Here you can solve the system for free linear equations Gauss method online large sizes in complex numbers with a very detailed solution. Our calculator can solve online both the usual definite and indefinite system of linear equations by the Gauss method, which has infinite set decisions. In this case, in the answer you will receive the dependence of some variables through other, free ones. You can also check the system of equations for consistency online using the Gaussian solution.
About the method
When solving a system of linear equations online method Gauss the following steps are performed.
- We write the extended matrix.
- In fact, the solution is divided into direct and reverse stroke Gauss method. The direct step of the Gaussian method is the reduction of a matrix to a stepwise form. The reverse of the Gaussian method is the reduction of a matrix to a special stepwise form. But in practice, it is more convenient to immediately zero out what is located both above and below the element in question. Our calculator uses exactly this approach.
- It is important to note that when solving using the Gaussian method, the presence in the matrix of at least one zero row with a NON-zero right-hand side (column of free terms) indicates the incompatibility of the system. In this case, a solution to the linear system does not exist.
To best understand how the Gaussian algorithm works online, enter any example, select "very detailed solution" and look up his solution online.
One of the simplest ways to solve a system of linear equations is a technique based on the calculation of determinants ( Cramer's rule). Its advantage is that it allows you to immediately record the solution; it is especially convenient in cases where the coefficients of the system are not numbers, but some parameters. Its disadvantage is the cumbersomeness of calculations in the case large number equations; moreover, Cramer's rule is not directly applicable to systems in which the number of equations does not coincide with the number of unknowns. In such cases, it is usually used Gaussian method.
Systems of linear equations having the same set of solutions are called equivalent. Obviously, the set of solutions of a linear system will not change if any equations are swapped, or if one of the equations is multiplied by some non-zero number, or if one equation is added to another.
Gauss method (method sequential elimination unknown) is that with the help of elementary transformations the system is reduced to an equivalent system of a step type. First, using the 1st equation, we eliminate x 1 of all subsequent equations of the system. Then, using the 2nd equation, we eliminate x 2 from the 3rd and all subsequent equations. This process, called direct Gaussian method, continues until there is only one unknown left on the left side of the last equation x n. After this it is done inverse of the Gaussian method– solving the last equation, we find x n; after that, using this value, from the penultimate equation we calculate x n–1, etc. We find the last one x 1 from the first equation.
It is convenient to carry out Gaussian transformations by performing transformations not with the equations themselves, but with the matrices of their coefficients. Consider the matrix:
called expanded matrix of the system, because, in addition to the main matrix of the system, it includes a column of free terms. The Gaussian method is based on reducing the main matrix of the system to a triangular form (or trapezoidal form in the case of non-square systems) using elementary row transformations (!) of the extended matrix of the system.
Example 5.1. Solve the system using the Gaussian method:
Solution. Let's write out the extended matrix of the system and, using the first row, after that we will reset the remaining elements:
we get zeros in the 2nd, 3rd and 4th rows of the first column:
Now we need all elements in the second column below the 2nd row to be equal to zero. To do this, you can multiply the second line by –4/7 and add it to the 3rd line. However, in order not to deal with fractions, let's create a unit in the 2nd row of the second column and only
Now, to get a triangular matrix, you need to reset the element of the fourth row of the 3rd column; to do this, you can multiply the third row by 8/54 and add it to the fourth. However, in order not to deal with fractions, we will swap the 3rd and 4th rows and the 3rd and 4th columns and only after that we will reset the specified element. Note that when rearranging the columns, the corresponding variables change places and this must be remembered; other elementary transformations with columns (addition and multiplication by a number) cannot be performed!
The last simplified matrix corresponds to a system of equations equivalent to the original one:
From here, using the inverse of the Gaussian method, we find from the fourth equation x 3 = –1; from the third x 4 = –2, from the second x 2 = 2 and from the first equation x 1 = 1. In matrix form, the answer is written as
We considered the case when the system is definite, i.e. when there is only one solution. Let's see what happens if the system is inconsistent or uncertain.
