In this article, the method is considered as a way to solve systems linear equations(SLAU). The method is analytical, that is, it allows you to write a solution algorithm in general view, 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 outline. 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
None hidden meaning not 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 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 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 joint systems are additionally 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 shorten big numbers in the 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 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. Therefore, 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 in last time the algorithm was performed 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 infinite number decisions. 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. That's what it is common decision SLAU.
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 common fraction, 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 can be done here is to remove from the third line overall coefficient "-1/7".
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 highest order of the determinant-square is 4. This means that there are solutions infinite set, and we 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 the matrix the following type:
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
Solution incompatible systems equations by the Gaussian method - 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 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 a mistake, it is more advisable to use 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.
Educational institution "Belarusian State
Agricultural Academy"
Department of Higher Mathematics
to study the topic “Gauss method for solving systems of linear
equations" by students of the accounting faculty of correspondence education (NISPO)
Gorki, 2013
Gauss method for solving systems of linear equations
Equivalent systems of equations
Two systems of linear equations are said to be equivalent if each solution of one of them is a solution of the other. The process of solving a system of linear equations consists of sequentially transforming it into an equivalent system using the so-called elementary transformations , which are:
1) rearrangement of any two equations of the system;
2) multiplying both sides of any equation of the system by a nonzero number;
3) adding to any equation another equation multiplied by any number;
4) crossing out an equation consisting of zeros, i.e. equations of the form
Gaussian elimination
Consider the system m linear equations with n unknown:
The essence of the Gauss method or method sequential elimination the unknowns are as follows.
First, using elementary transformations, the unknown is eliminated from all equations of the system except the first. Such system transformations are called Gaussian elimination step . The unknown is called enabling variable at the first step of transformation. The coefficient is called resolution factor , the first equation is called resolving equation , and the column of coefficients at permission column .
When performing one step of Gaussian elimination, you need to use the following rules:
1) the coefficients and the free term of the resolving equation remain unchanged;
2) the coefficients of the resolution column located below the resolution coefficient become zero;
3) all other coefficients and free terms when performing the first step are calculated according to the rectangle rule:
, Where i=2,3,…,m; j=2,3,…,n.
We will perform similar transformations on the second equation of the system. This will lead to a system in which the unknown will be eliminated in all equations except the first two. As a result of such transformations over each of the equations of the system (direct progression of the Gaussian method), the original system is reduced to an equivalent step system of one of the following types.
Reverse Gaussian Method
Step system
has a triangular appearance and that's it 
(i=1,2,…,n). Such a system has a unique solution. The unknowns are determined starting from the last equation (reverse of the Gaussian method).
The step system has the form
where, i.e. the number of equations of the system is less than or equal to the number of unknowns. This system has no solutions, since the last equation will not be satisfied for any values of the variable.
Step type system
has countless solutions. From the last equation, the unknown is expressed through the unknowns 
. Then, in the penultimate equation, instead of the unknown, its expression is substituted through the unknowns . Continuing the reverse of the Gaussian method, the unknowns 
can be expressed in terms of unknowns . In this case, the unknowns are called free
and can take any values, and unknown 
basic.
When solving systems in practice, it is convenient to perform all transformations not with a system of equations, but with an extended matrix of the system, consisting of coefficients for unknowns and a column of free terms.
Example 1. Solve system of equations 
Solution. Let's create an extended matrix of the system and perform elementary transformations:
.
In the extended matrix of the system, the number 3 (it is highlighted) is the resolution coefficient, the first row is the resolution row, and the first column is the resolution column. When moving to the next matrix, the resolution row does not change; all elements of the resolution column below the resolution element are replaced by zeros. And all other elements of the matrix are recalculated according to the quadrilateral rule. Instead of element 4 in the second line we write 
, instead of element -3 in the second line it will be written 
etc. Thus, the second matrix will be obtained. The resolution element of this matrix will be the number 18 in the second row. To form the next (third matrix), we leave the second row unchanged, in the column under the resolving element we write zero and recalculate the remaining two elements: instead of the number 1 we write 
, and instead of the number 16 we write .
As a result, the original system was reduced to an equivalent system
From the third equation we find 
. Let's substitute this value into the second equation: 
y=3. Let’s substitute the found values into the first equation y And z: 
, x=2.
Thus, the solution to this system of equations is x=2, y=3, 
.
