Here you can solve the system for free linear equations Gauss method online large sizes in complex numbers with a very detailed solution. Our calculator can solve online both the usual definite and indefinite system of linear equations by the Gauss method, which has infinite set decisions. In this case, in the answer you will receive the dependence of some variables through other, free ones. You can also check the system of equations for consistency online using the Gaussian solution.
About the method
When solving a system of linear equations online using the Gaussian method, the following steps are performed.
- We write the extended matrix.
- In fact, the solution is divided into direct and reverse stroke Gauss method. The direct approach of the Gaussian method is the reduction of a matrix to a stepwise form. The reverse of the Gaussian method is the reduction of a matrix to a special stepwise form. But in practice, it is more convenient to immediately zero out what is located both above and below the element in question. Our calculator uses exactly this approach.
- It is important to note that when solving using the Gaussian method, the presence in the matrix of at least one zero row with a NON-zero right-hand side (column of free terms) indicates the inconsistency of the system. In this case, a solution to the linear system does not exist.
To best understand how the Gaussian algorithm works online, enter any example, select "very detailed solution" and look up his solution online.
1. System of linear algebraic equations
1.1 The concept of a system of linear algebraic equations
A system of equations is a condition consisting of simultaneous execution of several equations with respect to several variables. A system of linear algebraic equations (hereinafter referred to as SLAE) containing m equations and n unknowns is called a system of the form:
where numbers a ij are called system coefficients, numbers b i are called free terms, a ij And b i(i=1,…, m; b=1,…, n) represent some known numbers, and x 1 ,…, x n– unknown. In the designation of coefficients a ij the first index i denotes the number of the equation, and the second j is the number of the unknown at which this coefficient stands. The numbers x n must be found. It is convenient to write such a system in a compact matrix form: AX=B. Here A is the matrix of system coefficients, called the main matrix;
– column vector of unknowns xj. is a column vector of free terms bi.
The product of matrices A*X is defined, since there are as many columns in matrix A as there are rows in matrix X (n pieces).
The extended matrix of a system is the matrix A of the system, supplemented by a column of free terms
1.2 Solving a system of linear algebraic equations
The solution to a system of equations is an ordered set of numbers (values of variables), when substituting them instead of variables, each of the equations of the system turns into a true equality.
A solution to a system is n values of the unknowns x1=c1, x2=c2,…, xn=cn, upon substitution of which all equations of the system become true equalities. Any solution to the system can be written as a column matrix
A system of equations is called consistent if it has at least one solution, and inconsistent if it does not have any solution.
A consistent system is said to be determinate if it has a single solution, and indefinite if it has more than one solution. In the latter case, each of its solutions is called a particular solution of the system. The set of all particular solutions is called the general solution.
Solving a system means finding out whether it is compatible or inconsistent. If the system is consistent, find it common decision.
Two systems are called equivalent (equivalent) if they have the same general solution. In other words, systems are equivalent if every solution of one of them is a solution of the other, and vice versa.
A transformation, the application of which turns a system into a new system equivalent to the original one, is called an equivalent or equivalent transformation. Examples of equivalent transformations include the following transformations: interchanging two equations of a system, interchanging two unknowns along with the coefficients of all equations, multiplying both sides of any equation of a system by a nonzero number.
A system of linear equations is called homogeneous if all free terms are equal to zero:
A homogeneous system is always consistent, since x1=x2=x3=…=xn=0 is a solution of the system. This solution is called zero or trivial.
2. Gaussian elimination method
2.1 The essence of the Gaussian elimination method
The classical method for solving systems of linear algebraic equations is the method of sequential elimination of unknowns - Gaussian method(it is also called the Gaussian elimination method). This is a method of sequential elimination of variables, when, using elementary transformations, a system of equations is reduced to an equivalent system of a step (or triangular) form, from which all other variables are found sequentially, starting with the last (by number) variables.
The solution process using the Gaussian method consists of two stages: forward and backward moves.
1. Direct stroke.
At the first stage, the so-called direct move is carried out, when, through elementary transformations over the rows, the system is brought to a stepped or triangular shape, or it is established that the system is incompatible. Namely, among the elements of the first column of the matrix, select a non-zero one, move it to the uppermost position by rearranging the rows, and subtract the resulting first row from the remaining rows after the rearrangement, multiplying it by a value equal to the ratio of the first element of each of these rows to the first element of the first row, zeroing thus the column below it.
After these transformations have been completed, the first row and first column are mentally crossed out and continued until a zero-size matrix remains. If at any iteration there is no non-zero element among the elements of the first column, then go to the next column and perform a similar operation.
At the first stage (direct stroke), the system is reduced to a stepped (in particular, triangular) form.
The system below has a stepwise form:
,
Coefficients aii are called the main (leading) elements of the system.
(if a11=0, rearrange the rows of the matrix so that a 11 was not equal to 0. This is always possible, because otherwise the matrix contains a zero column, its determinant is equal to zero and the system is inconsistent).Let's transform the system by eliminating the unknown x1 in all equations except the first (using elementary transformations of the system). To do this, multiply both sides of the first equation by
and add term by term with the second equation of the system (or from the second equation subtract term by term by the first, multiplied by ). Then we multiply both sides of the first equation by and add them to the third equation of the system (or from the third we subtract the first one multiplied by ). Thus, we sequentially multiply the first line by a number and add to i th line, for i= 2, 3, …,n.Continuing this process, we obtain an equivalent system:
– new values of coefficients for unknowns and free terms in the last m-1 equations of the system, which are determined by the formulas: Thus, at the first step, all coefficients lying under the first leading element a 11 are destroyed
0, in the second step the elements lying under the second leading element a 22 (1) are destroyed (if a 22 (1) 0), etc. Continuing this process further, we finally, at the (m-1) step, reduce the original system to a triangular system.If, in the process of reducing the system to a stepwise form, zero equations appear, i.e. equalities of the form 0=0, they are discarded. If an equation of the form appears
then this indicates the incompatibility of the system. This is where the direct progression of Gauss's method ends.
