Gauss elimination and gauss jordan elimination
Download
1 / 29

GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION - PowerPoint PPT Presentation


  • 901 Views
  • Updated On :

GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION. An m  n Matrix. If m and n are positive integers, then an m  n matrix is a rectangular array in which each entry a ij of the matrix is a number. The matrix has m rows and n columns. Terminology.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION' - Mia_John


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

An m n matrix l.jpg
An m  n Matrix

  • If m and n are positive integers, then an m  n matrixis a rectangular array in which each entryaij of the matrix is a number. The matrix has m rows and n columns.


Terminology l.jpg
Terminology

  • A real matrix is a matrix all of whose entries are real numbers.

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

  • An mn matrix is said to be of size (or dimension) mn.

  • If m=n the matrix is square of order n.

  • If m=n , then the ai,i’s are the diagonal entries


Augmented matrix for a system of equations l.jpg
Augmented Matrix for a System of Equations

  • Given a system of equations we can talk about its coefficient 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.


Row echelon form l.jpg
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


Gauss elimination l.jpg
Gauss Elimination

  • The operations in the Gauss elimination are called elementary 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


Gauss elimination for solving a system of equations l.jpg
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.


Example 1 gauss elimination l.jpg
Example 1: Gauss Elimination

Let us consider the set of linearly independent equations.

Augmented matrix for the set is:


Example 1 gauss elimination9 l.jpg
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.


Example 1 gauss elimination10 l.jpg
Example 1: Gauss Elimination

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


Example 1 gauss elimination11 l.jpg

13R’2+R’3

R’’3

0.5R’1

R’1

-R’2

R’’2

(1/168)R’’3

R’’’3

Example 1: Gauss Elimination

Step 2: Eliminate y from the 3rd equation.

Step 3:


Example 1 gauss elimination12 l.jpg
Example 1: Gauss Elimination

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


Example 2 gauss elimination l.jpg
Example 2: Gauss Elimination

Let us consider another set of linearly independent equations.

The augmented matrix for this set is:


Example 2 gauss elimination14 l.jpg
Example 2: Gauss Elimination

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


Example 2 gauss elimination15 l.jpg
Example 2: Gauss Elimination

Step 2: Eliminate y from the 3rd equation.

From Row 3, therefore, z = ?

From Row 2, ? From Row 1, ?


Example 3 gauss elimination l.jpg
Example 3: Gauss Elimination

When would you interchange two equations (rows)?

Let us consider the following set of equations.

The corresponding augmented matrix is:


Example 3 gauss elimination17 l.jpg
Example 3: Gauss Elimination

  • 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.

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


Example 4 gauss elimination l.jpg
Example 4: Gauss Elimination

  • 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 of Wild Thing seed mix contains 500 grams of wild flower seed, 250 grams of Echinacea seed and 250 grams of Chrysanthemum seed.

2) Mommy Dearest mix 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 Chest mix has gained a lot of attention lately, with the interest 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…


Example 4 gauss elimination19 l.jpg
Example 4: Gauss Elimination

Cont’d

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 on hand.

Use matrices and complete Gauss-Jordan Eliminationto determine how much of each mixture the store can prepare.


Example 4 gauss elimination20 l.jpg
Example 4: Gauss Elimination

  • 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 of Wild Thing

Let Y = Amount of Mommy Dearest

Let Z = Amount of Medicine Chest

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

Echinacea 0.25X + 0Y + 0.9Z = 15g

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

In matrix form, this can be written as


Example 4 gauss elimination21 l.jpg
Example 4: Gauss Elimination

Before the matrices are populated, it is (sometimes) helpful to re-arrange the equations to reduce the number of steps in the Gauss Elimination. 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 = 15g

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

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


Example 4 gauss elimination22 l.jpg
Example 4: Gauss Elimination

  • Apply the Gauss Elimination:

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


Gauss jordan elimination l.jpg
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)


Gauss jordan elimination24 l.jpg
Gauss-Jordan Elimination

Let us consider the set of linearly independent equations.

Augmented matrix for the set is:


Gauss jordan elimination25 l.jpg
Gauss-Jordan Elimination

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


Gauss jordan elimination26 l.jpg

13R’2+R’3

R’’3

0.5R’1

R’1

-R’2

R’’2

(1/168)R’’3

R’’’3

Gauss-Jordan Elimination

Step 2: Eliminate y from the 3rd equation.

Step 3:


Gauss jordan elimination27 l.jpg
Gauss-Jordan Elimination

Step 4: Eliminate z from the 2nd equation


Gauss jordan elimination28 l.jpg
Gauss-Jordan Elimination

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


Gauss jordan elimination29 l.jpg
Gauss-Jordan Elimination

Step 5-2: Eliminate z from the 1st equation


ad