Example 2. Solve system of equations 
Solution. Let us perform elementary transformations on the extended matrix of the system:
In the second matrix, each element of the third row is divided by 2.
In the fourth matrix, each element of the third and fourth rows was divided by 11.
. The resulting matrix corresponds to the system of equations 
Solving this system, we find 
, 
, .
Example 3. Solve system of equations
Solution. Let's write the extended matrix of the system and perform elementary transformations:
.
In the second matrix, each element of the second, third and fourth rows was divided by 7.
As a result, a system of equations was obtained
equivalent to the original one.
Since there are two fewer equations than unknowns, then from the second equation 
. Let's substitute the expression for into the first equation: , 
.
Thus, the formulas 
give a general solution to this system of equations. Unknowns are free and can take any value.
Let, for example, 
Then 
And 
. Solution 
is one of the particular solutions of the system, of which there are countless.
Questions for self-control of knowledge
1) What transformations of linear systems are called elementary?
2) What transformations of the system are called the Gaussian elimination step?
3) What is a resolving variable, resolving coefficient, resolving column?
4) What rules should be used when performing one step of Gaussian elimination?
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 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 
.
One of the simplest ways to solve a system of linear equations is a technique based on the calculation of determinants ( Cramer's rule). Its advantage is that it allows you to immediately record the solution; it is especially convenient in cases where the coefficients of the system are not numbers, but some parameters. Its disadvantage is the cumbersomeness of calculations in the case large number equations; moreover, Cramer's rule is not directly applicable to systems in which the number of equations does not coincide with the number of unknowns. In such cases, it is usually used Gaussian method.
Systems of linear equations having the same set of solutions are called equivalent. Obviously, many solutions linear system does not change if any equations are swapped, or if one of the equations is multiplied by some non-zero number, or if one equation is added to another.
Gauss method (method of sequential elimination of unknowns) is that with the help of elementary transformations the system is reduced to an equivalent system of a step type. First, using the 1st equation, we eliminate x 1 of all subsequent equations of the system. Then, using the 2nd equation, we eliminate x 2 from the 3rd and all subsequent equations. This process, called direct Gaussian method, continues until there is only one unknown left on the left side of the last equation x n. After this it is done inverse of the Gaussian method– solving the last equation, we find x n; after that, using this value, from the penultimate equation we calculate x n–1, etc. We find the last one x 1 from the first equation.
It is convenient to carry out Gaussian transformations by performing transformations not with the equations themselves, but with the matrices of their coefficients. Consider the matrix:
called extended matrix of the system, because, in addition to the main matrix of the system, it includes a column of free terms. The Gaussian method is based on reducing the main matrix of the system to a triangular form (or trapezoidal form in the case of non-square systems) using elementary row transformations (!) of the extended matrix of the system.
Example 5.1. Solve the system using the Gaussian method:
Solution. Let's write out the extended matrix of the system and, using the first row, after that we will reset the remaining elements:
we get zeros in the 2nd, 3rd and 4th rows of the first column:
Now we need all elements in the second column below the 2nd row to be equal to zero. To do this, you can multiply the second line by –4/7 and add it to the 3rd line. However, in order not to deal with fractions, let's create a unit in the 2nd row of the second column and only
Now, to get a triangular matrix, you need to reset the element of the fourth row of the 3rd column; to do this, you can multiply the third row by 8/54 and add it to the fourth. However, in order not to deal with fractions, we will swap the 3rd and 4th rows and the 3rd and 4th columns and only after that we will reset the specified element. Note that when rearranging the columns, the corresponding variables change places and this must be remembered; other elementary transformations with columns (addition and multiplication by a number) cannot be performed!
The last simplified matrix corresponds to a system of equations equivalent to the original one:
From here, using the inverse of the Gaussian method, we find from the fourth equation x 3 = –1; from the third x 4 = –2, from the second x 2 = 2 and from the first equation x 1 = 1. In matrix form, the answer is written as
We considered the case when the system is definite, i.e. when there is only one solution. Let's see what happens if the system is inconsistent or uncertain.