2. Reverse stroke.
At the second stage, the so-called reverse move is carried out, the essence of which is to express all the resulting basic variables in terms of non-basic ones and construct fundamental system solutions, or, if all variables are basic, then express numerically the unique solution to the system of linear equations.
This procedure begins with the last equation, from which the corresponding basic variable is expressed (there is only one in it) and substituted into the previous equations, and so on, going up the “steps”.
Each line corresponds to exactly one basis variable, so at every step except the last (topmost), the situation exactly repeats the case of the last line.
Note: in practice, it is more convenient to work not with the system, but with its extended matrix, performing all the elementary transformations on its rows. It is convenient for the coefficient a11 to be equal to 1 (rearrange the equations, or divide both sides of the equation by a11).
2.2 Examples of solving SLAEs using the Gaussian method
In this section there are three various examples Let us show how the Gaussian method can solve SLAE.
Example 1. Solve a 3rd order SLAE.
Let's reset the coefficients at
in the second and third lines. To do this, multiply them by 2/3 and 1, respectively, and add them to the first line:
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 ones appeared (or exist) in the matrix (as special case– identical) lines, then it follows 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, which is 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 in this example it turned out to be 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 easy to program and does not take into account specific features 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.
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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:
$\begin(cases) m = 0 \\ g = 2\\ y + m = 2\ \ x + 3y + 2g + m = 11\end(cases)$
Two systems of linear equations are called equivalent if the set of all their solutions coincides.
Elementary transformations of a system of equations are:
- Deleting trivial equations from the system, i.e. those for which all coefficients are equal to zero;
- Multiplying any equation by a number other than zero;
- Adding to any i-th equation any j-th equation multiplied by any number.
A variable x i is called free if this variable is not allowed, but the entire system of equations is allowed.
Theorem. Elementary transformations transform a system of equations into an equivalent one.
The meaning of the Gaussian method is to transform the original system of equations and obtain an equivalent resolved or equivalent inconsistent system.
So, the Gaussian method consists of the following steps:
- Let's look at the first equation. Let's choose the first non-zero coefficient and divide the entire equation by it. We obtain an equation in which some variable x i enters with a coefficient of 1;
- Let's subtract this equation from all the others, multiplying it by such numbers that the coefficients of the variable x i in the remaining equations are zeroed. We obtain a system resolved with respect to the variable x i and equivalent to the original one;
- If trivial equations arise (rarely, but it happens; for example, 0 = 0), we cross them out of the system. As a result, there are one fewer equations;
- We repeat the previous steps no more than n times, where n is the number of equations in the system. Each time we select a new variable for “processing”. If inconsistent equations arise (for example, 0 = 8), the system is inconsistent.
As a result, after a few steps we will obtain either a resolved system (possibly with free variables) or an inconsistent one. Allowed systems fall into two cases:
- The number of variables is equal to the number of equations. This means that the system is defined;
- Number of variables more number equations. We collect all the free variables on the right - we get formulas for the allowed variables. These formulas are written in the answer.
That's all! System of linear equations solved! This is a fairly simple algorithm, and to master it you do not have to contact a higher mathematics tutor. Let's look at an example:
Task. Solve the system of equations:
Description of steps:
- Subtract the first equation from the second and third - we get the allowed variable x 1;
- We multiply the second equation by (−1), and divide the third equation by (−3) - we get two equations in which the variable x 2 enters with a coefficient of 1;
- We add the second equation to the first, and subtract from the third. We get the allowed variable x 2 ;
- Finally, we subtract the third equation from the first - we get the allowed variable x 3;
- We have received an approved system, write down the response.
The general solution to a simultaneous system of linear equations is new system, equivalent to the original one, in which all allowed variables are expressed in terms of free ones.
When might a general solution be needed? If you have to do fewer steps than k (k is how many equations there are). However, the reasons why the process ends at some step l< k , может быть две:
- After the lth step, we obtained a system that does not contain an equation with number (l + 1). In fact, this is good, because... the authorized system is still obtained - even a few steps earlier.
- After the lth step, we obtained an equation in which all coefficients of the variables are equal to zero, and the free coefficient is different from zero. This is a contradictory equation, and, therefore, the system is inconsistent.
It is important to understand that the emergence of an inconsistent equation using the Gaussian method is a sufficient basis for inconsistency. At the same time, we note that as a result of the lth step, no trivial equations can remain - all of them are crossed out right in the process.
Description of steps:
- Subtract the first equation, multiplied by 4, from the second. We also add the first equation to the third - we get the allowed variable x 1;
- Subtract the third equation, multiplied by 2, from the second - we get the contradictory equation 0 = −5.
So, the system is inconsistent because an inconsistent equation has been discovered.
Task. Explore compatibility and find a general solution to the system:
Description of steps:
- We subtract the first equation from the second (after multiplying by two) and the third - we get the allowed variable x 1;
- Subtract the second equation from the third. Since all the coefficients in these equations are the same, the third equation will become trivial. At the same time, multiply the second equation by (−1);
- Subtract the second from the first equation - we get the allowed variable x 2. The entire system of equations is now also resolved;
- Since the variables x 3 and x 4 are free, we move them to the right to express the allowed variables. This is the answer.
So, the system is consistent and indeterminate, since there are two allowed variables (x 1 and x 2) and two free ones (x 3 and x 4).