CHEMICAL ENGINEERING

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(2140505)GAUSS ELIMINATION AND GAUSS JORDAN METHOD

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GAUSS ELIMINATION METHOD

In linear algebra, Gaussian elimination (also known

as row reduction) is an algorithm for solving systems

of linear equations.

It is usually understood as a sequence of operations

performed on the associated matrix of coefficients.

This method can also be used to find the rank of a

matrix, to calculate the determinant of a matrix, and to

calculate the inverse of an invertible square matrix.

The method is named after Carl Friedric Gauss(1777–

1855), although it was known to Chinese

mathematicians as early as 179 AD.

To perform row reduction on a matrix, one uses a

sequence of elementary row operations to modify the

matrix until the lower left-hand corner of the matrix is

filled with zeros, as much as possible. There are three

types of elementary row operations:

1) Swapping two rows

2) Multiplying a row by a non-zero number

3) Adding a multiple of one row to another row.

ALGORITHM OF GAUSS ELIMINATION METHOD

Start

Declare the variables and read the order of the matrix n.

Take the coefficients of the linear equation as:

Do for k=1 to n

Do for j=1 to n+1

Read a[k][j] End for j

End for k

Do for k=1 to n-1

Do for i=k+1 to n

Do for j=k+1 to n+1

a[i][j] = a[i][j] – a[i][k] /a[k][k] * a[k][j] End for j

End for i

End for k

Compute x[n] = a[n][n+1]/a[n][n]

Do for k=n-1 to 1

sum = 0

Do for j=k+1 to n

sum = sum + a[k][j] * x[j] End for j

x[k] = 1/a[k][k] * (a[k][n+1] – sum)

End for k

Display the result x[k]

Stop

AN EXAMPLE OF GAUSS ELIMINATION METHOD OF

CHEMICAL ENGINEERING APPLICATIONS

Example 1

A liquid-liquid extraction process conducted in the Electrochemical Materials Laboratory involved the extraction of nickel from the aqueous phase into an organic phase. A typical set of experimental data from the laboratory is given below.

Ni aqueous phase, 2 , 2.5, 3

Ni organic phase, 8.57 , 10 , 12

Assuming g is the amount of Ni in the organic phase and a is the amount of Ni in the aqueous phase, the quadratic interpolant that estimates is given by g= x1a

2

+ x2a + x3 , where 2 < a < 3

Find the values of both the using naïve Gauss elimination.

The solution of x1, x2, x3 is given by

Forward Elimination of Unknowns

Since there are three equations, there will be two steps of

forward elimination of unknowns.

First step

Divide Row 1 by 4 and then multiply it by 6.25, that is,

multiply Row 1 by .

12

10

57.8

139

15.225.6

124

3

2

1

x

x

x

Subtract the result from Row 2 to get

391.135625.1125.325.65625.11 Row

12

3906.3

57.8

139

5625.0625.00

124

3

2

1

x

x

x

Second step

We now divide Row 2 by −0.625 and then multiply it by

−1.5, that is, multiply Row 2 by .

Subtract the result from Row 3 to get

1375.835.15.104.22 Row

855.0

3906.3

57.8

1.000

5625.0625.00

124

3

2

1

x

x

x

Back Substitution

From the third equation,

Substituting the value of in the second equation,

855.01.0 3 x

1.0

0.8553 x

55.8

3906.35625.0625.0 32 xx

625.0

5625.03906.3 3

2

xx

625.0

55.85625.03906.3

27.2

Hence the solution vector is

Hence, g= 1.14a2 - 2.27a +8.55, where g is the amount

of nickel in organic phase and a is the amount of nickel

in aqueous phase.

Thus, if the value of g is given by 2.3g/l then the value

of a will be , g= 1.14(2.3)2 - 2.27(2.3) + 8.55

= 9.3596 g/l

55.8

27.2

14.1

3

2

1

x

x

x

GAUSS JORDAN METHOD

Some authors use the term Gaussian elimination to refer only to the procedure until the matrix is in echelon form, and use the term Gauss-Jordan elimination to refer to the procedure which ends in reduced echelon form.

In linear algebra, Gauss–Jordan elimination is an algorithm for getting matrices in reduced row echelon form using elementary row operations. It

is a variation of Gaussian elimination.

HISTORY ABOUT GAUSS JORDAN

METHOD

it is a variation of Gaussian elimination as

described by Wilhelm Jordan in 1887.

However, the method also appears in an article

by Clasen published in the same year. Jordan

and Clasen probably discovered Gauss–Jordan

elimination independently.

HERE ARE THE STEPS TO GAUSS-JORDAN

ELIMINATION:

Turn the equations into an augmented matrix.

Use elementary row operations on matrix [A|b] to

transform A into diagonal form. Make sure there are

no zeros in the diagonal.

Divide the diagonal element and the right-hand

element (of b) for that diagonal element's row so

that the diagonal element is equal to one.

HERE ARE THE STEPS TO GAUSS-JORDAN

ELIMINATION:

1. Write the augmented matrix of the system.

2. Use row operations to transform the augmented

matrix in the form described below, which is called

the reduced row echelon form (RREF).

(a) The rows (if any) consisting entirely of zeros are

grouped together at the bottom of the matrix.

(b) In each row that does not consist entirely of

zeros, the leftmost nonzero element is a 1 (called a

leading 1 or a pivot).

(c) Each column that contains a leading 1 has

zeros in all other entries.

(d) The leading 1 in any row is to the left of any

leading 1’s in the rows below it.

o 3. Stop process in step 2 if you obtain a row

whose elements are all zeros except the last one

on the right. In that case, the system is

inconsistent and has no solutions. Otherwise,

finish step 2 and read the solutions of the system

from the final matrix.

ALGORITHM OF GAUSS JORDAN

METHOD

Start

Read the order of matrix ’n’ and read the coefficient of the linear eq.

Do for k=1 to n.

Do for l=k+1 to n+1

a[k][l]=a[k][l]/a[k][k] end for l set a[k][k]=1

Do for i=1 to n.

If (i is not equal to k) then ,

do for j=k+1 to n+1

a[i][j]=a[i][j]-(a[k][j]*a[i][k])

End for j

End for i

End for k.

Do for m=1 to n.

x[m]=a[m][n+1],display x[m],[n] for m

Stop.

Example 1. Solve the following system by

using the Gauss-Jordan elimination method.

x + y + z = 5

2x + 3y + 5z = 8

4x + 5z = 2

Solution: The augmented matrix of the system is the following.

1 1 1|52 3 5|84 0 5|2

We will now perform row operations until we obtain a matrix in reduced row echelon form.

1 1 1|52 3 5|84 0 5|2

1 1 1| 50 1 3| − 24 0 5| 2

1 1 1| 50 1 3| − 20 −4 1| − 18

R3-4R1

1 1 1| 50 1 3| − 20 0 13| − 26

R3+4R2

(1/13)R31 1 1| 50 1 3| − 20 0 1| − 2

R3+4R2 1 1 1| 50 1 0| 40 0 1| − 2

R1-R3

1 1 0| 50 1 0| 40 0 1| − 2

1 0 0| 30 1 0| 40 0 1| − 2

From this final matrix, we can read the solution of the system. It is

R1-R2