In this article, the method is considered as a method for solving systems of linear equations (SLAEs). 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 an infinite number of solutions. Or they don't have it at all.
What does it mean to solve using the Gaussian method?
First, we need to write our system of equations in It looks like this. Take the system:
The coefficients are written in the form of a table, and the free terms are written in a separate column on the right. The column with free terms is separated for convenience. The matrix that includes this column is called extended.
Next, the main matrix with coefficients must be reduced to an upper triangular form. This is the main point of solving the system using the Gaussian method. Simply put, after certain manipulations, the matrix should look so that its lower left part contains only zeros:
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 Gaussian method in the most general terms. What happens if suddenly the system has no solution? Or are there infinitely many of them? To answer these and many other questions, it is necessary to consider separately all the elements used in solving the Gaussian method.
Matrices, their properties
There is no hidden meaning in the matrix. This is simply a convenient way to record data for subsequent operations with it. Even schoolchildren do not need to be afraid of them.
The matrix is always rectangular, because it is more convenient. Even in the Gauss method, where everything comes down to constructing a matrix of a triangular form, a rectangle appears in the entry, only with zeros in the place where there are no numbers. Zeros may not be written, but they are implied.
The matrix has a size. Its “width” is the number of rows (m), “length” is the number of columns (n). Then the size of the matrix A (capital Latin letters are usually used to denote them) 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 matrix A can be denoted by its row and column numbers: a xy ; x - row number, changes, y - column number, changes.
B is not the main point of the decision. In principle, all operations can be performed directly with the equations themselves, but the notation will 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 characteristic. There is no need to find out its meaning now; 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 from 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 at the intersection of the selected columns and rows will form a new square matrix. If the determinant of such a matrix is a non-zero number, it is called the basis minor of the original rectangular matrix.
Before you start solving a system of equations using the Gaussian method, it doesn’t 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 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 (if we remember about the basis minor, we can say that the rank of a matrix is the order of the basis minor).
Based on the situation with rank, SLAE can be divided into:
- Joint. U In joint systems, the rank of the main matrix (consisting only of coefficients) coincides with the rank of the extended matrix (with a column of free terms). Such systems have a solution, but not necessarily one, therefore, additionally joint systems are divided into:
- - certain- having a single 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;
- - undefined - with an infinite number of solutions. The rank of matrices in such systems is less than the number of unknowns.
- Incompatible. U In such systems, the ranks of the main and extended matrices do not coincide. Incompatible systems have no solution.
The Gauss method is good because during the solution it allows one to obtain either an unambiguous proof of the inconsistency of the system (without calculating the determinants of large matrices), or a solution in general form for a system with an infinite number of solutions.
Elementary transformations
Before proceeding directly to solving the system, you can 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 given elementary transformations are valid only for matrices, the source of which was the SLAE. Here is a list of these transformations:
- Rearranging lines. Obviously, if you change the order of the equations in the system record, this will not affect the solution in any way. Consequently, rows in the matrix of this system can also be swapped, not forgetting, of course, the column of free terms.
- Multiplying all elements of a string by a certain coefficient. Very helpful! It can be used to reduce large numbers in a matrix or remove zeros. Many decisions, as usual, will not change, but further operations will become more convenient. The main thing is that the coefficient is not equal to zero.
- Removing rows with proportional factors. This partly follows from the previous paragraph. If two or more rows in a matrix have proportional coefficients, then when one of the rows is multiplied/divided by the proportionality coefficient, two (or, again, more) absolutely identical rows are obtained, and the extra ones can be removed, leaving only one.
- Removing a null line. If, during the transformation, a row is obtained somewhere in which all elements, including the free term, are zero, then such a row 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 unobvious 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 breaking down 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
Let's say you need to add the first to the second, multiplied by the coefficient "-2".
a" 21 = a 21 + -2×a 11
a" 22 = a 22 + -2×a 12
a" 2n = a 2n + -2×a 1n
Then the second row in the matrix is replaced with a new one, and the first remains unchanged.
a 11 a 12 ... a 1n | b1
a" 21 a" 22 ... a" 2n | b 2
It should be noted that the multiplication coefficient can be selected in such a way that, as a result of adding two rows, one of the elements of the new row is equal to zero. Consequently, it is possible to obtain an equation in a 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 contain two fewer unknowns. And if each time you turn one coefficient of all rows that are below the original one to zero, then you can, like stairs, 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 as follows:
The main matrix is compiled from the system coefficients. A column of free terms is added to the extended matrix and, for convenience, separated by a line.
- 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, at each step of the algorithm, 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 zero. Now you need to forget about line number one and perform the same algorithm, starting from line two:
- coefficient k = (-a 32 /a 22);
- the second modified line is added to the “current” line;
- the result of the addition is substituted into 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. In the bottom line there is 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 line 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 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 happen that in the given triangular matrix there are no rows with one coefficient element of the equation and one free term. There are only lines 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 ones are those that stand “on the edge” of the rows in the step matrix. The rest are free. In the general solution, the basic variables are written through free ones.
