Explanation Given: Input-output matrix, A = [ .25 .08 .33 .11 ] and the demand matrix, D = [ 690 920 ] . Here the first row represents wheat and the second row represented oil. Formula used Production matrix, X X = ( I − A ) − 1 D A is Input-output matrix and D is demand matrix Calculation: First evaluate ( I − A ) as follows: I − A = [ 1 0 0 1 ] − [ .25 .08 .33 .11 ] = [ 0.75 − 0.08 − 0.33 0.89 ] The inverse of the matrix ( I − A ) is found by Gauss Jordan method as follows: Augmented matrix is, [ 0.75 − 0.08 1 0 − 0.33 0.89 0 1 ] Proceed with the following row operations: R 1 → 1 0.75 R 1 gives, [ 1 − 8 75 1.33 0 − 0.33 0.89 0 1 ] R 2 → R 2 + 0

Mathematics With Applications, Loose-leaf Edition Plus Mylab Math With Pearson Etext -- 18-week Access Card Package (12th Edition)

12th Edition

ISBN: 9780135998304

Author: Margaret L. Lial, Tom Hungerford, John P. Holcomb, Bernadette Mullins

Publisher: PEARSON

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Chapter 6.6, Problem 25E

To determine

To calculate: The amount of each quantity to be produced.

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Chapter 6.6, Problem 24E

Chapter 6.6, Problem 26E

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Chapter 6 Solutions

Mathematics With Applications, Loose-leaf Edition Plus Mylab Math With Pearson Etext -- 18-week Access Card Package (12th Edition)

Ch. 6.1 - Checkpoint1 Use the substitution method tosolve...Ch. 6.1 - Checkpoint 2 Use the elimination method to solve...Ch. 6.1 - Checkpoint 3 Solve the system of equations . Draw...Ch. 6.1 - Checkpoint 4 Solve the following system: Ch. 6.1 - Checkpoint 5 Solve the system Draw the graph of...Ch. 6.1 - Prob. 6CP Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Use substitution to solve each system. (See...Ch. 6.1 - Use substitution to solve each system. (See...

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