Three options for completing the forward move of the Gauss method. Gauss method online

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), 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.

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 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 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 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 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 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 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. 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 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, a more complex example for an 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.

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

Wish you luck!

Solutions and answers:

Example 2: Decision : 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: Decision : 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.
(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:

In this article, the method is considered as a way to solve systems of linear equations (SLAE). The method is analytical, that is, it allows you to write a solution algorithm in a general form, and then substitute values from specific examples there. Unlike the matrix method or Cramer's formulas, when solving a system of linear equations using the Gauss method, you can also work with those that have infinitely many solutions. Or they don't have it at all.

What does Gauss mean?

First you need to write down our system of equations in It looks like this. The system is taken:

The coefficients are written in the form of a table, and on the right in a separate column - free members. The column with free members is separated for convenience. The matrix that includes this column is called extended.

Further, the main matrix with coefficients must be reduced to the upper triangular shape. This is the main point of solving the system by the Gauss method. Simply put, after certain manipulations, the matrix should look like this, so that there are only zeros in its lower left part:

Then, if you write the new matrix again as a system of equations, you will notice that the last row already contains the value of one of the roots, which is then substituted into the equation above, another root is found, and so on.

This is a description of the solution by the Gauss method in the most general terms. And what happens if suddenly the system does not have a solution? Or are there an infinite number of them? To answer these and many more questions, it is necessary to consider separately all the elements used in the solution by the Gauss method.

Matrices, their properties

There is no hidden meaning in the matrix. It's just a convenient way to record data for later operations. Even schoolchildren should not be afraid of them.

The matrix is always rectangular, because it is more convenient. Even in the Gauss method, where everything boils down to building a triangular matrix, a rectangle appears in the entry, only with zeros in the place where there are no numbers. Zeros can be omitted, but they are implied.

The matrix has a size. Its "width" is the number of rows (m), its "length" is the number of columns (n). Then the size of the matrix A (capital Latin letters are usually used for their designation) will be denoted as A m×n . If m=n, then this matrix is square, and m=n is its order. Accordingly, any element of the matrix A can be denoted by the number of its row and column: a xy ; x - row number, changes , y - column number, changes .

B is not the main point of the solution. In principle, all operations can be performed directly with the equations themselves, but the notation will turn out to be much more cumbersome, and it will be much easier to get confused in it.

Determinant

The matrix also has a determinant. This is a very important feature. Finding out its meaning now is not worth it, you can simply show how it is calculated, and then tell what properties of the matrix it determines. The easiest way to find the determinant is through diagonals. Imaginary diagonals are drawn in the matrix; the elements located on each of them are multiplied, and then the resulting products are added: diagonals with a slope to the right - with a "plus" sign, with a slope to the left - with a "minus" sign.

It is extremely important to note that the determinant can only be calculated for a square matrix. For a rectangular matrix, you can do the following: choose the smallest of the number of rows and the number of columns (let it be k), and then randomly mark k columns and k rows in the matrix. The elements located at the intersection of the selected columns and rows will form a new square matrix. If the determinant of such a matrix is a number other than zero, then it is called the basis minor of the original rectangular matrix.

Before proceeding with the solution of the system of equations by the Gauss method, it does not hurt to calculate the determinant. If it turns out to be zero, then we can immediately say that the matrix has either an infinite number of solutions, or there are none at all. In such a sad case, you need to go further and find out about the rank of the matrix.

System classification

There is such a thing as the rank of a matrix. This is the maximum order of its non-zero determinant (remembering the basis minor, we can say that the rank of a matrix is the order of the basis minor).

According to how things are with the rank, SLAE can be divided into:

Joint. At of joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended one (with a column of free members). Such systems have a solution, but not necessarily one, therefore, joint systems are additionally divided into:
- certain- having a unique solution. In certain systems, the rank of the matrix and the number of unknowns (or the number of columns, which is the same thing) are equal;
- indefinite - with an infinite number of solutions. The rank of matrices for such systems is less than the number of unknowns.
Incompatible. At such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.

The Gauss method is good in that it allows one to obtain either an unambiguous proof of the inconsistency of the system (without calculating the determinants of large matrices) or a general solution for a system with an infinite number of solutions.

