is represented as , and the right side of the second equation as , then the system will take the form:
We have a proportion. Let's multiply its extreme and middle terms. Then the system will take the form:
Let's multiply the first equation by −3, and open the brackets in the second:
Now let's add both equations. As a result of adding these equations, we get an equality with zero on both sides:
It turns out that the system has countless solutions.
But we can’t just take arbitrary values from the sky for x And y. We can specify one of the values, and the other will be determined depending on the value we specify. For example, let x= 2 . Let's substitute this value into the system:
As a result of solving one of the equations, the value for y, which will satisfy both equations:
The resulting pair of values (2; −2) will satisfy the system:
Let's find another pair of values. Let x= 4. Let's substitute this value into the system:
You can tell by eye that the value y equals zero. Then we get a pair of values (4; 0) that satisfies our system:
Example 8. Solve the following system of equations using the addition method:
Multiply the first equation by 6, and the second by 12
Let's rewrite what's left:
Let's multiply the first equation by −1. Then the system will take the form:
Now let's add both equations. As a result of addition, equation 6 is formed b= 48, whose root is 8. Substitute b into the first equation and find a
System of linear equations with three variables
A linear equation with three variables includes three variables with coefficients, as well as an intercept term. In canonical form it can be written as follows:
ax + by + cz = d
This equation has countless solutions. By giving two variables different values, a third value can be found. The solution in this case is a triple of values ( x; y; z) which turns the equation into an identity.
If the variables x, y, z are interconnected by three equations, then a system of three linear equations with three variables is formed. To solve such a system, you can use the same methods that apply to linear equations with two variables: the substitution method and the addition method.
Example 1. Solve the following system of equations using the substitution method:
Let us express in the third equation x. Then the system will take the form:
Now let's do the substitution. Variable x is equal to the expression 3 − 2y − 2z . Let's substitute this expression into the first and second equations:
Let's open the brackets in both equations and present similar terms:
We have arrived at a system of linear equations with two variables. In this case, it is convenient to use the addition method. As a result, the variable y will disappear and we can find the value of the variable z
Now let's find the value y. To do this, it is convenient to use the equation − y+ z= 4. Substitute the value into it z
Now let's find the value x. To do this, it is convenient to use the equation x= 3 − 2y − 2z . Let's substitute the values into it y And z
Thus, the triple of values (3; −2; 2) is a solution to our system. By checking we make sure that these values satisfy the system:
Example 2. Solve the system using the addition method
Let's add the first equation with the second, multiplied by −2.
If the second equation is multiplied by −2, it takes the form −6x+ 6y − 4z = −4 . Now let's add it to the first equation:
We see that as a result of elementary transformations, the value of the variable was determined x. It is equal to one.
Let's return to the main system. Let's add the second equation with the third, multiplied by −1. If the third equation is multiplied by −1, it takes the form −4x + 5y − 2z = −1 . Now let's add it to the second equation:
We got the equation x− 2y= −1 . Let's substitute the value into it x which we found earlier. Then we can determine the value y
Now we know the meanings x And y. This allows you to determine the value z. Let's use one of the equations included in the system:
Thus, the triple of values (1; 1; 1) is the solution to our system. By checking we make sure that these values satisfy the system:
Problems on composing systems of linear equations
The task of composing systems of equations is solved by entering several variables. Next, equations are compiled based on the conditions of the problem. From the compiled equations they form a system and solve it. Having solved the system, it is necessary to check whether its solution satisfies the conditions of the problem.
Problem 1. A Volga car drove out of the city to the collective farm. She returned back along another road, which was 5 km shorter than the first. In total, the car traveled 35 km round trip. How many kilometers is the length of each road?
Solution
Let x— length of the first road, y- length of the second. If the car traveled 35 km round trip, then the first equation can be written as x+ y= 35. This equation describes the sum of the lengths of both roads.
It is said that the car returned along a road that was 5 km shorter than the first one. Then the second equation can be written as x− y= 5. This equation shows that the difference between the road lengths is 5 km.
Or the second equation can be written as x= y+ 5. We will use this equation.
Because the variables x And y in both equations denote the same number, then we can form a system from them:
Let's solve this system using some of the previously studied methods. In this case, it is convenient to use the substitution method, since in the second equation the variable x already expressed.
Substitute the second equation into the first and find y
Let's substitute the found value y in the second equation x= y+ 5 and we'll find x
The length of the first road was designated through the variable x. Now we have found its meaning. Variable x is equal to 20. This means that the length of the first road is 20 km.
And the length of the second road was indicated by y. The value of this variable is 15. This means the length of the second road is 15 km.
Let's check. First, let's make sure that the system is solved correctly:
Now let’s check whether the solution (20; 15) satisfies the conditions of the problem.