For convenience, the matrix is first rewritten back into a system of equations. Then in the last of them, where exactly there is only one basic variable left, it remains on one side, and everything else is transferred to the other. This is done for every equation with one basic variable. Then, in the remaining equations, where possible, the expression obtained for it is substituted instead of the basic variable. If the result is again an expression containing only one basic variable, it is again expressed from there, 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 specific case calculate the values of the basic variables. There are an infinite number of particular solutions that can be given.
Solution with specific examples
Here is a system of equations.
For convenience, it is better to immediately create its matrix
It is known that when solved by the Gaussian 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 row in place of the first one.
second line: k = (-a 21 /a 11) = (-3/1) = -3
a" 21 = a 21 + k×a 11 = 3 + (-3)×1 = 0
a" 22 = a 22 + k×a 12 = -1 + (-3)×2 = -7
a" 23 = a 23 + k×a 13 = 1 + (-3)×4 = -11
b" 2 = b 2 + k×b 1 = 12 + (-3)×12 = -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 = b 3 + k×b 1 = 3 + (-5)×12 = -57
Now, in order not to get confused, you need to write down a matrix with the intermediate results of the transformations.
Obviously, such a matrix can be made more convenient for perception using certain 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 line all elements are multiples of three. Then you can shorten 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 the first line alone and work with the second and third. The task is to add the second line to the third line, multiplied by such a coefficient that the element a 32 becomes equal to zero.
k = (-a 32 /a 22) = (-3/7) = -3/7 (if during some transformations the answer does not turn out to be an integer, it is recommended to maintain the accuracy of the calculations to leave it “as is”, in the form of an ordinary fractions, and only then, when the answers are received, decide whether to round and convert to another form of recording)
a" 32 = a 32 + k×a 22 = 3 + (-3/7)×7 = 3 + (-3) = 0
a" 33 = a 33 + k×a 23 = 6 + (-3/7)×11 = -9/7
b" 3 = b 3 + k×b 2 = 19 + (-3/7)×24 = -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 using the Gaussian method are not required. What you can do here is to remove the overall coefficient "-1/7" from the third line.
Now everything is beautiful. All that’s left to do is 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 Gaussian method. Equation (3) contains the z value:
y = (24 - 11×(61/9))/7 = -65/9
And the first equation allows us to find x:
x = (12 - 4z - 2y)/1 = 12 - 4×(61/9) - 2×(-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 answer is written in the following form:
x 1 = -2/3, y = -65/9, z = 61/9.
An example of an uncertain system
The variant of solving a certain system using the Gauss method has been analyzed; now it is necessary to consider the case if the system is uncertain, 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 appearance of the system is already alarming, because the number of unknowns is n = 5, and the rank of the system matrix is already exactly less than this number, because the number of rows is m = 4, that is, the largest order of the determinant-square is 4. This means that there are an infinite number of solutions, and you need to look for its general appearance. The Gauss method for linear equations allows you to do this.
First, as usual, an extended 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 is. Fourth line: k = (-a 4 1 /a 11) = -5
By multiplying the elements of the first row by each of their coefficients in turn and adding them to the required rows, we obtain a matrix of the following form:
As you can see, the second, third and fourth rows consist of elements proportional to each other. The second and fourth are generally identical, so one of them can be removed immediately, and the remaining one can be multiplied by the coefficient “-1” and get line number 3. And again, out of two identical lines, leave one.
The result is a matrix like this. While the system has not yet been written down, it is necessary to determine the basic variables here - those standing at the coefficients a 11 = 1 and a 22 = 1, and free ones - all the rest.
In the second equation there is only one basic variable - x 2. This means that it can be expressed from there by writing it through the variables x 3 , x 4 , x 5 , which are free.
We substitute the resulting expression into the first equation.
The result is 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 we can write the answer in general form.
You can also specify one of the particular solutions of the system. For such cases, zeros are usually chosen as values for free variables. Then the answer will be:
16, 23, 0, 0, 0.
An example of a non-cooperative system
Solving incompatible systems of equations using the Gauss method is the fastest. It ends immediately as soon as at one of the stages an equation is obtained that has no solution. That is, the stage of calculating the roots, which is quite long and tedious, is eliminated. 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 stepwise form:
k 1 = -2k 2 = -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
without a solution. Consequently, the system is inconsistent, and the answer will be the empty set.
Advantages and disadvantages of the method
If you choose which method to solve SLAEs on paper with a pen, then the method that was discussed in this article looks the most attractive. It is much more difficult to get confused in elementary transformations than if you have to manually search 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 mistakes, it is more advisable 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 actually 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 put the method into is spreadsheets, for example, Excel. Again, any SLAE entered into 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, multiplication of matrices (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 possible to determine the rank of the matrix much more quickly and, therefore, establish its compatibility or incompatibility.
Let a system of linear algebraic equations be given that needs to be solved (find such values of the unknowns xi 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 non-joint).
2) Have infinitely many solutions.
