Gauss method detailed explanation. Gauss method: description of the algorithm for solving a system of linear equations, examples, solutions
Ever since the beginning of the 16th-18th centuries, mathematicians began to intensively study the functions, thanks to which so much has changed in our lives. Computer technology without this knowledge would simply not exist. For solutions challenging tasks, linear equations and functions were created various concepts, theorems and methods of solution. One of such universal and rational ways and methods of solving linear equations and their systems became the Gaussian method. Matrices, their rank, determinant - everything can be calculated without using complex operations.
What is SLAU
In mathematics, there is the concept of SLAE - a system of linear algebraic equations. What does she represent? This is a set of m equations with the required n unknowns, usually denoted as x, y, z, or x 1 , x 2 ... x n, or other symbols. Solve by Gauss method this system- means to find all the required unknowns. If the system has the same number unknowns and equations, then it is called an n-th order system.
The most popular methods for solving SLAE
V educational institutions secondary education are studying various techniques for solving such systems. Most often this simple equations, consisting of two unknowns, so any existing method it won't take long to find answers to them. It can be like a substitution method, when another equation is derived from one equation and substituted into the original one. Or term by term subtraction and addition. But the Gauss method is considered the easiest and most universal. It makes it possible to solve equations with any number of unknowns. Why is this technique considered rational? Everything is simple. The matrix method is good because it does not require several times to rewrite unnecessary characters in the form of unknowns, it is enough to do arithmetic operations on the coefficients - and you will get a reliable result.
Where are SLAEs used in practice?
The solution of SLAE are the points of intersection of lines on the graphs of functions. In our high-tech computer age, people who are closely involved in the development of games and other programs need to know how to solve such systems, what they represent and how to check the correctness of the resulting result. Most often, programmers develop special linear algebra calculators, this includes a system of linear equations. The Gauss method allows you to calculate all existing solutions. Other simplified formulas and techniques are also used.
SLAE compatibility criterion
Such a system can only be solved if it is compatible. For clarity, we present the SLAE in the form Ax=b. It has a solution if rang(A) equals rang(A,b). In this case, (A,b) is an extended form matrix that can be obtained from matrix A by rewriting it with free terms. It turns out that solving linear equations using the Gaussian method is quite easy.
Perhaps some notation is not entirely clear, so it is necessary to consider everything with an example. Let's say there is a system: x+y=1; 2x-3y=6. It consists of only two equations in which there are 2 unknowns. The system will have a solution only if the rank of its matrix is equal to the rank of the augmented matrix. What is a rank? This is the number of independent lines of the system. In our case, the rank of the matrix is 2. Matrix A will consist of the coefficients located near the unknowns, and the coefficients behind the “=” sign will also fit into the expanded matrix.
Why SLAE can be represented in matrix form
Based on the compatibility criterion according to the proven Kronecker-Capelli theorem, the system of linear algebraic equations can be represented in matrix form. Using the Gaussian cascade method, you can solve the matrix and get the only reliable answer for the entire system. If the rank of an ordinary matrix is equal to the rank of its extended matrix, but less than the number of unknowns, then the system has an infinite number of answers.
Matrix transformations
Before moving on to solving matrices, it is necessary to know what actions can be performed on their elements. There are several elementary transformations:
- By rewriting the system into a matrix form and solving it, it is possible to multiply all the elements of the series by the same coefficient.
- In order to convert a matrix to canonical form, two parallel rows can be swapped. The canonical form implies that all elements of the matrix that are located along the main diagonal become ones, and the remaining ones become zeros.
- The corresponding elements of the parallel rows of the matrix can be added one to the other.
Jordan-Gauss method
The essence of solving systems of linear homogeneous and inhomogeneous equations Gaussian method is to gradually eliminate the unknowns. Let's say we have a system of two equations in which there are two unknowns. To find them, you need to check the system for compatibility. The Gaussian equation is solved very simply. It is necessary to write out the coefficients located near each unknown in a matrix form. To solve the system, you need to write out the augmented matrix. If one of the equations contains a smaller number of unknowns, then "0" must be put in place of the missing element. All known transformation methods are applied to the matrix: multiplication, division by a number, adding the corresponding elements of the rows to each other, and others. It turns out that in each row it is necessary to leave one variable with the value "1", the rest should be reduced to zero. For a more accurate understanding, it is necessary to consider the Gauss method with examples.
A simple example of solving a 2x2 system
To begin with, let's take a simple system of algebraic equations, in which there will be 2 unknowns.
Let's rewrite it in an augmented matrix.