It was said that the car traveled a total of 35 km round trip. We add the lengths of both roads and make sure that the solution (20; 15) satisfies this condition: 20 km + 15 km = 35 km
The following condition: the car returned back along another road, which was 5 km shorter than the first . We see that solution (20; 15) also satisfies this condition, since 15 km is shorter than 20 km by 5 km: 20 km − 15 km = 5 km
When composing a system, it is important that the variables represent the same numbers in all equations included in this system.
So our system contains two equations. These equations in turn contain variables x And y, which represent the same numbers in both equations, namely road lengths of 20 km and 15 km.
Problem 2. Oak and pine sleepers were loaded onto the platform, 300 sleepers in total. It is known that all oak sleepers weighed 1 ton less than all pine sleepers. Determine how many oak and pine sleepers there were separately, if each oak sleeper weighed 46 kg, and each pine sleeper 28 kg.
Solution
Let x oak and y pine sleepers were loaded onto the platform. If there were 300 sleepers in total, then the first equation can be written as x+y = 300 .
All oak sleepers weighed 46 x kg, and the pine ones weighed 28 y kg. Since oak sleepers weighed 1 ton less than pine sleepers, the second equation can be written as 28y − 46x= 1000 . This equation shows that the difference in mass between oak and pine sleepers is 1000 kg.
Tons were converted to kilograms since the mass of oak and pine sleepers was measured in kilograms.
As a result, we obtain two equations that form the system
Let's solve this system. Let us express in the first equation x. Then the system will take the form:
Substitute the first equation into the second and find y
Let's substitute y into the equation x= 300 − y and find out what it is x
This means that 100 oak and 200 pine sleepers were loaded onto the platform.
Let's check whether the solution (100; 200) satisfies the conditions of the problem. First, let's make sure that the system is solved correctly:
It was said that there were 300 sleepers in total. We add up the number of oak and pine sleepers and make sure that the solution (100; 200) satisfies this condition: 100 + 200 = 300.
The following condition: all oak sleepers weighed 1 ton less than all pine sleepers . We see that the solution (100; 200) also satisfies this condition, since 46 × 100 kg of oak sleepers is lighter than 28 × 200 kg of pine sleepers: 5600 kg − 4600 kg = 1000 kg.
Problem 3. We took three pieces of copper-nickel alloy in ratios of 2: 1, 3: 1 and 5: 1 by weight. A piece weighing 12 kg was fused from them with a ratio of copper and nickel content of 4: 1. Find the mass of each original piece if the mass of the first is twice the mass of the second.
A system of linear equations is a set of several linear equations considered together.
A system can have any number of equations with any number of unknowns.
A solution to a system of equations is a set of values of unknowns that satisfies all the equations of the system, that is, turning them into identities.
A system that has a solution is called consistent; otherwise, it is called inconsistent.
Various methods are used to solve the system.
Let 
(the number of equations is equal to the number of unknowns).
Cramer method
Consider solving a system of three linear equations with three unknowns:
(7)
To find unknowns 
Let's apply Cramer's formula:
(8)
Where 
- determinant of the system, the elements of which are the coefficients of the unknowns:
.
obtained by replacing the first column of the determinant 
column of free members:
.
Likewise:
;
.
Example 1. Solve the system using Cramer's formula:
.
Solution: Let's use formulas (8):
;
;
;
;
Answer: 
.
For any system 
linear equations with 
the unknowns can be stated:
Matrix solution
Let us consider solving system (7) of three linear equations with three unknowns using a matrix method.
Using the rules of matrix multiplication, this system of equations can be written as: 
, Where
.
Let the matrix 
non-degenerate, i.e. 
. Multiplying both sides of the matrix equation on the left by the matrix 
, inverse of the matrix 
, we get: 
.
Considering that 
, we have
(9)
Example 2. Solve the system using the matrix method:
.
Solution: Let's introduce the matrices:
- from the coefficients of the unknowns;
- column of free members.
Then the system can be written as a matrix equation: 
.
Let's use formula (9). Let's find the inverse matrix 
according to formula (6):
;
.
Hence,
Got:
.
Answer: 
.
Method of sequential elimination of unknowns (Gauss method)
The main idea of the method used is to sequentially eliminate unknowns. Let us explain the meaning of this method using a system of three equations with three unknowns:
.
Let's assume that 
(If 
, then we change the order of the equations, choosing as the first equation the one in which the coefficient at 
not equal to zero).
First step: a) divide the equation 
on 
; b) multiply the resulting equation by 
and subtract from 
; c) then multiply the result by 
and subtract from 
. As a result of the first step we will have the system:
,
Second step: we deal with the equation 
And 
exactly the same as with equations 
.