Example 5.2. Explore the system using the Gaussian method:
Solution. We write out and transform the extended matrix of the system
We write a simplified system of equations:
Here, in the last equation it turns out that 0=4, i.e. contradiction. Consequently, the system has no solution, i.e. she incompatible. à
Example 5.3. Explore and solve the system using the Gaussian method:
Solution. We write out and transform the extended matrix of the system:
As a result of the transformations, the last line contains only zeros. This means that the number of equations has decreased by one:
Thus, after simplifications, there are two equations left, and four unknowns, i.e. two unknown "extra". Let them be "superfluous", or, as they say, free variables, will x 3 and x 4. Then
Believing x 3 = 2a And x 4 = b, we get x 2 = 1–a And x 1 = 2b–a; or in matrix form
A solution written in this way is called general, because, giving parameters a And b different meanings, it is possible to describe all possible solutions of the system. a
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, if you feel like a teapot, then I recommend starting with the basics on the page Next, it is useful to study the lesson.
The Gaussian method is easy! Why? The famous German mathematician Johann Carl Friedrich Gauss received recognition during his lifetime greatest mathematician of all times, a genius and even nicknamed “the king of mathematics.” And everything ingenious, as you know, is simple! By the way, not only suckers get money, but also geniuses - Gauss’s portrait was on the 10 Deutschmark banknote (before the introduction of the euro), and Gauss still smiles mysteriously at Germans from ordinary postage stamps.
The Gauss method is simple in that the KNOWLEDGE OF A FIFTH-GRADE STUDENT IS ENOUGH to master it. You must know how to add and multiply! It is no coincidence that teachers often consider the method of sequential exclusion of unknowns in school mathematics electives. It’s a paradox, but students find the Gaussian method the most difficult. Nothing surprising - it’s all about the methodology, and I will try to talk about the algorithm of the method in an accessible form.
First, let's systematize a little knowledge about systems of linear equations. A system of linear equations can:
1) Have a unique solution. 2) Have infinitely many solutions. 3) Have no solutions (be non-joint).
The Gauss method is the most powerful and universal 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. And the method of sequential elimination of unknowns Anyway will 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 devoted to the situations of points No. 2-3. I note that the algorithm of the method itself works the same in all three cases.
Let's return 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 down extended system matrix: . I think everyone can see by what principle the coefficients are written. The vertical line inside the matrix does not have any mathematical meaning - it is simply a strikethrough for ease of design.
Reference : I recommend you 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 – this is the same matrix of the system plus a column of free terms, in this case: . For brevity, any of the matrices can be simply called a matrix.
After the extended system matrix is written, it is necessary to perform some actions with it, which are also called elementary transformations.
The following elementary transformations exist:
1) Strings matrices Can rearrange in some places. For example, in the matrix under consideration, you can painlessly rearrange the first and second rows: 
2) If the matrix has (or has appeared) proportional (like special case– identical) lines, then it follows delete All these rows are from the matrix 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 appears in the matrix during transformations, then it should also be delete. I won’t draw, of course, the zero line is the line in which all zeros.
4) The matrix row can be multiply (divide) to 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 because it simplifies further transformations of the matrix.
5) This transformation causes the most difficulties, but in fact there is nothing complicated either. To a row of a matrix you can add another string multiplied by a number, different from zero. Let's look at our matrix from a practical example: . First I'll describe the transformation in great detail. Multiply the first line 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 TO WHICH IS ADDED changes UT.
In practice, of course, they don’t write it in such detail, but write it briefly: 
Once again: to the second line added the first line multiplied by –2. A line is usually multiplied orally or on a draft, with the mental calculation process going something like this:
“I rewrite the matrix and rewrite the first line: 
»
“First column. At the bottom I need to get zero. Therefore, I multiply the one at the top by –2: , and add the first one to the second line: 2 + (–2) = 0. I write the result in the second line: 
»
“Now the second column. At the top, I multiply -1 by -2: . I add the first to the second line: 1 + 2 = 3. I write the result in the second line: 
»
“And the third column. At the top I multiply -5 by -2: . I add the first to the second line: –7 + 10 = 3. I write the result in the second line: 
»
Please carefully understand this example and understand the sequential calculation algorithm, if you understand this, then the Gaussian method is practically in your pocket. But, of course, we will still work on this transformation.
