home biology tutor How to check the uniqueness of the solutions of systems. Examples of systems of linear equations: solution method. General solution of a system of linear equations

How to check the uniqueness of the solutions of systems. Examples of systems of linear equations: solution method. General solution of a system of linear equations

To investigate a system of linear agebraic equations (SLAE) for compatibility means to find out whether this system has solutions or not. Well, if there are solutions, then indicate how many of them.

We will need information from the topic "System of linear algebraic equations. Basic terms. Matrix notation". In particular, such concepts as the matrix of the system and the extended matrix of the system are needed, since the formulation of the Kronecker-Capelli theorem is based on them. As usual, the matrix of the system will be denoted by the letter $A$, and the extended matrix of the system by the letter $\widetilde(A)$.

Kronecker-Capelli theorem

Linear system algebraic equations is consistent if and only if the rank of the system matrix is equal to the rank of the extended matrix of the system, i.e. $\rank A=\rang\widetilde(A)$.

Let me remind you that a system is called joint if it has at least one solution. The Kronecker-Capelli theorem says this: if $\rang A=\rang\widetilde(A)$, then there is a solution; if $\rang A\neq\rang\widetilde(A)$, then this SLAE has no solutions (is inconsistent). The answer to the question about the number of these solutions is given by a corollary of the Kronecker-Capelli theorem. The statement of the corollary uses the letter $n$, which is equal to the number of variables in the given SLAE.

Corollary from the Kronecker-Capelli theorem

If $\rang A\neq\rang\widetilde(A)$, then the SLAE is inconsistent (has no solutions).
If $\rang A=\rang\widetilde(A)< n$, то СЛАУ является неопределённой (имеет бесконечное количество решений).
If $\rang A=\rang\widetilde(A) = n$, then the SLAE is definite (it has exactly one solution).

Note that the formulated theorem and its corollary do not indicate how to find the solution to the SLAE. With their help, you can only find out whether these solutions exist or not, and if they exist, then how many.

Example #1

Explore SLAE $ \left \(\begin(aligned) & -3x_1+9x_2-7x_3=17;\\ & -x_1+2x_2-4x_3=9;\\ & 4x_1-2x_2+19x_3=-42. \end(aligned )\right.$ for consistency If the SLAE is consistent, indicate the number of solutions.

To find out the existence of solutions to a given SLAE, we use the Kronecker-Capelli theorem. We need the matrix of the system $A$ and the extended matrix of the system $\widetilde(A)$, we write them down:

$$ A=\left(\begin(array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \end(array) \right);\; \widetilde(A)=\left(\begin(array) (ccc|c) -3 & 9 &-7 & 17 \\ -1 & 2 & -4 & 9\\ 4 & -2 & 19 & -42 \end(array)\right). $$

We need to find $\rang A$ and $\rang\widetilde(A)$. There are many ways to do this, some of which are listed in the Matrix Rank section. Usually, two methods are used to study such systems: "Calculation of the rank of a matrix by definition" or "Calculation of the rank of a matrix by the method of elementary transformations".

Method number 1. Calculation of ranks by definition.

According to the definition, the rank is the highest order of the minors of the matrix , among which there is at least one other than zero. Usually, the study begins with the first-order minors, but here it is more convenient to proceed immediately to the calculation of the third-order minor of the matrix $A$. The elements of the third-order minor are at the intersection of three rows and three columns of the matrix under consideration. Since the matrix $A$ contains only 3 rows and 3 columns, the third order minor of the matrix $A$ is the determinant of the matrix $A$, i.e. $\DeltaA$. To calculate the determinant, we apply formula No. 2 from the topic "Formulas for calculating second and third order determinants":

$$ \Delta A=\left| \begin(array) (ccc) -3 & 9 & -7 \\ -1 & 2 & -4 \\ 4 & -2 & 19 \end(array) \right|=-21. $$

So, there is a third-order minor of the matrix $A$, which is not equal to zero. A 4th-order minor cannot be composed, since it requires 4 rows and 4 columns, and the matrix $A$ has only 3 rows and 3 columns. So, the highest order of minors of the matrix $A$, among which there is at least one non-zero one, is equal to 3. Therefore, $\rang A=3$.

We also need to find $\rang\widetilde(A)$. Let's look at the structure of the $\widetilde(A)$ matrix. Up to the line in the matrix $\widetilde(A)$ there are elements of the matrix $A$, and we found out that $\Delta A\neq 0$. Therefore, the matrix $\widetilde(A)$ has a third-order minor that is not equal to zero. We cannot compose fourth-order minors of the matrix $\widetilde(A)$, so we conclude: $\rang\widetilde(A)=3$.

Since $\rang A=\rang\widetilde(A)$, according to the Kronecker-Capelli theorem, the system is consistent, i.e. has a solution (at least one). To indicate the number of solutions, we take into account that our SLAE contains 3 unknowns: $x_1$, $x_2$ and $x_3$. Since the number of unknowns is $n=3$, we conclude: $\rang A=\rang\widetilde(A)=n$, therefore, according to the corollary of the Kronecker-Capelli theorem, the system is definite, i.e. has a unique solution.

Problem solved. What are the disadvantages and advantages of this method? First, let's talk about the pros. First, we needed to find only one determinant. After that, we immediately made a conclusion about the number of solutions. Usually, in standard typical calculations, systems of equations are given that contain three unknowns and have a single solution. For such systems this method very convenient, because we know in advance that there is a solution (otherwise there would be no example in a typical calculation). Those. it only remains for us to show that there is a solution to the most fast way. Secondly, the calculated value of the system matrix determinant (i.e. $\Delta A$) will come in handy later: when we start solving the given system using the Cramer method or using the inverse matrix .

However, by definition, the method of calculating the rank is undesirable if the system matrix $A$ is rectangular. In this case, it is better to apply the second method, which will be discussed below. Besides, if $\Delta A=0$, then we will not be able to say anything about the number of solutions for a given inhomogeneous SLAE. Maybe SLAE has an infinite number of solutions, or maybe none. If $\Delta A=0$ then it is required additional research, which is often cumbersome.

Summarizing what has been said, I note that the first method is good for those SLAEs whose system matrix is square. At the same time, the SLAE itself contains three or four unknowns and is taken from standard standard calculations or control works.

Method number 2. Calculation of the rank by the method of elementary transformations.

This method is described in detail in the corresponding topic. We will calculate the rank of the matrix $\widetilde(A)$. Why matrices $\widetilde(A)$ and not $A$? The point is that the matrix $A$ is a part of the matrix $\widetilde(A)$, so by calculating the rank of the matrix $\widetilde(A)$ we will simultaneously find the rank of the matrix $A$.

\begin(aligned) &\widetilde(A) =\left(\begin(array) (ccc|c) -3 & 9 &-7 & 17 \\ -1 & 2 & -4 & 9\\ 4 & - 2 & 19 & -42 \end(array) \right) \rightarrow \left|\text(swap first and second lines)\right| \rightarrow \\ &\rightarrow \left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ -3 & 9 &-7 & 17\\ 4 & -2 & 19 & - 42 \end(array) \right) \begin(array) (l) \phantom(0) \\ r_2-3r_1\\ r_3+4r_1 \end(array) \rightarrow \left(\begin(array) (ccc| c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 6 & 3 & -6 \end(array) \right) \begin(array) (l) \phantom(0 ) \\ \phantom(0)\\ r_3-2r_2 \end(array)\rightarrow\\ &\rightarrow \left(\begin(array) (ccc|c) -1 & 2 & -4 & 9 \\ 0 & 3 &5 & -10\\ 0 & 0 & -7 & 14 \end(array) \right) \end(aligned)

We have reduced the matrix $\widetilde(A)$ to a stepped form . The resulting step matrix has three non-zero rows, so its rank is 3. Therefore, the rank of the matrix $\widetilde(A)$ is 3, i.e. $\rank\widetilde(A)=3$. Making transformations with the elements of the matrix $\widetilde(A)$, we simultaneously transformed the elements of the matrix $A$ located before the line. The $A$ matrix is also stepped: $\left(\begin(array) (ccc) -1 & 2 & -4 \\ 0 & 3 &5 \\ 0 & 0 & -7 \end(array) \right )$. Conclusion: the rank of the matrix $A$ is also equal to 3, i.e. $\rank A=3$.