As a result, the original system is transformed into the so-called stepwise form:
From the transformed system, all unknowns are determined sequentially without difficulty.
Comment. In practice, it is more convenient to reduce to a stepwise form not the system of equations itself, but a matrix of coefficients, unknowns, and free terms.
Example 3. Solve the system using the Gaussian method:
.
We will write the transition from one matrix to another using the equivalence sign ~.
~
~
~
~
~
.
Using the resulting matrix, we write out the transformed system:
.
Answer: 
.
Note: If the system has a unique solution, then the step system is reduced to a triangular one, that is, to one in which the last equation will contain one unknown. In the case of an uncertain system, that is, one in which the number of unknowns is greater than the number of linearly independent equations, there will be no triangular system, since the last equation will contain more than one unknown (the system has an infinite number of solutions). When the system is inconsistent, then, after reducing it to stepwise form, it will contain at least one 
value of the form 
, that is, an equation in which all unknowns have zero coefficients and the right-hand side is nonzero (the system has no solutions). The Gauss method is applicable to an arbitrary system of linear equations (for any 
And 
).
Existence theorem for a solution to a system of linear equations
When solving a system of linear equations using the Gaussian method, the answer to the question whether this system is compatible or inconsistent can be given only at the end of the calculations. However, it is often important to solve the question of compatibility or incompatibility of a system of equations without finding the solutions themselves. The answer to this question is given by the following Kronecker-Capelli theorem.
Let the system be given 
linear equations with 
unknown:
(10)
In order for system (10) to be consistent, it is necessary and sufficient that the rank of the system matrix
.
was equal to the rank of its extended matrix
.
Moreover, if 
, then system (10) has a unique solution; if 
, then the system has an infinite number of solutions.
Consider a homogeneous system (all free terms are equal to zero) of linear equations:
.
This system is always consistent since it has a zero solution.
The following theorem gives conditions under which the system also has solutions other than zero.
Terema. In order for a homogeneous system of line equations to have a zero solution, it is necessary and sufficient that its determinant 
was equal to zero:
.
Thus, if 
, then the solution is the only one. If 
, then there are an infinite number of other non-zero solutions. Let us indicate one of the ways to find solutions for a homogeneous system of three linear equations with three unknowns in the case 
.
It can be proven that if 
, and the first and second equations are disproportionate (linearly independent), then the third equation is a consequence of the first two. The solution of a homogeneous system of three equations with three unknowns is reduced to the solution of two equations with three unknowns. A so-called free unknown appears, to which arbitrary values can be assigned.
Example 4. Find all solutions of the system:
.
Solution. Determinant of this system
.
Therefore, the system has zero solutions. You can notice that the first two equations, for example, are not proportional, therefore, they are linearly independent. The third is a consequence of the first two (it turns out if you add twice the second to the first equation). Rejecting it, we obtain a system of two equations with three unknowns:
.
Assuming, for example, 
, we get
.
Solving a system of two linear equations, we express 
And 
through 
:
. Therefore, the solution to the system can be written as: 
, Where 
- arbitrary number.
Example 5. Find all solutions of the system:
.
Solution. It is easy to see that in this system there is only one independent equation (the other two are proportional to it). A system of three equations with three unknowns has been reduced to one equation with three unknowns. Two free unknowns appear. Finding, for example, from the first equation 
for arbitrary 
And 
, we obtain solutions to this system. The general form of the solution can be written, where 
And 
- arbitrary numbers.
Self-test questions
Formulate Cramer's rule for solving the system 
linear equations with 
unknown.
What is the essence of the matrix method of solving systems?
What is Gauss' method for solving a system of linear equations?
State the Kronecker-Capelli theorem.
Formulate a necessary and sufficient condition for the existence of nonzero solutions to a homogeneous system of linear equations.
Examples for self-solution
Find all solutions of the systems:
1.
; 2.
;
3.
; 4.
;
5.
; 6.
;
7.
; 8.
;
9.
; 10.
;
11.
; 12.
;
13.
;
14.
;
15.
.
Determine at what values 
And 
system of equations
a) has a unique solution;
b) has no solution;
c) has infinitely many solutions.
16.
; 17.
;
Find all solutions of the following homogeneous systems:
18.
; 19.
;
20.
; 21.
;
22.
; 23.
;
Answers to examples
1.
; 2.
; 3. Ǿ; 4. Ǿ;
5.
- arbitrary number.
6.
, Where 
- arbitrary number.
7.
; 8.
; 9. Ǿ; 10. Ǿ;
11.
, Where 
- arbitrary number.
12. , where 
And 
- arbitrary numbers.
13.
; 14.
Where 
And 
- arbitrary numbers.
15. Ǿ; 16. a) 
; b) 
; V) 
.