Elementary transformations do not change the solution of the system of equations
! ATTENTION: considered manipulations cannot be used, if you are offered a task where the matrices are given “by themselves.” For example, with “classical” operations with matrices Under no circumstances should you rearrange anything inside the matrices! Let's return to our system. It is practically taken to pieces.
Let us write down the extended matrix of the system and, using elementary transformations, reduce it to stepped view:
(1) The first line was added to the second line, multiplied by –2. And again: why do we multiply the first line 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 line by 3.
The purpose of elementary transformations
–
reduce the matrix to stepwise form: 
. In the design of the task, they just mark out the “stairs” 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 scientific and educational literature it is often called trapezoidal view or triangular view.
As a result of elementary transformations, we obtained equivalent original system of equations:
Now the system needs to be “unwinded” in the opposite direction - from bottom to top, this process is called inverse of the Gaussian method.
In the lower equation we already have a ready-made result: .
Let's consider the first equation of the system and substitute the already known value of “y” into it:
Let's consider the most common situation, when the Gaussian method requires solving a system of three linear equations with three unknowns.
Example 1
Solve the system of equations using the Gauss method: 
Let's write the extended matrix of the system: 
Now I will immediately draw the result that we will come to during the solution: 
And I repeat, our goal is to bring the matrix to a stepwise form using elementary transformations. Where to start?
First, look at the top left number: 
Should almost always be here unit. Generally speaking, –1 (and sometimes other numbers) will do, but somehow it has traditionally happened that one 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. It's already easier.
The unit in the top left corner is organized. Now you need to get zeros in these places: 
We get zeros using 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, 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:
We write the result in the second line: 
We deal with the third line in the same way (3, 2, –5, –1). To get a zero in the first position, you need to the third line add the first line multiplied by –3. Mentally or on a draft, multiply the first line by –3: (–3, –6, 3, –27). AND to the third line we add the first line multiplied by –3:
We write the result in the third line:
In practice, these actions are usually performed orally and written down in one step: 
No need to count everything at once and at the same time. The order of calculations and “writing in” the results consistent and usually it’s like this: first we rewrite the first line, and we puff on ourselves little by little - CONSISTENTLY and ATTENTIVELY:
And I have already discussed the mental process 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 numbers, the simpler the solution:
At the final stage of elementary transformations, you need to get another zero here: 
For this to the third line we add the second line multiplied by –2:
Try to figure out this action yourself - mentally multiply the second line by –2 and perform 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 system of linear equations was obtained: 
Cool.
Now the reverse of the Gaussian method comes into play. The equations “unwind” from bottom to top.
In the third equation we already have a ready result:
Let's look at the second equation: . The meaning of "zet" is already known, thus:
And finally, the first equation: . “Igrek” and “zet” are known, it’s just a matter of little things:
Answer: 
As has been repeatedly noted, for any system of equations it is possible and necessary to check the solution found, fortunately, this is easy and quick.
Example 2
This is an example for an independent solution, a sample of the final design and an answer at the end of the lesson.
It should be noted that your progress of the decision may not coincide with my decision process, 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”. We should have one there. The problem is that there are no units in the first column at all, so rearranging the rows will not solve anything. 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 added the first and second lines, while the second line did not change.
Now at the top left there is “minus one”, which suits us quite well. Anyone who wants to get +1 can perform an additional movement: multiply the first line by –1 (change its sign).
(2) The first line multiplied by 5 was added to the second line. The first line multiplied by 3 was added to the third line.
(3) The first line was multiplied by –1, in principle, this is for beauty. The sign of the third line was also changed and it was moved to second place, so that on the second “step” we had the required unit.