Since $\rang A=\rang\widetilde(A)$, according to the Kronecker-Capelli theorem, the system is consistent, i.e. has a solution. To indicate the number of solutions, we take into account that our SLAE contains 3 unknowns: $x_1$, $x_2$ and $x_3$. Since the number of unknowns is $n=3$, we conclude: $\rang A=\rang\widetilde(A)=n$, therefore, according to the corollary of the Kronecker-Capelli theorem, the system is defined, i.e. has a unique solution.

What are the advantages of the second method? The main advantage is its versatility. It doesn't matter to us whether the matrix of the system is square or not. In addition, we have actually carried out transformations of the Gauss method forward. There are only a couple of steps left, and we could get the solution of this SLAE. To be honest, I like the second way more than the first, but the choice is a matter of taste.

Answer: The given SLAE is consistent and defined.

Example #2

Explore SLAE $ \left\( \begin(aligned) & x_1-x_2+2x_3=-1;\\ & -x_1+2x_2-3x_3=3;\\ & 2x_1-x_2+3x_3=2;\\ & 3x_1- 2x_2+5x_3=1;\\ & 2x_1-3x_2+5x_3=-4.\end(aligned) \right.$ for compatibility.

We will find the ranks of the system matrix and the extended matrix of the system by the method of elementary transformations. Extended system matrix: $\widetilde(A)=\left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ -1 & 2 & -3 & 3 \\ 2 & -1 & 3 & 2 \\ 3 & -2 & 5 & 1 \\ 2 & -3 & 5 & -4 \end(array) \right)$. Let's find the required ranks by transforming the augmented matrix of the system:

$$ \left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ -1 & 2 & -3 & 3 \\ 2 & -3 & 5 & -4 \\ 3 & -2 & 5 & 1 \\ 2 & -1 & 3 & 2 \end(array) \right) \begin(array) (l) \phantom(0)\\r_2+r_1\\r_3-2r_1\\ r_4 -3r_1\\r_5-2r_1\end(array)\rightarrow \left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ 0 & 1 & -1 & 2 \\ 0 & -1 & 1 & -2 \\ 0 & 1 & -1 & 4 \\ 0 & 1 & -1 & 4 \end(array) \right) \begin(array) (l) \phantom(0)\\ \phantom(0)\\r_3-r_2\\ r_4-r_2\\r_5+r_2\end(array)\rightarrow\\ $$ $$ \rightarrow\left(\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \end(array) \ right) \begin(array) (l) \phantom(0)\\\phantom(0)\\\phantom(0)\\ r_4-r_3\\\phantom(0)\end(array)\rightarrow \left (\begin(array) (ccc|c) 1 & -1 & 2 & -1\\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end(array) \right) $$

The extended matrix of the system is reduced to a stepped form. The rank of a step matrix is equal to the number of its nonzero rows, so $\rang\widetilde(A)=3$. The matrix $A$ (up to the line) is also reduced to a stepped form, and its rank is equal to 2, $\rang(A)=2$.

Since $\rang A\neq\rang\widetilde(A)$, then, according to the Kronecker-Capelli theorem, the system is inconsistent (ie, has no solutions).

Answer: The system is inconsistent.

Example #3

Explore SLAE $ \left\( \begin(aligned) & 2x_1+7x_3-5x_4+11x_5=42;\\ & x_1-2x_2+3x_3+2x_5=17;\\ & -3x_1+9x_2-11x_3-7x_5=-64 ;\\ & -5x_1+17x_2-16x_3-5x_4-4x_5=-90;\\ & 7x_1-17x_2+23x_3+15x_5=132. \end(aligned) \right.$ for compatibility.

We bring the augmented matrix of the system to a stepped form:

$$ \left(\begin(array)(ccccc|c) 2 & 0 & 7 & -5 & 11 & 42\\ 1 & -2 & 3 & 0 & 2 & 17 \\ -3 & 9 & -11 & 0 & -7 & -64 \\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \end(array) \right) \overset (r_1\leftrightarrow(r_3))(\rightarrow) $$ $$ \rightarrow\left(\begin(array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 2 & 0 & 7 & -5 & 11 & 42\\ -3 & 9 & -11 & 0 & -7 & -64\\ -5 & 17 & -16 & -5 & -4 & -90 \\ 7 & -17 & 23 & 0 & 15 & 132 \end(array) \right) \begin(array) (l) \phantom(0)\\ r_2-2r_1 \\r_3+3r_1 \\ r_4+5r_1 \\ r_5-7r_1 \end( array) \rightarrow \left(\begin(array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 0 & 4 & 1 & -5 & 7 & 8\\ 0 & 3 & - 2 & 0 & -1 & -13\\ 0 & 7 & -1 & -5 & 6 & -5 \\ 0 & -3 & 2 & 0 & 1 & 13 \end(array) \right) \begin( array) (l) \phantom(0)\\ \phantom(0)\\4r_3+3r_2 \\ 4r_4-7r_2 \\ 4r_5+3r_2 \end(array) \rightarrow $$ $$ \rightarrow\left(\begin (array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 0 & 4 & 1 & -5 & 7 & 8\\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & -11 & 15 & -25 & -76 \\ 0 & 0 & 11 & -15 & 25 & 76 \end(array) \right) \begin(array) (l) \phantom(0)\\ \phantom(0)\\\phantom(0) \\ r_4 -r_3 \\ r_5+r_2 \end(array) \rightarrow \left(\begin(array)(ccccc|c) 1 & -2 & 3 & 0 & 2 & 17\\ 0 & 4 & 1 & -5 & 7 & 8\\ 0 & 0 & -11 & 15 & -25 & -76\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end(array) \right) $$

We have reduced the extended matrix of the system and the matrix of the system itself to a stepped form. The rank of the extended matrix of the system is equal to three, the rank of the matrix of the system is also equal to three. Since the system contains $n=5$ unknowns, i.e. $\rang\widetilde(A)=\rang(A)\lt(n)$, then, according to the corollary of the Kronecker-Capelli theorem, this system is indeterminate, i.e. has an infinite number of solutions.

Answer: the system is indeterminate.

In the second part, we will analyze examples that are often included in typical calculations or test papers in higher mathematics: a study on the compatibility and solution of SLAE depending on the values of the parameters included in it.

Solving systems of linear algebraic equations (SLAE) is undoubtedly the most important topic of the linear algebra course. A huge number of problems from all branches of mathematics are reduced to solving systems linear equations. These factors explain the reason for creating this article. The material of the article is selected and structured so that with its help you can

choose the optimal method for solving your system of linear algebraic equations,
study the theory of the chosen method,
solve your system of linear equations, having considered in detail the solutions of typical examples and problems.

Brief description of the material of the article.

Let's give it all first necessary definitions, concepts, and introduce notation.

Next, we consider methods for solving systems of linear algebraic equations in which the number of equations is equal to the number of unknown variables and which have a unique solution. First, let's focus on the Cramer method, secondly, we will show the matrix method for solving such systems of equations, and thirdly, we will analyze the Gauss method (the method of successive elimination of unknown variables). To consolidate the theory, we will definitely solve several SLAEs in various ways.

After that, we turn to solving systems of linear algebraic equations of a general form, in which the number of equations does not coincide with the number of unknown variables or the main matrix of the system is degenerate. We formulate the Kronecker-Capelli theorem, which allows us to establish the compatibility of SLAEs. Let us analyze the solution of systems (in the case of their compatibility) using the concept of the basis minor of a matrix. We will also consider the Gauss method and describe in detail the solutions of the examples.

