Problem 1

Solve a system of linear equations in two ways: using Cramer’s formulas and Gauss’s method

1) solve the inhomogeneous system of linear algebraic equations Ax = B using the Cramer method

The determinant of system D is not equal to zero. Let's find the auxiliary determinants D 1, D 2, D 3, if they are not equal to zero, then there are no solutions, if they are equal, then there are an infinite number of solutions

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: we got the solution:

2) solve the inhomogeneous system of linear algebraic equations Ax = B using the Gauss method

Let's create an extended matrix of the system

Let's take the first line as a guide, and the element a 11 = 1 as a guide. Using the guide line we get zeros in the first column.

corresponds to the set of solutions of the system of linear equations

Answer: we got the solution:

Problem 2

Given the coordinates of the vertices of triangle ABC

Find:

1) length of side AB;

4) equation of median AE;

Construct the given triangle and all lines in the coordinate system.

A(1; -1), B(4; 3). C(5; 1).

1) Distance between points A( x 1; at 1) and B( x 2; at 2) is determined by the formula

using which we find the length of side AB;

2) equations of sides AB and BC and their angular coefficients;

Equation of a straight line passing through two given points of the plane A( x 1; at 1) and B( x 2; at 2) has the form

Substituting the coordinates of points A and B into (2), we obtain the equation of side AB:

We find the angle coefficient k AB of straight line AB by transforming the resulting equation to the form of an equation of a straight line with an angle coefficient y =kx - b.

, that is, from where

Similarly, we obtain the equation of straight line BC and find its angular coefficient.

Substituting the coordinates of points B and C into (2), we obtain the equation for side BC:

We find the angle coefficient k of the BC of the straight BC by transforming the resulting equation to the form of the equation of a straight line with an angular coefficient y =kx - b.

, that is

3) internal angle at vertex B in radians with an accuracy of 0.01

To find the internal angle of our triangle, we use the formula:

Note that the procedure for calculating the difference between the angular coefficients in the numerator of this fraction depends on the relative position of the straight lines AB and BC.

Substituting the previously calculated values ​​of k BC and k AB into (3), we find:

Now, using the tables with an engineering microcalculator, we get B » 1.11 rad.

4) equation of median AE;

To compile the equation of the median AE, we first find the coordinates of point E, which lies in the middle of the segment BC

Substituting the coordinates of points A and E into equation (2), we obtain the median equation:

5) equation and length of height CD;

To compile the equation for height CD, we use the equation of a straight line passing through a given point M( x 0 ; y 0)with a given slope k, which has the form

and the condition of perpendicularity of straight lines AB and CD, which is expressed by the relation k AB k CD = -1, whence k CD = -1/k AB = - 3/4

Substituting in (4) instead of k the value k C D = -3/4, and instead of x 0 , y 0 the corresponding coordinates of point C, we obtain the equation for the height CD

To calculate the length of the height CD, we use the formula for finding the distance d from a given point M( x 0 ; y 0) to a given straight line with the equation Ax+ By + C = 0, which has the form:

Substituting in (5) instead x 0 ; y 0 coordinates of point C, and instead of A, B, C the coefficients of the equation of straight line AB, we get

6) the equation of a straight line passing through point E parallel to side AB and the point M of its intersection with height CD;

Since the desired straight line EF is parallel to the straight line AB, then k EF = k AB = 4/3. Substituting into equation (4) instead x 0 ; y 0 coordinates of point E, and instead of k the value k EF we obtain the equation of the straight line EF".

To find the coordinates of point M, we solve jointly the equations of lines EF and CD.

Thus, M(5.48, 0.64).

7) equation of a circle with center at point E passing through vertex B

Since the circle has a center at point E(4.5; 2) and passes through the vertex B(4; 3), then its radius

Canonical equation of a circle of radius R with center at point M 0 ( x 0 ; y 0) has the form

Triangle ABC, height CD, median AE, straight line EF, point M and a circle constructed in the x0y coordinate system in Fig. 1.

Problem 3

Draw up an equation of a line, for each point of which its distance to point A (2; 5) is equal to the distance to the straight line y = 1. Plot the resulting curve in the coordinate system

Solution

Let M ( x, y) - current point of the desired curve. Let us drop the perpendicular MB from point M to the straight line y = 1 (Fig. 2). Then B(x; 1). Since MA = MB, then

We compose the main determinant for the system

and calculate it.

