Solution of systems by Cramer's method examples. Linear equations

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

Cramer's method.

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

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

Concerning Variables X and at.
Solution:
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's 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 action, 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 be used to solve systems of higher dimensions.

Comment: If it turns out that , and it is impossible to divide by zero, then they say that the system does not have a unique solution. In this case, the system has either infinitely many solutions or no solutions at all.

Example 2(an infinite number of solutions):

Solve the system of equations:

regarding variables X and at.
Solution:
Find the determinant of the matrix, composed of the coefficients of the system:

Solving systems by the substitution method.

The first of the equations of the system is an equality that is true for any values of the variables (because 4 is always equal to 4). So there is only one equation left. This is a relationship equation between variables.
We got that the solution of the system is any pair of values of variables related by 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 relationship equation.

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

Example 3(no solutions, the system is inconsistent):

Solve the system of equations:

Solution:
Find the determinant of the matrix, composed of the coefficients of the system:

You can't use Cramer's formulas. Let's solve this system by the substitution method

The second equation of the system is an equality that is not valid 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 whole system has no solutions.
Answer: no solutions

Consider a system of 3 equations with three unknowns

Using third-order determinants, the solution of 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 is Cramer's rule solutions of the system of three linear equations with 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, you can apply Cramer's rule, but first calculate three more determinants:

Examination:

Therefore, the solution is 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 takes place

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

(2.5)

where  – main matrix determinant,  i – matrix determinant, derived from the main, replacementith column free members column.

Note that if =0, then Cramer's rule is not applicable. 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 higher-order determinants.

2.4. nth order determinants

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

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

We get



Using the concept of algebraic complement, we can formulate the 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 some row (or column) and 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 in any row or column, we get n determinants ( n–1)-th order. In order to have fewer such determinants, it is advisable to choose 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 expanding them in any row or column. Usually in such cases, choose the column or row that has the most zeros. The selected row or column will be marked with an arrow.



2.5. Basic properties of determinants

Expanding the determinant in 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 determinants of the 1st order, i.e. to the elements of the matrix whose determinant is being calculated. So, to calculate the 2nd order determinants, you will have to calculate the sum of two terms, for the 3rd order determinants - the sum of 6 terms, for the 4th order determinants - 24 terms. The number of terms will increase sharply as the order of the determinant increases. This means that the calculation of determinants of very high orders becomes a rather laborious task, beyond the power 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 rows and columns are swapped in it, i.e. when transposing a matrix:

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

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

Consequence . If a determinant has two identical rows (columns), then it zero.

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

For instance,

Consequence . If all elements of some row (column) of the 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 some number.

For instance,

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

With the number of equations the same as the number of unknowns with the main determinant of the matrix, which is not equal to zero, the coefficients of the system (there is a solution for such equations and it is only one).

Cramer's theorem.

When the determinant of the matrix of a square system is non-zero, then the system is compatible and it has one solution and it can be found by Cramer's formulas:

where Δ - system matrix determinant,

Δ i- determinant of the matrix of the system, in which instead of i th column is the column of right parts.

When the determinant of the system is zero, then the system can become consistent or inconsistent.

This method is usually used for small systems with volume calculations and if when it is necessary to determine 1 of the unknowns. The complexity of the method is that it is necessary to calculate many determinants.

Description of Cramer's method.

There is a system of equations:

A system of 3 equations can be solved by Cramer's method, which was discussed above for a system of 2 equations.

We compose the determinant from the coefficients of the unknowns:

It will be system qualifier. When D≠0, so the system is compatible. Now we will compose 3 additional determinants:

We solve the system by Cramer's formulas:

Examples of solving systems of equations by Cramer's method.

Example 1.

Given system:

Let's solve it by Cramer's method.

First you need to calculate the determinant of the matrix of the system:

Because Δ≠0, hence, from Cramer's theorem, the system is compatible and it has one solution. We calculate additional determinants. The determinant Δ 1 is obtained from the determinant Δ by replacing its first column with a column of free coefficients. We get:

In the same way, we obtain the determinant Δ 2 from the determinant of the matrix of the system, replacing the second column with a column of free coefficients:

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

We first consider Cramer's rule in detail for a system of two linear equations in two unknowns. What for? - After all the simplest system can be solved by the school method, by term addition!

The fact is that even if sometimes, but there is such a task - 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 it is advisable to solve exactly according to Cramer's rule!

Consider the system of equations

At the first step, we calculate the determinant , it is called the 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 the Latin letter.

The roots of the equation are found by 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 surely get terrible fancy fractions, which are extremely inconvenient to work with, and the design of the solution will look just awful. You can multiply the second equation by 6 and subtract term by term, but the same fractions will appear here.

