home Treatment and prevention 5 Cramer method for solving systems of linear equations. Linear equations

5 Cramer method for solving systems of linear equations. Linear equations

2. Solving systems of equations using the matrix method (using an inverse matrix).
3. Gauss method for solving systems of equations.

Cramer's method.

The Cramer method is used to solve systems of linear algebraic equations ( SLAU).

Formulas using the example of a system of two equations with two variables.
Given: Solve the system using Cramer's method

Regarding variables X And at.
Solution:
Let's find the determinant of the matrix, composed of the coefficients of the system Calculation of determinants. :

Let's apply Cramer's formulas and find the values of the variables:
And .
Example 1:
Solve the system of equations:

regarding variables X And at.
Solution:

Let us replace the first column in this determinant with a column of coefficients from the right side of the system and find its value:

Let's do a similar thing, replacing the second column in the first determinant:

Applicable Cramer's formulas and find the values of the variables:
And .
Answer:
Comment: This method can solve systems of higher dimensions.

Comment: If it turns out that , but cannot be divided by zero, then they say that the system does not have a unique solution. In this case, the system either has infinitely many solutions or has no solutions at all.

Example 2(infinite number of solutions):

Solve the system of equations:

regarding variables X And at.
Solution:
Let us find the determinant of the matrix, composed of the coefficients of the system:

Solving systems using the substitution method.

The first of the system's equations is an equality that is true for any values of the variables (because 4 is always equal to 4). This means there is only one equation left. This is an equation for the relationship between variables.
We found that the solution to the system is any pair of values of variables related to each other by the equality .
Common decision will be written like this:
Particular solutions can be determined by choosing an arbitrary value of y and calculating x from this connection equality.

etc.
There are infinitely many such solutions.
Answer: common decision
Private solutions:

Example 3(no solutions, system is incompatible):

Solve the system of equations:

Solution:
Let us find the determinant of the matrix, composed of the coefficients of the system:

Cramer's formulas cannot be used. Let's solve this system using the substitution method

The second equation of the system is an equality that is not true for any values of the variables (of course, since -15 is not equal to 2). If one of the equations of the system is not true for any values of the variables, then the entire system has no solutions.
Answer: no solutions

In order to master this paragraph, you must be able to reveal the determinants “two by two” and “three by three”. If you are bad with qualifiers, please study the lesson How to calculate the determinant?

First we will look in detail at Cramer's rule for the system of two linear equations with two unknowns. For what? – After all, the simplest system can be solved using the school method, the method of term-by-term addition!

The fact is that, albeit sometimes, such a task occurs - to solve a system of two linear equations with two unknowns using Cramer's formulas. Secondly, a simpler example will help you understand how to use Cramer's rule to more complex case– systems of three equations with three unknowns.

In addition, there are systems of linear equations with two variables, which are advisable to solve using Cramer’s rule!

Consider the system of equations

At the first step, we calculate the determinant, it is called main determinant of the system.

Gauss method.

If , then the system has a unique solution, and to find the roots we must calculate two more determinants:
And

In practice, the above qualifiers can also be denoted by a Latin letter.

We find the roots of the equation using the formulas:
,

Example 7

Solve a system of linear equations

Solution: We see that the coefficients of the equation are quite large, on the right side there are decimals with a comma. The comma is a rather rare guest in practical tasks in mathematics; I took this system from an econometric problem.

How to solve such a system? You can try to express one variable in terms of another, but in this case you will probably end up with terrible fancy fractions that are extremely inconvenient to work with, and the design of the solution will look simply terrible. You can multiply the second equation by 6 and subtract term by term, but the same fractions will arise here too.

What to do? In such cases, Cramer's formulas come to the rescue.

;

Answer: ,

Both roots have infinite tails and are found approximately, which is quite acceptable (and even commonplace) for econometrics problems.