(4) The second line was added to the third line, multiplied by 2.
(5) The third line was divided by 3.
A bad sign that indicates an error in calculations (more rarely, a typo) is a “bad” bottom line. That is, if we got something like , below, and, accordingly, 
, then with a high degree of probability we can say that an error was made during elementary transformations.
We charge the reverse, in the design of examples they often do not rewrite the system itself, but the equations are “taken directly from the given matrix.” The reverse stroke, I remind you, works from bottom to top. Yes, here is a gift:
Answer: 
.
Example 4
Solve a system of linear equations using the Gauss method 
This is an example for you to solve on your own, it is somewhat more complicated. It's okay if someone gets confused. Full solution and sample design at the end of the lesson. Your solution may be different from my solution.
In the last part we will look at some features of the Gaussian algorithm. The first feature is that sometimes some variables are missing from the system equations, for example: 
How to correctly write the extended system matrix? I already talked about this point in class. Cramer's rule. Matrix method. In the extended matrix of the system, we put zeros in place of missing variables: 
By the way, that's pretty easy example, since there is already one zero in the first column, and there are fewer elementary conversions 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 there? In some cases they can. Consider the system: 
.
Here on the upper left “step” we have a two. But we notice the fact that all the numbers in the first column are divisible by 2 without a remainder - and the other is two and six. And the two at the top left will suit us! In 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. This way we will get the required zeros in the first column.
Or something like this conditional example: 
. Here the three on the second “step” 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: add the second line to the third line, multiplied by –4, as a result of which the zero we need will be obtained.
Gauss's method is universal, but there is one peculiarity. Confidently learn to solve systems using other methods (Cramer’s method, matrix method) you can literally the first time - there is a very strict algorithm. But in order to feel confident in the Gaussian method, you should “get your teeth into” and solve at least 5-10 ten systems. Therefore, at first there may be confusion and errors in calculations, and there is nothing unusual or tragic about this.
Rainy autumn weather outside the window.... Therefore, for everyone who wants a more complex example to solve on their own:
Example 5
Solve a system of 4 linear equations with four unknowns using the Gauss method. 
Such a task is not so rare in practice. I think even a teapot who has thoroughly studied this page will understand the algorithm for solving such a system intuitively. Fundamentally, everything is the same - there are just more actions.
Cases when the system has no solutions (inconsistent) or has infinitely many solutions are discussed in the lesson Incompatible systems and systems with a common solution. There you can fix the considered algorithm of the Gaussian method.
I 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 stepwise form.
Elementary transformations performed:
(1) The first line was added to the second line, multiplied by –2. The first line was added to the third line, multiplied by –1.
Attention!
Here you may be tempted to subtract the first from the third line; I highly recommend not subtracting it - the risk of error greatly increases. Just fold it!
(2) The sign of the second line was changed (multiplied by –1). The second and third lines have been swapped.
Please note
, that on the “steps” we are satisfied not only with one, but also with –1, which is even more convenient.
(3) The second line was added to the third line, multiplied by 5.
(4) The sign of the second line was changed (multiplied by –1). The third line was divided by 14.
Reverse:
Answer
:
.
Example 4:
Solution
:
Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:
Conversions performed: (1) A second line was added to the first line. Thus, the desired unit is organized on the upper left “step”. (2) The first line multiplied by 7 was added to the second line. The first line multiplied by 6 was added to the third line.
With the second “step” everything gets 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 was added to the second line, multiplied by –3. The required item on the second step has been received. . (5) The second line was added to the third line, multiplied by 6. (6) The second line was multiplied by –1, the third line was divided by -83.
Reverse:
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 line was added to the second line, 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 was added to the third line, multiplied by 4. The second line was added to the fourth line, multiplied by –1. (4) The sign of the second line was changed. The fourth line was divided by 3 and placed in place of the third line. (5) The third line was added to the fourth line, multiplied by –5.