Elementary transformations

Before proceeding directly to the solution of the system, it is possible to make it less cumbersome and more convenient for calculations. This is achieved through elementary transformations - such that their implementation does not change the final answer in any way. It should be noted that some of the above elementary transformations are valid only for matrices, the source of which was precisely the SLAE. Here is a list of these transformations:

String permutation. It is obvious that if we change the order of the equations in the system record, then this will not affect the solution in any way. Consequently, it is also possible to interchange rows in the matrix of this system, not forgetting, of course, about the column of free members.
Multiplying all elements of a string by some factor. Very useful! With it, you can reduce large numbers in the matrix or remove zeros. The set of solutions, as usual, will not change, and it will become more convenient to perform further operations. The main thing is that the coefficient is not equal to zero.
Delete rows with proportional coefficients. This partly follows from the previous paragraph. If two or more rows in the matrix have proportional coefficients, then when multiplying / dividing one of the rows by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and you can remove the extra ones, leaving only one.
Removing the null line. If in the course of transformations a string is obtained somewhere in which all elements, including the free member, are zero, then such a string can be called zero and thrown out of the matrix.
Adding to the elements of one row the elements of another (in the corresponding columns), multiplied by a certain coefficient. The most obscure and most important transformation of all. It is worth dwelling on it in more detail.

Adding a string multiplied by a factor

For ease of understanding, it is worth disassembling this process step by step. Two rows are taken from the matrix:

a 11 a 12 ... a 1n | b1

a 21 a 22 ... a 2n | b 2

Suppose you need to add the first to the second, multiplied by the coefficient "-2".

a" 21 \u003d a 21 + -2 × a 11

a" 22 \u003d a 22 + -2 × a 12

a" 2n \u003d a 2n + -2 × a 1n

Then in the matrix the second row is replaced with a new one, and the first one remains unchanged.

a 11 a 12 ... a 1n | b1

a" 21 a" 22 ... a" 2n | b 2

It should be noted that the multiplication factor can be chosen in such a way that, as a result of the addition of two strings, one of the elements of the new string is equal to zero. Therefore, it is possible to obtain an equation in the system, where there will be one less unknown. And if you get two such equations, then the operation can be done again and get an equation that will already contain two less unknowns. And if each time we turn to zero one coefficient for all rows that are lower than the original one, then we can, like steps, go down to the very bottom of the matrix and get an equation with one unknown. This is called solving the system using the Gaussian method.

In general

Let there be a system. It has m equations and n unknown roots. You can write it down like this:

The main matrix is compiled from the coefficients of the system. A column of free members is added to the extended matrix and separated by a bar for convenience.

the first row of the matrix is multiplied by the coefficient k = (-a 21 / a 11);
the first modified row and the second row of the matrix are added;
instead of the second row, the result of the addition from the previous paragraph is inserted into the matrix;
now the first coefficient in the new second row is a 11 × (-a 21 /a 11) + a 21 = -a 21 + a 21 = 0.

Now the same series of transformations is performed, only the first and third rows are involved. Accordingly, in each step of the algorithm, the element a 21 is replaced by a 31 . Then everything is repeated for a 41 , ... a m1 . The result is a matrix where the first element in the rows is equal to zero. Now we need to forget about line number one and execute the same algorithm starting from the second line:

coefficient k \u003d (-a 32 / a 22);
the second modified line is added to the "current" line;
the result of the addition is substituted in the third, fourth, and so on lines, while the first and second remain unchanged;
in the rows of the matrix, the first two elements are already equal to zero.

The algorithm must be repeated until the coefficient k = (-a m,m-1 /a mm) appears. This means that the last time the algorithm was executed was only for the lower equation. Now the matrix looks like a triangle, or has a stepped shape. The bottom line contains the equality a mn × x n = b m . The coefficient and free term are known, and the root is expressed through them: x n = b m /a mn. The resulting root is substituted into the top row to find x n-1 = (b m-1 - a m-1,n ×(b m /a mn))÷a m-1,n-1 . And so on by analogy: in each next line there is a new root, and, having reached the "top" of the system, you can find many solutions. It will be the only one.

