Gaussian matrices. Gauss method for dummies: examples of solutions. Problems for using the Gauss rule
Let a system of linear algebraic equations be given, which must be solved (find such values \u200b\u200bof 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 inconsistent).
2) Have infinitely many solutions.
3) Have a unique solution.
As we remember, Cramer's rule and the matrix method are inapplicable 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 the in every casewill lead us to the answer! The algorithm of the method itself works the same in all three cases. If the knowledge of determinants is required in the Cramer and matrix methods, then for the application of the Gauss method, knowledge of only arithmetic operations is necessary, which makes it accessible even for primary school students.
Extended matrix transformations ( this is the matrix of the system - a matrix composed only of the coefficients of unknowns, plus a column of free terms)systems of linear algebraic equations in the Gauss method:
1) from strings matrices can rearrangeplaces.
2) if the matrix contains (or is) proportional (as a special case - the same) rows, then it follows delete from the matrix all these rows except one.
3) if a zero row appeared in the matrix during the transformations, then it also follows delete.
4) the row of the matrix can be multiply (divide)to any number other than zero.
5) the row of the matrix can be add another string multiplied by a numbernonzero.
In the Gauss method, elementary transformations do not change the solution of the system of equations.
Gauss method consists of two stages:
- “Direct move” - using elementary transformations to reduce the extended matrix of the system of linear algebraic equations to a “triangular” stepwise form: the elements of the extended matrix located below the main diagonal are equal to zero (“top-down” move). For example, to this form:
To do this, perform the following actions:
1) Suppose we consider the first equation of a system of linear algebraic equations and the coefficient at x 1 is K. The second, third, etc. the equations are transformed as follows: each equation (coefficients for unknowns, including free terms) is divided by the coefficient for the unknown x 1, standing in each equation, and multiplied by K. After that, we subtract the first from the second equation (coefficients for unknowns and free terms). We get the coefficient 0 for x 1 in the second equation. Subtract the first equation from the third transformed equation until all equations, except for the first one, for unknown x 1 have a coefficient of 0.
2) Go to the next equation. Let this be the second equation and the coefficient at x 2 is equal to M. With all the "lower" equations we proceed as described above. Thus, "under" the unknown x 2 in all equations will be zeros.
3) Go to the next equation and so on until there is one last unknown and the transformed free term.
- "Reverse" of the Gauss method - obtaining a solution to a system of linear algebraic equations ("bottom-up" move). From the last "lower" equation we get one first solution - the unknown x n. To do this, we solve the elementary equation A * x n \u003d B. In the example above, x 3 \u003d 4. Substitute the found value into the "upper" next equation and solve it with respect to the next unknown. For example, x 2 - 4 \u003d 1, i.e. x 2 \u003d 5. And so on until we find all the unknowns.
Example.
Let's solve the system of linear equations by 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 a unit there. The problem is that there are no ones in the first column at all, so rearranging the rows will not solve anything. In such cases, the unit needs to be organized using an elementary transformation. This can usually be done in several ways. Let's do this:
1 step
... To the first line, 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 is "minus one", which is fine for us. 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 the second place, thus, on the second “step, we have the required unit.
Step 4 ... The third line was added to the second line multiplied by 2.
Step 5 ... The third line was split by 3.
A sign that indicates an error in calculations (less often - a typo) is the "bad" bottom line. That is, if at the bottom we got something like (0 0 11 | 23), and, accordingly, 11x 3 \u003d 23, x 3 \u003d 23/11, then with a high degree of probability it can be argued that an error was made during elementary transformations.
We carry out the reverse move, in the design of examples, the system itself is often not rewritten, and the equations are "taken directly from the given matrix." The reverse move, I remind you, works from the bottom up. In this example, we got a gift:
x 3 \u003d 1
x 2 \u003d 3
x 1 + x 2 - x 3 \u003d 1, therefore x 1 + 3 - 1 \u003d 1, x 1 \u003d –1
Answer: x 1 \u003d –1, x 2 \u003d 3, x 3 \u003d 1.
Let's solve the same system according to 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
Subtracting 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 "stepwise" 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 get x 3 \u003d 0.96 or approximately 1.
x 2 \u003d 3 and x 1 \u003d –1.
Solving in this way, you will never get confused in the calculations and, despite the calculation errors, you will get the result.