Reverse:
Answer :
Educational institution "Belarusian State
Agricultural Academy"
Department of Higher Mathematics
to study the topic “Gauss method for solving systems of linear
equations" by students of the accounting faculty of correspondence education (NISPO)
Gorki, 2013
Gauss method for solving systems of linear equations
Equivalent systems of equations
Two systems of linear equations are said to be equivalent if each solution of one of them is a solution of the other. The process of solving a system of linear equations consists of sequentially transforming it into an equivalent system using the so-called elementary transformations , which are:
1) rearrangement of any two equations of the system;
2) multiplying both sides of any equation of the system by a nonzero number;
3) adding to any equation another equation multiplied by any number;
4) crossing out an equation consisting of zeros, i.e. equations of the form
Gaussian elimination
Consider the system m linear equations with n unknown:
The essence of the Gaussian method or the method of sequential elimination of unknowns is as follows.
First, using elementary transformations, the unknown is eliminated from all equations of the system except the first. Such system transformations are called Gaussian elimination step . The unknown is called enabling variable at the first step of transformation. The coefficient is called resolution factor , the first equation is called resolving equation , and the column of coefficients at permission column .
When performing one step of Gaussian elimination, you need to use the following rules:
1) the coefficients and the free term of the resolving equation remain unchanged;
2) the coefficients of the resolution column located below the resolution coefficient become zero;
3) all other coefficients and free terms when performing the first step are calculated according to the rectangle rule:
, Where i=2,3,…,m; j=2,3,…,n.
We will perform similar transformations on the second equation of the system. This will lead to a system in which the unknown will be eliminated in all equations except the first two. As a result of such transformations over each of the equations of the system (direct progression of the Gaussian method), the original system is reduced to an equivalent step system of one of the following types.
Reverse Gaussian Method
Step system
has triangular view and that's all 
(i=1,2,…,n). Such a system has a unique solution. The unknowns are determined starting from the last equation (reverse of the Gaussian method).
The step system has the form
where, i.e. the number of equations of the system is less than or equal to the number of unknowns. This system has no solutions, since the last equation will not be satisfied for any values of the variable.
Step type system
has countless solutions. From the last equation, the unknown is expressed through the unknowns 
. Then, in the penultimate equation, instead of the unknown, its expression is substituted through the unknowns . Continuing the reverse of the Gaussian method, the unknowns 
can be expressed in terms of unknowns . In this case, the unknowns are called free
and can take any values, and unknown 
basic.
When solving systems in practice, it is convenient to perform all transformations not with a system of equations, but with an extended matrix of the system, consisting of coefficients for unknowns and a column of free terms.
Example 1. Solve system of equations 
Solution. Let's create an extended matrix of the system and perform elementary transformations:
.
In the extended matrix of the system, the number 3 (it is highlighted) is the resolution coefficient, the first row is the resolution row, and the first column is the resolution column. When moving to the next matrix, the resolution row does not change; all elements of the resolution column below the resolution element are replaced by zeros. And all other elements of the matrix are recalculated according to the quadrilateral rule. Instead of element 4 in the second line we write 
, instead of element -3 in the second line it will be written 
etc. Thus, the second matrix will be obtained. The resolution element of this matrix will be the number 18 in the second row. To form the next (third matrix), we leave the second row unchanged, in the column under the resolving element we write zero and recalculate the remaining two elements: instead of the number 1 we write 
, and instead of the number 16 we write .
As a result, the original system was reduced to an equivalent system
From the third equation we find 
. Let's substitute this value into the second equation: 
y=3. Let's substitute the found values into the first equation y And z: 
, x=2.
Thus, the solution to this system of equations is x=2, y=3, 
.
Example 2. Solve system of equations 
Solution. Let us perform elementary transformations on the extended matrix of the system:
In the second matrix, each element of the third row is divided by 2.
In the fourth matrix, each element of the third and fourth rows was divided by 11.
. The resulting matrix corresponds to the system of equations 
Solving this system, we find 
, 
, .
Example 3. Solve system of equations
Solution. Let's write the extended matrix of the system and perform elementary transformations:
.