When there are no solutions

If in one of the matrix rows all elements, except for the free term, are equal to zero, then the equation corresponding to this row looks like 0 = b. It has no solution. And since such an equation is included in the system, then the set of solutions of the entire system is empty, that is, it is degenerate.

When there are an infinite number of solutions

It may turn out that in the given triangular matrix there are no rows with one element-the coefficient of the equation, and one - a free member. There are only strings that, when rewritten, would look like an equation with two or more variables. This means that the system has an infinite number of solutions. In this case, the answer can be given in the form of a general solution. How to do it?

All variables in the matrix are divided into basic and free. Basic - these are those that stand "on the edge" of the rows in the stepped matrix. The rest are free. In the general solution, the basic variables are written in terms of the free ones.

For convenience, the matrix is first rewritten back into a system of equations. Then in the last of them, where exactly only one basic variable remained, it remains on one side, and everything else is transferred to the other. This is done for each equation with one basic variable. Then, in the rest of the equations, where possible, instead of the basic variable, the expression obtained for it is substituted. If the result is again an expression containing only one basic variable, it is expressed from there again, and so on, until each basic variable is written as an expression with free variables. This is the general solution of SLAE.

You can also find the basic solution of the system - give the free variables any values, and then for this particular case calculate the values of the basic variables. There are infinitely many particular solutions.

Solution with specific examples

Here is the system of equations.

For convenience, it is better to immediately create its matrix

It is known that when solving by the Gauss method, the equation corresponding to the first row will remain unchanged at the end of the transformations. Therefore, it will be more profitable if the upper left element of the matrix is the smallest - then the first elements of the remaining rows after the operations will turn to zero. This means that in the compiled matrix it will be advantageous to put the second in place of the first row.

second line: k = (-a 21 / a 11) = (-3/1) = -3

a" 21 \u003d a 21 + k × a 11 \u003d 3 + (-3) × 1 \u003d 0

a" 22 \u003d a 22 + k × a 12 \u003d -1 + (-3) × 2 \u003d -7

a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11

b "2 \u003d b 2 + k × b 1 \u003d 12 + (-3) × 12 \u003d -24

third line: k = (-a 3 1 /a 11) = (-5/1) = -5

a" 3 1 = a 3 1 + k×a 11 = 5 + (-5)×1 = 0

a" 3 2 = a 3 2 + k×a 12 = 1 + (-5)×2 = -9

a" 3 3 = a 33 + k×a 13 = 2 + (-5)×4 = -18

b "3 \u003d b 3 + k × b 1 \u003d 3 + (-5) × 12 \u003d -57

Now, in order not to get confused, it is necessary to write down the matrix with the intermediate results of the transformations.

It is obvious that such a matrix can be made more convenient for perception with the help of some operations. For example, you can remove all "minuses" from the second line by multiplying each element by "-1".

It is also worth noting that in the third row all elements are multiples of three. Then you can reduce the string by this number, multiplying each element by "-1/3" (minus - at the same time to remove negative values).

Looks much nicer. Now we need to leave alone the first line and work with the second and third. The task is to add the second row to the third row, multiplied by such a coefficient that the element a 32 becomes equal to zero.

k = (-a 32 / a 22) = (-3/7) = -3/7 fractions, and only then, when the answers are received, decide whether to round up and translate into another form of notation)

a" 32 = a 32 + k × a 22 = 3 + (-3/7) × 7 = 3 + (-3) = 0

a" 33 \u003d a 33 + k × a 23 \u003d 6 + (-3/7) × 11 \u003d -9/7

b "3 \u003d b 3 + k × b 2 \u003d 19 + (-3/7) × 24 \u003d -61/7

The matrix is written again with new values.

1	2	4	12
0	7	11	24
0	0	-9/7	-61/7

As you can see, the resulting matrix already has a stepped form. Therefore, further transformations of the system by the Gauss method are not required. What can be done here is to remove the overall coefficient "-1/7" from the third line.

