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GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION

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.

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GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION

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  1. GAUSS ELIMINATION AND GAUSS-JORDAN ELIMINATION

  2. 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.

  3. 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

  4. 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.

  5. 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

  6. 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

  7. 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.

  8. Example 1: Gauss Elimination Let us consider the set of linearly independent equations. Augmented matrix for the set is:

  9. 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.

  10. Example 1: Gauss Elimination Step 1: Eliminate x from the 2nd and 3rd equation.

  11. 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:

  12. 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

  13. Example 2: Gauss Elimination Let us consider another set of linearly independent equations. The augmented matrix for this set is:

  14. Example 2: Gauss Elimination Step 1: Eliminate x from the 2nd and 3rd equation.

  15. Example 2: Gauss Elimination Step 2: Eliminate y from the 3rd equation. From Row 3, therefore, z = ? From Row 2, ? From Row 1, ?

  16. Example 3: Gauss Elimination When would you interchange two equations (rows)? Let us consider the following set of equations. The corresponding augmented matrix is:

  17. 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

  18. 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…

  19. 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.

  20. 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

  21. 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

  22. Example 4: Gauss Elimination • Apply the Gauss Elimination: Z = 10, X = 24, and Y = 20

  23. 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)

  24. Gauss-Jordan Elimination Let us consider the set of linearly independent equations. Augmented matrix for the set is:

  25. Gauss-Jordan Elimination Step 1: Eliminate x from the 2nd and 3rd equation.

  26. 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:

  27. Gauss-Jordan Elimination Step 4: Eliminate z from the 2nd equation

  28. Gauss-Jordan Elimination Step 5-1: Eliminate y from the 1st equation

  29. Gauss-Jordan Elimination Step 5-2: Eliminate z from the 1st equation

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