Be sure to dwell on the structure of the general solution of homogeneous and inhomogeneous systems of linear algebraic equations. Let us give the concept of a fundamental system of solutions and show how the general solution of the SLAE is written using the vectors of the fundamental system of solutions. For a better understanding, let's look at a few examples.

In conclusion, we consider systems of equations that are reduced to linear ones, as well as various problems, in the solution of which SLAEs arise.

Page navigation.

Definitions, concepts, designations.

We will consider systems of p linear algebraic equations with n unknown variables (p may be equal to n ) of the form

Unknown variables, - coefficients (some real or complex numbers), - free members (also real or complex numbers).

This form of SLAE is called coordinate.

AT matrix form this system of equations has the form ,
where - the main matrix of the system, - the matrix-column of unknown variables, - the matrix-column of free members.

If we add to the matrix A as the (n + 1)-th column the matrix-column of free terms, then we get the so-called expanded matrix systems of linear equations. Usually, the augmented matrix is denoted by the letter T, and the column of free members is separated by a vertical line from the rest of the columns, that is,

By solving a system of linear algebraic equations called a set of values of unknown variables , which turns all the equations of the system into identities. Matrix equation for the given values of the unknown variables also turns into an identity .

If a system of equations has at least one solution, then it is called joint.

If the system of equations has no solutions, then it is called incompatible.

If a SLAE has a unique solution, then it is called certain; if there is more than one solution, then - uncertain.

If the free terms of all equations of the system are equal to zero , then the system is called homogeneous, otherwise - heterogeneous.

Solution of elementary systems of linear algebraic equations.

If the number of system equations is equal to the number of unknown variables and the determinant of its main matrix is not equal to zero, then we will call such SLAEs elementary. Such systems of equations have a unique solution, and in the case of a homogeneous system, all unknown variables are equal to zero.

We began to study such SLAEs in high school. When solving them, we took one equation, expressed one unknown variable in terms of others and substituted it into the remaining equations, then took the next equation, expressed the next unknown variable and substituted it into other equations, and so on. Or they used the addition method, that is, they added two or more equations to eliminate some unknown variables. We will not dwell on these methods in detail, since they are essentially modifications of the Gauss method.

The main methods for solving elementary systems of linear equations are the Cramer method, the matrix method and the Gauss method. Let's sort them out.

Solving systems of linear equations by Cramer's method.

Let us need to solve a system of linear algebraic equations

in which the number of equations is equal to the number of unknown variables and the determinant of the main matrix of the system is different from zero, that is, .

Let be the determinant of the main matrix of the system, and are determinants of matrices that are obtained from A by replacing 1st, 2nd, …, nth column respectively to the column of free members:

With such notation, the unknown variables are calculated by the formulas of Cramer's method as . This is how the solution of a system of linear algebraic equations is found by the Cramer method.

Example.

Cramer method .

Decision.

The main matrix of the system has the form . Calculate its determinant (if necessary, see the article):

Since the determinant of the main matrix of the system is different from zero, the system has a unique solution that can be found by Cramer's method.

Compose and calculate the necessary determinants (the determinant is obtained by replacing the first column in matrix A with a column of free members, the determinant - by replacing the second column with a column of free members, - by replacing the third column of matrix A with a column of free members):

Finding unknown variables using formulas :

Answer:

The main disadvantage of Cramer's method (if it can be called a disadvantage) is the complexity of calculating the determinants when the number of system equations is more than three.

Solving systems of linear algebraic equations by the matrix method (using the inverse matrix).

Let the system of linear algebraic equations be given in matrix form , where the matrix A has dimension n by n and its determinant is nonzero.

Since , then the matrix A is invertible, that is, there is an inverse matrix . If we multiply both sides of the equality by on the left, then we get a formula for finding the column matrix of unknown variables. So we got the solution to the system of linear algebraic equations matrix method.

Example.

Solve System of Linear Equations matrix method.

Decision.

Let's rewrite the system of equations in matrix form:

As

then the SLAE can be solved by the matrix method. Using the inverse matrix, the solution to this system can be found as .

Let's build an inverse matrix using a matrix of algebraic complements of the elements of matrix A (if necessary, see the article):

It remains to calculate - the matrix of unknown variables by multiplying the inverse matrix on the matrix-column of free members (if necessary, see the article):

Answer:

or in another notation x 1 = 4, x 2 = 0, x 3 = -1.

The main problem in finding solutions to systems of linear algebraic equations by the matrix method is the complexity of finding the inverse matrix, especially for square matrices of order higher than the third.

Solving systems of linear equations by the Gauss method.

Suppose we need to find a solution to a system of n linear equations with n unknown variables
the determinant of the main matrix of which is different from zero.

The essence of the Gauss method consists in the successive exclusion of unknown variables: first, x 1 is excluded from all equations of the system, starting from the second, then x 2 is excluded from all equations, starting from the third, and so on, until only the unknown variable x n remains in the last equation. Such a process of transforming the equations of the system for the successive elimination of unknown variables is called direct Gauss method. After the forward run of the Gauss method is completed, x n is found from the last equation, x n-1 is calculated from the penultimate equation using this value, and so on, x 1 is found from the first equation. The process of calculating unknown variables when moving from the last equation of the system to the first is called reverse Gauss method.

Let us briefly describe the algorithm for eliminating unknown variables.

We will assume that , since we can always achieve this by rearranging the equations of the system. We exclude the unknown variable x 1 from all equations of the system, starting from the second one. To do this, add the first equation multiplied by to the second equation of the system, add the first multiplied by to the third equation, and so on, add the first multiplied by to the nth equation. The system of equations after such transformations will take the form

where , a .

We would come to the same result if we expressed x 1 in terms of other unknown variables in the first equation of the system and substituted the resulting expression into all other equations. Thus, the variable x 1 is excluded from all equations, starting from the second.

Next, we act similarly, but only with a part of the resulting system, which is marked in the figure

To do this, add the second equation multiplied by to the third equation of the system, add the second multiplied by to the fourth equation, and so on, add the second multiplied by to the nth equation. The system of equations after such transformations will take the form

where , a . Thus, the variable x 2 is excluded from all equations, starting from the third.

Next, we proceed to the elimination of the unknown x 3, while acting similarly with the part of the system marked in the figure

So we continue the direct course of the Gauss method until the system takes the form

From this moment, we begin the reverse course of the Gauss method: we calculate x n from the last equation as , using the obtained value of x n we find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.

Example.

Solve System of Linear Equations Gaussian method.

Decision.

Let's exclude the unknown variable x 1 from the second and third equations of the system. To do this, to both parts of the second and third equations, we add the corresponding parts of the first equation, multiplied by and by, respectively:

Now we exclude x 2 from the third equation by adding to its left and right parts the left and right parts of the second equation, multiplied by:

On this, the forward course of the Gauss method is completed, we begin the reverse course.

From the last equation of the resulting system of equations, we find x 3:

From the second equation we get .

From the first equation we find the remaining unknown variable and this completes the reverse course of the Gauss method.

Answer:

X 1 \u003d 4, x 2 \u003d 0, x 3 \u003d -1.

Solving systems of linear algebraic equations of general form.

In the general case, the number of equations of the system p does not coincide with the number of unknown variables n:

Such SLAEs may have no solutions, have a single solution, or have infinitely many solutions. This statement also applies to systems of equations whose main matrix is square and degenerate.

Kronecker-Capelli theorem.

Before finding a solution to a system of linear equations, it is necessary to establish its compatibility. The answer to the question when SLAE is compatible, and when it is incompatible, gives Kronecker–Capelli theorem:
for a system of p equations with n unknowns (p can be equal to n ) to be consistent it is necessary and sufficient that the rank of the main matrix of the system is equal to the rank of the extended matrix, that is, Rank(A)=Rank(T) .