Now everything is beautiful. The point is small - write the matrix again in the form of a system of equations and calculate the roots

x + 2y + 4z = 12(1)

7y + 11z = 24 (2)

The algorithm by which the roots will now be found is called the reverse move in the Gauss method. Equation (3) contains the value of z:

y = (24 - 11×(61/9))/7 = -65/9

And the first equation allows you to find x:

x = (12 - 4z - 2y)/1 = 12 - 4x(61/9) - 2x(-65/9) = -6/9 = -2/3

We have the right to call such a system joint, and even definite, that is, having a unique solution. The response is written in the following form:

x 1 \u003d -2/3, y \u003d -65/9, z \u003d 61/9.

An example of an indefinite system

The variant of solving a certain system by the Gauss method has been analyzed, now it is necessary to consider the case if the system is indefinite, that is, infinitely many solutions can be found for it.

x 1 + x 2 + x 3 + x 4 + x 5 = 7 (1)

3x 1 + 2x 2 + x 3 + x 4 - 3x 5 = -2 (2)

x 2 + 2x 3 + 2x 4 + 6x 5 = 23 (3)

5x 1 + 4x 2 + 3x 3 + 3x 4 - x 5 = 12 (4)

The very form of the system is already alarming, because the number of unknowns is n = 5, and the rank of the matrix of the system is already exactly less than this number, because the number of rows is m = 4, that is, the largest order of the square determinant is 4. This means that there are an infinite number of solutions, and it is necessary to look for its general form. The Gauss method for linear equations makes it possible to do this.

First, as usual, the augmented matrix is compiled.

Second line: coefficient k = (-a 21 / a 11) = -3. In the third line, the first element is before the transformations, so you don't need to touch anything, you need to leave it as it is. Fourth line: k = (-a 4 1 /a 11) = -5

Multiplying the elements of the first row by each of their coefficients in turn and adding them to the desired rows, we obtain a matrix of the following form:

As you can see, the second, third and fourth rows consist of elements that are proportional to each other. The second and fourth are generally the same, so one of them can be removed immediately, and the rest multiplied by the coefficient "-1" and get line number 3. And again, leave one of two identical lines.

It turned out such a matrix. The system has not yet been written down, it is necessary here to determine the basic variables - standing at the coefficients a 11 \u003d 1 and a 22 \u003d 1, and free - all the rest.

The second equation has only one basic variable - x 2 . Hence, it can be expressed from there, writing through the variables x 3 , x 4 , x 5 , which are free.

We substitute the resulting expression into the first equation.

It turned out an equation in which the only basic variable is x 1. Let's do the same with it as with x 2 .

All basic variables, of which there are two, are expressed in terms of three free ones, now you can write the answer in a general form.

You can also specify one of the particular solutions of the system. For such cases, as a rule, zeros are chosen as values for free variables. Then the answer will be:

16, 23, 0, 0, 0.

An example of an incompatible system

The solution of inconsistent systems of equations by the Gauss method is the fastest. It ends as soon as at one of the stages an equation is obtained that has no solution. That is, the stage with the calculation of the roots, which is quite long and dreary, disappears. The following system is considered:

x + y - z = 0 (1)

2x - y - z = -2 (2)

4x + y - 3z = 5 (3)

As usual, the matrix is compiled:

1	1	-1	0
2	-1	-1	-2
4	1	-3	5

And it is reduced to a stepped form:

k 1 \u003d -2k 2 \u003d -4

1	1	-1	0
0	-3	1	-2
0	0	0	7

After the first transformation, the third line contains an equation of the form

having no solution. Therefore, the system is inconsistent, and the answer is the empty set.

Advantages and disadvantages of the method

If you choose which method to solve SLAE on paper with a pen, then the method that was considered in this article looks the most attractive. In elementary transformations, it is much more difficult to get confused than it happens if you have to manually look for a determinant or some tricky inverse matrix. However, if you use programs for working with data of this type, for example, spreadsheets, then it turns out that such programs already contain algorithms for calculating the main parameters of matrices - determinant, minors, inverse, and so on. And if you are sure that the machine will calculate these values itself and will not make a mistake, it is more expedient to use the matrix method or Cramer's formulas, because their application begins and ends with the calculation of determinants and inverse matrices.