To solve this system of linear equations, only two operations are required. We need to bring the matrix to the canonical form so that there are units along the main diagonal. So, translating from the matrix form back into the system, we get the equations: 1x+0y=b1 and 0x+1y=b2, where b1 and b2 are the answers obtained in the process of solving.
- The first step in solving the augmented matrix will be as follows: the first row must be multiplied by -7 and the corresponding elements added to the second row, respectively, in order to get rid of one unknown in the second equation.
- Since the solution of equations by the Gauss method implies bringing the matrix to the canonical form, then it is necessary to do the same operations with the first equation and remove the second variable. To do this, we subtract the second line from the first and get the necessary answer - the solution of the SLAE. Or, as shown in the figure, we multiply the second row by a factor of -1 and add the elements of the second row to the first row. This is the same.
As you can see, our system is solved by the Jordan-Gauss method. We rewrite it in the required form: x=-5, y=7.
An example of solving SLAE 3x3
Suppose we have a more complex system of linear equations. The Gauss method makes it possible to calculate the answer even for the most seemingly confusing system. Therefore, in order to delve deeper into the calculation methodology, we can move on to a more complex example with three unknowns.
As in the previous example, we rewrite the system in the form of an expanded matrix and begin to bring it to the canonical form.
To solve this system, you will need to perform much more actions than in the previous example.
- First you need to make in the first column one single element and the rest zeros. To do this, multiply the first equation by -1 and add the second equation to it. It is important to remember that we rewrite the first line in original form, and the second - already in the modified.
- Next, we remove the same first unknown from the third equation. To do this, we multiply the elements of the first row by -2 and add them to the third row. Now the first and second lines are rewritten in their original form, and the third - already with changes. As you can see from the result, we got the first one at the beginning of the main diagonal of the matrix and the rest are zeros. A few more actions, and the system of equations by the Gauss method will be reliably solved.
- Now you need to do operations on other elements of the rows. The third and fourth steps can be combined into one. We need to divide the second and third lines by -1 to get rid of the negative ones on the diagonal. We have already brought the third line to the required form.
- Next, we canonicalize the second line. To do this, we multiply the elements of the third row by -3 and add them to the second line of the matrix. It can be seen from the result that the second line is also reduced to the form we need. It remains to do a few more operations and remove the coefficients of the unknowns from the first row.
- In order to make 0 from the second element of the row, you need to multiply the third row by -3 and add it to the first row.
- The next decisive step is to add to the first line necessary elements second row. So we get the canonical form of the matrix, and, accordingly, the answer.
As you can see, the solution of equations by the Gauss method is quite simple.
An example of solving a 4x4 system of equations
Some more complex systems of equations can be solved by the Gaussian method using computer programs. It is necessary to drive coefficients for unknowns into existing empty cells, and the program will calculate the required result step by step, describing each action in detail.
Described below step-by-step instruction solutions to this example.
In the first step, free coefficients and numbers for unknowns are entered into empty cells. Thus, we get the same augmented matrix that we write by hand.
And all the necessary arithmetic operations are performed to bring the extended matrix to the canonical form. It must be understood that the answer to a system of equations is not always integers. Sometimes the solution can be from fractional numbers.
Checking the correctness of the solution
The Jordan-Gauss method provides for checking the correctness of the result. In order to find out whether the coefficients are calculated correctly, you just need to substitute the result into the original system of equations. The left side of the equation must match the right side, which is behind the equals sign. If the answers do not match, then you need to recalculate the system or try to apply another method of solving SLAE known to you, such as substitution or term-by-term subtraction and addition. After all, mathematics is a science that has a huge number of various techniques solutions. But remember: the result should always be the same, no matter what solution method you used.
Gauss method: the most common errors in solving SLAE
During the solution of linear systems of equations, errors most often occur, such as incorrect transfer of coefficients to a matrix form. There are systems in which some unknowns are missing in one of the equations, then, transferring the data to the expanded matrix, they can be lost. As a result, when solving this system, the result may not correspond to the real one.
Another of the main mistakes can be incorrect writing out the final result. It must be clearly understood that the first coefficient will correspond to the first unknown from the system, the second - to the second, and so on.
The Gauss method describes in detail the solution of linear equations. Thanks to him, it is easy to perform the necessary operations and find the right result. In addition, this is a universal tool for finding a reliable answer to equations of any complexity. Maybe that is why it is so often used in solving SLAE.