In the second matrix, each element of the second, third and fourth rows was divided by 7.
As a result, a system of equations was obtained
equivalent to the original one.
Since there are two fewer equations than unknowns, then from the second equation 
. Let's substitute the expression for into the first equation: , 
.
Thus, the formulas 
give general solution of this system of equations. Unknowns are free and can take any value.
Let, for example, 
Then 
And 
. Solution 
is one of the particular solutions of the system, of which there are countless.
Questions for self-control of knowledge
1) What transformations linear systems are called elementary?
2) What transformations of the system are called the Gaussian elimination step?
3) What is a resolving variable, resolving coefficient, resolving column?
4) What rules should be used when performing one step of Gaussian elimination?
Given online calculator finds a solution to a system of linear equations (SLE) using the Gaussian method. A detailed solution is given. To calculate, select the number of variables and the number of equations. Then enter the data into the cells and click on the "Calculate" button.
|
Number representation:
Integers and/or Common fractionsWhole Numbers and/or Decimals
Number of places after decimal separator
×
Warning
Clear all cells?
Close Clear
Data entry instructions. Numbers are entered as integers (examples: 487, 5, -7623, etc.), decimals (ex. 67., 102.54, etc.) or fractions. The fraction must be entered in the form a/b, where a and b (b>0) are integers or decimal numbers. Examples 45/5, 6.6/76.4, -7/6.7, etc.
Gauss method
The Gauss method is a method of transition from the original system of linear equations (using equivalent transformations) to a system that is easier to solve than the original system.
Equivalent transformations of a system of linear equations are:
- swapping two equations in the system,
- multiplying any equation in the system by a non-zero real number,
- adding to one equation another equation multiplied by an arbitrary number.
Consider a system of linear equations:
| (1) |
Let us write system (1) in matrix form:
| Ax=b | (2) |
  | (3) |
A- called the coefficient matrix of the system, b− right side of the restrictions, x− vector of variables to be found. Let rank( A)=p.
Equivalent transformations do not change the rank of the coefficient matrix and the rank of the extended matrix of the system. The set of solutions of the system also does not change under equivalent transformations. The essence of the Gauss method is to reduce the matrix of coefficients A to diagonal or stepped.
Let's build an extended matrix of the system:
At the next stage, we reset all elements of column 2, below the element. If this element is zero, then this row is swapped with the row lying below this row and having a non-zero element in the second column. Next, reset all elements of column 2 below the leading element a 22. To do this, add lines 3, ... m with string 2 multiplied by − a 32 /a 22 , ..., −a m2/ a 22, respectively. Continuing the procedure, we obtain a matrix of diagonal or stepped form. Let the resulting extended matrix have the form:
| (7) |
 |
Because rangA=rang(A|b), then the set of solutions (7) is ( n−p)− variety. Hence n−p the unknowns can be chosen arbitrarily. The remaining unknowns from system (7) are calculated as follows. From the last equation we express x p through the remaining variables and insert into the previous expressions. Next, from the penultimate equation we express x p−1 through the remaining variables and insert into the previous expressions, etc. Let's consider the Gauss method using specific examples.
Examples of solving a system of linear equations using the Gauss method
Example 1. Find a general solution to a system of linear equations using the Gauss method:
Let us denote by a ij elements i-th line and j th column.
a 1 1 . To do this, add lines 2,3 with line 1, multiplied by -2/3,-1/2, respectively:
Matrix recording type: Ax=b, Where
Let us denote by a ij elements i-th line and j th column.
Let's exclude the elements of the 1st column of the matrix below the element a 11. To do this, add lines 2,3 with line 1, multiplied by -1/5,-6/5, respectively:
We divide each row of the matrix by the corresponding leading element (if the leading element exists):
Where x 3 , x
Substituting the upper expressions into the lower ones, we obtain the solution.
Then the vector solution can be represented as follows:
 |
Where x 3 , x 4 are arbitrary real numbers.