Application

Since the Gaussian solution is an algorithm, and the matrix is, in fact, a two-dimensional array, it can be used in programming. But since the article positions itself as a guide "for dummies", it should be said that the easiest place to shove the method into is spreadsheets, for example, Excel. Again, any SLAE entered in a table in the form of a matrix will be considered by Excel as a two-dimensional array. And for operations with them, there are many nice commands: addition (you can only add matrices of the same size!), Multiplication by a number, matrix multiplication (also with certain restrictions), finding the inverse and transposed matrices and, most importantly, calculating the determinant. If this time-consuming task is replaced by a single command, it is much faster to determine the rank of a matrix and, therefore, to establish its compatibility or inconsistency.

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.

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.

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 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:

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.

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: »

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

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

(2) Divide the second row by 3.

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:

The result is written in the second line:

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.

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.

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

(5) The third row was divided by 3.

Answer: .

Example 4

Solve a system of linear equations using the Gauss method

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: .

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 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.

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 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: Decision : 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: Decision : 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 :

Let a system of linear algebraic equations be given, which must be solved (find such values of the unknowns хi that turn each equation of the system into an equality).

We know that a system of linear algebraic equations can:

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

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 primary school students.

Extended matrix transformations ( this is the matrix of the system - a matrix composed only of the coefficients of the unknowns, plus a column of free terms) systems of linear algebraic equations in the Gauss method:

1) with troky matrices can rearrange places.

2) if there are (or are) proportional (as a special case - identical) rows in the matrix, then it follows delete from the matrix, all these rows except one.

3) if a zero row appeared in the matrix during the transformations, then it also follows delete.

4) the row of the matrix can multiply (divide) to any number other than zero.

5) to the row of the matrix, you can add another string multiplied by a number, different from zero.

In the Gauss method, elementary transformations do not change the solution of the system of equations.

The Gauss method consists of two stages:

"Direct move" - using elementary transformations, bring the extended matrix of the system of linear algebraic equations to a "triangular" stepped form: the elements of the extended matrix located below the main diagonal are equal to zero (top-down move). For example, to this kind:

To do this, perform the following steps:

1) Let us consider the first equation of a system of linear algebraic equations and the coefficient at x 1 is equal to K. The second, third, etc. we transform the equations as follows: we divide each equation (coefficients for unknowns, including free terms) by the coefficient for unknown x 1, which is in each equation, and multiply by K. After that, subtract the first from the second equation (coefficients for unknowns and free terms). We get at x 1 in the second equation the coefficient 0. From the third transformed equation we subtract the first equation, so until all equations, except the first, with unknown x 1 will not have a coefficient 0.

2) Move on to the next equation. Let this be the second equation and the coefficient at x 2 is equal to M. With all the "subordinate" equations, we proceed as described above. Thus, "under" the unknown x 2 in all equations will be zeros.

3) We pass to the next equation and so on until one last unknown and transformed free term remains.

The "reverse move" of the Gauss method is to obtain a solution to a system of linear algebraic equations (the "bottom-up" move). From the last "lower" equation we get one first solution - the unknown x n. To do this, we solve the elementary equation A * x n \u003d B. In the example above, x 3 \u003d 4. We substitute the found value in the “upper” next equation and solve it with respect to the next unknown. For example, x 2 - 4 \u003d 1, i.e. x 2 \u003d 5. And so on until we find all the unknowns.

Example.

We solve the system of linear equations using the Gauss method, as some authors advise:

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. Let's do it like this:
1 step . 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. Whoever wants to get +1 can perform an additional action: multiply the first line by -1 (change its sign).

2 step . 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 step . 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 step . To the third line, add the second line, multiplied by 2.

5 step . The third line is divided by 3.

A sign that indicates an error in calculations (less often a typo) is a “bad” bottom line. That is, if we got something like (0 0 11 | 23) below, and, accordingly, 11x 3 = 23, x 3 = 23/11, then with a high degree of probability we can say that a mistake was made during elementary transformations.

We perform a 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." In this example, the gift turned out:

x 3 = 1
x 2 = 3
x 1 + x 2 - x 3 \u003d 1, therefore x 1 + 3 - 1 \u003d 1, x 1 \u003d -1

Answer:x 1 \u003d -1, x 2 \u003d 3, x 3 \u003d 1.