Example 5.2. Explore the system using the Gaussian method:
Solution. We write out and transform the extended matrix of the system
We write a simplified system of equations:
Here, in the last equation it turned out that 0=4, i.e. contradiction. Consequently, the system has no solution, i.e. she incompatible. à
Example 5.3. Explore and solve the system using the Gaussian method:
Solution. We write out and transform the extended matrix of the system:
As a result of the transformations, the last line contains only zeros. This means that the number of equations has decreased by one:
Thus, after simplifications, there are two equations left, and four unknowns, i.e. two unknown "extra". Let them be "superfluous", or, as they say, free variables, will x 3 and x 4 . Then
Believing x 3 = 2a And x 4 = b, we get x 2 = 1–a And x 1 = 2b–a; or in matrix form
A solution written in this way is called general, because, giving parameters a And b different meanings, it is possible to describe all possible solutions of the system. a
Definition and description of the Gaussian method
The Gaussian transformation method (also known as the method of sequential elimination of unknown variables from an equation or matrix) for solving systems of linear equations is a classical method for solving systems of algebraic equations (SLAE). This classical method is also used to solve problems such as obtaining inverse matrices and determining the rank of the matrix.
Transformation using the Gaussian method consists of making small (elementary) sequential changes to a system of linear algebraic equations, leading to the elimination of variables from it from top to bottom with the formation of a new triangular system of equations that is equivalent to the original one.
Definition 1
This part of the solution is called the forward Gaussian solution, since the entire process is carried out from top to bottom.
After reducing the original system of equations to a triangular one, we find all system variables from bottom to top (that is, the first variables found occupy exactly the last lines of the system or matrix). This part of the solution is also known as the inverse of the Gaussian solution. His algorithm is as follows: first, the variables closest to the bottom of the system of equations or matrix are calculated, then the resulting values are substituted higher and thus another variable is found, and so on.
Description of the Gaussian method algorithm
The sequence of actions for the general solution of a system of equations using the Gaussian method consists in alternately applying the forward and backward strokes to the matrix based on the SLAE. Let the initial system of equations have the following form:
$\begin(cases) a_(11) \cdot x_1 +...+ a_(1n) \cdot x_n = b_1 \\ ... \\ a_(m1) \cdot x_1 + a_(mn) \cdot x_n = b_m \end(cases)$
To solve SLAEs using the Gaussian method, it is necessary to write the original system of equations in the form of a matrix:
$A = \begin(pmatrix) a_(11) & … & a_(1n) \\ \vdots & … & \vdots \\ a_(m1) & … & a_(mn) \end(pmatrix)$, $b =\begin(pmatrix) b_1 \\ \vdots \\ b_m \end(pmatrix)$
The matrix $A$ is called the main matrix and represents the coefficients of the variables written in order, and $b$ is called the column of its free terms. The matrix $A$, written through a bar with a column of free terms, is called an extended matrix:
$A = \begin(array)(ccc|c) a_(11) & … & a_(1n) & b_1 \\ \vdots & … & \vdots & ...\\ a_(m1) & … & a_( mn) & b_m \end(array)$
Now it is necessary, using elementary transformations on the system of equations (or on the matrix, since this is more convenient), to bring it to the following form:
$\begin(cases) α_(1j_(1)) \cdot x_(j_(1)) + α_(1j_(2)) \cdot x_(j_(2))...+ α_(1j_(r)) \cdot x_(j_(r)) +... α_(1j_(n)) \cdot x_(j_(n)) = β_1 \\ α_(2j_(2)) \cdot x_(j_(2)). ..+ α_(2j_(r)) \cdot x_(j_(r)) +... α_(2j_(n)) \cdot x_(j_(n)) = β_2 \\ ...\\ α_( rj_(r)) \cdot x_(j_(r)) +... α_(rj_(n)) \cdot x_(j_(n)) = β_r \\ 0 = β_(r+1) \\ … \ \ 0 = β_m \end(cases)$ (1)
The matrix obtained from the coefficients of the transformed system of equation (1) is called a step matrix; this is what step matrices usually look like:
$A = \begin(array)(ccc|c) a_(11) & a_(12) & a_(13) & b_1 \\ 0 & a_(22) & a_(23) & b_2\\ 0 & 0 & a_(33) & b_3 \end(array)$
These matrices are characterized by the following set of properties:
- All its zero lines come after non-zero lines
- If some row of a matrix with number $k$ is non-zero, then the previous row of the same matrix has fewer zeros than this one with number $k$.
