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Gauss - Jordan
Elimination

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This is a variation of Gaussian elimination. Gaussian elimination gives us tools to solve large linear systems numerically. It is done by manipulating the given matrix using the elementary row operations to put the matrix into row echelon form. To be in row echelon form, a matrix must conform to the following criteria:
If a row does not consist entirely of zeros, then the first non zero number in the row is a 1.(the leading 1)
If there are any rows entirely made up of zeros, then they are grouped at the bottom of the matrix.
In any two successive rows that do not consist entirely of zeros, the leading 1 in the lower row occurs farther to the right that the leading 1 in the higher row.
From this form, the solution is easily(relatively) derived. The variation made in the Gauss-Jordan method is called back substitution. Back substitution consists of taking a row echelon matrix and operating on it in reverse order. Normally the matrix is simplified from top to bottom to achieve row echelon form. When Gauss-Jordan has finished, all that remains in the matrix is a main diagonal of ones and the augmentation, this matrix is now in reduced row echelon form. For a matrix to be in reduced row echelon form, it must be in row echelon form and submit to one added criteria:

Each column that contains a leading 1 has zeros everywhere else.
Since the matrix is representing the coefficients of the given variables in the system, the augmentation now represents the values of each of those variables. The solution to the system can now be found by inspection and no additional work is required. Consider the following example:

Start With: Elementary Row
Operation(S) Product

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Place into augmented matrix
R2 - (-1)R1 --> R2
R3 - ( 3)R1 --> R3

(-1)R2 --> R2
R3 - (-10)R2 --> R3

(-1/52)R3 --> R3

In Row Echelon Form --->

R2 - (-5)R3 --> R2
R1 - (2)R3 --> R1

R1 - (1)R2 --> R1

Reduced Row Echelon Form --->



It is now obvious, by inspection, that the solution to this linear system is x=3, y=1, and z=2. Again, by solution, it is meant the x, y, and z required to satisfy all the equations simultaneously.

There are some problems that could arise while searching for these solutions. If the lines are parallel then they will not intersect and thus provide no solution. In three dimensions the problem of skewing is possible. Lines are skewed if they lie in parallel planes yet have different slopes. If this problem occurs, it will made evident in the matrix by a row (or more than one) of zeros being present when the matrix is in row echelon form. Another problem that may arise is a division by zero. If a zero is placed in the main diagonal of the row being operated on, when you divide that row by the diagonal number the division by zero error will occur. To trap this error, simply check the diagonal number being worked with. If it is zero, exchange that row with the row below it. Exchanging rows is a legal elementary row operation.

If a matrix is ill-conditioned, bad round off errors may occur. Since a large number of multiplications and divisions are performed, if the numbers are not of relative size then round off error will be apparent. If the matrix is small then the error won't have time to propagate; but if the matrix is large, the round off error could deem the output solution unreliable.


Gauss-Jordan Elimination Algorithm In a Nut****l
Read in a matrix of size mxn.(m rows and n columns)
Begin working down the main diagonal starting at row one, column one.
Check for a zero in the working position.
If working position is zero, interchange the working row with the row directly beneath it. If you are on row m (last row) then the system is unsolvable.
Repeat step 3.
Divide the working row by the number in the working position. This will force the diagonal element to have a value equal to one.
Replace the rows beneath the working row by the following:
Store the value in the working column and next row.
Multiply working position by the stored number.
Subtract the product(step 2) from the from the next row entry.
Move column one to the right.
Multiply entry in working row new column by stored number.
Repeat steps 2 - 5 until last column have been operated on.
Keeping the working row constant, repeat steps 1-6 coupling the next row in line with the working row until all rows have been operated on. If this is done correctly, all the entries below the working position will be zero.
Move to the next position in the diagonal.
Repeat steps 3-7 until the last element in the diagonal has been operated on.
Your matrix is now in row echelon form.
Now begin the back substitution.
For the sake of not sounding redundant, back substitution is doing steps 3-7 going back up the diagonal.

 

 

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http://aspire.cs.uah.edu/****book/gauss.html
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http://www.3rbe.com/books/index.php?act=category&id=32

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حبيبي .. طريقة جاوس تعتمد على طرق عديدة
الطريقة الي استخمتها انا .. هو الحصول على القطر الأساسي الي تتكون من 1

بمعنى اخر ..
لو كانت عندنا مصفوفة 3 *3
100
010
001

فأنا احاول اوصل الى هذا المصفوفة ..
ومن جهة اخرى ينتج لي مصفوفة ثانية (( من الناحية الأخرى ))

كل ما في الموضوع اني اسعى للحصول على 1 .. حتى ينتج لي المصفوفة المحايدة