This method of solving a system of linear algebraic equations is easily programmable and does not take into account the specific features of the coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.
Wish you success! See you in class! Tutor.
blog. site, with full or partial copying of the material, a link to the source is required.
Let a system of linear algebraic equations be given, which must be solved (find such values \u200b\u200bof 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 inconsistent).
2) Have infinitely many solutions.
3) Have a unique solution.
As we remember, Cramer's rule and the matrix method are inapplicable 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 the in every casewill lead us to the answer! The algorithm of the method itself works the same in all three cases. If the knowledge of determinants is required in the Cramer and matrix methods, then for the application of the Gauss method, knowledge of only arithmetic operations is necessary, which makes it accessible even for primary school students.
Extended matrix transformations ( this is the matrix of the system - a matrix composed only of the coefficients of unknowns, plus a column of free terms)systems of linear algebraic equations in the Gauss method:
1) from strings matrices can rearrangeplaces.
2) if the matrix contains (or is) proportional (as a special case - the same) rows, then it follows delete from the matrix all these rows except one.
3) if a zero row appeared in the matrix during the transformations, then it also follows delete.
4) the row of the matrix can be multiply (divide)to any number other than zero.
5) the row of the matrix can be add another string multiplied by a numbernonzero.
In the Gauss method, elementary transformations do not change the solution of the system of equations.
Gauss method consists of two stages:
- “Direct move” - using elementary transformations to reduce the extended matrix of the system of linear algebraic equations to a “triangular” stepwise form: the elements of the extended matrix located below the main diagonal are equal to zero (“top-down” move). For example, to this form:
To do this, perform the following actions:
1) Suppose we consider the first equation of a system of linear algebraic equations and the coefficient at x 1 is K. The second, third, etc. the equations are transformed as follows: each equation (coefficients for unknowns, including free terms) is divided by the coefficient for the unknown x 1, standing in each equation, and multiplied by K. After that, we subtract the first from the second equation (coefficients for unknowns and free terms). We get the coefficient 0 for x 1 in the second equation. Subtract the first equation from the third transformed equation until all equations, except for the first one, for unknown x 1 have a coefficient of 0.
2) Go to the next equation. Let this be the second equation and the coefficient at x 2 is equal to M. With all the "lower" equations we proceed as described above. Thus, "under" the unknown x 2 in all equations will be zeros.
3) Go to the next equation and so on until there is one last unknown and the transformed free term.
- "Reverse" of the Gauss method - obtaining a solution to a system of linear algebraic equations ("bottom-up" move). From the last "lower" equation we get one first solution - the unknown x n. To do this, we solve the elementary equation A * x n \u003d B. In the example above, x 3 \u003d 4. Substitute the found value into the "upper" next equation and solve it with respect to the next unknown. For example, x 2 - 4 \u003d 1, i.e. x 2 \u003d 5. And so on until we find all the unknowns.
Example.
Let's solve the system of linear equations by 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 a unit there. The problem is that there are no ones in the first column at all, so rearranging the rows will not solve anything. In such cases, the unit needs to be organized using an elementary transformation. This can usually be done in several ways. Let's do this:
1 step
... To the first line, 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 is "minus one", which is fine for us. 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 the second place, thus, on the second “step, we have the required unit.
Step 4 ... The third line was added to the second line multiplied by 2.
Step 5 ... The third line was split by 3.
A sign that indicates an error in calculations (less often - a typo) is the "bad" bottom line. That is, if at the bottom we got something like (0 0 11 | 23), and, accordingly, 11x 3 \u003d 23, x 3 \u003d 23/11, then with a high degree of probability it can be argued that an error was made during elementary transformations.
We carry out the reverse move, in the design of examples, the system itself is often not rewritten, and the equations are "taken directly from the given matrix." The reverse move, I remind you, works from the bottom up. In this example, we got a gift:
x 3 \u003d 1
x 2 \u003d 3
x 1 + x 2 - x 3 \u003d 1, therefore x 1 + 3 - 1 \u003d 1, x 1 \u003d –1
Answer: x 1 \u003d –1, x 2 \u003d 3, x 3 \u003d 1.
Let's solve the same system according to 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
Subtracting 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 "stepwise" 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 get x 3 \u003d 0.96 or approximately 1.
x 2 \u003d 3 and x 1 \u003d –1.