3) Have a single solution.
As we remember, Cramer's rule and the matrix method are not suitable 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 will lead us to the answer! The method algorithm itself works the same in all three cases. If the Cramer and matrix methods require knowledge of determinants, then to apply the Gauss method you only need knowledge of arithmetic operations, which makes it accessible even to primary school students.
Augmented 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 troki matrices Can rearrange in some places.
2) if proportional (as a special case – identical) rows appear (or exist) in the matrix, then you should delete from the matrix all these rows except one.
3) if a zero row appears in the matrix during transformations, then it should also be delete.
4) a row of the matrix can be multiply (divide) to any number other than zero.
5) to a 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 a system of linear algebraic equations to a “triangular” step form: the elements of the extended matrix located below the main diagonal are equal to zero (top-down move). For example, to this type:
To do this, perform the following steps:
1) Let us consider the first equation of a system of linear algebraic equations and the coefficient for x 1 is equal to K. The second, third, etc. we transform the equations as follows: we divide each equation (coefficients of the unknowns, including free terms) by the coefficient of the unknown x 1 in each equation, and multiply by K. After this, we subtract the first from the second equation (coefficients of unknowns and free terms). For x 1 in the second equation we obtain the coefficient 0. From the third transformed equation we subtract the first equation until all equations except the first, for unknown x 1, have a coefficient 0.
2) Let's move on to the next equation. Let this be the second equation and the coefficient for x 2 equal to M. We proceed with all “lower” equations as described above. Thus, “under” the unknown x 2 there will be zeros in all equations.
3) Move on to the next equation and so on until one last unknown and the transformed free term remain.
- 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 obtain one first solution - the unknown x n. To do this, we solve the elementary equation A * x n = B. In the example given above, x 3 = 4. We substitute the found value into the “upper” next equation and solve it with respect to the next unknown. For example, x 2 – 4 = 1, i.e. x 2 = 5. And so on until we find all the unknowns.
Example.
Let's solve the system of linear equations using the Gauss method, as some authors advise:
Let us write down the extended matrix of the system and, using elementary transformations, bring it to a stepwise form:
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. Let's do 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 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 action: multiply the first line by –1 (change its sign).
Step 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.
Step 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.
Step 4 . The third line was added to the second line, multiplied by 2.
Step 5 . The third line was divided by 3.
A sign that indicates an error in calculations (more rarely, 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 an error was made during elementary transformations.
Let’s do the reverse; in the design of examples, the system itself is often not rewritten, but the equations are “taken directly from the given matrix.” The reverse move, I remind you, works from the bottom up. In this example, the result was a gift:
x 3 = 1
x 2 = 3
x 1 + x 2 – x 3 = 1, therefore x 1 + 3 – 1 = 1, x 1 = –1
Answer:x 1 = –1, x 2 = 3, x 3 = 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
Multiplying 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
Subtracting the second from the third equation, we obtain a “stepped” extended matrix:
4 2 –1 1
0 0.4 –0.6 0.6
0 0 –1.275 –1.225
Thus, since the error accumulated during the calculations, we obtain x 3 = 0.96 or approximately 1.
x 2 = 3 and x 1 = –1.
By 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 coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.
I wish you success! See you in class! Tutor Dmitry Aystrakhanov.
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Since the beginning of the 16th-18th centuries, mathematicians have intensively begun to study functions, thanks to which so much in our lives has changed. Computer technology simply would not exist without this knowledge. Various concepts, theorems, and solution techniques have been created to solve complex problems, linear equations, and functions. One of such universal and rational methods and techniques for solving linear equations and their systems was the Gauss method. Matrices, their rank, determinant - everything can be calculated without using complex operations.
What is SLAU
In mathematics, there is the concept of SLAE - a system of linear algebraic equations. What is she like? This is a set of m equations with the required n unknown quantities, usually denoted as x, y, z, or x 1, x 2 ... x n, or other symbols. Solving a given system using the Gaussian method means finding all the unknown unknowns. If a system has the same number of unknowns and equations, then it is called an nth order system.
The most popular methods for solving SLAEs
In educational institutions of secondary education, various methods for solving such systems are studied. Most often these are simple equations consisting of two unknowns, so any existing method for finding the answer to them will not take much time. This can be like a substitution method, when another is derived from one equation and substituted into the original one. Or the method of term-by-term subtraction and addition. But the Gauss method is considered the easiest and most universal. It makes it possible to solve equations with any number of unknowns. Why is this particular technique considered rational? It's simple. The good thing about the matrix method is that it does not require rewriting unnecessary symbols several times as unknowns; it is enough to perform arithmetic operations on the coefficients - and you will get a reliable result.
Where are SLAEs used in practice?
The solution to SLAEs are the points of intersection of lines on the graphs of functions. In our high-tech computer age, people who are closely associated with the development of games and other programs need to know how to solve such systems, what they represent and how to check the correctness of the resulting result. Most often, programmers develop special linear algebra calculator programs, which also includes a system of linear equations. The Gauss method allows you to calculate all existing solutions. Other simplified formulas and techniques are also used.
