Examples of systems of linear equations: solution method. Solving Linear Equations with Examples How to Solve a System of Equations in 3 Variables
2.3.1. Definition.
Let linear equations be given:
a 1 x + b 1 y + c 1 z = d 1 , (2.3.1)
a 2 x + b 2 y + c 2 z = d 2 , (2.3.2)
a 3 x + b 3 y + c 3 z = d 3 . (2.3.3)
If it is required to find a general solution of equations (2.3.1) ¾ (2.3.3), then they say that they form system . The system consisting of equations (2.3.1) ¾ (2.3.3) is denoted as follows:
The general solution of the equations that make up the system is called system solution . Solve the system (2.3.4) ¾ this means either finding the set of all its solutions or proving that there are none.
As in the previous cases, below we will find conditions under which the system (2.3.4) has a unique solution, has more than one solution, and has no solution.
2.3.2. Definition. Let system (2.3.4) of linear equations be given. matrices
are called respectively ( basic )matrix and expanded matrix systems.
2.3.3. Definitions of equivalent systems of the form (2.3.4), as well as elementary transformations of the 1st and 2nd types, are introduced in the same way as for systems of two equations with two and three unknowns.
Elementary transformation The 3rd type of system (2.3.4) is the interchange of some two equations of this system. Similar to the previous cases of systems of 2 equations under elementary transformations of the system, a system is obtained,equivalent to this.
2.3.4. An exercise. Solve systems of equations:
Solution. a)
(1) Swapped the first and second equations of the system (transformation of the 3rd type).
(2) The first equation multiplied by 4 is subtracted from the second, and the first equation multiplied by 6 is subtracted from the third (type 2 transformation); thus, the unknown was excluded from the second and third equations x .
(3) The second equation multiplied by 14 is subtracted from the third; unknown was excluded from the third y .
(4) From the last equation we find z = 1, substituting which into the second, we find y = 0. Finally, substituting y = 0 and z = 1 in the first equation, we find x = -2.с
(1) Swapped the first and second equations of the system.
(2) The first equation times 4 is subtracted from the second, and the first equation times 6 is subtracted from the third.
(3) The second and third equations coincided. We exclude one of them from the system (or, in other words, if we subtract the second equation from the third equation, then the third equation turns into the identity 0 = 0; it is excluded from the system. We assume z = a .
(4) Substitute z = a into the second and first equations.
(5) Substituting y = 12 - 12a into the first equation, we find x .
c) If the first equation is divided by 4, and the third ¾ by 6, then we arrive at an equivalent system
which is equivalent to the equation x - 2y - z = -3. Solutions to this equation are known (see Example 2.2.3 b))
The last equality in the resulting system is contradictory. Therefore, the system has no solutions.
Transformations (1) and (2) ¾ are exactly the same as the corresponding transformations of the system b))
(3) Subtract the second equation from the last equation.
Answer: a) (-2; 0; 1);
b) (21 - 23 a ; 12 - 12a ; a ), a Î R;
c) ((-3 + 2 a + b ; a ; b )|a , b Î R};
d) The system has no solutions.
2.3.5. It follows from the previous examples that system with three unknowns, as well as a system with two unknowns, may have only one solution, an infinite number of solutions and not having a single solution. Below we will analyze all possible cases. But first we introduce some notation.
Denote by D the determinant of the matrix of the system:
Denote by D 1 the determinant obtained from D by replacing the first column with the column of free members:
Similarly, let's put
D 2 = and D 3 = .
2.3.6. Theorem. If a D¹0, then the system(2.3.4)has the only solution
, , . (2.3.5)
Formulas (2.3.5) are called formulas = = 0 for all i ¹ j and at least one of the determinants , , not equal to zero, then the solution system does not have.
4) If a = = = = = = 0 for all i ¹ j , then the system has an infinite number of solutions, depending on two parameters.
For the system we compose the main determinant
and calculate it.
Then we make additional determinants
and calculate them.
According to Cramer's rule, the solution of the system is found by the formulas
; 
;
,if 
1)
Let's calculate:
By Cramer's formulas we find:
Answer: (1; 2; 3)
2)
Let's calculate:
Since the main determinant 
, and at least one additional is not equal to zero (in our case 
), then the system has no solution.