Solving in this way, you will never get confused in the calculations and, despite the calculation errors, you will get the result.
This method of solving a system of linear algebraic equations is easily programmable and does not take into account the specific features of the coefficients for unknowns, because in practice (in economic and technical calculations) one has to deal with non-integer coefficients.
Wish you success! See you in class! Tutor Dmitry Aistrakhanov.
site, with full or partial copying of the material, a link to the source is required.
Solution of systems of linear equations by the Gauss method.Let us need to find a solution to the system from n linear equations with n unknown variables
the determinant of the main matrix of which is nonzero.
The essence of the Gauss method consists in the successive elimination of unknown variables: first, x 1 from all equations of the system, starting with the second, further exclude x 2of all equations, starting with the third, and so on, until only the unknown variable remains in the last equation x n... Such a process of transforming the equations of the system for the successive elimination of unknown variables is called by the direct Gauss method... After completing the forward run of the Gauss method, from the last equation, we find x n, using this value from the penultimate equation is calculated x n-1, and so on, from the first equation we find x 1... The process of calculating unknown variables when moving from the last equation of the system to the first is called backward Gaussian method.
Let us briefly describe the algorithm for eliminating unknown variables.
We will assume that, since we can always achieve this by rearranging the equations of the system. Eliminate the unknown variable x 1 of all equations of the system, starting with the second. To do this, to the second equation of the system we add the first one multiplied by, to the third equation we add the first one multiplied by, and so on, to nthto the equation we add the first, multiplied by. The system of equations after such transformations takes the form
where, a.
We would arrive at the same result if we expressed x 1 through other unknown variables in the first equation of the system and the resulting expression was substituted into all other equations. So the variable x 1 excluded from all equations, starting with the second.
For this, to the third equation of the system we add the second multiplied by, to the fourth equation we add the second multiplied by, and so on, to nthto the equation we add the second, multiplied by. The system of equations after such transformations takes the form
where, a. So the variable x 2 excluded from all equations starting with the third.
So we continue the direct course of the Gauss method until the system takes the form
From this point on, we start the reverse course of the Gauss method: calculate x n from the last equation as, using the obtained value x n find x n-1 from the penultimate equation, and so on, we find x 1 from the first equation.
Example.
Solve the system of linear equations using the Gaussian method. ...
Answer:
x 1 \u003d 4, x 2 \u003d 0, x 3 \u003d -1.
KOSTROM BRANCH OF THE MILITARY UNIVERSITY OF RHB PROTECTION
Department of "Automation of command and control of troops"
For teachers only
"I approve"
Head of department number 9
colonel A.B. YAKOVLEV
"____" ______________ 2004
associate Professor A.I. SMIRNOVA
"MATRIXES. GAUSS'S METHOD"
LECTURE No. 2/3
Discussed at the meeting of the department number 9
"____" ___________ 2003
Minutes No. ___________
Kostroma, 2003
Cobsession
Introduction
1. Actions on matrices.
2. Solution of systems of linear equations by the Gauss method.
Conclusion
Literature
1. V.E. Schneider et al., Short Course in Higher Mathematics, Volume I, Ch. 2, §6, 7.
2.V.S. Shchipachev, Higher Mathematics, Ch. 10, § 1, 7.
INTRODUCTION
The lecture discusses the concept of a matrix, actions on matrices, as well as the Gauss method for solving systems of linear equations. For a special case, the so-called square matrices, one can calculate the determinants, the concept of which was discussed in the previous lecture. The Gauss method is more general than the previously considered Cramer's method for solving linear systems. The questions discussed at the lecture are used in various branches of mathematics and in applied questions.
1st study question ACTIONS ON MATRICES
DEFINITION 1. Rectangular table fromm, n numbers containingm - lines andn - columns, type:
called size matrix m ´ n
The numbers that make up the matrix are called matrix elements.
Item position and i j in the matrix are characterized by a double index:
the first i - line number;
second j - the number of the column at the intersection of which the element stands.
In abbreviated form, matrices are denoted in capital letters: A, B, C ...
Briefly, you can write like this:
DEFINITION 2.A matrix with the number of rows equal to the number of columns, i.e.m = n is called square.
The number of rows (columns) of a square matrix is \u200b\u200bcalled the order of the matrix.
EXAMPLE.
REMARK 1. We will consider matrices whose entries are numbers. In mathematics and its applications, there are matrices whose elements are other objects, for example, functions, vectors.