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 according to ready-made formulas, however, there is one caveat. When use this method, compulsory The fragment of the assignment is the following fragment: "so the system has a unique solution". Otherwise, the reviewer may punish you for disrespecting Cramer's theorem.

It will not be superfluous to check, which is convenient to carry out on a calculator: we substitute the approximate values \u200b\u200bin the left side of each equation of the system. As a result, with a small error, numbers that are on the right side should be obtained.

Example 8

Express your answer in ordinary improper fractions. Make a check.

This is an example for independent solution(example of finishing and answer at the end of the lesson).

We turn to the consideration of 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 by 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.

, so the system has a unique solution.

Answer: .

Actually, there is nothing special to comment here again, in view of the fact that the decision is made according to ready-made formulas. But there are a couple of notes.

It happens that as a result of calculations, “bad” irreducible fractions are obtained, for example: .
I recommend the following "treatment" algorithm. If there is no computer at hand, we do this:

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

2) If no errors were found as a result of the check, then most likely a typo was made in the condition of the assignment. In this case, calmly and CAREFULLY solve the task to the end, and then make sure to check and draw it up on a clean copy after the decision. Of course, checking a fractional answer is an unpleasant task, but it will be a disarming argument for the teacher, who, well, really likes to put a minus for any bad thing like. How to deal with fractions is detailed in the answer for Example 8.

If you have a computer at hand, then use an automated program to check it, which can be downloaded for free at the very beginning of the lesson. By the way, it is most advantageous to use the program right away (even before starting the solution), you will immediately see the intermediate step at which you made a mistake! The same calculator automatically calculates the solution of 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 put in place of missing variables.
By the way, it is rational to open determinants with zeros in the row (column) in which zero is located, since there are noticeably fewer calculations.

Example 10

Solve the system using Cramer's formulas.

This is an example for self-solving (finishing sample and 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 Determinant Properties lesson. 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.

Solution of the system using the inverse matrix

Method inverse matrix is, in fact, a special case matrix equation(See Example No. 3 of the specified lesson).

To study this section, you need to be able to expand the determinants, find the inverse matrix and perform matrix multiplication. Relevant links will be given as the explanation progresses.

Example 11

Solve the system with the matrix method

Solution: We write the system in matrix form:
, where

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

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

First, let's deal with the determinant:

Here the determinant is expanded by the first line.

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

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

Reference: It is useful to know the meaning of double subscripts in linear algebra. The first digit is the line number 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, while, for example, the element is in the 3rd row, 2nd column

In the course of solving, it is better to describe the calculation of minors in detail, although, with a certain experience, they can be adjusted to count with errors orally.

Cramer's method or the so-called Cramer's rule is a way to search for unknown quantities from systems of equations. It can be used only if the number of required 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 of such a system is calculated using 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 the Cramer method is as follows:

To find a solution to the system by 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 the Cramer method, turned 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 apply the Gaussian method.
Then you need to replace the last column of the main matrix with the column of free members and calculate the determinant $D_1$.
Repeat the same for all columns, getting the determinants from $D_1$ to $D_n$, where $n$ is the number of the rightmost column.
After all determinants of $D_1$...$D_n$ are 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, several methods can be used:

The rule of triangles, or the rule of Sarrus, resembling the same rule. The essence of the triangle method is that when calculating the determinant of the product of all numbers connected in the figure by a red line on the right, they are written with a plus sign, and all numbers connected in a similar way in the figure on the left - with a minus sign. Both rules are suitable for 3 x 3 matrices. 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 the secondary diagonal or parallel to it are written with a minus sign.

Figure 1. Rule of triangles for calculating the determinant for the Cramer method

With a method known as the Gaussian method, this method is also sometimes referred to as determinant reduction. In this case, the matrix is transformed and brought to a triangular form, and then all the numbers on the main diagonal are multiplied. It should be remembered that in such a search for a determinant, one cannot multiply or divide rows or columns by numbers without taking them out as a factor 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, with each permutation of the rows or columns of the matrix, one should remember the need to change the final sign of the matrix.
When solving Cramer's SLAE with 4 unknowns, it is best to use the Gaussian method to search and find determinants or determine the determinant through the search for minors.

Solving systems of equations by Cramer's method

We apply the Cramer 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 an expanded form for convenience:

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

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 by the Cramer 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 members:

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

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

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 a SLAE with a 3rd order (3 x 3) main matrix and three desired ones.

Solve the system of equations:

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

We calculate the main determinant of the matrix using the above rule under paragraph 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 \u003d 120 - 63 - 36 - 168 + 60 + 27 \u003d - $ 60

Let's find the required values:

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

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

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