gaus jordan

8/9/2019 Gaus Jordan

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Gaus

Jordan


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IntroductionSeveral methods exist for the numerical solution of systems of linear

equations, between them they are: Elimination Gaussiana and Gauss-

Jordn. This practice focuses on the above mentioned methods for what next

e it gives a small explanation of each one of they;


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Elimination Gaussiana

This method equations system proposes the variables phasing down in e,

handling a matriz squared up to leaving it in his triangular form, this way to

find the value of a variable and to allow other equations with 2, 3, 4., n

variables, for of equal form to be finding consecutively e value of the second,

third one. er.sima variable doing uua replacement backwards.

This process is realized by means of the use of such basic operations like

the sum, subtraction. multiplication and division; the multiplication of a line is

very frequent for a constant, and the sum of this one with another line, since

this way it is possible to come up to obtaining an equation where the

coefficient of a variable is 0.

To the line that is selected to realize this process there is called he a linepivot


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METHODOF GAUS-JORDANThe Gauss-Jordan method consists of the consecutive elimination of the

unknowns with the intention of coming to a staggered system. To carry out

the above mentioned transformation one resorts to the elementary

transformations, considering them on a matriz that it should represent to the

equations system only across his coefficients. This matriz is known as amatriz of the System; if also this one contains the independent terms itnames him by Developing matriz of the system.


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To illustrate the central idea of the method, let's consider the problem of

solving the following equations system:

Which is represented by the following arrangement matricial:


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1er step will be to obtain a line 'pivot', which first element is 1, normalizing a

line, in this case the 2nd one.

On having begun the gradation, we exchange the reglones 1 y2

Now, for comenzara to stagger, we will multiply 1 er line for 3


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Now we will add 1 er line to the second one, and

we will replace the last one with the result. Later, we normalize the line again

1.

Just like earlier, we will multiply the first line, for the first element of 3 er linechanged sign, in this case by 2.

once again we will add lines, now 1st and el3ro, replacing the last one withthe result and later we normalize again the pivot.


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Once the first column has filled zeros, we change pivot to the immediate line

down and fill with zeros the second column, down.

andste algorithm recurs up to turning to: to matril in a top diagonal matriz,

with alone zeros below his diagonal reign!.

This first part of the method is known as an Elimination Gaussiana. Since it is

appreciated, the system has been transformed into the equivalent one which

equations are 1,2 and 3 variables. From aqu \, with: to elimination Gaussiana

the values of the unknowns are obtained to turn of one replacementbackwards, how does it appear:


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A more efficient method to obtain the values of the unknowns is of

completing the diagonalizacin of the matriz. This part is known like

elimination of Jordan and as a whole with the elimination Gaussiana the

Gauss-Jordan method agrees. This continuation consists of the

escalonacin of the matriz, now up. Continuing with the example we have

that:

We take the first reglon as as a pivot so we will normalize it

it saves we repeat the previous steps but up We Multiply the pivot by the last

element of the 2nd line of the matriz of the system, with the changed sign

"To begin to do zero the third column, we will add the lines 3 and 2, replacingthe last one with the result and re-normalizing the pivot.


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Ending with the third column, I will multiply the pivot for 2, we it add to 1er

line, change this one with the result and re-normalize the pivot.

now we use 2do line as pivot, in case it is already not necessary to normalize

it. To do zeros the 2nd column we continue the previous steps.

Once the matriz stays diagonalizada, the solution of the system appears in

the Developing matriz of the System


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Example

To solve the following equations system

The system will be solved I happen or happen:


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It is necessary to write the system in form matricial.

One thinks about how to simplify the equation, for which the second line will

split between 3.7 (the highest value).


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The 2nd line joins the 1st line.

Now the 2nd line will multiply for-0.3.

The 2nd line will join the 3rd line.


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The Io will divide line between 3.759 v the 2nd line will split between-0.3

The 1st line multiplies for 0.538

The 1st line joins the 3rd line


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The Io multiplies line by 0.541 and one adds him to the 2nd line

They split it line between 0.541 and 30 line between-0.578

The 3rd line multiplies for-0.861 and one adds him to the 1st line


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The 3rd line splits between-0.861 and immediately later it turns to multiply

the same line now by-0.601 and joins the 2nd line.

The 3rd line splits between-0.601.

The matriz re-makes itself comfortable.


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