Let us consider the application of the Kronecker-Cappelli theorem for determining the compatibility of a system of linear equations as an example.

Example.

Find out if the system of linear equations has solutions.

Decision.

. Let us use the method of bordering minors. Minor of the second order different from zero. Let's go over the third-order minors surrounding it:

Since all bordering third-order minors are equal to zero, the rank of the main matrix is two.

In turn, the rank of the augmented matrix is equal to three, since the minor of the third order

different from zero.

Thus, Rang(A) , therefore, according to the Kronecker-Capelli theorem, we can conclude that the original system of linear equations is inconsistent.

Answer:

There is no solution system.

So, we have learned to establish the inconsistency of the system using the Kronecker-Capelli theorem.

But how to find the solution of the SLAE if its compatibility is established?

To do this, we need the concept of the basis minor of a matrix and the theorem on the rank of a matrix.

The highest order minor of the matrix A, other than zero, is called basic.

It follows from the definition of the basis minor that its order is equal to the rank of the matrix. For a non-zero matrix A, there can be several basic minors; there is always one basic minor.

For example, consider the matrix .

All third-order minors of this matrix are equal to zero, since the elements of the third row of this matrix are the sum of the corresponding elements of the first and second rows.

The following minors of the second order are basic, since they are nonzero

Minors are not basic, since they are equal to zero.

Matrix rank theorem.

If the rank of a matrix of order p by n is r, then all elements of the rows (and columns) of the matrix that do not form the chosen basis minor are linearly expressed in terms of the corresponding elements of the rows (and columns) that form the basis minor.

What does the matrix rank theorem give us?

If, by the Kronecker-Capelli theorem, we have established the compatibility of the system, then we choose any basic minor of the main matrix of the system (its order is equal to r), and exclude from the system all equations that do not form the chosen basic minor. The SLAE obtained in this way will be equivalent to the original one, since the discarded equations are still redundant (according to the matrix rank theorem, they are a linear combination of the remaining equations).

As a result, after discarding the excessive equations of the system, two cases are possible.

If the number of equations r in the resulting system is equal to the number of unknown variables, then it will be definite and the only solution can be found by the Cramer method, the matrix method or the Gauss method.

Example.

Decision.

Rank of the main matrix of the system is equal to two, since the minor of the second order different from zero. Extended matrix rank is also equal to two, since the only minor of the third order is equal to zero

and the minor of the second order considered above is different from zero. Based on the Kronecker-Capelli theorem, one can assert the compatibility of the original system of linear equations, since Rank(A)=Rank(T)=2 .

As a basis minor, we take . It is formed by the coefficients of the first and second equations:

The third equation of the system does not participate in the formation of the basis minor, so we exclude it from the system based on the matrix rank theorem:

Thus we have obtained an elementary system of linear algebraic equations. Let's solve it by Cramer's method:

Answer:

x 1 \u003d 1, x 2 \u003d 2.

If the number of equations r in the resulting SLAE less than number unknown variables n, then on the left-hand side of the equations we leave the terms that form the basis minor, and transfer the remaining terms to the right-hand side of the equations of the system with the opposite sign.

The unknown variables (there are r of them) remaining on the left-hand sides of the equations are called main.

Unknown variables (there are n - r of them) that ended up on the right side are called free.

Now we assume that the free unknown variables can take arbitrary values, while the r main unknown variables will be expressed in terms of the free unknown variables in a unique way. Their expression can be found by solving the resulting SLAE by the Cramer method, the matrix method, or the Gauss method.

Let's take an example.

Example.

Solve System of Linear Algebraic Equations .

Decision.

Find the rank of the main matrix of the system by the bordering minors method. Let us take a 1 1 = 1 as a non-zero first-order minor. Let's start searching for a non-zero second-order minor surrounding this minor:

So we found a non-zero minor of the second order. Let's start searching for a non-zero bordering minor of the third order:

Thus, the rank of the main matrix is three. The rank of the augmented matrix is also equal to three, that is, the system is consistent.

The found non-zero minor of the third order will be taken as the basic one.

For clarity, we show the elements that form the basis minor:

We leave the terms participating in the basic minor on the left side of the equations of the system, and transfer the rest with opposite signs to the right sides:

We give free unknown variables x 2 and x 5 arbitrary values, that is, we take , where are arbitrary numbers. In this case, the SLAE takes the form

We solve the obtained elementary system of linear algebraic equations by the Cramer method:

Hence, .

In the answer, do not forget to indicate free unknown variables.

Answer:

Where are arbitrary numbers.

Summarize.

To solve a system of linear algebraic equations of a general form, we first find out its compatibility using the Kronecker-Capelli theorem. If the rank of the main matrix is not equal to the rank of the extended matrix, then we conclude that the system is inconsistent.

If the rank of the main matrix is equal to the rank of the extended matrix, then we choose the basic minor and discard the equations of the system that do not participate in the formation of the chosen basic minor.

If the order of the basis minor is equal to the number unknown variables, then the SLAE has a unique solution that can be found by any method known to us.

If the order of the basis minor is less than the number of unknown variables, then on the left side of the equations of the system we leave the terms with the main unknown variables, transfer the remaining terms to the right sides and assign arbitrary values to the free unknown variables. From the resulting system of linear equations, we find the main unknown variables by the Cramer method, the matrix method or the Gauss method.

Gauss method for solving systems of linear algebraic equations of general form.

Using the Gauss method, one can solve systems of linear algebraic equations of any kind without their preliminary investigation for compatibility. The process of successive exclusion of unknown variables makes it possible to draw a conclusion about both the compatibility and inconsistency of the SLAE, and if a solution exists, it makes it possible to find it.

From the point of view of computational work, the Gaussian method is preferable.

Watch it detailed description and analyzed examples in the article Gauss method for solving systems of linear algebraic equations of general form.

Recording the general solution of homogeneous and inhomogeneous linear algebraic systems using the vectors of the fundamental system of solutions.

In this section, we will focus on joint homogeneous and inhomogeneous systems of linear algebraic equations that have an infinite number of solutions.

Let's deal with homogeneous systems first.

Fundamental decision system of a homogeneous system of p linear algebraic equations with n unknown variables is a set of (n – r) linearly independent solutions of this system, where r is the order of the basis minor of the main matrix of the system.

If we designate linearly independent solutions of a homogeneous SLAE as X (1) , X (2) , …, X (n-r) (X (1) , X (2) , …, X (n-r) are matrices columns of dimension n by 1 ) , then the general solution of this homogeneous system is represented as a linear combination of vectors of the fundamental system of solutions with arbitrary constant coefficients С 1 , С 2 , …, С (n-r), that is, .

What does the term general solution of a homogeneous system of linear algebraic equations (oroslau) mean?

The meaning is simple: the formula sets everything possible solutions the original SLAE, in other words, taking any set of values of arbitrary constants С 1 , С 2 , …, С (n-r) , according to the formula we get one of the solutions of the original homogeneous SLAE.

Thus, if we find a fundamental system of solutions, then we can set all solutions of this homogeneous SLAE as .

Let us show the process of constructing a fundamental system of solutions for a homogeneous SLAE.

We choose the basic minor of the original system of linear equations, exclude all other equations from the system, and transfer to the right-hand side of the equations of the system with opposite signs all terms containing free unknown variables. Let's give the free unknown variables the values 1,0,0,…,0 and calculate the main unknowns by solving the resulting elementary system of linear equations in any way, for example, by the Cramer method. Thus, X (1) will be obtained - the first solution of the fundamental system. If we give the free unknowns the values 0,1,0,0,…,0 and calculate the main unknowns, then we get X (2) . Etc. If we give the free unknown variables the values 0,0,…,0,1 and calculate the main unknowns, then we get X (n-r) . This is how the fundamental system of solutions of the homogeneous SLAE will be constructed and its general solution can be written in the form .

For inhomogeneous systems of linear algebraic equations, the general solution is represented as

Let's look at examples.

Example.