Then we compose additional determinants

and calculate them.

According to Cramer's rule, the solution to the system is found using the formulas

;
;
,If

1)

Let's calculate:

Using Cramer's formulas we find:

Answer: (1; 2; 3)

2)

Let's calculate:

Since the main determinant
, and at least one additional one 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 number of solutions, which can be found as follows:

Solve the systems yourself:

A)
b)

Answer: a) (1; 2; 5) b) ;;

Practical lesson No. 3 on the topic:

Dot product of two vectors and its application

1. If given
And
, then we find the scalar product using the formula:

∙

2.If, then the scalar product of these two vectors is found by the formula

1. Given two vectors
And

We find their scalar product as follows:

.

2. Two vectors are given:

={2;3;–4}
={1; –5; 6}

The scalar product is found like this:

3.
,

3.1 Finding the work of a constant force on a straight section of path

1) Under the influence of a force of 15 N, the body moved in a straight line 2 meters. The angle between the force and the direction of movement =60 0. Calculate the work done by a force to move a body.

Given:

Solution:

2) Given:

Solution:

3) A body moved from point M(1; 2; 3) to point N(5; 4; 6) under the influence of a force of 60 N. The angle between the direction of the force and the displacement vector =45 0. Calculate the work done by this force.

Solution: find the displacement vector

Finding the module of the displacement vector:

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 at what  the vectors
And
mutually orthogonal.

Because
, That
, Means

Decide for yourself:

A)

. Find their scalar product.

b) Calculate how much work the force produces
, if the point of its application, moving rectilinearly, has moved from point M (5; -6; 1) to point N (1; -2; 3)

c) Determine whether the vectors are orthogonal
And

Answers: a) 1 b) 16 c) yes

3.3. Finding the angle between vectors

1)

. Find .

We find

substitute into the formula:

.

1). Given are the vertices of the triangle A(3; 2; –3), B(5; 1; –1), C(1; –2; 1). Find the angle at vertex A.

Let's put it into the formula:

Decide for yourself:

Given are the vertices of the triangle A(3; 5; -2), B(5; 7; -1), C(4; 3; 0). Determine the interior angle at vertex A.

Answer: 90 o

Practical lesson No. 4 on the topic:

VECTOR PRODUCT OF TWO VECTORS AND ITS APPLICATION.

Formula for finding the cross product of two vectors:

	looks like

1) Find the modulus of the vector product:

Let's compose a determinant and calculate it (using Sarrus's rule or the theorem on the expansion of the determinant into the elements of the first row).

1st method: according to Sarrus's rule

Method 2: expand the determinant into the elements of the first row.

2) Find the modulus of the vector 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 vector product and its modulus

4.2. CALCULATING THE AREA OF A TRIANGLE

Example: given are 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 emanating from 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) Given vectors::
,
.

They are collinear because
And

after reducing each fraction we get the ratio

b) Given vectors:

.

They are not collinear because
or

Decide for yourself:

a) At what values ​​m and n of the vector
collinear?

Answer:
;

b) Find the vector product and its modulus
,
.

Answer:
,
.

Practical lesson No. 5 on the topic:

STRAIGHT LINE ON A PLANE

Problem No. 1. Find the equation of a line passing through point A(-2; 3) parallel to the line

1. Find the slope of the line
.

is the equation of a straight line with an angular coefficient and an initial ordinate (
). That's why
.

2. Since the lines MN and AC are parallel, their angular coefficients are equal, i.e.
.

3. To find the equation of straight line AC, we use the equation of a straight line passing through a point with a given angular coefficient:

. In this formula instead And substitute the coordinates of point A(-2; 3), instead Let’s substitute – 3. As a result of the substitution we get:

Answer:

Task No. 2. Find the equation of a line passing through the point K(1; –2) parallel to the line.

1. Let's find the slope of the line.

This is the general equation of a line, which in general form is given by the formula. Comparing the equations, we find that A = 2, B = –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 angular coefficients are equal:
.