17. a) 
; b) 
; V) 
;
18.
; 19.
; 20., where 
- arbitrary number.
21. , where 
- arbitrary number.
22. , where 
- arbitrary number.
23. , where 
And 
- arbitrary numbers.
Linear equations (first degree equations) with two unknowns
Definition 1. Linear equation (first degree equation) with two unknowns x and y name an equation of the form
Solution . Let us express from equality (2) the variable y through the variable x:
From formula (3) it follows that solutions to equation (2) are all pairs of numbers of the form
where x is any number.
Note. As can be seen from the solution to Example 1, equation (2) has infinitely many solutions. However, it is important to note that not any pair of numbers (x; y) is a solution to this equation. In order to obtain any solution to equation (2), the number x can be taken as any, and the number y can then be calculated using formula (3).
Systems of two linear equations in two unknowns
Definition 3. A system of two linear equations with two unknowns x and y call a system of equations of the form
Where a 1 , b 1 , c 1 , a 2 , b 2 , c 2 – given numbers.
Definition 4. In the system of equations (4) the numbers a 1 , b 1 , a 2 , b 2 called , and numbers c 1 , c 2 – free members.
Definition 5. By solving the system of equations (4) call a pair of numbers ( x; y) , which is a solution to both one and the other equation of system (4).
Definition 6. The two systems of equations are called equivalent (equivalent), if all solutions of the first system of equations are solutions of the second system, and all solutions of the second system are solutions of the first system.
The equivalence of systems of equations is indicated using the symbol “”
Systems of linear equations are solved using , which we will illustrate with examples.
Example 2. Solve system of equations
Solution . In order to solve system (5) eliminate the unknown from the second equation of the system X .
To this end, we first transform system (5) to a form in which the coefficients for unknown x in the first and second equations of the system become the same.
If the first equation of system (5) is multiplied by the coefficient at x in the second equation (number 7), and the second equation is multiplied by the coefficient at x in the first equation (number 2), then system (5) will take the form
Now let us perform the following transformations on system (6):
- from the second equation we subtract the first equation and replace the second equation of the system with the resulting difference.
As a result, system (6) is transformed into an equivalent system
From the second equation we find y= 3, and substituting this value into the first equation, we get
Answer . (-2 ; 3) .
Example 3. Find all values of the parameter p for which the system of equations
A) has a unique solution;
b) has infinitely many solutions;
V) has no solutions.
Solution . Expressing x through y from the second equation of system (7) and substituting the resulting expression instead of x into the first equation of system (7), we obtain
Let us study solutions to system (8) depending on the values of the parameter p. To do this, first consider the first equation of system (8):
| y (2 - p) (2 + p) = 2 + p | (9) |
If 
, then equation (9) has a unique solution
Thus, in the case when , system (7) has a unique solution
If p= - 2, then equation (9) takes the form
and its solution is any number 
. Therefore, the solution to system (7) is infinite set everyone pairs of numbers
,
where y is any number.
If p= 2, then equation (9) takes the form
and has no solutions, which implies that system (7) has no solutions.
Systems of three linear equations in three unknowns
Definition 7. A system of three linear equations with three unknowns x, y and z call a system of equations having the form
Where a 1 , b 1 , c 1 , d 1 , a 2 , b 2 , c 2 , d 2 , a 3 , b 3 , c 3 , d 3 – given numbers.
Definition 8. In the system of equations (10) the numbers a 1 , b 1 , c 1 , a 2 , b 2 , c 2 , a 3 , b 3 , c 3 called coefficients for unknowns, and the numbers d 1 , d 2 , d 3 – free members.
Definition 9. By solving the system of equations (10) name three numbers (x; y ; z) , when substituting them into each of the three equations of system (10), the correct equality is obtained.
Example 4. Solve system of equations
Solution . We will solve system (11) using method of sequential elimination of unknowns.
To do this first we exclude the unknown from the second and third equations of the system y by performing the following transformations on system (11):
- We will leave the first equation of the system unchanged;
- to the second equation we add the first equation and replace the second equation of the system with the resulting sum;
- from the third equation we subtract the first equation and replace the third equation of the system with the resulting difference.
As a result, system (11) is transformed into an equivalent system
Now eliminate the unknown from the third equation of the system x by performing the following transformations on system (12):
- We will leave the first and second equations of the system unchanged;
- from the third equation we subtract the second equation and replace the third equation of the system with the resulting difference.
As a result, system (12) is transformed into an equivalent system
From the system (13) we consistently find
z = - 2 ; x = 1 ; y = 2 .
Answer . (1; 2; -2) .
Example 5. Solve system of equations
Solution . Note that from this system one can obtain a convenient consequence, adding all three equations of the system:
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