Comments are not needed here, since the task is solved using ready-made formulas, however, there is one caveat. When to use this method, compulsory A fragment of the task design is the following fragment: “This means that the system has a unique solution”. Otherwise, the reviewer may punish you for disrespect for Cramer's theorem.

It would not be superfluous to check, which is convenient to carry out on a calculator: we substitute approximate values in left side each equation of the system. As a result, with a small error, you should get numbers that are on the right sides.

Example 8

Present the answer in ordinary improper fractions. Do a check.

This is an example for you to solve on your own (an example of the final design and the answer at the end of the lesson).

Let's move on to consider Cramer's rule for a system of three equations with three unknowns:

We find the main determinant of the system:

If , then the system has infinitely many solutions or is inconsistent (has no solutions). In this case, Cramer's rule will not help; you need to use the Gauss method.

If , then the system has a unique solution and to find the roots we must calculate three more determinants:
, ,

And finally, the answer is calculated using the formulas:

As you can see, the “three by three” case is fundamentally no different from the “two by two” case; the column of free terms sequentially “walks” from left to right along the columns of the main determinant.

Example 9

Solve the system using Cramer's formulas.

Solution: Let's solve the system using Cramer's formulas.

, which means the system has a unique solution.

Answer: .

Actually, here again there is nothing special to comment on, due to the fact that the solution follows ready-made formulas. But there are a couple of comments.

It happens that as a result of calculations, “bad” irreducible fractions are obtained, for example: .
I recommend the following “treatment” algorithm. If you don’t have a computer at hand, do this:

1) There may be an error in the calculations. As soon as you encounter a “bad” fraction, you immediately need to check Is the condition rewritten correctly?. If the condition is rewritten without errors, then you need to recalculate the determinants using expansion in another row (column).

2) If no errors are identified as a result of checking, then most likely there was a typo in the task conditions. In this case, calmly and CAREFULLY work through the task to the end, and then be sure to check and we draw it up on a clean sheet after the decision. Of course, checking a fractional answer is an unpleasant task, but it will be a disarming argument for the teacher, who really likes to give a minus for any bullshit like . How to handle fractions is described in detail in the answer to Example 8.

If you have a computer at hand, then use an automated program to check, which can be downloaded for free at the very beginning of the lesson. By the way, it is most profitable to use the program right away (even before starting the solution); you will immediately see the intermediate step where you made a mistake! The same calculator automatically calculates the solution to the system matrix method.

Second remark. From time to time there are systems in the equations of which some variables are missing, for example:

Here in the first equation there is no variable , in the second there is no variable . In such cases, it is very important to correctly and CAREFULLY write down the main determinant:
– zeros are placed in place of missing variables.
By the way, it is rational to open determinants with zeros according to the row (column) in which the zero is located, since there are noticeably fewer calculations.

Example 10

Solve the system using Cramer's formulas.

This is an example for an independent solution (a sample of the final design and the answer at the end of the lesson).

For the case of a system of 4 equations with 4 unknowns, Cramer’s formulas are written according to similar principles. You can see a live example in the lesson Properties of Determinants. Reducing the order of the determinant - five 4th order determinants are quite solvable. Although the task is already very reminiscent of a professor’s shoe on the chest of a lucky student.

Solving the system using an inverse matrix

Method inverse matrix- this is essentially special case matrix equation(See Example No. 3 of the specified lesson).

To study this section, you must be able to expand determinants, find the inverse of a matrix, and perform matrix multiplication. Relevant links will be provided as the explanations progress.

Example 11

Solve the system using the matrix method

Solution: Let's write the system in matrix form:
, Where

Please look at the system of equations and matrices. I think everyone understands the principle by which we write elements into matrices. The only comment: if some variables were missing from the equations, then zeros would have to be placed in the corresponding places in the matrix.

We find the inverse matrix using the formula:
, where is the transposed matrix algebraic additions corresponding matrix elements.

First, let's look at the determinant:

Here the determinant is expanded on the first line.