3)
Let's calculate:
Since all determinants are equal to zero, the system has an infinite set of solutions, which can be found as 
Solve your own systems:
a) 
b) 
Answer: a) (1; 2; 5) b) 
;
;
Practical lesson number 3 on the topic:
The scalar product of two vectors and its application
1. If given 
and 
, then the scalar product is found by the formula: 
∙
2. If, then the scalar product of these two vectors is found by the formula
1. Two vectors are given 
and 
We find their scalar product as follows:
.
2. Two vectors are given:
={2;3;–4}
={1;
–5; 6}
dot product is found like this:
3.
,
3.1 Finding the work of a constant force on a straight section of the path
1) Under the action of a force of 15N, the body has moved in a straight line by 2 meters. The angle between the force and the direction of movement =60 0 . Calculate the work done by the force to move the body.
Given: 
Solution:
2) Given: 
Solution:
3) A body moved from point M(1; 2; 3) to point N(5; 4; 6) under the action of a force of 60N. Angle between force direction and displacement vector =45 0 . Calculate the work done by this force.
Solution: find the displacement vector 
Find the displacement vector modulus:
According to the formula 
find a job:
3.2 Determining the orthogonality of two vectors
Two vectors are orthogonal if 
, that is
because 
1) 
– not orthogonal
2) 
-orthogonal
3) Determine for which the vectors 
and 
mutually orthogonal.
Because 
, then 
, means
Decide for yourself:
a) 
. Find their scalar product.
b) Calculate how much work the force does 
, if the point of its application, moving in a straight line, has moved from point M (5; -6; 1) to point N (1; -2; 3)
c) Determine if the vectors are orthogonal 
and
Answers: a) 1 b) 16 c) yes
3.3 Finding the angle between vectors
1)
. Find 
.
We find 
plug into the formula:
.
one). The vertices of the triangle A(3; 2; -3), B(5; 1; -1), C(1; -2; 1) are given. Find the angle at vertex A.
Substitute in the formula: 
Decide for yourself:
The vertices of the triangle A(3; 5; -2), B(5; 7; -1), C(4; 3; 0) are given. Determine the interior angle at vertex A.
Answer: 90 o
Practical lesson number 4 on the topic:
VECTOR PRODUCT OF TWO VECTORS AND ITS APPLICATION.
The formula for finding the cross product of two vectors:
|
|
|
|
|
has the form |
||
1) Find the vector product module:
We compose the determinant and calculate it (according to the Sarrus rule or the theorem on the expansion of the determinant in terms of the elements of the first row).
1st method: according to the Sarrus rule
2nd way: expand the determinant by the elements of the first row.
2) Find the module of the cross product:
4.1. CALCULATION OF THE AREA OF A PARALLELOGRAM BUILT ON TWO VECTORS.
1) Calculate the area of a parallelogram built on vectors
2). Find the cross product and its modulus
4.2. CALCULATION OF THE AREA OF A TRIANGLE
Example: given the vertices of the triangle A(1; 0; -1), B(1; 2; 0), C(3; -1; 1). Calculate the area of the triangle.
First, let's find the coordinates of two vectors coming out of the same vertex.
Let's find their vector product
4.3. DETERMINATION OF COLLINEARITY OF TWO VECTORS
If the vector 
and 
are collinear, then
, i.e. the coordinates of the vectors must be proportional.
a) Vector data:: 
,
.
They are collinear because 
and
after reducing each fraction, the ratio is obtained 
b) Vector data: 
.
They are not collinear because 
or 
Decide for yourself:
a) For what values of m and n of the vector 
collinear?
Answer: 
;
b) Find the cross product and its modulus 
,
.
Answer: 
,
.
Practical lesson number 5 on the topic:
STRAIGHT LINE ON THE PLANE
Task number 1. Find the equation of a straight line passing through the point A (-2; 3) parallel to the straight line 
1. Find the slope of the straight line 
.
is the equation of a straight line with slope and initial ordinate ( 
). That's why 
.
2. Since the lines MN and AC are parallel, their slopes are equal, i.e. 
.
3. To find the equation of the straight line AC, we use the equation of a straight line passing through a point with a given slope:
. In this formula, instead of 
and 
we substitute the coordinates of the point A (-2; 3), instead of 
let's substitute - 3. As a result of the substitution, we get:
Answer: 
Task number 2. Find the equation of a straight line passing through the point K (1; -2) parallel to the straight line.