Let's solve the same system using the proposed algorithm. We get

4 2 –1 1
5 3 –2 2
3 2 –3 0

Divide the second equation by 5 and the third by 3. We get:

4 2 –1 1
1 0.6 –0.4 0.4
1 0.66 –1 0

Multiply the second and third equations by 4, we get:

4 2 –1 1
4 2,4 –1.6 1.6
4 2.64 –4 0

Subtract the first equation from the second and third equations, we have:

4 2 –1 1
0 0.4 –0.6 0.6
0 0.64 –3 –1

Divide the third equation by 0.64:

4 2 –1 1
0 0.4 –0.6 0.6
0 1 –4.6875 –1.5625

Multiply the third equation by 0.4

4 2 –1 1
0 0.4 –0.6 0.6
0 0.4 –1.875 –0.625

Subtract the second equation from the third equation, we get the “stepped” augmented matrix:

4 2 –1 1
0 0.4 –0.6 0.6
0 0 –1.275 –1.225

Thus, since an error accumulated in the process of calculations, we get x 3 \u003d 0.96, or approximately 1.

x 2 \u003d 3 and x 1 \u003d -1.

Solving in this way, you will never get confused in the calculations and, despite the calculation errors, you will get the result.

This method of solving a system of linear algebraic equations is easily programmable and does not take into account the specific features of the coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.

Wish you luck! See you in class! Tutor.

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What does Gauss mean?

First you need to write down our system of equations in It looks like this. The system is taken:

Matrices, their properties

There is no hidden meaning in the matrix. It's just a convenient way to record data for later operations. Even schoolchildren should not be afraid of them.

Determinant

System classification

According to how things are with the rank, SLAE can be divided into:

Joint. At of joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended one (with a column of free members). Such systems have a solution, but not necessarily one, therefore, joint systems are additionally divided into:
- certain- having a unique solution. In certain systems, the rank of the matrix and the number of unknowns (or the number of columns, which is the same thing) are equal;
- indefinite - with an infinite number of solutions. The rank of matrices for such systems is less than the number of unknowns.
Incompatible. At such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.

Elementary transformations

String permutation. It is obvious that if we change the order of the equations in the system record, then this will not affect the solution in any way. Consequently, it is also possible to interchange rows in the matrix of this system, not forgetting, of course, about the column of free members.
Multiplying all elements of a string by some factor. Very useful! With it, you can reduce large numbers in the matrix or remove zeros. The set of solutions, as usual, will not change, and it will become more convenient to perform further operations. The main thing is that the coefficient is not equal to zero.
Delete rows with proportional coefficients. This partly follows from the previous paragraph. If two or more rows in the matrix have proportional coefficients, then when multiplying / dividing one of the rows by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and you can remove the extra ones, leaving only one.
Removing the null line. If in the course of transformations a string is obtained somewhere in which all elements, including the free member, are zero, then such a string can be called zero and thrown out of the matrix.
Adding to the elements of one row the elements of another (in the corresponding columns), multiplied by a certain coefficient. The most obscure and most important transformation of all. It is worth dwelling on it in more detail.

Adding a string multiplied by a factor

For ease of understanding, it is worth disassembling this process step by step. Two rows are taken from the matrix:

a 11 a 12 ... a 1n | b1

a 21 a 22 ... a 2n | b 2

Suppose you need to add the first to the second, multiplied by the coefficient "-2".

a" 21 \u003d a 21 + -2 × a 11

a" 22 \u003d a 22 + -2 × a 12

a" 2n \u003d a 2n + -2 × a 1n

Then in the matrix the second row is replaced with a new one, and the first one remains unchanged.

a 11 a 12 ... a 1n | b1

a" 21 a" 22 ... a" 2n | b 2

In general

Let there be a system. It has m equations and n unknown roots. You can write it down like this:

The main matrix is compiled from the coefficients of the system. A column of free members is added to the extended matrix and separated by a bar for convenience.

the first row of the matrix is multiplied by the coefficient k = (-a 21 / a 11);
the first modified row and the second row of the matrix are added;
instead of the second row, the result of the addition from the previous paragraph is inserted into the matrix;
now the first coefficient in the new second row is a 11 × (-a 21 /a 11) + a 21 = -a 21 + a 21 = 0.

coefficient k \u003d (-a 32 / a 22);
the second modified line is added to the "current" line;
the result of the addition is substituted in the third, fourth, and so on lines, while the first and second remain unchanged;
in the rows of the matrix, the first two elements are already equal to zero.