Attention! If , then the inverse matrix does not exist, and it is impossible to solve the system using the matrix method. In this case, the system is solved by the method of eliminating unknowns (Gauss method).

Now we need to calculate 9 minors and write them into the minors matrix

Reference: It is useful to know the meaning of double subscripts in linear algebra. The first digit is the number of the line in which the element is located. The second digit is the number of the column in which the element is located:

That is, a double subscript indicates that the element is in the first row, third column, and, for example, the element is in 3 row, 2 column

During the solution, it is better to describe the calculation of minors in detail, although with some experience you can get used to calculating them with errors orally.

In the first part we looked a little theoretical material, the substitution method, as well as the method of term-by-term addition of system equations. I recommend everyone who accessed the site through this page to read the first part. Perhaps some visitors will find the material too simple, but in the process of solving systems of linear equations, I made a number of very important comments and conclusions regarding the solution of mathematical problems in general.

Now we will analyze Cramer’s rule, as well as solving a system of linear equations using an inverse matrix (matrix method). All materials are presented simply, in detail and clearly; almost all readers will be able to learn how to solve systems using the above methods.

First, we will take a closer look at Cramer's rule for a system of two linear equations in two unknowns. For what? – After all, the simplest system can be solved using the school method, the method of term-by-term addition!

The fact is that, albeit sometimes, such a task occurs - to solve a system of two linear equations with two unknowns using Cramer's formulas. Secondly, a simpler example will help you understand how to use Cramer's rule for a more complex case - a system of three equations with three unknowns.

In addition, there are systems of linear equations with two variables, which are advisable to solve using Cramer’s rule!

Consider the system of equations

At the first step, we calculate the determinant, it is called main determinant of the system.

Gauss method.

If , then the system has a unique solution, and to find the roots we must calculate two more determinants:
And

In practice, the above qualifiers can also be denoted by a Latin letter.

We find the roots of the equation using the formulas:
,

Example 7

Solve a system of linear equations

Solution: We see that the coefficients of the equation are quite large; on the right side there are decimal fractions with a comma. The comma is a rather rare guest in practical tasks in mathematics; I took this system from an econometric problem.

What to do? In such cases, Cramer's formulas come to the rescue.

;

Answer: ,

Both roots have infinite tails and are found approximately, which is quite acceptable (and even commonplace) for econometrics problems.

Comments are not needed here, since the task is solved using ready-made formulas, however, there is one caveat. When using this method, compulsory A fragment of the task design is the following fragment: “This means that the system has a unique solution”. Otherwise, the reviewer may punish you for disrespect for Cramer's theorem.

It would not be superfluous to check, which can be conveniently carried out on a calculator: we substitute approximate values into the left side of each equation of the system. As a result, with a small error, you should get numbers that are on the right sides.

Example 8

Present the answer in ordinary improper fractions. Do a check.

This is an example for you to solve on your own (an example of the final design and the answer at the end of the lesson).

Let's move on to consider Cramer's rule for a system of three equations with three unknowns:

We find the main determinant of the system:

If , then the system has infinitely many solutions or is inconsistent (has no solutions). In this case, Cramer's rule will not help; you need to use the Gauss method.

If , then the system has a unique solution and to find the roots we must calculate three more determinants:
, ,

And finally, the answer is calculated using the formulas:

Example 9

Solve the system using Cramer's formulas.

Solution: Let's solve the system using Cramer's formulas.

, which means the system has a unique solution.

Answer: .

Actually, here again there is nothing special to comment on, due to the fact that the solution follows ready-made formulas. But there are a couple of comments.

Example 10

Solve the system using Cramer's formulas.

This is an example for an independent solution (a sample of the final design and the answer at the end of the lesson).

Solving the system using an inverse matrix

The inverse matrix method is essentially a special case matrix equation(See Example No. 3 of the specified lesson).

To study this section, you must be able to expand determinants, find the inverse of a matrix, and perform matrix multiplication. Relevant links will be provided as the explanations progress.