SLAU compatibility criterion
Such a system can only be solved if it is compatible. For clarity, let us represent the SLAE in the form Ax=b. It has a solution if rang(A) equals rang(A,b). In this case, (A,b) is an extended form matrix that can be obtained from matrix A by rewriting it with free terms. It turns out that solving linear equations using the Gaussian method is quite easy.
Perhaps some of the symbols are not entirely clear, so it is necessary to consider everything with an example. Let's say there is a system: x+y=1; 2x-3y=6. It consists of only two equations, in which there are 2 unknowns. The system will have a solution only if the rank of its matrix is equal to the rank of the extended matrix. What is rank? This is the number of independent lines of the system. In our case, the rank of the matrix is 2. Matrix A will consist of coefficients located near the unknowns, and the coefficients located behind the “=” sign also fit into the extended matrix.
Why can SLAEs be represented in matrix form?
Based on the compatibility criterion according to the proven Kronecker-Capelli theorem, a system of linear algebraic equations can be represented in matrix form. Using the Gaussian cascade method, you can solve the matrix and get a single reliable answer for the entire system. If the rank of an ordinary matrix is equal to the rank of its extended matrix, but is less than the number of unknowns, then the system has an infinite number of answers.
Matrix transformations
Before moving on to solving matrices, you need to know what actions can be performed on their elements. There are several elementary transformations:
- By rewriting the system in matrix form and solving it, you can multiply all elements of the series by the same coefficient.
- In order to transform the matrix into canonical form, you can swap two parallel rows. The canonical form implies that all matrix elements that are located along the main diagonal become ones, and the remaining ones become zeros.
- The corresponding elements of parallel rows of the matrix can be added to one another.
Jordan-Gauss method
The essence of solving systems of linear homogeneous and inhomogeneous equations using the Gaussian method is to gradually eliminate the unknowns. Let's say we have a system of two equations in which there are two unknowns. To find them, you need to check the system for compatibility. The equation is solved very simply by the Gauss method. It is necessary to write down the coefficients located near each unknown in matrix form. To solve the system, you will need to write out the extended matrix. If one of the equations contains a smaller number of unknowns, then “0” must be put in place of the missing element. All known transformation methods are applied to the matrix: multiplication, division by a number, adding the corresponding elements of the series to each other, and others. It turns out that in each row it is necessary to leave one variable with the value “1”, the rest should be reduced to zero. For a more precise understanding, it is necessary to consider the Gauss method with examples.
A simple example of solving a 2x2 system
To begin with, let's take a simple system of algebraic equations, in which there will be 2 unknowns.
Let's rewrite it into an extended matrix.
To solve this system of linear equations, only two operations are required. We need to bring the matrix to canonical form so that there are ones along the main diagonal. So, transferring from the matrix form back to the system, we get the equations: 1x+0y=b1 and 0x+1y=b2, where b1 and b2 are the resulting answers in the solution process.
- The first action when solving an extended matrix will be this: the first row must be multiplied by -7 and added corresponding elements to the second row in order to get rid of one unknown in the second equation.
- Since solving equations using the Gauss method involves reducing the matrix to canonical form, then it is necessary to perform the same operations with the first equation and remove the second variable. To do this, we subtract the second line from the first and get the required answer - the solution of the SLAE. Or, as shown in the figure, we multiply the second row by a factor of -1 and add the elements of the second row to the first row. It is the same.
As we can see, our system was solved by the Jordan-Gauss method. We rewrite it in the required form: x=-5, y=7.
An example of a 3x3 SLAE solution
Suppose we have a more complex system of linear equations. The Gauss method makes it possible to calculate the answer even for the most seemingly confusing system. Therefore, in order to delve deeper into the calculation methodology, you can move on to a more complex example with three unknowns.
As in the previous example, we rewrite the system in the form of an extended matrix and begin to bring it to its canonical form.
To solve this system, you will need to perform much more actions than in the previous example.
- First you need to make the first column one unit element and the rest zeros. To do this, multiply the first equation by -1 and add the second equation to it. It is important to remember that we rewrite the first line in its original form, and the second in a modified form.
- Next, we remove this same first unknown from the third equation. To do this, multiply the elements of the first row by -2 and add them to the third row. Now the first and second lines are rewritten in their original form, and the third - with changes. As you can see from the result, we got the first one at the beginning of the main diagonal of the matrix and the remaining zeros. A few more steps, and the system of equations by the Gaussian method will be reliably solved.
- Now you need to perform operations on other elements of the rows. The third and fourth actions can be combined into one. We need to divide the second and third lines by -1 to get rid of the minus ones on the diagonal. We have already brought the third line to the required form.
- Next we bring the second line to canonical form. To do this, we multiply the elements of the third row by -3 and add them to the second row of the matrix. From the result it is clear that the second line is also reduced to the form we need. It remains to perform a few more operations and remove the coefficients of the unknowns from the first line.
