gaussian gaussianjordan

7/30/2019 Gaussian GaussianJordan

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GAUSS ELIMINATION AND

GAUSS-JORDAN ELIMINATION


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An m n Matrix

If m and n are positive integers, then an

m nmatrix

is arectangular array in which each entryaij of the matrix is a

number. The matrix has mrows and ncolumns.

nmmmm

n

n

n

aaaa

aaaa

aaaa

aaaa

,3,2,1,

,33,32,31,3

,23,22,21,2

,13,12,11,1


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Terminology

A real matrix is a matrix all of whose entries arereal numbers.

i (j) is called the row (column) subscript.

An mn matrix is said to be ofsize (ordimension)

mn.

Ifm=n the matrix is square oforder n.

Ifm=n , then the ai,is are the diagonal entries


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Augmented Matrix for a System of Equations

Given a system of equations we can talk about itscoefficient matrix and its augmented matrix.

To solve the system we can now use row operations

instead of equation operations to put the augmented

matrix in row echelon form.


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Row-Echelon Form

A matrix is in row-echelon form if:

The lower left quadrant of the matrix has all zero entries.

In each row that is not all zeros the first entry is a 1.

The diagonal elements of the coefficient matrix are all 1


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Gauss Elimination

The operations in the Gauss elimination are calledelementary operations.

Elementary operations for rows are:

Interchange of two rows.

Multiplication of a row by a nonzero constant.

Addition of a constant multiple of one row to another

row.

Two matrices are said to be row equivalent if one

matrix can be obtained from the other using

elementary row operations


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Gauss Elimination for Solving A

System of Equations

1. Write the augmented matrix of the system.

2. Use elementary row operations to construct a row

equivalent matrix in row-echelon form.

3. Write the system of equations corresponding to the

matrix in row-echelon form.

4. Use back-substitution to find the solutions to this

system.


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Example 1: Gauss Elimination

Augmented matrix: For Gauss Elimination, the Augmented Matrix (A) is

used so that both A and b can be manipulated together.


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31835

7753

36542

31835

7753

36542

zyx

zyx

zyx

Step 1: Eliminate x from the 2nd and 3rd equation.

1215.2013

615.14

36542

1215.20130

615.1410

36542

zy

zy

zyx


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67216800

615.1410

36542

672168

615.14

36542

z

zy

zyx

4100

615.1410

185.221

Step 2: Eliminate y from the 3rd equation.

13R2+R3 R3

Step 3:

0.5R1 R1

-R2 R2

(1/168)R3 R3

4

615.14

185.22

z

zy

zyx


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From Row 3, z = 4

From Row 2, y -14.5z = -61 or, y - 14.5 (4) = 61 or, y = - 3

From Row 1, x 2y + 2.5z = 18 or, x 2 (- 3) + 2.55 (4) = 18

or, x = 2

4100

615.1410

185.221

4

615.14

185.22

z

zy

zyx


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Let us consider another set of linearly independent equations.

The augmented matrix for this set is:


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From Row 3, therefore, z = ?

From Row 2, ? From Row 1, ?



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When would yo u interchange two equat ions (rows )?

Let us consider the following set of equations.

The corresponding augmented matrix is:


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The solution is: x = - 3, y = 4, z = 2

Eqn. (1) (Row 1) cannot be used to eliminate x from Eqns. (2)

and (3) (Rows 2 and 3).

Interchange Row 1 with Row 2. The augmented matrix

becomes:

Now follow the steps mentioned earlier to solve for the

unknowns.


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Problem:

A garden supply centre buys flower seed in bulk then mixes and packages

the seeds for home garden use. The supply center provides 3 different

mixes of flower seeds: Wild Thing, Mommy Dearest and Medicine

Chest.

1) One kilogram ofWild Thingseed mix contains 500 grams of wild flowerseed, 250 grams of Echinacea seed and 250 grams of Chrysanthemum

seed.

2) Mommy Dearestmix is a product that is commonly purchased through

the gift store and consists of 75% Chrysanthemum seed and 25% wild

flower seed.

3) The Medicine Chestmix has gained a lot of attention lately, with theinterest in medicinal plants, and contains only Echinacea seed, but the mix

must include some vermiculite (10% by weight of the total mixture) for ease

of planting.

To be continued


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Contd

In a single order, the store received 17 grams of wild flower seed, 15

grams of Echinacea seed and 21 grams of Chrysanthemum seed.

Assume that the garden center has an ample supply of vermiculite onhand.

Use matrices and complete Gauss-Jordan Elimination to determine

how much of each mixture the store can prepare.


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Solution:

Assign variables to the

amount of each mix that will

be produced.

Perform a balance on each

of the components that are

available.

Let X = Amount ofWild Thing

Let Y = Amount ofMommy Dearest

Let Z = Amount ofMedicine Chest

Wild flower 0.5X + 0.25Y + 0Z = 17g

Echinacea 0.25X + 0Y + 0.9Z = 15gChrysanthemum 0.25X + 0.75Y + 0Z = 21g

In matrix form, this can be written as bAx


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Before the matrices are populated, it is (sometimes) helpful to re-

arrange the equations to reduce the number of steps in the GaussElimination. To do this (if there seems like an easy solution), attempt

to move zeros to the bottom left, and try to maintain the first row with

non-zeros except for the last entry, since row 1 is used to reduce

other rows.

By moving the last column (Z) to the front, and switching the first and

second row, the new set of equations becomes:

Echinacea 0.9Z + 0.25X + 0Y = 15gWild flower 0Z + 0.5X + 0.25Y = 17g

Chrysanthemum 0Z + 0.25X + 0.75Y = 21g


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Apply the Gauss

Elimination:

Z = 10, X = 24, and Y = 20

214

3

4

10

174

1

2

1

0

1504

1

10

9


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Gauss-Jordan Elimination

In Gauss-Jordan elimination, we continue the reduction of

the augmented matrix until we get a row equivalent matrix

in reduced row-echelon form. (r-e form where every

column with a leading 1 has rest zeros)

cb

a

100010

001


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31835

7753

36542

zyx

zyx

zyx

Let us consider the set of linearly independent equations.

Augmented matrix for the set is:

31835

7753

36542


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31835

7753

36542

31835

7753

36542

zyx

zyx

zyx

Step 1: Eliminate x from the 2nd and 3rd equation.

1215.2013

615.14

36542

1215.20130

615.1410

36542

zy

zy

zyx


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67216800

615.1410

36542

672168

615.14

36542

z

zy

zyx

4100

615.1410

185.221


13R2+R3 R3

Step 3:

0.5R1 R1

-R2 R2

(1/168)R3 R3

4

615.14

185.22

z

zy

zyx


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)2()5.14()3( RowRow

4100

615.1410

185.221

4

615.14

185.22

z

zy

zyx

Step 4: Eliminate z from the 2nd

equation

4100

3010185.221

4

3185.22

z

yzyx


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1)1()2()2( RowNewRowRow

4100

3010

185.221

4

3

185.22

z

y

zyx

Step 5-1: Eliminate y from the 1st equation

4100

3010125.201

4

3125.2

z

yzx


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