1. System of linear algebraic equations
1.1 The concept of a system of linear algebraic equations
A system of equations is a condition consisting in the simultaneous execution of several equations in several variables. A system of linear algebraic equations (hereinafter referred to as SLAE) containing m equations and n unknowns is a system of the form:
where the numbers a ij are called the coefficients of the system, the numbers b i are free members, aij and b i(i=1,…, m; b=1,…, n) are some known numbers, and x 1 ,…, x n- unknown. In the notation of the coefficients aij the first index i denotes the number of the equation, and the second index j is the number of the unknown at which this coefficient stands. Subject to finding the number x n . It is convenient to write such a system in a compact matrix form: AX=B. Here A is the matrix of coefficients of the system, called the main matrix;
is a column vector of unknown xj. is a column vector of free members bi.
The product of matrices A * X is defined, since there are as many columns in matrix A as there are rows in matrix X (n pieces).
The extended matrix of the system is the matrix A of the system, supplemented by a column of free terms
1.2 Solution of a system of linear algebraic equations
The solution of a system of equations is an ordered set of numbers (values of variables), when substituting them instead of variables, each of the equations of the system turns into a true equality.
The solution of the system is n values of the unknowns x1=c1, x2=c2,…, xn=cn, substituting which all equations of the system turn into true equalities. Any solution of the system can be written as a matrix-column
A system of equations is called consistent if it has at least one solution, and inconsistent if it has no solutions.
A joint system is called definite if it has only decision, and indefinite if it has more than one solution. In the latter case, each of its solutions is called a particular solution of the system. The set of all particular solutions is called the general solution.
To solve a system means to find out whether it is consistent or inconsistent. If the system is compatible, find it common decision.
Two systems are called equivalent (equivalent) if they have the same general solution. In other words, systems are equivalent if every solution to one of them is a solution to the other, and vice versa.
A transformation, the application of which turns the system into new system, equivalent to the original one, is called an equivalent or equivalent transformation. The following transformations can serve as examples of equivalent transformations: swapping two equations of the system, swapping two unknowns together with the coefficients of all equations, multiplying both parts of any equation of the system by a non-zero number.
A system of linear equations is called homogeneous if all free terms are equal to zero:
A homogeneous system is always consistent, since x1=x2=x3=…=xn=0 is a solution to the system. This solution is called null or trivial.
2. Gaussian elimination method
2.1 The essence of the Gaussian elimination method
The classical method for solving systems of linear algebraic equations is the method sequential exclusion unknown - Gauss method(It is also called the Gaussian elimination method). This is a method of successive elimination of variables, when, with the help of elementary transformations, a system of equations is reduced to an equivalent system of a stepped (or triangular) form, from which all other variables are found sequentially, starting from the last (by number) variables.
The Gaussian solution process consists of two stages: forward and backward moves.
1. Direct move.
At the first stage, the so-called direct move is carried out, when, by means of elementary transformations over rows, the system is brought to a stepped or triangular form, or it is established that the system is inconsistent. Namely, among the elements of the first column of the matrix, a non-zero one is chosen, it is moved to the uppermost position by permuting the rows, and the first row obtained after the permutation is subtracted from the remaining rows, multiplying it by a value equal to the ratio of the first element of each of these rows to the first element of the first row, zeroing thus the column below it.
After the indicated transformations have been made, the first row and the first column are mentally crossed out and continue until a zero-size matrix remains. If at some of the iterations among the elements of the first column there was not found a non-zero one, then go to the next column and perform a similar operation.
At the first stage (forward run), the system is reduced to a stepped (in particular, triangular) form.
The system below is stepwise:
,
The coefficients aii are called the main (leading) elements of the system.
(if a11=0, rearrange the rows of the matrix so that a 11 was not equal to 0. This is always possible, because otherwise the matrix contains a zero column, its determinant zero and the system is inconsistent).We transform the system by eliminating the unknown x1 in all equations except the first one (using elementary transformations systems). To do this, multiply both sides of the first equation by
and add term by term with the second equation of the system (or from the second equation we subtract term by term the first multiplied by ). Then we multiply both parts of the first equation by and add it to the third equation of the system (or subtract the first one multiplied by the third term by term). Thus, we successively multiply the first row by a number and add to i-th line, for i= 2, 3, …,n.Continuing this process, we get the equivalent system:
– new values of the coefficients for unknowns and free terms in the last m-1 equations of the system, which are determined by the formulas: Thus, at the first step, all coefficients under the first leading element a 11 are destroyed
0, the second step destroys the elements under the second leading element a 22 (1) (if a 22 (1) 0), and so on. Continuing this process further, we will finally reduce the original system to a triangular system at the (m-1) step.If, in the process of reducing the system to a stepwise form, zero equations appear, i.e. equalities of the form 0=0, they are discarded. If there is an equation of the form
This indicates the incompatibility of the system. This completes the direct course of the Gauss method.