When there are no solutions

When there are an infinite number of solutions

Solution with specific examples

Here is the system of equations.

For convenience, it is better to immediately create its matrix

second line: k = (-a 21 / a 11) = (-3/1) = -3

a" 21 \u003d a 21 + k × a 11 \u003d 3 + (-3) × 1 \u003d 0

a" 22 \u003d a 22 + k × a 12 \u003d -1 + (-3) × 2 \u003d -7

a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11

b "2 \u003d b 2 + k × b 1 \u003d 12 + (-3) × 12 \u003d -24

third line: k = (-a 3 1 /a 11) = (-5/1) = -5

a" 3 1 = a 3 1 + k×a 11 = 5 + (-5)×1 = 0

a" 3 2 = a 3 2 + k×a 12 = 1 + (-5)×2 = -9

a" 3 3 = a 33 + k×a 13 = 2 + (-5)×4 = -18

b "3 \u003d b 3 + k × b 1 \u003d 3 + (-5) × 12 \u003d -57

Now, in order not to get confused, it is necessary to write down the matrix with the intermediate results of the transformations.

k = (-a 32 / a 22) = (-3/7) = -3/7 fractions, and only then, when the answers are received, decide whether to round up and translate into another form of notation)

a" 32 = a 32 + k × a 22 = 3 + (-3/7) × 7 = 3 + (-3) = 0

a" 33 \u003d a 33 + k × a 23 \u003d 6 + (-3/7) × 11 \u003d -9/7

b "3 \u003d b 3 + k × b 2 \u003d 19 + (-3/7) × 24 \u003d -61/7

The matrix is written again with new values.

1	2	4	12
0	7	11	24
0	0	-9/7	-61/7

Now everything is beautiful. The point is small - write the matrix again in the form of a system of equations and calculate the roots

x + 2y + 4z = 12(1)

7y + 11z = 24 (2)

The algorithm by which the roots will now be found is called the reverse move in the Gauss method. Equation (3) contains the value of z:

y = (24 - 11×(61/9))/7 = -65/9

And the first equation allows you to find x:

x = (12 - 4z - 2y)/1 = 12 - 4x(61/9) - 2x(-65/9) = -6/9 = -2/3

We have the right to call such a system joint, and even definite, that is, having a unique solution. The response is written in the following form:

x 1 \u003d -2/3, y \u003d -65/9, z \u003d 61/9.

An example of an indefinite system

x 1 + x 2 + x 3 + x 4 + x 5 = 7 (1)

3x 1 + 2x 2 + x 3 + x 4 - 3x 5 = -2 (2)

x 2 + 2x 3 + 2x 4 + 6x 5 = 23 (3)

5x 1 + 4x 2 + 3x 3 + 3x 4 - x 5 = 12 (4)

First, as usual, the augmented matrix is compiled.

Multiplying the elements of the first row by each of their coefficients in turn and adding them to the desired rows, we obtain a matrix of the following form:

The second equation has only one basic variable - x 2 . Hence, it can be expressed from there, writing through the variables x 3 , x 4 , x 5 , which are free.

We substitute the resulting expression into the first equation.

It turned out an equation in which the only basic variable is x 1. Let's do the same with it as with x 2 .

All basic variables, of which there are two, are expressed in terms of three free ones, now you can write the answer in a general form.

You can also specify one of the particular solutions of the system. For such cases, as a rule, zeros are chosen as values for free variables. Then the answer will be:

16, 23, 0, 0, 0.

An example of an incompatible system

x + y - z = 0 (1)

2x - y - z = -2 (2)

4x + y - 3z = 5 (3)

As usual, the matrix is compiled:

1	1	-1	0
2	-1	-1	-2
4	1	-3	5

And it is reduced to a stepped form:

k 1 \u003d -2k 2 \u003d -4

1	1	-1	0
0	-3	1	-2
0	0	0	7

After the first transformation, the third line contains an equation of the form

having no solution. Therefore, the system is inconsistent, and the answer is the empty set.