- To make 0 from the second element of a row, you need to multiply the third row by -3 and add it to the first row.
- The next decisive step will be to add the necessary elements of the second row to the first row. This way we get the canonical form of the matrix, and, accordingly, the answer.
As you can see, solving equations using the Gauss method is quite simple.
An example of solving a 4x4 system of equations
Some more complex systems of equations can be solved using the Gaussian method using computer programs. It is necessary to enter the coefficients for the unknowns into the existing empty cells, and the program itself will step by step calculate the required result, describing in detail each action.
Step-by-step instructions for solving such an example are described below.
In the first step, free coefficients and numbers for unknowns are entered into empty cells. Thus, we get the same extended matrix that we write manually.
And all the necessary arithmetic operations are performed to bring the extended matrix to its canonical form. It is necessary to understand that the answer to a system of equations is not always integers. Sometimes the solution may be from fractional numbers.
Checking the correctness of the solution
The Jordan-Gauss method provides for checking the correctness of the result. In order to find out whether the coefficients are calculated correctly, you just need to substitute the result into the original system of equations. The left side of the equation must match the right side behind the equal sign. If the answers do not match, then you need to recalculate the system or try to apply to it another method of solving SLAEs known to you, such as substitution or term-by-term subtraction and addition. After all, mathematics is a science that has a huge number of different solution methods. But remember: the result should always be the same, no matter what solution method you used.
Gauss method: the most common errors when solving SLAEs
When solving linear systems of equations, errors most often occur such as incorrect transfer of coefficients into matrix form. There are systems in which some unknowns are missing from one of the equations; then, when transferring data to an extended matrix, they can be lost. As a result, when solving this system, the result may not correspond to the actual one.
Another major mistake may be incorrectly writing out the final result. It is necessary to clearly understand that the first coefficient will correspond to the first unknown from the system, the second - to the second, and so on.
The Gauss method describes in detail the solution of linear equations. Thanks to it, it is easy to carry out the necessary operations and find the right result. In addition, this is a universal tool for finding a reliable answer to equations of any complexity. Maybe that's why it is so often used when solving SLAEs.
Carl Friedrich Gauss, the greatest mathematician, hesitated for a long time, choosing between philosophy and mathematics. Perhaps it was precisely this mindset that allowed him to make such a noticeable “legacy” in world science. In particular, by creating the "Gauss Method" ...
For almost 4 years, articles on this site dealt with school education, mainly from the point of view of philosophy, the principles of (mis)understanding introduced into the minds of children. The time is coming for more specifics, examples and methods... I believe that this is exactly the approach to the familiar, confusing and important areas of life gives better results.
We people are designed in such a way that no matter how much we talk about abstract thinking, But understanding Always happens through examples. If there are no examples, then it is impossible to grasp the principles... Just as it is impossible to get to the top of a mountain except by walking the entire slope from the foot.
Same with school: for now living stories It is not enough that we instinctively continue to regard it as a place where children are taught to understand.
For example, teaching the Gaussian method...
Gauss method in 5th grade school
I’ll make a reservation right away: the Gauss method has a much wider application, for example, when solving systems of linear equations. What we will talk about takes place in 5th grade. This started, having understood which, it is much easier to understand the more “advanced options”. In this article we are talking about Gauss's method (method) for finding the sum of a series
Here is an example that my youngest son, who attends 5th grade at a Moscow gymnasium, brought from school.
School demonstration of the Gauss method
A mathematics teacher using an interactive whiteboard (modern teaching methods) showed children a presentation of the history of the “creation of the method” by little Gauss.
The school teacher whipped little Karl (an outdated method, not used in schools these days) because he
instead of sequentially adding numbers from 1 to 100, find their sum noticed that pairs of numbers equally spaced from the edges of an arithmetic progression add up to the same number. for example, 100 and 1, 99 and 2. Having counted the number of such pairs, little Gauss almost instantly solved the problem proposed by the teacher. For which he was executed in front of an astonished public. So that others would be discouraged from thinking.
What did little Gauss do? developed number sense? Noticed some feature number series with a constant step (arithmetic progression). AND exactly this later made him a great scientist, those who know how to notice, having feeling, instinct of understanding.
This is why mathematics is valuable, developing ability to see general in particular - abstract thinking. Therefore, most parents and employers instinctively consider mathematics an important discipline ...
“Then you need to learn mathematics, because it puts your mind in order.
M.V.Lomonosov".
However, the followers of those who flogged future geniuses with rods turned the Method into something the opposite. As my supervisor said 35 years ago: “The question has been learned.” Or as my youngest son said yesterday about Gauss’s method: “Maybe it’s not worth making a big science out of this, huh?”
The consequences of the creativity of the “scientists” are visible in the level of current school mathematics, the level of its teaching and the understanding of the “Queen of Sciences” by the majority.
However, let's continue...
Methods for explaining the Gauss method in 5th grade school
A mathematics teacher at a Moscow gymnasium, explaining the Gauss method according to Vilenkin, complicated the task.