After obtaining the step matrix, it is necessary to substitute the resulting variables into the remaining equations (starting from the end) and obtain the remaining values of the variables.
Basic rules and permitted transformations when using the Gauss method
When simplifying a matrix or system of equations using this method, you need to use only elementary transformations.
Such transformations are considered to be operations that can be applied to a matrix or system of equations without changing its meaning:
- rearrangement of several lines,
- adding or subtracting from one row of a matrix another row from it,
- multiplying or dividing a string by a constant not equal to zero,
- a line consisting of only zeros, obtained in the process of calculating and simplifying the system, must be deleted,
- You also need to remove unnecessary proportional lines, choosing for the system the only one with coefficients that are more suitable and convenient for further calculations.
All elementary transformations are reversible.
Analysis of the three main cases that arise when solving linear equations using the method of simple Gaussian transformations
There are three cases that arise when using the Gaussian method to solve systems:
- When a system is inconsistent, that is, it does not have any solutions
- The system of equations has a solution, and a unique one, and the number of non-zero rows and columns in the matrix is equal to each other.
- The system has a certain number or set of possible solutions, and the number of rows in it is less than the number of columns.
Outcome of a solution with an inconsistent system
For this option, when solving matrix equation The Gauss method is characterized by obtaining some line with the impossibility of fulfilling the equality. Therefore, if at least one incorrect equality occurs, the resulting and original systems do not have solutions, regardless of the other equations they contain. An example of an inconsistent matrix:
$\begin(array)(ccc|c) 2 & -1 & 3 & 0 \\ 1 & 0 & 2 & 0\\ 0 & 0 & 0 & 1 \end(array)$
In the last line an impossible equality arose: $0 \cdot x_(31) + 0 \cdot x_(32) + 0 \cdot x_(33) = 1$.
A system of equations that has only one solution
These systems, after being reduced to a step matrix and removing rows with zeros, have the same number of rows and columns in the main matrix. Here simplest example such a system:
$\begin(cases) x_1 - x_2 = -5 \\ 2 \cdot x_1 + x_2 = -7 \end(cases)$
Let's write it in the form of a matrix:
$\begin(array)(cc|c) 1 & -1 & -5 \\ 2 & 1 & -7 \end(array)$
To bring the first cell of the second row to zero, we multiply the top row by $-2$ and subtract it from the bottom row of the matrix, and leave the top row in its original form, as a result we have the following:
$\begin(array)(cc|c) 1 & -1 & -5 \\ 0 & 3 & 10 \end(array)$
This example can be written as a system:
$\begin(cases) x_1 - x_2 = -5 \\ 3 \cdot x_2 = 10 \end(cases)$
The lower equation yields the following value for $x$: $x_2 = 3 \frac(1)(3)$. Substitute this value into the upper equation: $x_1 – 3 \frac(1)(3)$, we get $x_1 = 1 \frac(2)(3)$.
A system with many possible solutions
This system is characterized by a smaller number of significant rows than the number of columns in it (the rows of the main matrix are taken into account).
Variables in such a system are divided into two types: basic and free. When transforming such a system, the main variables contained in it must be left in the left area up to the “=” sign, and the remaining variables must be moved to the right side of the equality.
Such a system has only a certain general solution.
Let us analyze the following system of equations:
$\begin(cases) 2y_1 + 3y_2 + x_4 = 1 \\ 5y_3 - 4y_4 = 1 \end(cases)$
Let's write it in the form of a matrix:
$\begin(array)(cccc|c) 2 & 3 & 0 & 1 & 1 \\ 0 & 0 & 5 & 4 & 1 \\ \end(array)$
Our task is to find a general solution to the system. For this matrix, the basis variables will be $y_1$ and $y_3$ (for $y_1$ - since it comes first, and in the case of $y_3$ - it is located after the zeros).
As basis variables, we choose exactly those that are the first in the row and are not equal to zero.
The remaining variables are called free; we need to express the basic ones through them.
Using the so-called reverse stroke, we analyze the system from bottom to top; to do this, we first express $y_3$ from the bottom line of the system:
$5y_3 – 4y_4 = 1$
$5y_3 = 4y_4 + 1$
$y_3 = \frac(4/5)y_4 + \frac(1)(5)$.