Find the fundamental system of solutions and the general solution of a homogeneous system of linear algebraic equations .

Decision.

The rank of the main matrix of homogeneous systems of linear equations is always equal to the rank of the extended matrix. Let us find the rank of the main matrix by the method of fringing minors. As a nonzero minor of the first order, we take the element a 1 1 = 9 of the main matrix of the system. Find the bordering non-zero minor of the second order:

A minor of the second order, different from zero, is found. Let's go through the third-order minors bordering it in search of a non-zero one:

All bordering minors of the third order are equal to zero, therefore, the rank of the main and extended matrix is two. Let's take the basic minor. For clarity, we note the elements of the system that form it:

The third equation of the original SLAE does not participate in the formation of the basic minor, therefore, it can be excluded:

We leave the terms containing the main unknowns on the right-hand sides of the equations, and transfer the terms with free unknowns to the right-hand sides:

Let us construct a fundamental system of solutions to the original homogeneous system of linear equations. The fundamental system of solutions of this SLAE consists of two solutions, since the original SLAE contains four unknown variables, and the order of its basic minor is two. To find X (1), we give the free unknown variables the values x 2 \u003d 1, x 4 \u003d 0, then we find the main unknowns from the system of equations
.

Let's solve it by Cramer's method:

Thus, .

Now let's build X (2) . To do this, we give the free unknown variables the values x 2 \u003d 0, x 4 \u003d 1, then we find the main unknowns from the system of linear equations
.

Let's use Cramer's method again:

We get .

So we got two vectors of the fundamental system of solutions and , now we can write down the general solution of a homogeneous system of linear algebraic equations:

, where C 1 and C 2 are arbitrary numbers., are equal to zero. We also take the minor as the basic one, exclude the third equation from the system, and transfer the terms with free unknowns to the right-hand sides of the system equations:

To find, we give the free unknown variables the values x 2 \u003d 0 and x 4 \u003d 0, then the system of equations takes the form , from which we find the main unknown variables using the Cramer method:

We have , hence,

where C 1 and C 2 are arbitrary numbers.

It should be noted that the solutions of an indefinite homogeneous system of linear algebraic equations generate linear space

Decision.

The canonical equation of an ellipsoid in a rectangular Cartesian coordinate system has the form . Our task is to determine the parameters a , b and c . Since the ellipsoid passes through points A, B and C, then when substituting their coordinates into the canonical equation of the ellipsoid, it should turn into an identity. So we get a system of three equations:

Denote , then the system becomes a system of linear algebraic equations .

Let us calculate the determinant of the main matrix of the system:

Since it is non-zero, we can find the solution by Cramer's method:
). Obviously, x = 0 and x = 1 are the roots of this polynomial. quotient from division on the is an . Thus, we have a decomposition and the original expression will take the form .

Let's use the method of indefinite coefficients.

Equating the corresponding coefficients of the numerators, we arrive at a system of linear algebraic equations . Its solution will give us the desired indefinite coefficients A, B, C and D.

We solve the system using the Gauss method:

In the reverse course of the Gauss method, we find D = 0, C = -2, B = 1, A = 1 .

We get

Answer:

Decision. A= . Find r(A). As matrix A has order 3x4, then the highest order of minors is 3. In this case, all minors of the third order are equal to zero (check it yourself). Means, r(A)< 3. Возьмем главный basic minor = -5-4 = -9 ≠ 0. Hence r(A) =2.

Consider matrix With = .

Minor third order ≠ 0. Hence, r(C) = 3.

Since r(A) ≠ r(C) , then the system is inconsistent.

Example 2 Determine the compatibility of the system of equations

Solve this system if it is compatible.

Decision.

A = , C = . Obviously, r(А) ≤ 3, r(C) ≤ 4. Since detC = 0, then r(C)< 4. Consider minor third order, located in the upper left corner of the matrix A and C: = -23 ≠ 0. Hence, r(A) = r(C) = 3.

Number unknown in the system n=3. So the system has a unique solution. In this case, the fourth equation is the sum of the first three and can be ignored.

According to Cramer's formulas we get x 1 = -98/23, x 2 = -47/23, x 3 = -123/23.

2.4. Matrix method. Gauss method

system n linear equations with n unknowns can be solved matrix method according to the formula X \u003d A -1 B (for Δ ≠ 0), which is obtained from (2) by multiplying both parts by A -1 .

Example 1. Solve a system of equations

by the matrix method (in Section 2.2 this system was solved using the Cramer formulas)

Decision. Δ=10 ≠ 0 A = - nonsingular matrix.

= (verify this for yourself by doing the necessary calculations).

A -1 \u003d (1 / Δ) x \u003d .

X \u003d A -1 B \u003d x= .

Answer: .

From a practical point of view matrix method and formulas Kramer are associated with a large amount of computation, so preference is given to Gauss method, which consists in the successive elimination of unknowns. To do this, the system of equations is reduced to an equivalent system with a triangular augmented matrix (all elements below the main diagonal are equal to zero). These actions are called direct move. From the resulting triangular system, the variables are found using successive substitutions (backward).

Example 2. Solve the system using the Gauss method

(This system was solved above using the Cramer formula and the matrix method).

Decision.

Direct move. We write the augmented matrix and use elementary transformations Let's bring it to a triangular form:

~ ~ ~ ~ .

Get system

Reverse move. From the last equation we find X 3 = -6 and substitute this value into the second equation:

X 2 = - 11/2 - 1/4X 3 = - 11/2 - 1/4(-6) = - 11/2 + 3/2 = -8/2 = -4.

X 1 = 2 -X 2 + X 3 = 2+4-6 = 0.

Answer: .

2.5. General solution of a system of linear equations

Let a system of linear equations be given = b i(i=). Let r(A) = r(C) = r, i.e. the system is collaborative. Any non-zero minor of order r is basic minor. Without loss of generality, we will assume that the basis minor is located in the first r (1 ≤ r ≤ min(m,n)) rows and columns of the matrix A. Discarding last m-r equations of the system, we write a shortened system:

which is equivalent to the original. Let's name the unknowns x 1 ,….x r basic , and x r +1 ,…, x r free and move the terms containing the free unknowns to the right side of the equations of the truncated system. We get the system with respect to the basic unknowns:

which for each set of values of free unknowns x r +1 \u003d C 1, ..., x n \u003d C n-r has the only solution x 1 (C 1, ..., C n-r), ..., x r (C 1, ..., C n-r), found by Cramer's rule.

Appropriate solution shortened, and hence the original system has the form:

Х(С 1 ,…, С n-r) = - general solution of the system.

If we give some numerical values to the free unknowns in the general solution, then we get the solution linear system, called private .

Example. Establish compatibility and find the overall solution of the system

Decision. A = , С = .

So as r(A)= r(C) = 2 (see for yourself), then the original system is compatible and has an infinite number of solutions (since r< 4).

As appears from Cramer's theorems, when solving a system of linear equations, three cases may occur:

First case: the system of linear equations has a unique solution

(the system is consistent and definite)

Second case: the system of linear equations has an infinite number of solutions

(the system is consistent and indeterminate)

** ,

those. the coefficients of the unknowns and the free terms are proportional.

Third case: the system of linear equations has no solutions

(system inconsistent)

So the system m linear equations with n variables is called incompatible if it has no solutions, and joint if it has at least one solution. A joint system of equations that has only one solution is called certain, and more than one uncertain.

Examples of solving systems of linear equations by the Cramer method

Let the system

Based on Cramer's theorem

………….
,

where
-

system identifier. The remaining determinants are obtained by replacing the column with the coefficients of the corresponding variable (unknown) with free members:

Example 2

Therefore, the system is definite. To find its solution, we calculate the determinants

By Cramer's formulas we find:

So, (1; 0; -1) is the only solution to the system.

To check the solutions of the systems of equations 3 X 3 and 4 X 4, you can use the online calculator, decisive method Kramer.