3. To find the equation of the straight line KS, we use the formula for the equation of the straight line passing through a point with a given angular coefficient
. In this formula instead And let's substitute the coordinates of the point K(–2; 3), instead of

Problem No. 3. Find the equation of the line passing through the point K(–1; –3) perpendicular to the line.

1. is a general equation of a straight line, which in general form is 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 angular coefficients 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 the straight line passing through the point with a given angular coefficient

. In this formula instead And substitute the coordinates of the point K(–1;–3), instead let's substitute As a result of the substitution we get:

Decide for yourself:

1. Find the equation of the line passing through the point K(–4; 1) parallel to the line
.

Answer:
.

2. Find the equation of the line passing through the point K(5; –2) parallel to the line
.

3. Find the equation of the line passing through the point K(–2, –6) perpendicular to the line
.

4. Find the equation of the line passing through the point K(7; –2) perpendicular to the line
.

Answer:
.

5. Find the equation of the perpendicular dropped from the point K(–6; 7) to the straight line
.

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 necessary to find a general solution to 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 previous cases, below we will find conditions under which system (2.3.4) has a unique solution, has more than one solution, and has no solution.

2.3.2. Definition. Let the system (2.3.4) of linear equations be given. Matrices

are called accordingly ( basic )matrix And extended 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 called the interchange of some two equations of this system. Similar to the previous cases of systems of 2 equations with elementary transformations of the system, the system is obtained,equivalent to this.

2.3.4. Exercise. Solve systems of equations:

Solution. A)

(1) We swapped the first and second equations of the system (type 3 transformation).

(2) The first equation multiplied by 4 was subtracted from the second, and the first equation multiplied by 6 was 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, was subtracted from the third; the 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 into the first equation, we find x = -2.ñ

(1) We swapped the first and second equations of the system.

(2) The first equation multiplied by 4 was subtracted from the second, and the first equation multiplied by 6 was 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 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. The 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 system b))

(3) Subtract the second 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. From the previous examples it follows that system with three unknowns, like 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.

Let D denote the determinant of the system matrix:

Let D 1 denote the determinant obtained from D by replacing the first column with a column of free terms:

Similarly, let us put

D 2 = and D 3 = .

2.3.6. Theorem. If D¹0, then the system(2.3.4)has a unique 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 system has no solutions.

4) If = = = = = = 0 for all i ¹ j , then the system has an infinite number of solutions, depending on two parameters.

PRACTICAL LESSON No. 7

SOLUTION OF A SYSTEM OF 3 LINEAR EQUATIONS

WITH THREE VARIABLES

Target:

Develop the ability to transform matrices;

Develop system solving skills3 linear equations in three variables using Cramer's method;

Consolidate knowledge about the properties of 2nd and 3rd order determinants;

Material and technical support: guidelines for performing the work;

Lead time: 2 academic hours;

Progress of the lesson:

Study brief theoretical information;

Complete tasks;

Draw a conclusion on the work;

Prepare a defense of your work on test questions.

Brief theoretical information:

A matrix is ​​a square or rectangular table, filled with numbers. These numbers are called matrix elements.

Matrix elements, horizontally located, form the rows of the matrix. Matrix elements, arranged vertically, form the matrix columns.

Lines are numbered from left to right, starting from number1, columns are numbered from top to bottom, starting from number1.

MatrixA , havingm lines andn columns, called a matrixsizem onn and is designatedA m∙n . Elementa i j matricesA = { a ij } stands at the intersectioni - oh lines andj- th column.

The main diagonal of a square matrix is ​​the diagonal leading from the upper left corner of the matrix to the lower right corner.The side diagonal of a square matrix is ​​the diagonal leading from the lower left corner of the matrix to the upper right corner.

Two matrices are considered equal if they have the same dimension and their corresponding elements are equal.

Each matrix can be multiplied by any number, and ifk – number, thenk ∙ A ={ k ∙ a ij }.

Matrices of the same sizeA m∙n AndB m∙n can be folded, andA m∙n + B m∙n = { a ij + b i j }.

The matrix addition operation has the propertiesA + B = B + A , A +( B + C ) = ( A + B ) + C .

Example 1. After performing operations on matrices, find the matrix C= 2A - B, where, .

Solution.

Let's calculate matrix 2A of dimension 3x3:

Let's calculate the matrix C = 2A - In dimension 3x3:

C = 2 A - B .