REMARK 2. Matrix is \u200b\u200ba special mathematical concept. With the help of matrices, it is convenient to write various transformations, linear systems, etc., therefore matrices are often found in mathematical and technical literature.
DEFINITION 3.Size Matrix1 none line is called matrix - string.
T-size matrix1 consisting of one column is called matrix - column.
DEFINITION 4. Zero Matrix called a matrix, all elements of which are equal to zero.
Consider a square matrix of order n:
side diagonalmain diagonal
The diagonal of a square matrix going from the upper left element of the table to the lower right is called the main diagonal of the matrix (the main diagonal contains elements of the form and i i).
The diagonal going from the top right element to the bottom left is called by the side diagonal of the matrix.
Let's consider some special types of square matrices.
1) A square matrix is \u200b\u200bcalled diagonalif all elements not on the main diagonal are equal to zero.
2) A diagonal matrix in which all elements of the main diagonal are equal to one is called single... It is indicated:
3) A square matrix is \u200b\u200bcalled triangular, if all elements to one side of the main diagonal are zero:
upper lower
triangular matrix triangular matrix
For a square matrix, the concept is introduced: determinant of a matrix... It is a determinant composed of matrix elements. It is indicated:
It is clear that the determinant of the identity matrix is \u200b\u200bequal to 1: 1 E½ \u003d 1
COMMENT. A non-square matrix has no determinant.
If the determinant of a quadratic matrix is \u200b\u200bnonzero, then the matrix is \u200b\u200bcalled non-degenerate, if the determinant is zero, then the matrix is \u200b\u200bcalled degenerate.
DEFINITION 5. The matrix obtained from this by replacing its rows with columns with the same numbers is called transposed to the given one.
Matrix transposed to AND, denote A T.
EXAMPLE.
3 3 2DEFINITION.Two matrices of the same size are called equal, if all their corresponding elements are equal .
Let's consider operations on matrices.
ADD OF MATRIXES.
The addition operation is introduced only for matrices of the same size.
DEFINITION 7. The sum of two matrices A \u003d (a i j ) and B \u003d ( b i j ) the same size the matrix С \u003d (with i j) of the same size, whose elements are equal to the sums of the corresponding elements of the matrix terms, i.e. from i j \u003d a i j + b i j
The sum of matrices is denoted A + B.
EXAMPLE.
REAL MULTIPLICATION OF MATRICES
DEFINITION 8.To multiply a matrix by a numberk, you need to multiply each element of the matrix by this number:
if a A \u003d(and i j )then k · A= (k · a i j )
EXAMPLE.
PROPERTIES OF MATRIX ADD AND MULTIPLICATION BY NUMBER
1. Displacement property: A + B \u003d B + A
2. Combination property: (A + B) + C \u003d A + (B + C)
3. Distribution property: k · (A + B) = k A + k Bwhere k – number
MATRIX MULTIPLICATION
The matrix ANDwill be called a globule with a matrix INif the number of matrix columns AND is equal to the number of rows of the matrix IN, i.e. for consistent matrices the matrix AND has a size m ´ n , matrix IN has a size n ´ k . Square matrices are consistent if they are of the same order.
DEFINITION 9.Product of matrix A of sizem ´ n per matrix B sizen ´ k called a matrix C of sizem ´ kwhose element a i j located ini -Th line andj - th column, is equal to the sum of products of elementsi - th row of matrix A to the corresponding elementsj - column of matrix B, i.e.
c i j = a i 1 b 1 j + a i 2 b 2 j +……+ a i n b n j
We denote: C \u003d A· IN.
thenComposition IN´ AND does not make sense, because matrices
not agreed.NOTE 1. If AND´ IN makes sense then IN´ AND may not make sense.
REMARK 2. If it makes sense AND´ IN and IN´ AND, then, generally speaking
AND´ IN ¹ IN´ AND, i.e. matrix multiplication does not have a transposition law.
NOTE 3. If ANDIs a square matrix and EIs the identity matrix of the same order, then AND´ E= E´ A \u003d A.
It follows from this that the identity matrix plays the role of unity during multiplication.
EXAMPLES... Find, if possible, AND´ IN and IN´ AND.
Decision: Square matrices of the same second order are matched in the same order and therefore AND´ IN and IN´ AND exist.