If there are no variables in the system of linear equations in one or more equations, then in the determinant the elements corresponding to them are equal to zero! This is the next example.

Example 3 Solve the system of linear equations by Cramer's method:

Decision. We find the determinant of the system:

Look carefully at the system of equations and at the determinant of the system and repeat the answer to the question in which cases one or more elements of the determinant are equal to zero. So, the determinant is not equal to zero, therefore, the system is definite. To find its solution, we calculate the determinants for the unknowns

By Cramer's formulas we find:

So, the solution of the system is (2; -1; 1).

6. General system linear algebraic equations. Gauss method.

As we remember, Cramer's rule and the matrix method are unsuitable in cases where the system has infinitely many solutions or is inconsistent. Gauss method – the most powerful and versatile tool for finding solutions to any system of linear equations, which in every case lead us to the answer! The algorithm of the method in all three cases works the same way. If the Cramer and matrix methods require knowledge of determinants, then the application of the Gauss method requires knowledge of only arithmetic operations, which makes it accessible even to schoolchildren primary school.

First, we systematize the knowledge about systems of linear equations a little. A system of linear equations can:

1) Have a unique solution.
2) Have infinitely many solutions.
3) Have no solutions (be incompatible).

The Gauss method is the most powerful and versatile tool for finding a solution any systems of linear equations. As we remember Cramer's rule and matrix method are unsuitable in cases where the system has infinitely many solutions or is inconsistent. A method of successive elimination of unknowns anyway lead us to the answer! In this lesson, we will again consider the Gauss method for case No. 1 (the only solution to the system), the article is reserved for the situations of points No. 2-3. I note that the method algorithm itself works in the same way in all three cases.

Back to the simplest system from the lesson How to solve a system of linear equations?
and solve it using the Gaussian method.

The first step is to write extended matrix system:
. By what principle the coefficients are recorded, I think everyone can see. The vertical line inside the matrix does not carry any mathematical meaning - it's just a strikethrough for ease of design.

Reference:I recommend to remember terms linear algebra. System Matrix is a matrix composed only of the coefficients of the unknowns, in this example system matrix: . Extended System Matrix is the same matrix of the system plus a column of free terms, in this case: . Any of the matrices can be called simply a matrix for brevity.

After the extended matrix of the system is written, it is necessary to perform some actions with it, which are also called elementary transformations.

There are the following elementary transformations:

1) Strings matrices can be rearranged places. For example, in the matrix under consideration, you can safely rearrange the first and second rows:

2) If there are (or appeared) proportional (as a special case - identical) rows in the matrix, then it follows delete from the matrix, all these rows except one. Consider, for example, the matrix . In this matrix, the last three rows are proportional, so it is enough to leave only one of them: .

3) If a zero row appeared in the matrix during the transformations, then it also follows delete. I will not draw, of course, the zero line is the line in which only zeros.

4) The row of the matrix can be multiply (divide) for any number non-zero. Consider, for example, the matrix . Here it is advisable to divide the first line by -3, and multiply the second line by 2: . This action is very useful, as it simplifies further transformations of the matrix.

5) This transformation causes the most difficulties, but in fact there is nothing complicated either. To the row of the matrix, you can add another string multiplied by a number, different from zero. Consider our matrix from a practical example: . First, I will describe the transformation in great detail. Multiply the first row by -2: , and to the second line we add the first line multiplied by -2: . Now the first line can be divided "back" by -2: . As you can see, the line that is ADDED LI – hasn't changed. Always the line is changed, TO WHICH ADDED UT.

In practice, of course, they don’t paint in such detail, but write shorter:

Once again: to the second line added the first row multiplied by -2. The line is usually multiplied orally or on a draft, while the mental course of calculations is something like this:

“I rewrite the matrix and rewrite the first row: »

First column first. Below I need to get zero. Therefore, I multiply the unit above by -2:, and add the first to the second line: 2 + (-2) = 0. I write the result in the second line: »

“Now the second column. Above -1 times -2: . I add the first to the second line: 1 + 2 = 3. I write the result to the second line: »

“And the third column. Above -5 times -2: . I add the first line to the second line: -7 + 10 = 3. I write the result in the second line: »

Please think carefully about this example and understand the sequential calculation algorithm, if you understand this, then the Gauss method is practically "in your pocket". But, of course, we are still working on this transformation.

Elementary transformations do not change the solution of the system of equations

! ATTENTION: considered manipulations can not use, if you are offered a task where the matrices are given "by themselves". For example, with "classic" matrices in no case should you rearrange something inside the matrices!

Let's return to our system. She's practically broken into pieces.

Let us write the augmented matrix of the system and, using elementary transformations, reduce it to stepped view:

(1) The first row was added to the second row, multiplied by -2. And again: why do we multiply the first row by -2? In order to get zero at the bottom, which means getting rid of one variable in the second line.

(2) Divide the second row by 3.

The purpose of elementary transformations – convert the matrix to step form: . In the design of the task, they directly draw out the “ladder” with a simple pencil, and also circle the numbers that are located on the “steps”. The term "stepped view" itself is not entirely theoretical; in the scientific and educational literature, it is often called trapezoidal view or triangular view.

As a result of elementary transformations, we have obtained equivalent original system of equations:

Now the system needs to be "untwisted" in the opposite direction - from the bottom up, this process is called reverse Gauss method.

In the lower equation, we already have the finished result: .

Consider the first equation of the system and substitute the already known value of “y” into it:

Let us consider the most common situation, when the Gaussian method is required to solve a system of three linear equations with three unknowns.

Example 1

Solve the system of equations using the Gauss method:

Let's write the augmented matrix of the system:

Now I will immediately draw the result that we will come to in the course of the solution:

And I repeat, our goal is to bring the matrix to a stepped form using elementary transformations. Where to start taking action?

First, look at the top left number:

Should almost always be here unit. Generally speaking, -1 (and sometimes other numbers) will also suit, but somehow it has traditionally happened that a unit is usually placed there. How to organize a unit? We look at the first column - we have a finished unit! Transformation one: swap the first and third lines:

Now the first line will remain unchanged until the end of the solution. Now fine.

The unit in the top left is organized. Now you need to get zeros in these places:

Zeros are obtained just with the help of a "difficult" transformation. First, we deal with the second line (2, -1, 3, 13). What needs to be done to get zero in the first position? Need to the second line add the first line multiplied by -2. Mentally or on a draft, we multiply the first line by -2: (-2, -4, 2, -18). And we consistently carry out (again mentally or on a draft) addition, to the second line we add the first line, already multiplied by -2:

The result is written in the second line:

Similarly, we deal with the third line (3, 2, -5, -1). To get zero in the first position, you need to the third line add the first line multiplied by -3. Mentally or on a draft, we multiply the first line by -3: (-3, -6, 3, -27). And to the third line we add the first line multiplied by -3:

The result is written in the third line:

In practice, these actions are usually performed verbally and written down in one step:

No need to count everything at once and at the same time. The order of calculations and "insertion" of results consistent and usually like this: first we rewrite the first line, and puff ourselves quietly - CONSISTENTLY and ATTENTIVELY:

And I have already considered the mental course of the calculations themselves above.

In this example, this is easy to do, we divide the second line by -5 (since all numbers there are divisible by 5 without a remainder). At the same time, we divide the third line by -2, because the smaller the number, the easier solution:

On the final stage elementary conversions need to get one more zero here:

For this to the third line we add the second line, multiplied by -2:

Try to parse this action yourself - mentally multiply the second line by -2 and carry out the addition.

The last action performed is the hairstyle of the result, divide the third line by 3.

As a result of elementary transformations, an equivalent initial system of linear equations was obtained:

Cool.

Now the reverse course of the Gaussian method comes into play. The equations "unwind" from the bottom up.