What if the difference (step) of an arithmetic progression is not one, but another number? For example, 20.
The problem he gave to the fifth graders:
20+40+60+80+ ... +460+480+500
Before getting acquainted with the gymnasium method, let’s take a look at the Internet: how do school teachers and math tutors do it?..
Gaussian method: explanation No. 1
A well-known tutor on his YOUTUBE channel gives the following reasoning:
"Let's write the numbers from 1 to 100 as follows:
first a series of numbers from 1 to 50, and strictly below it another series of numbers from 50 to 100, but in the reverse order"
1, 2, 3, ... 48, 49, 50
100, 99, 98 ... 53, 52, 51
"Please note: the sum of each pair of numbers from the top and bottom rows is the same and equals 101! Let's count the number of pairs, it is 50 and multiply the sum of one pair by the number of pairs! Voila: The answer is ready!"
“If you couldn’t understand, don’t be upset!” the teacher repeated three times during the explanation. "You will take this method in 9th grade!"
Gaussian method: explanation No. 2
Another tutor, less well-known (judging by the number of views), takes a more scientific approach, offering a solution algorithm of 5 points that must be completed sequentially.
For the uninitiated, 5 is one of the Fibonacci numbers traditionally considered magical. A 5 step method is always more scientific than a 6 step method, for example. ...And this is hardly an accident, most likely, the Author is a hidden adherent of the Fibonacci theory
Given an arithmetic progression: 4, 10, 16 ... 244, 250, 256 .
Algorithm for finding the sum of numbers in a series using the Gauss method:
4, 10, 16 ... 244, 250, 256
256, 250, 244 ... 16, 10, 4
At the same time, you need to remember plus one rule : we must add one to the resulting quotient: otherwise we will get a result that is less by one than the true number of pairs: 42 + 1 = 43.
This is the required sum of the arithmetic progression from 4 to 256 with a difference of 6!
Gauss method: explanation in 5th grade at a Moscow gymnasium
Here's how to solve the problem of finding the sum of a series:
20+40+60+ ... +460+480+500
in the 5th grade of a Moscow gymnasium, Vilenkin’s textbook (according to my son).
After showing the presentation, the math teacher showed a couple of examples using the Gaussian method and gave the class a task of finding the sum of the numbers in a series in increments of 20.
This required the following:
As you can see, this is a more compact and effective technique: the number 3 is also a member of the Fibonacci sequence
My comments on the school version of the Gauss method
The great mathematician would definitely have chosen philosophy if he had foreseen what his “method” would be turned into by his followers German teacher, who flogged Karl with rods. He would have seen the symbolism, the dialectical spiral and the undying stupidity of the “teachers”, trying to measure the harmony of living mathematical thought with the algebra of misunderstanding ....
By the way: did you know. that our education system is rooted in the German school of the 18th and 19th centuries?
But Gauss chose mathematics.
What is the essence of his method?
IN simplification. IN observing and grasping simple patterns of numbers. IN turning dry school arithmetic into interesting and exciting activity , activating in the brain the desire to continue, rather than blocking high-cost mental activity.
Is it possible to use one of the given “modifications of Gauss’s method” to calculate the sum of the numbers of an arithmetic progression almost instantly? According to the “algorithms”, little Karl would be guaranteed to avoid spanking, develop an aversion to mathematics and suppress his creative impulses in the bud.
Why did the tutor so persistently advise fifth-graders “not to be afraid of misunderstanding” of the method, convincing them that they would solve “such” problems as early as 9th grade? Psychologically illiterate action. It was a good move to note: "See? You already in 5th grade you can solve problems that you will complete only in 4 years! What a great fellow you are!”
To use the Gaussian method, a level of class 3 is sufficient, when normal children already know how to add, multiply and divide 2-3 digit numbers. Problems arise due to the inability of adult teachers who are “out of touch” to explain the simplest things in normal human language, not to mention mathematical... They are unable to get people interested in mathematics and completely discourage even those who are “capable.”
Or, as my son commented: “making a big science out of it.”
Gauss method, my explanations
My wife and I explained this “method” to our child, it seems, even before school...
Simplicity instead of complexity or a game of questions and answers
"Look, here are the numbers from 1 to 100. What do you see?"
The point is not what exactly the child sees. The trick is to get him to look.
"How can you put them together?" The son realized that such questions are not asked “just like that” and you need to look at the question “somehow differently, differently than he usually does”
It doesn't matter if the child sees the solution right away, it's unlikely. It is important that he stopped being afraid to look, or as I say: “moved the task”. This is the beginning of the journey to understanding
“Which is easier: adding, for example, 5 and 6 or 5 and 95?” A leading question... But any training comes down to “guiding” a person to the “answer” - in any way acceptable to him.
At this stage, guesses may already arise about how to “save” on calculations.
All we did was hint: the “frontal, linear” method of counting is not the only possible one. If a child understands this, then later he will come up with many more such methods, because it's interesting!!! And he will definitely avoid “misunderstanding” mathematics and will not feel disgusted with it. He got the win!
