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GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATIONPowerPoint Presentation

GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION

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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.

GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION

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GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION

- 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.

- A real matrix is a matrix all of whose entries are real numbers.
- i (j) is called the row (column) subscript.
- An mn matrix is said to be of size (or dimension) mn.
- If m=n the matrix is square of order n.
- If m=n , then the ai,i’s are the diagonal entries

- 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.

- 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

- 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

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.

Let us consider the set of linearly independent equations.

Augmented matrix for the set is:

- Augmented matrix:
- For Gauss Elimination, the Augmented Matrix (A) is used so that both A and b can be manipulated together.

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

13R’2+R’3

R’’3

0.5R’1

R’1

-R’2

R’’2

(1/168)R’’3

R’’’3

Step 2: Eliminate y from the 3rd equation.

Step 3:

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

Let us consider another set of linearly independent equations.

The augmented matrix for this set is:

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

Step 2: Eliminate y from the 3rd equation.

From Row 3, therefore, z = ?

From Row 2, ? From Row 1, ?

When would you interchange two equations (rows)?

Let us consider the following set of equations.

The corresponding augmented matrix is:

- 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

- 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…

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.

- 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 flower0.5X + 0.25Y + 0Z = 17g

Echinacea0.25X + 0Y + 0.9Z = 15g

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

In matrix form, this can be written as

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:

Echinacea0.9Z + 0.25X + 0Y = 15g

Wild flower0Z + 0.5X + 0.25Y = 17g

Chrysanthemum0Z + 0.25X + 0.75Y = 21g

- Apply the Gauss Elimination:

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

- 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)

Let us consider the set of linearly independent equations.

Augmented matrix for the set is:

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

13R’2+R’3

R’’3

0.5R’1

R’1

-R’2

R’’2

(1/168)R’’3

R’’’3

Step 2: Eliminate y from the 3rd equation.

Step 3:

Step 4: Eliminate z from the 2nd equation

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

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