2. Reverse move.
At the second stage, the so-called reverse move is carried out, the essence of which is to express all the resulting basic variables in terms of non-basic ones and construct fundamental system solutions, or, if all variables are basic, then express in numerical form the only solution of the system of linear equations.
This procedure begins with the last equation, from which the corresponding basic variable is expressed (there is only one in it) and substituted into the previous equations, and so on, going up the "steps".
Each line corresponds to exactly one basic variable, so at each step, except for the last (topmost), the situation exactly repeats the case of the last line.
Note: in practice, it is more convenient to work not with the system, but with its extended matrix, performing all elementary transformations on its rows. It is convenient that the coefficient a11 be equal to 1 (rearrange the equations, or divide both sides of the equation by a11).
2.2 Examples of solving SLAE by the Gauss method
In this section, three various examples Let us show how the SLAE can be solved by the Gauss method.
Example 1. Solve SLAE of the 3rd order.
Set the coefficients to zero at
in the second and third lines. To do this, multiply them by 2/3 and 1, respectively, and add them to the first line:
We continue to consider systems of linear equations. This lesson is the third on the topic. If you have a vague idea of what a system of linear equations is in general, you feel like a teapot, then I recommend starting with the basics on the Next page, it is useful to study the lesson.
Gauss method is easy! Why? The famous German mathematician Johann Carl Friedrich Gauss, during his lifetime, received recognition as the greatest mathematician of all time, a genius, and even the nickname "King of Mathematics". And everything ingenious, as you know, is simple! By the way, not only suckers, but also geniuses get into the money - the portrait of Gauss flaunted on a bill of 10 Deutschmarks (before the introduction of the euro), and Gauss still mysteriously smiles at the Germans from ordinary postage stamps.
The Gauss method is simple in that it IS ENOUGH THE KNOWLEDGE OF A FIFTH-GRADE STUDENT to master it. Must be able to add and multiply! It is no coincidence that the method of successive elimination of unknowns is often considered by teachers at school mathematical electives. It is a paradox, but the Gauss method causes the greatest difficulties for students. Nothing surprising - it's all about the methodology, and I will try to tell in an accessible form about the algorithm of the method.
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), an 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 coefficients for unknowns, in this example, the matrix of the system: . Extended System Matrix is the same matrix of the system plus a column of free members, 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 rearrange 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 case study: . 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. Is 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 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 CAREFULLY:
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 simpler the solution:
At the final stage of elementary transformations, one more zero must be obtained 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 self-solving, a sample of finishing 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 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. Full solution and design sample 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 else like this conditional 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. Confidently learn to solve systems by other methods (Cramer's method, matrix method) can be literally the first time - there is a very strict algorithm. But in order to feel confident in the Gauss method, you should “fill your hand” and solve at least 5-10 ten 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, a more complex example for an independent solution:
Example 5
Solve a system of 4 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 success!
Solutions and answers:
Example 2:
Solution
:
Let us write down the extended matrix of the system and, using elementary transformations, 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:
Solution
:
We write the extended matrix of the system and, using elementary transformations, bring it to a 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. (6) The second row was multiplied by -1, the third row was divided by -83.
Reverse move:
Answer :
Example 5:
Solution
:
Let us write down the matrix of the system and, using elementary transformations, bring it to a stepwise form:
Conversions performed: (1) The first and second lines have been swapped. (2) The first row was added to the second row, multiplied by -2. The first line was added to the third line, multiplied by -2. The first line was added to the fourth line, multiplied by -3. (3) The second line multiplied by 4 was added to the third line. The second line multiplied by -1 was added to the fourth line. (4) The sign of the second line has been changed. The fourth line was divided by 3 and placed instead of the third line. (5) The third line was added to the fourth line, multiplied by -5.
Reverse move:
Answer :
Here you can solve a system of linear equations for free Gauss method online large sizes in complex numbers with a very detailed solution. Our calculator can solve online both conventional definite and indefinite systems of linear equations using the Gaussian method, which has an infinite number of solutions. In this case, in the answer you will receive the dependence of some variables through others, free ones. You can also check the system of equations for compatibility online using the Gaussian solution.
About method
When solving a system of linear equations online method Gauss performs the following steps.
- We write the augmented matrix.
- In fact, the solution is divided into the forward and backward steps of the Gaussian method. The direct move of the Gauss method is called the reduction of the matrix to a stepped form. The reverse move of the Gauss method is the reduction of a matrix to a special stepped form. But in practice, it is more convenient to immediately zero out what is both above and below the element in question. Our calculator uses exactly this approach.