Example 11

Solve the system using the matrix method

Solution: Let's write the system in matrix form:
, Where

We find the inverse matrix using the formula:
, where is the transposed matrix of algebraic complements of the corresponding elements of the matrix.

First, let's look at the determinant:

Here the determinant is expanded on the first line.

Now we need to calculate 9 minors and write them into the minors matrix

Cramer's method or the so-called Cramer's rule is a method of searching for unknown quantities from systems of equations. It can be used only if the number of sought values is equivalent to the number of algebraic equations in the system, that is, the main matrix formed from the system must be square and not contain zero rows, and also if its determinant must not be zero.

Theorem 1

Cramer's theorem If the main determinant $D$ of the main matrix, compiled on the basis of the coefficients of the equations, is not equal to zero, then the system of equations is consistent, and it has a unique solution. The solution to such a system is calculated through the so-called Cramer formulas for solving systems of linear equations: $x_i = \frac(D_i)(D)$

What is the Cramer method?

The essence of Cramer's method is as follows:

To find a solution to the system using Cramer's method, first of all we calculate the main determinant of the matrix $D$. When the calculated determinant of the main matrix, when calculated by Cramer's method, turns out to be equal to zero, then the system does not have a single solution or has an infinite number of solutions. In this case, to find a general or some basic answer for the system, it is recommended to use the Gaussian method.
Then you need to replace the outermost column main matrix to the column of free terms and calculate the determinant $D_1$.
Repeat the same for all columns, obtaining determinants from $D_1$ to $D_n$, where $n$ is the number of the rightmost column.
After all determinants $D_1$...$D_n$ have been found, the unknown variables can be calculated using the formula $x_i = \frac(D_i)(D)$.

Techniques for calculating the determinant of a matrix

To calculate the determinant of a matrix with a dimension greater than 2 by 2, you can use several methods:

The rule of triangles, or Sarrus's rule, reminiscent of the same rule. The essence of the triangle method is that when calculating the determinant, the products of all numbers connected in the figure by the red line on the right are written with a plus sign, and all numbers connected in a similar way in the figure on the left are written with a minus sign. Both rules are suitable for matrices of size 3 x 3. In the case of the Sarrus rule, the matrix itself is first rewritten, and next to it its first and second columns are rewritten again. Diagonals are drawn through the matrix and these additional columns; matrix members lying on the main diagonal or parallel to it are written with a plus sign, and elements lying on or parallel to the secondary diagonal are written with a minus sign.

Figure 1. Triangle rule for calculating the determinant for Cramer's method

Using a method known as the Gaussian method, this method is also sometimes called reducing the order of the determinant. In this case, the matrix is transformed and reduced to triangular view, and then all the numbers on the main diagonal are multiplied. It should be remembered that when searching for a determinant in this way, you cannot multiply or divide rows or columns by numbers without taking them out as a multiplier or divisor. In the case of searching for a determinant, it is only possible to subtract and add rows and columns to each other, having previously multiplied the subtracted row by a non-zero factor. Also, whenever you rearrange the rows or columns of the matrix, you should remember the need to change the final sign of the matrix.
When solving a SLAE with 4 unknowns using the Cramer method, it is best to use the Gauss method to search and find determinants or determine the determinant by searching for minors.

Solving systems of equations using Cramer's method

Let's apply Cramer's method for a system of 2 equations and two required quantities:

$\begin(cases) a_1x_1 + a_2x_2 = b_1 \\ a_3x_1 + a_4x_2 = b_2 \\ \end(cases)$

Let's display it in expanded form for convenience:

$A = \begin(array)(cc|c) a_1 & a_2 & b_1 \\ a_3 & a_4 & b_1 \\ \end(array)$

Let's find the determinant of the main matrix, also called the main determinant of the system:

$D = \begin(array)(|cc|) a_1 & a_2 \\ a_3 & a_4 \\ \end(array) = a_1 \cdot a_4 – a_3 \cdot a_2$

If the main determinant is not equal to zero, then to solve the slough using Cramer’s method it is necessary to calculate a couple more determinants from two matrices with the columns of the main matrix replaced by a row of free terms:

$D_1 = \begin(array)(|cc|) b_1 & a_2 \\ b_2 & a_4 \\ \end(array) = b_1 \cdot a_4 – b_2 \cdot a_4$

$D_2 = \begin(array)(|cc|) a_1 & b_1 \\ a_3 & b_2 \\ \end(array) = a_1 \cdot b_2 – a_3 \cdot b_1$

Now let's find the unknowns $x_1$ and $x_2$:

$x_1 = \frac (D_1)(D)$

$x_2 = \frac (D_2)(D)$

Example 1

Cramer's method for solving SLAEs with a main matrix of 3rd order (3 x 3) and three required ones.

Solve the system of equations:

$\begin(cases) 3x_1 – 2x_2 + 4x_3 = 21 \\ 3x_1 +4x_2 + 2x_3 = 9\\ 2x_1 – x_2 - x_3 = 10 \\ \end(cases)$

Let's calculate the main determinant of the matrix using the rule stated above under point number 1:

$D = \begin(array)(|ccc|) 3 & -2 & 4 \\3 & 4 & -2 \\ 2 & -1 & 1 \\ \end(array) = 3 \cdot 4 \cdot ( -1) + 2 \cdot (-2) \cdot 2 + 4 \cdot 3 \cdot (-1) – 4 \cdot 4 \cdot 2 – 3 \cdot (-2) \cdot (-1) - (- 1) \cdot 2 \cdot 3 = - 12 – 8 -12 -32 – 6 + 6 = - 64$

And now three other determinants:

$D_1 = \begin(array)(|ccc|) 21 & 2 & 4 \\ 9 & 4 & 2 \\ 10 & 1 & 1 \\ \end(array) = 21 \cdot 4 \cdot 1 + (- 2) \cdot 2 \cdot 10 + 9 \cdot (-1) \cdot 4 – 4 \cdot 4 \cdot 10 – 9 \cdot (-2) \cdot (-1) - (-1) \cdot 2 \ cdot 21 = - 84 – 40 – 36 – 160 – 18 + 42 = - $296

$D_2 = \begin(array)(|ccc|) 3 & 21 & 4 \\3 & 9 & 2 \\ 2 & 10 & 1 \\ \end(array) = 3 \cdot 9 \cdot (- 1) + 3 \cdot 10 \cdot 4 + 21 \cdot 2 \cdot 2 – 4 \cdot 9 \cdot 2 – 21 \cdot 3 \cdot (-1) – 2 \cdot 10 \cdot 3 = - 27 + 120 + 84 – 72 + 63 – 60 = $108

$D_3 = \begin(array)(|ccc|) 3 & -2 & 21 \\ 3 & 4 & 9 \\ 2 & 1 & 10 \\ \end(array) = 3 \cdot 4 \cdot 10 + 3 \cdot (-1) \cdot 21 + (-2) \cdot 9 \cdot 2 – 21 \cdot 4 \cdot 2 - (-2) \cdot 3 \cdot 10 - (-1) \cdot 9 \cdot 3 = 120 – 63 – 36 – 168 + 60 + 27 = - $60

Let's find the required quantities:

$x_1 = \frac(D_1) (D) = \frac(- 296)(-64) = 4 \frac(5)(8)$

$x_2 = \frac(D_1) (D) = \frac(108) (-64) = - 1 \frac (11) (16)$

$x_3 = \frac(D_1) (D) = \frac(-60) (-64) = \frac (15) (16)$

Consider a system of 3 equations with three unknowns

Using 3rd order determinants, the solution to such a system can be written in the same form as for a system of two equations, i.e.