Now we substitute the expressed $y_3$ into the upper equation of the system $2y_1 + 3y_2 + y_4 = 1$: $2y_1 + 3y_2 - (\frac(4)(5)y_4 + \frac(1)(5)) + y_4 = 1$
We express $y_1$ in terms of free variables $y_2$ and $y_4$:
$2y_1 + 3y_2 - \frac(4)(5)y_4 - \frac(1)(5) + y_4 = 1$
$2y_1 = 1 – 3y_2 + \frac(4)(5)y_4 + \frac(1)(5) – y_4$
$2y_1 = -3y_2 - \frac(1)(5)y_4 + \frac(6)(5)$
$y_1 = -1.5x_2 – 0.1y_4 + 0.6$
The solution is ready.
Example 1
Solve slough using the Gaussian method. Examples. An example of solving a system of linear equations given by a 3 by 3 matrix using the Gaussian method
$\begin(cases) 4x_1 + 2x_2 – x_3 = 1 \\ 5x_1 + 3x_2 - 2x^3 = 2\\ 3x_1 + 2x_2 – 3x_3 = 0 \end(cases)$
Let's write our system in the form of an extended matrix:
$\begin(array)(ccc|c) 4 & 2 & -1 & 1 \\ 5 & 3 & -2 & 2 \\ 3 & 2 & -3 & 0\\ \end(array)$
Now, for convenience and practicality, you need to transform the matrix so that $1$ is in the upper corner of the outermost column.
To do this, to the 1st line you need to add the line from the middle, multiplied by $-1$, and write the middle line itself as it is, it turns out:
$\begin(array)(ccc|c) -1 & -1 & 1 & -1 \\ 5 & 3 & -2 & 2 \\ 3 & 2 & -3 & 0\\ \end(array)$
$\begin(array)(ccc|c) -1 & -1 & 1 & -1 \\ 0 & -2 & 3 & -3 \\ 0 & -1 & 0 & -3\\ \end(array) $
Multiply the top and last lines by $-1$, and also swap the last and middle lines:
$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & -2 & 3 & -3\\ \end(array)$
$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 3 & 3\\ \end(array)$
And divide the last line by $3$:
$\begin(array)(ccc|c) 1 & 1 & -1 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 1\\ \end(array)$
We obtain the following system of equations, equivalent to the original one:
$\begin(cases) x_1 + x_2 – x_3 = 1\\ x_2 = 3 \\ x_3 = 1 \end(cases)$
From the upper equation we express $x_1$:
$x1 = 1 + x_3 – x_2 = 1 + 1 – 3 = -1$.
Example 2
An example of solving a system defined using a 4 by 4 matrix using the Gaussian method
$\begin(array)(cccc|c) 2 & 5 & 4 & 1 & 20 \\ 1 & 3 & 2 & 1 & 11 \\ 2 & 10 & 9 & 7 & 40\\ 3 & 8 & 9 & 2 & 37 \\ \end(array)$.
At the beginning, we swap the top lines following it to get $1$ in the upper left corner:
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 2 & 5 & 4 & 1 & 20 \\ 2 & 10 & 9 & 7 & 40\\ 3 & 8 & 9 & 2 & 37 \\ \end(array)$.
Now multiply the top line by $-2$ and add to the 2nd and 3rd. To the 4th we add the 1st line, multiplied by $-3$:
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & -1 & 0 & -1 & -2 \\ 0 & 4 & 5 & 5 & 18\\ 0 & - 1 & 3 & -1 & 4 \\ \end(array)$
Now to line number 3 we add line 2 multiplied by $4$, and to line 4 we add line 2 multiplied by $-1$.
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & -1 & 0 & -1 & -2 \\ 0 & 0 & 5 & 1 & 10\\ 0 & 0 & 3 & 0 & 6 \\ \end(array)$
We multiply line 2 by $-1$, and divide line 4 by $3$ and replace line 3.
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & 0 & 2\\ 0 & 0 & 5 & 1 & 10 \\ \end(array)$
Now we add to the last line the penultimate one, multiplied by $-5$.
$\begin(array)(cccc|c) 1 & 3 & 2 & 1 & 11 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 0 & 1 & 0 & 2\\ 0 & 0 & 0 & 1 & 0 \\ \end(array)$
We solve the resulting system of equations:
$\begin(cases) m = 0 \\ g = 2\\ y + m = 2\ \ x + 3y + 2g + m = 11\end(cases)$