1. Find the slope of the straight line.
This is the general equation of a straight line, which is generally given by the formula. Comparing the equations and we find that A \u003d 2, B \u003d -3. The slope of the straight line given by the equation is found by the formula 
. Substituting A = 2 and B = –3 into this formula, we obtain the slope of the straight line MN. So, 
.
2. Since the lines MN and KS are parallel, their slopes are equal: 
.
3. To find the equation of the straight line KS, we use the formula for the equation of a straight line passing through a point with a given slope 
. In this formula, instead of 
and 
we substitute the coordinates of the point K(–2; 3), instead of 
Task number 3. Find the equation of a straight line passing through the point K (–1; –3) perpendicular to the straight line.
1. is the general equation of a straight line, which is generally given by the formula.
and we find that A = 3, B = 4.
The slope of the straight line given by the equation is found by the formula: 
. Substituting A = 3 and B = 4 into this formula, we obtain the slope of the straight line MN: 
.
2. Since the lines MN and KD are perpendicular, their slopes are inversely proportional and opposite in sign:
.
3. To find the equation of the straight line KD, we use the formula for the equation of a straight line passing through a point with a given slope
. In this formula, instead of 
and 
we substitute the coordinates of the point K(–1; –3), instead of 
let's substitute . As a result of the substitution, we get:
Decide for yourself:
1. Find the equation of a straight line passing through the point K (–4; 1) parallel to the straight line 
.
Answer: 
.
2. Find the equation of a straight line passing through the point K (5; -2) parallel to the straight line 
.
3. Find the equation of a straight line passing through the point K (–2; –6) perpendicular to the straight line 
.
4. Find the equation of a straight line passing through the point K (7; -2) perpendicular to the straight line 
.
Answer: 
.
5. Find the equation of the perpendicular dropped from the point K (–6; 7) to the straight line 
.
Consider a system of three linear equations with three unknowns
a 11 , a 12 , …, a 33 are the coefficients for the unknowns,
b 1 , b 2 , b 3- free members.
To solve system (2.4) means to find such an ordered triple of numbers x 1 \u003d c 1, x 2 \u003d c 2, x 3 \u003d c 3, when substituting them into the equations of the system, the latter turn into identities.
A system of equations that has solutions (single or infinite set) is called joint, a system of equations that has no solutions, incompatible.
Let us present three methods for solving system (2.4).
Cramer's rule
Compose the determinant of the system from the coefficients of the unknowns
(2.5)
If , then system (2.4) has a unique solution, which is found by the Cramer formulas:
where , , are obtained from the determinant by replacing the first, second, and third columns, respectively, with a column of free terms of system (2.4).
(2.7)
Example 7 Solve the system 
We calculate the determinant of system (2.5) and the determinants , , (2.6).
hence the system has a unique solution.
By Cramer's formulas (2.6) we find:
You can make a check by substituting the values of the unknowns into the equations of the system.
So, x 1 \u003d x 2 \u003d x 3 \u003d 1 is the solution of the system.
Gauss method
Consider system (2.4):
The Gauss method, otherwise the method of sequential elimination of unknowns, is as follows. Let Exclude from the 2nd and 3rd equations of the system x 1. We get the system:
We get a triangular system. From the 3rd equation we find x 3, substituting it into the 2nd equation, we find x2, then from the 1st equation we find x 1, substituting into it x2 and x 3.
Example 8 Solve the system 
We rearrange the 3rd and 1st equations so that in the 1st equation the coefficient at x 1 was equal to 1.
Exclude x 1 from the 2nd and 3rd equations. To do this, multiply the 1st equation by (-4) and add it to the 2nd equation; then multiply the 1st equation by (-6) and add it to the 3rd equation. We get the system:
Exclude x2 from the 3rd equation. To do this, multiply the 2nd equation by (-13/10) and add it to the 3rd equation. We get the system:
From the last equation we find x 3= -1, we substitute into the 2nd equation:
10x2 - 13(-1) = -7, -10x2 = - 20, x2 = 2.
Substituting x2 and x 3 into the 1st equation, we get
So the solution to the system is: x 1 = 1, x2 = 2, x 3 = -1.
