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GAUSS AND GAUSS-JORDAN ELIMINATION

JONATHAN ORLANDO CELIS ARIASCOD: 2061226

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CONTENT

Gauss Elimination

Gauss Elimination example

Gauss-Jordan Elimination

Gauss-Jordan Elimination example

Problems with Gauss-Jordan elimination

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GAUSS ELIMINATION We will now explore a more versatile way than

the method of determinants to determine if a

system of equations has a solution.

We will indeed be able to use the results of this

method to find the actual solution(s) of the

system (if any).

It should be noted that this method can be

applied to systems of equations with an

unequal number of equations and unknowns.

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GAUSS ELIMINATION EXAMPLE

• Consider the system of equations:

• To solve for x, y, and z we must eliminate some of the unknowns from some of the equations. Consider adding -2 times the first equation to the second equation and also adding 6 times the first equation to the third equation. The result is:

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GAUSS ELIMINATION EXAMPLE

• We have now eliminated the x term from the last two equations. Now simplify the last two equations by dividing by 2 and 3, respectively:

• To eliminate the y term in the last equation, multiply the second equation by -5 and add it to the third equation:

  

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GAUSS ELIMINATION EXAMPLE

• The third equation says z=-2.

• Substituting this into the second equation yields y=-1.

• Using both of these results in the first equation gives x=3.

• The process of progressively solving for the unknowns is called back-substitution, this is the essence of Gaussian elimination.

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GAUSS-JORDAN ELIMINATION• This is a variation of Gaussian elimination.• 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:

1. If a row does not consist entirely of zeros, then the first non zero number in the row is a 1.(the leading 1)

2. If there are any rows entirely made up of zeros, then they are grouped at the bottom of the matrix.

3. 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.

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GAUSS-JORDAN ELIMINATION EXAMPLE

• Consider the system of equations:

R2 - (-1)R1 --> R2 R3 - ( 3)R1 --> R3

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

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GAUSS-JORDAN ELIMINATION EXAMPLE

(-1/52)R3 --> R3

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

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GAUSS-JORDAN ELIMINATION EXAMPLE

• 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

R1 - (1)R2 --> R1

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PROBLEMS WITH GAUSS-JORDAN ELIMINATION

• There are some problems that could arise while searching for these

solutions.

1. If the lines are parallel then they will not intersect and thus provide no

solution.

2. 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.

3. 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.

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