In the third equation, we already have the finished result:

Let's look at the second equation: . The meaning of "z" is already known, thus:

And finally, the first equation: . "Y" and "Z" are known, the matter is small:

Answer:

As has been repeatedly noted, for any system of equations, it is possible and necessary to check the found solution, fortunately, this is not difficult and fast.

Example 2

This is an example for independent decision, a sample finishing touch, and an answer at the end of the lesson.

It should be noted that your course of action may not coincide with my course of action, and this is a feature of the Gauss method. But the answers must be the same!

Example 3

Solve a system of linear equations using the Gauss method

We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

We look at the upper left "step". There we should have a unit. The problem is that there are no ones in the first column at all, so nothing can be solved by rearranging the rows. In such cases, the unit must be organized using an elementary transformation. This can usually be done in several ways. I did this:
(1) To the first line we add the second line, multiplied by -1. That is, we mentally multiplied the second line by -1 and performed the addition of the first and second lines, while the second line did not change.

Now at the top left "minus one", which suits us perfectly. Who wants to get +1 can perform an additional gesture: multiply the first line by -1 (change its sign).

(2) The first row multiplied by 5 was added to the second row. The first row multiplied by 3 was added to the third row.

(3) The first line was multiplied by -1, in principle, this is for beauty. The sign of the third line was also changed and moved to the second place, thus, on the second “step, we had the desired unit.

(4) The second line multiplied by 2 was added to the third line.

(5) The third row was divided by 3.

A bad sign that indicates a calculation error (less often a typo) is a “bad” bottom line. That is, if we got something like below, and, accordingly, , then with a high degree of probability it can be argued that an error was made in the course of elementary transformations.

We charge the reverse move, in the design of examples, the system itself is often not rewritten, and the equations are “taken directly from the given matrix”. The reverse move, I remind you, works from the bottom up. Yes, here is a gift:

Answer: .

Example 4

Solve a system of linear equations using the Gauss method

This is an example for an independent solution, it is somewhat more complicated. It's okay if someone gets confused. Complete Solution and sample design at the end of the lesson. Your solution may differ from mine.

In the last part, we consider some features of the Gauss algorithm.
The first feature is that sometimes some variables are missing in the equations of the system, for example:

How to correctly write the augmented matrix of the system? I already talked about this moment in the lesson. Cramer's rule. Matrix method. In the expanded matrix of the system, we put zeros in place of the missing variables:

By the way, this is a fairly easy example, since there is already one zero in the first column, and there are fewer elementary transformations to perform.

The second feature is this. In all the examples considered, we placed either –1 or +1 on the “steps”. Could there be other numbers? In some cases they can. Consider the system: .

Here on the upper left "step" we have a deuce. But we notice the fact that all the numbers in the first column are divisible by 2 without a remainder - and another two and six. And the deuce at the top left will suit us! At the first step, you need to perform the following transformations: add the first line multiplied by -1 to the second line; to the third line add the first line multiplied by -3. Thus, we will get the desired zeros in the first column.

Or another hypothetical example: . Here, the triple on the second “rung” also suits us, since 12 (the place where we need to get zero) is divisible by 3 without a remainder. It is necessary to carry out the following transformation: to the third line, add the second line, multiplied by -4, as a result of which the zero we need will be obtained.

The Gauss method is universal, but there is one peculiarity. You can confidently learn how to solve systems by other methods (Cramer's method, matrix method) literally from the first time - there is a very rigid algorithm. But in order to feel confident in the Gauss method, you should “fill your hand” and solve at least 5-10 systems. Therefore, at first there may be confusion, errors in calculations, and there is nothing unusual or tragic in this.

Rainy autumn weather outside the window .... Therefore, for everyone more complex example for independent solution:

Example 5

Solve a system of four linear equations with four unknowns using the Gauss method.

Such a task in practice is not so rare. I think that even a teapot who has studied this page in detail understands the algorithm for solving such a system intuitively. Basically the same - just more action.

The cases when the system has no solutions (inconsistent) or has infinitely many solutions are considered in the lesson. Incompatible systems and systems with a common solution. There you can fix the considered algorithm of the Gauss method.

Wish you luck!

Solutions and answers:

Example 2: Decision: Let's write the augmented matrix of the system and with the help of elementary transformations we will bring it to a stepped form.

Performed elementary transformations:
(1) The first row was added to the second row, multiplied by -2. The first line was added to the third line, multiplied by -1. Attention! Here it may be tempting to subtract the first from the third line, I strongly do not recommend subtracting - the risk of error greatly increases. We just fold!
(2) The sign of the second line was changed (multiplied by -1). The second and third lines have been swapped. note that on the “steps” we are satisfied not only with one, but also with -1, which is even more convenient.
(3) To the third line, add the second line, multiplied by 5.
(4) The sign of the second line was changed (multiplied by -1). The third line was divided by 14.

Reverse move:

Answer: .

Example 4: Decision: Let's write the augmented matrix of the system and with the help of elementary transformations we bring it to the step form:

Conversions performed:
(1) The second line was added to the first line. Thus, the desired unit is organized on the upper left “step”.
(2) The first row multiplied by 7 was added to the second row. The first row multiplied by 6 was added to the third row.

With the second "step" everything is worse, the "candidates" for it are the numbers 17 and 23, and we need either one or -1. Transformations (3) and (4) will be aimed at obtaining the desired unit

(3) The second line was added to the third line, multiplied by -1.
(4) The third line, multiplied by -3, was added to the second line.
The necessary thing on the second step is received .
(5) To the third line added the second, multiplied by 6.

Within the lessons Gauss method and Incompatible systems/systems with a common solution we considered inhomogeneous systems of linear equations, where free member(which is usually on the right) at least one of the equations was different from zero.
And now, after a good warm-up with matrix rank, we will continue to polish the technique elementary transformations on the homogeneous system linear equations.
According to the first paragraphs, the material may seem boring and ordinary, but this impression is deceptive. In addition to further development of technical methods, there will be many new information, so please try not to neglect the examples in this article.

However, two more cases are widespread in practice:

– The system is inconsistent (has no solutions);
The system is consistent and has infinitely many solutions.

Note : the term "consistency" implies that the system has at least some solution. In a number of tasks, it is required to preliminarily examine the system for compatibility, how to do this - see the article on matrix rank.

For these systems, the most universal of all solution methods is used - Gauss method. In fact, the "school" method will also lead to the answer, but in higher mathematics it is customary to use the Gaussian method of successive elimination of unknowns. Those who are not familiar with the Gauss method algorithm, please study the lesson first gauss method for dummies.

The elementary matrix transformations themselves are exactly the same, the difference will be in the end of the solution. First, consider a couple of examples where the system has no solutions (inconsistent).

Example 1

What immediately catches your eye in this system? The number of equations is less than the number of variables. If the number of equations is less than the number of variables, then we can immediately say that the system is either inconsistent or has infinitely many solutions. And it remains only to find out.

The beginning of the solution is quite ordinary - we write the extended matrix of the system and, using elementary transformations, we bring it to a stepwise form:

(1) On the upper left step, we need to get +1 or -1. There are no such numbers in the first column, so rearranging the rows will not work. The unit will have to be organized independently, and this can be done in several ways. I did this: To the first line, add the third line, multiplied by -1.

(2) Now we get two zeros in the first column. To the second line we add the first line multiplied by 3. To the third line we add the first line multiplied by 5.

(3) After the transformation is done, it is always advisable to see if it is possible to simplify the resulting strings? Can. We divide the second line by 2, at the same time getting the desired -1 on the second step. Divide the third line by -3.

(4) Add the second line to the third line.

Probably, everyone paid attention to the bad line, which turned out as a result of elementary transformations: . It is clear that this cannot be so. Indeed, we rewrite the resulting matrix back to the system of linear equations:

If, as a result of elementary transformations, a string of the form is obtained, where is a non-zero number, then the system is inconsistent (has no solutions) .

How to record the end of a task? Let's draw with white chalk: "as a result of elementary transformations, a line of the form is obtained, where" and give the answer: the system has no solutions (inconsistent).