Determinant of a third-order matrix is the number defined by the equality:

.

This number represents an algebraic sum consisting of six terms. Each term contains exactly one element from each row and each column of the matrix. Each term consists of the product of three factors.

Fig.1.1. Fig.1.2.

The signs with which the terms of the determinant are included in the formula for finding the third-order determinant can be determined using the given scheme, which is called the rule of triangles or Sarrus’s rule. The first three terms are taken with a plus sign and are determined from figure (1.1.), and the next three terms are taken with a minus sign and are determined from figure (1.2).

Example 2. Calculate the third-order determinant using Sarrus's rule:

Solution:

Example 3. Calculate the third-order determinant using the expansion method over the elements of the first row:

Solution:

We use the formula:

3 -2 +2 = 3(-5 + 16) – 2(1+32) + 2(2 +20) = 33 – 66 + 44 = 11.

Let's consider the main properties of determinants:

A determinant with a zero row (column) is equal to zero.

If you multiply any row (any column) of a matrix by any number, then the determinant of the matrix will be multiplied by this number.

The determinant does not change when the matrix is ​​transposed.

The determinant changes sign when any two rows (columns) of the matrix are rearranged.

The determinant of a matrix with two identical rows (columns) is equal to zero.

The determinant does not change if any other row is added to any row, multiplied by any number. A similar statement is true for columns.

The properties of matrices and determinants are widely used when solving a system of three linear equations with three unknowns:

,

where x 1 , X 2 , X 3 are variables, and 11 , A 12 ,…, A 33 - numerical coefficients. It should be remembered that when solving a system, one of three possible answers is possible:

1) the system has a unique solution – (x 1 ; X 2 ; X 3 );

2) the system has infinitely many solutions (undefined);

3) the system has no solutions (inconsistent).

Consider solving a system of three linear equations with three unknownsCramer's method, whichallows you to findthe only solution to the system, based on the ability to calculate third-order determinants:

Example 3. Find a solution to a system of three linear equations with three unknowns using Cramer’s formulas:

Solution. Find third order determinants usingSarrus's rule or expansion by elements of the first row:

We find the solution to the system using the formulas:

Answer: (- 152; 270; -254)

Tasks for independent completion:

I. Find the transformation matrix.

II. Compute determinantIIIorder.

III. Solve the system using Cramer's method.

Option 1.

1. C = A +3 B , If, . 2..

Option 2.

1. C =2 A - B ,If, . 2..

Option 3.

1. C = 3 A + B , If, . 2. .

Option 4.

1. C = A - 4 B , If, . 2..

Option 5.

1. C = 4 A - B , If, . 2..

Option 6.

1. C = A +2 B , If, . 2..

Option 7.

1. C =2 A + B , If, . 2..

Option 8.

1. C =3 A - B , If, . 2..

Option 9.

1. C = A - 3 B , If, . 2..

Option 10.

1. C = A - 2 B , If, . 2..

Option 11.

1. C = A +4 B , If, . 2..

Option 12.

1. C =4 A + B , If, . 2..

Option 13.

1. C = A +3 B , If, . 2..

Option 14.

1. C =2 A - B , If, . 2..

Option 15.

1. C =3 A + B , If, . 2..

Questions for self-control:

What is a matrix?

Rules for calculating third-order determinants?

Write Cramer's formulas for solving a system of three linear equations with three variables.

Systems of equations are widely used in the economic sector for mathematical modeling of various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.

Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.

A system of linear equations is two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.

Linear equation

Equations of the form ax+by=c are called linear. The designations x, y are the unknowns whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.

Types of systems of linear equations

The simplest examples are considered to be systems of linear equations with two variables X and Y.

F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.

Solve system of equations - this means finding values ​​(x, y) at which the system turns into a true equality or establishing that suitable values ​​of x and y do not exist.

A pair of values ​​(x, y), written as the coordinates of a point, is called a solution to a system of linear equations.

If systems have one common solution or no solution exists, they are called equivalent.

Homogeneous systems of linear equations are systems whose right-hand side is equal to zero. If the right part after the equal sign has a value or is expressed by a function, such a system is heterogeneous.

The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three or more variables.

When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.