Solution of the system using the inverse matrix
Given system: (2.8)
Let's make a matrix BUT from the coefficients of the unknowns, the column matrix X– from unknowns, matrix-column AT- from free members.
,
System (2.8) can be written in matrix form as follows:
Decision Matrix X is found according to the formula:
A -1 is the inverse of the matrix BUT, it is composed of algebraic complements of matrix elements BUT by formula (2.3):
– determinant or matrix determinant BUT, .
Example 9 Solve system:
We introduce matrices: 
,
The inverse matrix was calculated in Example 6. Using formula (2.9), we find a solution to the system
So, x 1=1, x2=1, x 3=1.
Elements of vector algebra
Vector- directed segment; denoted by or . BUT is the beginning of the vector, AT- the end.
Length or module vector is denoted by .
Rice. 21.
In the 0xyz coordinate space, the vector can be represented as
(3.1)
This formula gives expansion of a vector in terms of a basis vectors , , ; , , - rectangular Cartesian coordinates of the vector (otherwise, projections of the vector on the coordinate axes).
Formula (3.1) can be written as follows:
– vector has coordinates , , .
Length(modulus) of the vector is found by the formula:
. (3.2)
If the vector is given by the origin coordinates A(x1,y1,z1) and end B(x2,y2,z2), then the coordinates are found by the formulas:
If the expansions of vectors and along the coordinate axes are known, then when adding (subtracting) vectors, their coordinates of the same name are added (subtracted), when a vector is multiplied by a number, the coordinates of the vector are multiplied by this number, i.e.
(3.4)
Dot product vectors and , denoted by , is the number equal to the product of the lengths of these vectors and the cosine of the angle between them
. (3.5)
If , then
. (3.6)
If vectors and collinear(parallel), then
. (3.7)
If vectors and orthogonal(perpendicular), then
Or 
(3.8)
Example 10 Given points A 1(1,0,-1), A2(2,-1,1), A 3(0,1,-2). By means of vector algebra, given what to find:
1) coordinates of vectors and .
We use formula (3.3):
2) Vector coordinates
Using formulas (3.4) and (3.5), we obtain
Or 
1.2. According to the rule of triangles: , and the length of the vector . Answer: 
3. Points A(0,-2,3), B(2,1,4), C(3,4,5) are given. Find:
a) coordinates (projections) of vectors and
b) vector coordinates
c) vector length
4. Vectors are given Find the scalar product of vectors .
5. Prove that the vectors and are collinear.
6. Prove that the vectors are orthogonal.
Linear equations (first degree equations) with two unknowns
Definition 1 . Linear equation (first degree equation) with two unknowns x and y name an equation that looks like
Solution . Let us express from equality (2) the variable y in terms of the variable x :
It follows from formula (3) that all pairs of numbers of the form
where x is any number.
Remark . As can be seen from the solution of 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 some solution to equation (2), the number x can be taken as any number, 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 are called a system of equations having the form
where a 1 , b 1 , c 1 , a 2 , b 2 , c 2 are given numbers.
Definition 4 . In the system of equations (4), the numbers a 1 , b 1 , a 2 , b 2 are called and the numbers c 1 , c 2 – free members.
Definition 5 . By solving the system of equations (4) name a pair of numbers x; y) , which is a solution to both one and the other equations 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 denoted using the symbol ""
Systems of linear equations are solved with the help of which we will illustrate with examples.
Example 2 . Solve a system of equations
Solution . To solve system (5) we eliminate the unknown from the second equation of the system X .
To this end, we first transform system (5) into 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
Let us now perform the following transformations on system (6):
- subtract the first equation from the second 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;
in) has no solutions.
Solution . Expressing x in terms of 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 the solutions of system (8) depending on the values of the parameter p . To do this, we first consider the first equation of system (8):
| y (2 - p) (2 + p) = 2 + p | (9) |
If a 
, then equation (9) has a unique solution
Thus, in the case when , system (7) has the only solution
If a p= - 2 , then equation (9) takes the form
and its solution is any number 
. Therefore, the solution to system (7) is infinite set all pairs of numbers
,
where y is any number.
If a p= 2 , then equation (9) takes the form
and has no solutions, whence it follows 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 the 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 are 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 at unknown, and the numbers d 1 , d 2 , d 3 – free members.