- It is important to note that when solving by the Gauss method, the presence in the matrix of at least one zero row with a nonzero right side(column of free members) indicates the incompatibility of the system. Solution linear system in this case does not exist.
To better understand how the Gaussian algorithm works online, enter any example, select "very detailed solution and look up his solution online.
Let the system be given, ∆≠0. (one)Gauss method is a method of successive elimination of unknowns.
The essence of the Gauss method is to transform (1) to a system with a triangular matrix , from which then successively ( backwards) the values of all unknowns are obtained. Let's consider one of the computational schemes. This circuit is called the single division circuit. So let's take a look at this diagram. Let a 11 ≠0 (leading element) divide by a 11 the first equation. Get
(2)
Using equation (2), it is easy to exclude the unknowns x 1 from the remaining equations of the system (for this, it is enough to subtract equation (2) from each equation preliminarily multiplied by the corresponding coefficient at x 1), that is, at the first step we obtain
.
In other words, at step 1, each element of subsequent rows, starting from the second, is equal to the difference between the original element and the product of its “projection” on the first column and the first (transformed) row.
After that, leaving the first equation alone, over the rest of the equations of the system obtained at the first step, we will perform a similar transformation: we choose from among them an equation with a leading element and use it to exclude x 2 from the remaining equations (step 2).
After n steps, instead of (1) we get an equivalent system 
(3)
Thus, at the first stage, we will obtain a triangular system (3). This step is called forward.
At the second stage (reverse move) we sequentially find from (3) the values x n , x n -1 , …, x 1 .
Let's denote the obtained solution as x 0 . Then the difference ε=b-A x 0 is called residual.
If ε=0, then the found solution x 0 is correct.
Calculations by the Gauss method are performed in two stages:
- The first stage is called the direct course of the method. At the first stage, the original system is converted to a triangular form.
- The second stage is called reverse. At the second stage, a triangular system equivalent to the original one is solved.
At each step, it was assumed that the leading element is different from zero. If this is not the case, then any other element can be used as a leader, as if rearranging the equations of the system.
Purpose of the Gauss method
The Gauss method is intended for solving systems of linear equations. Refers to direct methods of solution.Types of Gauss method
- Classical Gauss method;
- Modifications of the Gauss method. One of the modifications of the Gaussian method is the circuit with the choice of the main element. A feature of the Gauss method with the choice of the main element is such a permutation of the equations so that at the k-th step the leading element is the largest element in the k-th column.
- Jordan-Gauss method;
Illustrate the difference Jordan-Gauss method from the Gauss method on examples.
Gauss solution example
Let's solve the system: 
For the convenience of calculations, we swap the lines: 
Multiply the 2nd row by (2). Add the 3rd line to the 2nd 
Multiply the 2nd row by (-1). Add the 2nd row to the 1st 
From the 1st line we express x 3:
From the 2nd line we express x 2:
From the 3rd line we express x 1: 
An example of a solution by the Jordan-Gauss method
We will solve the same SLAE using the Jordano-Gauss method. 
We will sequentially choose the resolving element of the RE, which lies on the main diagonal of the matrix.
The enabling element is equal to (1).
NE \u003d SE - (A * B) / RE
RE - enabling element (1), A and B - matrix elements forming a rectangle with elements of STE and RE.
Let's present the calculation of each element in the form of a table:
| x 1 | x2 | x 3 | B |
| 1 / 1 = 1 | 2 / 1 = 2 | -2 / 1 = -2 | 1 / 1 = 1 |
 |  |  |  |
The enabling element is equal to (3).
In place of the resolving element, we get 1, and in the column itself we write zeros.
All other elements of the matrix, including the elements of column B, are determined by the rectangle rule.
To do this, select four numbers that are located at the vertices of the rectangle and always include the enabling element of the RE.
| x 1 | x2 | x 3 | B |
 |  |
||
| 0 / 3 = 0 | 3 / 3 = 1 | 1 / 3 = 0.33 | 4 / 3 = 1.33 |
 |
The enabling element is (-4).
In place of the resolving element, we get 1, and in the column itself we write zeros.
All other elements of the matrix, including the elements of column B, are determined by the rectangle rule.
To do this, select four numbers that are located at the vertices of the rectangle and always include the enabling element of the RE.
Let's present the calculation of each element in the form of a table:
| x 1 | x2 | x 3 | B |
 |  |  |  |
 |  |
||
| 0 / -4 = 0 | 0 / -4 = 0 | -4 / -4 = 1 | -4 / -4 = 1 |
Answer: x 1 = 1, x 2 = 1, x 3 = 1