Simple and complex methods for solving systems of equations

There is no general analytical method for solving such systems; all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as graphical and matrix methods, solution by the Gaussian method.

The main task when teaching solution methods is to teach how to correctly analyze the system and find the optimal solution algorithm for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of using a particular method

Solving examples of systems of linear equations in the 7th grade general education curriculum is quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.

Solving systems using the substitution method

The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system

Let us give a solution to an example of a system of linear equations of class 7 using the substitution method:

As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Solving this example is easy and allows you to get the Y value. The last step is to check the obtained values.

It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, solving by substitution is also inappropriate.

Solution of an example of a system of linear inhomogeneous equations:

Solution using algebraic addition

When searching for solutions to systems using the addition method, equations are added term by term and multiplied by various numbers. The ultimate goal of mathematical operations is an equation in one variable.

Application of this method requires practice and observation. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.

Solution algorithm:

1. Multiply both sides of the equation by a certain number. As a result of the arithmetic operation, one of the coefficients of the variable should become equal to 1.
2. Add the resulting expression term by term and find one of the unknowns.
3. Substitute the resulting value into the 2nd equation of the system to find the remaining variable.

Method of solution by introducing a new variable

A new variable can be introduced if the system requires finding a solution for no more than two equations; the number of unknowns should also be no more than two.

The method is used to simplify one of the equations by introducing a new variable. The new equation is solved for the introduced unknown, and the resulting value is used to determine the original variable.

The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to a standard quadratic trinomial. You can solve a polynomial by finding the discriminant.

It is necessary to find the value of the discriminant using the well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. In the given example, a=1, b=16, c=39, therefore D=100. If the discriminant is greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant is less than zero, then there is one solution: x = -b / 2*a.

The solution for the resulting systems is found by the addition method.

Visual method for solving systems

Suitable for 3 equation systems. The method consists in constructing graphs of each equation included in the system on the coordinate axis. The coordinates of the intersection points of the curves will be the general solution of the system.

The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.

As can be seen from the example, for each line two points were constructed, the values ​​of the variable x were chosen arbitrarily: 0 and 3. Based on the values ​​of x, the values ​​for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.

The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.

The following example requires finding a graphical solution to a system of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.

As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.

The systems from examples 2 and 3 are similar, but when constructed it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether a system has a solution or not; it is always necessary to construct a graph.

The matrix and its varieties

Matrices are used to concisely write a system of linear equations. A matrix is ​​a special type of table filled with numbers. n*m has n - rows and m - columns.

A matrix is ​​square when the number of columns and rows are equal. A matrix-vector is a matrix of one column with an infinitely possible number of rows. A matrix with ones along one of the diagonals and other zero elements is called identity.

An inverse matrix is ​​a matrix when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.

Rules for converting a system of equations into a matrix

In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.

A matrix row is said to be nonzero if at least one element of the row is not zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.

The matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all elements of the matrix are sequentially multiplied by a number.

Options for finding the inverse matrix

The formula for finding the inverse matrix is ​​quite simple: K -1 = 1 / |K|, where K -1 is the inverse matrix, and |K| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements by each other. For the “three by three” option, there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the numbers of columns and rows of elements are not repeated in the work.

Solving examples of systems of linear equations using the matrix method

The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with a large number of variables and equations.

In the example, a nm are the coefficients of the equations, the matrix is ​​a vector x n are variables, and b n are free terms.

Solving systems using the Gaussian method

In higher mathematics, the Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variables of systems with a large number of linear equations.

The Gauss method is very similar to solutions by substitution and algebraic addition, but is more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce the system to the form of an inverted trapezoid. By means of algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, while 3 and 4 are, respectively, with 3 and 4 variables.

After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.

In school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:

As can be seen from the example, at step (3) two equations were obtained: 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. Solving any of the equations will allow you to find out one of the variables x n.

Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.

The Gaussian method is difficult for middle school students to understand, but it is one of the most interesting ways to develop the ingenuity of children enrolled in advanced learning programs in math and physics classes.

For ease of recording, calculations are usually done as follows:

The coefficients of the equations and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right. Roman numerals indicate the numbers of equations in the system.

First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is ​​written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.

The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is ​​reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.

This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.

The free use of any solution method will require care and some experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of ​​human activity, while others exist for educational purposes.