If child discovered that adding pairs of numbers that add up to a hundred is a piece of cake, then "arithmetic progression with difference 1"- a rather dreary and uninteresting thing for a child - suddenly found life for him . Order emerged from chaos, and this always causes enthusiasm: that's how we are made!
A question to answer: why, after the insight a child has received, should he again be driven into the framework of dry algorithms, which are also functionally useless in this case?!
Why force stupid rewrites? sequence numbers in a notebook: so that even the capable do not have a single chance of understanding? Statistically, of course, but mass education is geared towards “statistics”...
Where did the zero go?
And yet, adding numbers that add up to 100 is much more acceptable to the mind than those that add up to 101...
The "Gauss School Method" requires exactly this: mindlessly fold pairs of numbers equidistant from the center of the progression, Despite everything.
What if you look?
Still, zero is the greatest invention of mankind, which is more than 2,000 years old. And math teachers continue to ignore him.
It is much easier to transform a series of numbers starting with 1 into a series starting with 0. The sum will not change, will it? You need to stop “thinking in textbooks” and start looking... And see that pairs with a sum of 101 can be completely replaced by pairs with a sum of 100!
0 + 100, 1 + 99, 2 + 98 ... 49 + 51
How to abolish the "plus 1 rule"?
To be honest, I first heard about such a rule from that YouTube tutor...
What do I still do when I need to determine the number of members of a series?
I look at the sequence:
1, 2, 3, .. 8, 9, 10
and when you’re completely tired, then move on to a simpler row:
1, 2, 3, 4, 5
and I figure: if you subtract one from 5, you get 4, but I’m absolutely clear I see 5 numbers! Therefore, you need to add one! The number sense developed in elementary school suggests: even if there are a whole Google of members of the series (10 to the hundredth power), the pattern will remain the same.
What the hell are the rules?..
So that in a couple or three years you can fill all the space between your forehead and the back of your head and stop thinking? How to earn your bread and butter? After all, we are moving in even ranks into the era of the digital economy!
More about Gauss’s school method: “why make science out of this?..”
It was not for nothing that I posted a screenshot from my son’s notebook...
"What happened in class?"
“Well, I counted right away, raised my hand, but she didn’t ask. Therefore, while the others were counting, I began to do homework in Russian so as not to waste time. Then, when the others finished writing (???), she called me to the board. I said the answer."
“That’s right, show me how you solved it,” said the teacher. I showed it. She said: “Wrong, you need to count as I showed!”
“It’s good that she didn’t give a bad grade. And she made me write in their notebook “the course of the solution” in their own way. Why make a big science out of this?..”
The main crime of a math teacher
Hardly after that incident Carl Gauss experienced a high sense of respect for his school mathematics teacher. But if he knew how followers of that teacher will distort the very essence of the method... he would roar with indignation and, through the World Intellectual Property Organization WIPO, achieve a ban on the use of his good name in school textbooks!..
In what the main mistake of the school approach? Or, as I put it, a crime of school mathematics teachers against children?
Algorithm of misunderstanding
What do school methodologists do, the vast majority of whom don’t know how to think?
They create methods and algorithms (see). This a defensive reaction that protects teachers from criticism (“Everything is done according to...”) and children from understanding. And thus - from the desire to criticize teachers!(The second derivative of bureaucratic “wisdom”, a scientific approach to the problem). A person who does not grasp the meaning will rather blame his own misunderstanding, rather than the stupidity of the school system.
This is what happens: parents blame their children, and teachers... do the same for children who “don’t understand mathematics!”
Are you smart?
What did little Karl do?
A completely unconventional approach to a formulaic task. This is the essence of His approach. This the main thing that should be taught in school is to think not with textbooks, but with your head. Of course, there is also an instrumental component that can be used... in search of simpler and more efficient counting methods.
Gauss method according to Vilenkin
In school they teach that Gauss's method is to
What, if the number of elements of the series is odd, as in the problem that was assigned to my son?..
The "catch" is that in this case you should find an “extra” number in the series and add it to the sum of the pairs. In our example this number is 260.
How to detect? Copying all pairs of numbers into a notebook!(This is why the teacher made the kids do this stupid job of trying to teach "creativity" using the Gaussian method... And this is why such a "method" is practically inapplicable to large data series, AND this is why it is not the Gaussian method.)
A little creativity in the school routine...
The son acted differently.
(20 + 500, 40 + 480 ...).
0+500, 20+480, 40+460 ...
Not difficult, right?
But in practice it becomes even easier, which allows you to carve out 2-3 minutes for remote sensing in Russian, while the rest are “counting”. In addition, it retains the number of steps of the method: 5, which does not allow the approach to be criticized for being unscientific.
Obviously this approach is simpler, faster and more universal, in the style of the Method. But... the teacher not only did not praise, but also forced me to rewrite it “in the correct way” (see screenshot). That is, she made a desperate attempt to stifle the creative impulse and the ability to understand mathematics at the root! Apparently, so that she could later be hired as a tutor... She attacked the wrong person...