(2.4)

if 0. Here

It's there Cramer's rule solving a system of three linear equations in three unknowns.

Example 2.3. Solve a system of linear equations using Cramer's rule:

Solution . Finding the determinant of the main matrix of the system

Since 0, then to find a solution to the system we can apply Cramer’s rule, but first we calculate three more determinants:

Examination:

Therefore, the solution was found correctly. 

Cramer's rules derived for linear systems 2nd and 3rd order, suggest that the same rules can be formulated for linear systems of any order. Really happens

Cramer's theorem. Quadratic system of linear equations with a nonzero determinant of the main matrix of the system (0) has one and only one solution and this solution is calculated using the formulas

(2.5)

Where  – determinant of the main matrix,  i – matrix determinant, obtained from the main one, replacingith column column of free members.

Note that if =0, then Cramer’s rule does not apply. This means that the system either has no solutions at all or has infinitely many solutions.

Having formulated Cramer's theorem, the question naturally arises of calculating determinants of higher orders.

2.4. Determinants of nth order

Additional minor M ij element a ij is called the determinant obtained from given by deletions i th line and j th column. Algebraic complement A ij element a ij the minor of this element taken with the sign (–1) is called i + j, i.e. A ij = (–1) i + j M ij .

For example, let's find the minors and algebraic complements of the elements a 23 and a 31 qualifiers

We get



Using the concept of algebraic complement we can formulate determinant expansion theoremn-th order by row or column.

Theorem 2.1. Matrix determinantAis equal to the sum of the products of all elements of a certain row (or column) by their algebraic complements:

(2.6)

This theorem underlies one of the main methods for calculating determinants, the so-called. order reduction method. As a result of the expansion of the determinant n th order over any row or column, we get n determinants ( n–1)th order. To have fewer such determinants, it is advisable to select the row or column that has the most zeros. In practice, the expansion formula for the determinant is usually written as:

those. algebraic additions are written explicitly in terms of minors.

Examples 2.4. Calculate the determinants by first sorting them into some row or column. Typically, in such cases, select the column or row that has the most zeros. The selected row or column will be indicated by an arrow.



2.5. Basic properties of determinants

Expanding the determinant over any row or column, we get n determinants ( n–1)th order. Then each of these determinants ( n–1)th order can also be decomposed into a sum of determinants ( n–2)th order. Continuing this process, one can reach the 1st order determinants, i.e. to the elements of the matrix whose determinant is calculated. So, to calculate 2nd order determinants, you will have to calculate the sum of two terms, for 3rd order determinants - the sum of 6 terms, for 4th order determinants - 24 terms. The number of terms will increase sharply as the order of the determinant increases. This means that calculating determinants of very high orders becomes a rather labor-intensive task, beyond the capabilities of even a computer. However, determinants can be calculated in another way, using the properties of determinants.

Property 1 . The determinant will not change if the rows and columns in it are swapped, i.e. when transposing a matrix:

This property indicates the equality of the rows and columns of the determinant. In other words, any statement about the columns of a determinant is also true for its rows and vice versa.

Property 2 . The determinant changes sign when two rows (columns) are interchanged.

Consequence . If the determinant has two identical rows (columns), then it is equal to zero.

Property 3 . The common factor of all elements in any row (column) can be taken out of the determinant sign.

For example,

Consequence . If all elements of a certain row (column) of a determinant are equal to zero, then the determinant itself is equal to zero.

Property 4 . The determinant will not change if the elements of one row (column) are added to the elements of another row (column), multiplied by any number.

For example,

Property 5 . The determinant of the product of matrices is equal to the product of the determinants of matrices:

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5 Cramer method for solving systems of linear equations. Linear equations

Cramer's method.

Solving the system using an inverse matrix

Solving the system using an inverse matrix

What is the Cramer method?

Techniques for calculating the determinant of a matrix

Solving systems of equations using Cramer's method

2.4. Determinants of nth order

2.5. Basic properties of determinants