Definition 9 . By solving the system of equations (10) name a trio of numbers (x; y ; z) , when substituting them into each of the three equations of system (10), the correct equality is obtained.
Example 4 . Solve a system of equations
Solution . We will solve system (11) using method of successive elimination of unknowns.
For this, first we eliminate the unknown from the second and third equations of the system y by performing the following transformations on system (11):
- we leave the first equation of the system unchanged;
- add the first equation to the second equation and replace the second equation of the system with the resulting sum;
- subtract the first equation from the third equation and replace the third equation of the system with the resulting difference.
As a result, system (11) is transformed into
An equation with one unknown, which, after opening the brackets and reducing like terms, takes the form
ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.
For example, all equations:
2x + 3 \u003d 7 - 0.5x; 0.3x = 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.
The value of the unknown that turns the equation into a true equality is called decision or the root of the equation .
For example, if in the equation 3x + 7 \u003d 13 we substitute the number 2 instead of the unknown x, then we get the correct equality 3 2 + 7 \u003d 13. This means that the value x \u003d 2 is the solution or the root of the equation.
And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 2 + 7 ≠ 13. Therefore, the value x \u003d 3 is not a solution or a root of the equation.
The solution of any linear equations is reduced to the solution of equations of the form
ax + b = 0.
We transfer the free term from the left side of the equation to the right side, while changing the sign in front of b to the opposite, we get
If a ≠ 0, then x = – b/a .
Example 1 Solve the equation 3x + 2 =11.
We transfer 2 from the left side of the equation to the right, while changing the sign in front of 2 to the opposite, we get
3x \u003d 11 - 2.
Let's do the subtraction, then
3x = 9.
To find x, you need to divide the product by a known factor, that is,
x = 9:3.
So the value x = 3 is the solution or the root of the equation.
Answer: x = 3.
If a = 0 and b = 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0, we get 0, but b is also 0. The solution to this equation is any number.
Example 2 Solve the equation 5(x - 3) + 2 = 3 (x - 4) + 2x - 1.
Let's expand the brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.
5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.
Here are similar members:
0x = 0.
Answer: x is any number.
If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when multiplying any number by 0, we get 0, but b ≠ 0.
Example 3 Solve the equation x + 8 = x + 5.
Let us group the terms containing unknowns on the left side, and the free terms on the right side:
x - x \u003d 5 - 8.
Here are similar members:
0x = - 3.
Answer: no solutions.
On the figure 1 the scheme for solving the linear equation is shown
Let us compose a general scheme for solving equations with one variable. Consider the solution of example 4.
Example 4 Let's solve the equation
1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.
2) After reduction we get
4 (x - 4) + 3 2 (x + 1) - 12 = 6 5 (x - 3) + 24x - 2 (11x + 43)
3) To separate members containing unknown and free members, open the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) We group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) Here are similar members:
- 22x = - 154.
6) Divide by - 22 , We get
x = 7.
As you can see, the root of the equation is seven.
In general, such equations can be solved as follows:
a) bring the equation to an integer form;
b) open brackets;
c) group the terms containing the unknown in one part of the equation, and the free terms in the other;
d) bring similar members;
e) solve an equation of the form aх = b, which was obtained after bringing like terms.
However, this scheme is not required for every equation. When solving many simpler equations, one has to start not from the first, but from the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.
Example 5 Solve the equation 2x = 1/4.
We find the unknown x \u003d 1/4: 2,
x = 1/8 .
Consider the solution of some linear equations encountered in the main state exam.
Example 6 Solve equation 2 (x + 3) = 5 - 6x.
2x + 6 = 5 - 6x
2x + 6x = 5 - 6
Answer: - 0.125
Example 7 Solve the equation - 6 (5 - 3x) \u003d 8x - 7.
– 30 + 18x = 8x – 7
18x - 8x = - 7 +30
Answer: 2.3
Example 8 Solve the Equation
3(3x - 4) = 4 7x + 24
9x - 12 = 28x + 24
9x - 28x = 24 + 12
Example 9 Find f(6) if f (x + 2) = 3 7's
Solution
Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.
We solve the linear equation x + 2 = 6,
we get x \u003d 6 - 2, x \u003d 4.
If x = 4 then
f(6) = 3 7-4 = 3 3 = 27
Answer: 27.
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