Everything that I described so long and tediously can be explained to a normal child in a maximum of half an hour. Along with examples.
And in such a way that he will never forget it.
And it will be step towards understanding...not just mathematicians.
Admit it: how many times in your life have you added using the Gaussian method? And I never did!
But instinct of understanding, which develops (or is extinguished) in the process of studying mathematical methods at school... Oh!.. This is truly an irreplaceable thing!
Especially in the age of universal digitalization, which we have quietly entered under the strict leadership of the Party and the Government.
A few words in defense of teachers...
It is unfair and wrong to place all responsibility for this style of teaching solely on school teachers. The system is in effect.
Some teachers understand the absurdity of what is happening, but what to do? The Law on Education, Federal State Educational Standards, methods, lesson plans... Everything must be done “in accordance and on the basis” and everything must be documented. Step aside - stood in line to be fired. Let’s not be hypocrites: the salaries of Moscow teachers are very good... If they fire you, where to go?..
Therefore this site not about education. He's about individual education, the only possible way to get out of the crowd generation Z ...
One of the universal and effective methods for solving linear algebraic systems is Gaussian method , consisting in the sequential elimination of unknowns.
Recall that the two systems are called equivalent (equivalent) if the sets of their solutions coincide. In other words, systems are equivalent if every solution of one of them is a solution of the other and vice versa. Equivalent systems are obtained when elementary transformations equations of the system:
multiplying both sides of the equation by a number other than zero;
adding to some equation the corresponding parts of another equation, multiplied by a number other than zero;
rearranging two equations.
Let a system of equations be given
The process of solving this system using the Gaussian method consists of two stages. At the first stage (direct motion), the system, using elementary transformations, is reduced to stepwise , or triangular form, and at the second stage (reverse) there is a sequential, starting from the last variable number, determination of the unknowns from the resulting step system.
Let us assume that the coefficient of this system 
, otherwise in the system the first row can be swapped with any other row so that the coefficient at 
was different from zero.
Let's transform the system by eliminating the unknown 
in all equations except the first. To do this, multiply both sides of the first equation by 
and add term by term with the second equation of the system. Then multiply both sides of the first equation by 
and add it to the third equation of the system. Continuing this process, we obtain the equivalent system
Here 
– new values of coefficients and free terms that are obtained after the first step.
Similarly, considering the main element 
, exclude the unknown 
from all equations of the system, except the first and second. Let's continue this process as long as possible, and as a result we will get a stepwise system
,
Where 
,
,…,
– main elements of the system 
.
If, in the process of reducing the system to a stepwise form, equations appear, i.e., equalities of the form 
, they are discarded since they are satisfied by any set of numbers 
. If at 
If an equation of the form appears that has no solutions, this indicates the incompatibility of the system.
During the reverse stroke, the first unknown is expressed from the last equation of the transformed step system 
through all the other unknowns 
which are called free
.
Then the variable expression 
from the last equation of the system is substituted into the penultimate equation and the variable is expressed from it 
. Variables are defined sequentially in a similar way 
. Variables 
, expressed through free variables, are called basic
(dependent). The result is a general solution to the system of linear equations.
To find private solution
systems, free unknown 
in the general solution arbitrary values are assigned and the values of the variables are calculated 
.
It is technically more convenient to subject to elementary transformations not the system equations themselves, but the extended matrix of the system
.
The Gauss method is a universal method that allows you to solve not only square, but also rectangular systems in which the number of unknowns 
not equal to the number of equations 
.
The advantage of this method is also that in the process of solving we simultaneously examine the system for compatibility, since, having given the extended matrix 
to stepwise form, it is easy to determine the ranks of the matrix 
and extended matrix 
and apply Kronecker-Capelli theorem
.
Example 2.1 Solve the system using the Gauss method
Solution. Number of equations 
and the number of unknowns 
.
Let's create an extended matrix of the system by assigning coefficients to the right of the matrix 
free members column 
.
Let's present the matrix 
to a triangular view; To do this, we will obtain “0” below the elements located on the main diagonal using elementary transformations.
To get the "0" in the second position of the first column, multiply the first row by (-1) and add it to the second row.
We write this transformation as the number (-1) against the first line and denote it with an arrow going from the first line to the second line.
To get "0" in the third position of the first column, multiply the first row by (-3) and add to the third row; Let's show this action using an arrow going from the first line to the third.
.
In the resulting matrix, written second in the chain of matrices, we get “0” in the second column in the third position. To do this, we multiplied the second line by (-4) and added it to the third. In the resulting matrix, multiply the second row by (-1), and divide the third by (-8). All elements of this matrix lying below the diagonal elements are zeros.
Because , the system is collaborative and defined.
The system of equations corresponding to the last matrix has a triangular form:
From the last (third) equation 
. Substitute into the second equation and get 
.
Let's substitute 
And 
into the first equation, we find 
.