If, according to the condition, it is required to EXPLORE the system for compatibility, then it is necessary to issue a solution in a more solid style involving the concept matrix rank and the Kronecker-Capelli theorem.

Please note that there is no reverse motion of the Gaussian algorithm here - there are no solutions and there is simply nothing to find.

Example 2

Solve a system of linear equations

This is a do-it-yourself example. Full solution and answer at the end of the lesson. Again, I remind you that your solution path may differ from my solution path, the Gaussian algorithm does not have a strong “rigidity”.

Another one technical feature solutions: elementary transformations can be stopped At once, as soon as a line like , where . Consider a conditional example: suppose that after the first transformation we get a matrix . The matrix has not yet been reduced to a stepped form, but there is no need for further elementary transformations, since a line of the form has appeared, where . It should be immediately answered that the system is incompatible.

When a system of linear equations has no solutions, this is almost a gift, in view of the fact that it turns out short solution, sometimes literally in 2-3 actions.

But everything in this world is balanced, and the problem in which the system has infinitely many solutions is just longer.

Example 3

Solve a system of linear equations

There are 4 equations and 4 unknowns, so the system can either have a single solution, or have no solutions, or have infinitely many solutions. Whatever it was, but the Gauss method in any case will lead us to the answer. Therein lies its versatility.

The beginning is again standard. We write the extended matrix of the system and, using elementary transformations, bring it to a step form:

That's all, and you were afraid.

(1) Note that all the numbers in the first column are divisible by 2, so a 2 is fine on the top left rung. To the second line we add the first line, multiplied by -4. To the third line we add the first line, multiplied by -2. To the fourth line we add the first line, multiplied by -1.

Attention! Many may be tempted from the fourth line subtract first line. This can be done, but it is not necessary, experience shows that the probability of an error in calculations increases several times. Just add up: To the fourth line, add the first line, multiplied by -1 - exactly!

(2) The last three lines are proportional, two of them can be deleted.

Here again it is necessary to show increased attention, but are the lines really proportional? For reinsurance (especially for a teapot), it would not be superfluous to multiply the second row by -1, and divide the fourth row by 2, resulting in three identical rows. And only after that remove two of them.

As a result of elementary transformations, the extended matrix of the system is reduced to a stepped form:

When completing a task in a notebook, it is advisable to make the same notes in pencil for clarity.

We rewrite the corresponding system of equations:

The “usual” only solution of the system does not smell here. There is no bad line either. This means that this is the third remaining case - the system has infinitely many solutions. Sometimes, by condition, it is necessary to investigate the compatibility of the system (i.e., to prove that a solution exists at all), you can read about this in the last paragraph of the article How to find the rank of a matrix? But for now, let's break down the basics:

The infinite set of solutions of the system is briefly written in the form of the so-called general system solution .

We will find the general solution of the system using the reverse motion of the Gauss method.

First we need to determine what variables we have basic, and which variables free. It is not necessary to bother with the terms of linear algebra, it is enough to remember that there are such basis variables and free variables.

Basic variables always "sit" strictly on the steps of the matrix.
In this example, the basic variables are and

Free variables are everything remaining variables that did not get a step. In our case, there are two of them: – free variables.

Now you need all basis variables express only through free variables.

The reverse move of the Gaussian algorithm traditionally works from the bottom up.
From the second equation of the system, we express the basic variable:

Now look at the first equation: . First, we substitute the found expression into it:

It remains to express the basic variable in terms of free variables:

The result is what you need - all the basis variables ( and ) are expressed only through free variables :

Actually, the general solution is ready:

How to write down the general solution?
Free variables are written into the general solution "on their own" and strictly in their places. In this case, free variables should be written in the second and fourth positions:
.

The resulting expressions for the basic variables and obviously needs to be written in the first and third positions:

Giving free variables arbitrary values, there are infinitely many private decisions. The most popular values are zeros, since the particular solution is the easiest to obtain. Substitute in the general solution:

is a private decision.

Ones are another sweet couple, let's substitute into the general solution:

is another particular solution.

It is easy to see that the system of equations has infinitely many solutions(since we can give free variables any values)

Each a particular solution must satisfy to each system equation. This is the basis for a “quick” check of the correctness of the solution. Take, for example, a particular solution and substitute it into left side each equation of the original system:

Everything has to come together. And with any particular solution you get, everything should also converge.

But, strictly speaking, the verification of a particular solution sometimes deceives; some particular solution can satisfy each equation of the system, and the general solution itself is actually found incorrectly.

Therefore, the verification of the general solution is more thorough and reliable. How to check the resulting general solution ?

It's easy, but quite tedious. We need to take expressions basic variables, in this case and , and substitute them into the left side of each equation of the system.

To the left side of the first equation of the system:

To the left side of the second equation of the system:

The right side of the original equation is obtained.

Example 4

Solve the system using the Gauss method. Find a general solution and two private ones. Check the overall solution.

This is a do-it-yourself example. Here, by the way, again the number of equations is less than the number of unknowns, which means that it is immediately clear that the system will either be inconsistent or have an infinite number of solutions. What is important in the decision process itself? Attention, and again attention. Full solution and answer at the end of the lesson.

And a couple more examples to reinforce the material

Example 5

Solve a system of linear equations. If the system has infinitely many solutions, find two particular solutions and check the general solution

Decision: Let's write the augmented matrix of the system and with the help of elementary transformations we bring it to the step form:

(1) Add the first line to the second line. To the third line we add the first line multiplied by 2. To the fourth line we add the first line multiplied by 3.
(2) To the third line, add the second line, multiplied by -5. To the fourth line we add the second line, multiplied by -7.
(3) The third and fourth lines are the same, we delete one of them.

Here is such a beauty:

Basis variables sit on steps, so they are base variables.
There is only one free variable, which did not get a step:

Reverse move:
We express the basic variables in terms of the free variable:
From the third equation:

Consider the second equation and substitute the found expression into it:

Consider the first equation and substitute the found expressions and into it:

Yes, a calculator that counts ordinary fractions is still convenient.

So the general solution is:

Once again, how did it happen? The free variable sits alone in its rightful fourth place. The resulting expressions for the basic variables , also took their ordinal places.

Let us immediately check the general solution. Work for blacks, but I have already done it, so catch =)

We substitute three heroes , , into the left side of each equation of the system:

The corresponding right-hand sides of the equations are obtained, so the general solution is found correctly.

Now from the found general solution we get two particular solutions. The chef here is the only free variable . You don't need to break your head.

Let then is a private decision.
Let then is another particular solution.

Answer: Common decision: , particular solutions: , .

I shouldn't have remembered about blacks here ... ... because all sorts of sadistic motives came into my head and I remembered the well-known fotozhaba, in which Ku Klux Klansmen in white overalls run across the field after a black football player. I sit and smile quietly. You know how distracting….

A lot of math is harmful, so a similar final example for an independent solution.

Example 6

Find the general solution of the system of linear equations.

I have already checked the general solution, the answer can be trusted. Your solution may differ from my solution, the main thing is that the general solutions match.

Probably, many people noticed an unpleasant moment in the solutions: very often, during the reverse course of the Gauss method, we had to fiddle with ordinary fractions. In practice, this is true, cases where there are no fractions are much less common. Be prepared mentally, and most importantly, technically.

I will dwell on some features of the solution that were not found in the solved examples.

The general solution of the system can sometimes include a constant (or constants), for example: . Here one of the basic variables is equal to a constant number: . There is nothing exotic in this, it happens. Obviously, in this case, any particular solution will contain a five in the first position.

Rarely, but there are systems in which the number of equations is greater than the number of variables. The Gaussian method works in the most severe conditions; one should calmly bring the extended matrix of the system to a stepped form according to the standard algorithm. Such a system may be inconsistent, may have infinitely many solutions, and, oddly enough, may have a unique solution.