GAUSS ELIMINATION AND GAUSSJORDAN ELIMINATION

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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 ? n
  matrix is a rectangular array in which each entry
  aij of the matrix is a number. The matrix has m
  rows and n columns.

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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 mn the matrix is square of order n.
• If mn , 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 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.

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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
  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

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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
Let us consider the set of linearly independent
equations.
Augmented matrix for the set is
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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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Example 1 Gauss Elimination
Step 1 Eliminate x from the 2nd and 3rd
equation.
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Example 1 Gauss Elimination
Step 2 Eliminate y from the 3rd equation.
Step 3
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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
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Example 2 Gauss Elimination
Let us consider another set of linearly
independent equations.
The augmented matrix for this set is
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Example 2 Gauss Elimination
Step 1 Eliminate x from the 2nd and 3rd equation.
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Example 2 Gauss Elimination
Step 2 Eliminate y from the 3rd equation.
From Row 3, therefore, z ? From Row 2, ?
From Row 1, ?
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Example 3 Gauss Elimination
When would you interchange two equations (rows)?
Let us consider the following set of equations.
The corresponding augmented matrix is
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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
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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
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Example 4 Gauss Elimination
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 on hand. Use matrices and complete
Gauss-Jordan Elimination to determine how much of
each mixture the store can prepare.
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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
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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
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Example 4 Gauss Elimination

• Apply the Gauss Elimination

Z 10, X 24, and Y 20
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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)

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Gauss-Jordan Elimination
Let us consider the set of linearly independent
equations.
Augmented matrix for the set is
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Gauss-Jordan Elimination
Step 1 Eliminate x from the 2nd and 3rd
equation.
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Gauss-Jordan Elimination
Step 2 Eliminate y from the 3rd equation.
Step 3
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Gauss-Jordan Elimination
Step 4 Eliminate z from the 2nd equation
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Gauss-Jordan Elimination
Step 5-1 Eliminate y from the 1st equation
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Gauss-Jordan Elimination
Step 5-2 Eliminate z from the 1st equation