Explanation Strategy for finding the inverse of a matrix: If a sequence of a row operations performed on the n × n matrix A reduces A to the n × n identity matrix I , then those same row operations, performed in the same order on I , will transform I into A − 1 . If no sequence of row operations will reduce A to I , then A is not invertible. A notation for this process uses an n -row-by- 2 n -column matrix, with matrix A as the left half and matrix I as the right half. If A is invertible, the proper row operations on [ A | I ] will transfer it into [ I | A − 1 ] . Given: A clothing manufacturer makes coats, shirts, and slacks. The times required for cutting, sewing, and packaging each time shown in the table. Clothing manufacturer makes: Coats Shirts Slacks Cutting 20 min 15 min 10 min Sewing 60 min 30 min 24 min Packaging 5 min 12 min 6 min The times required: Time Available Cutting 115 hr Sewing 280 hr Packaging 65 hr Calculation: We can let x , y , and z represent the coats, shirts, and slacks to be manufactured respectively. We can set up the following system of equations: 20 min is needed for every coats manufacture 15 min is needed for every shirts manufacture, 20 min is needed for every slacks manufacture. 20 x + 15 y + 10 z = 115 60 min is needed for every coats manufacture 30 min is needed for every shirts manufacture, 24 min is needed for every slacks manufacture. 60 x + 30 y + 24 z = 280 5 min is needed for every coats manufacture 12 min is needed for every shirts manufacture, 6 min is needed for every slacks manufacture. 5 x + 12 y + 6 z = 65 A system of equation can always be written as a matrix equation A X = B , where A is the coefficient matrix of the system, X is a column matrix of variables, and B is the column matrix of constants. In matrix form, the system becomes, [ 20 15 10 60 30 24 5 12 6 ] [ x y z ] = [ 115 280 65 ] If matrix A is invertible, the matrix equation A X = B is easy to solve, as we see in the next step. First we find the inverse of matrix A . Let A = [ 20 15 10 60 30 24 5 12 6 ] We can set up a 3 × 6 matrix with A on the left and I on the right of the line, [ A | I ] = [ 20 15 10 60 30 24 5 12 6 | 1 0 0 0 1 0 0 0 1 ] We perform row operations on the entire matrix to transform the left half into I . [ A | I ] = [ 20 15 10 60 30 24 5 12 6 | 1 0 0 0 1 0 0 0 1 ] R 1 ↔ R 2 [ 60 30 24 20 15 10 5 12 6 | 0 1 0 1 0 0 0 0 1 ] R 2 → R 2 − 1 3 ⋅ R 1 [ 60 30 24 0 5 2 5 12 6 | 0 1 0 1 − 1 3 0 0 0 1 ] R 3 → R 3 − 1 12 ⋅ R 1 [ 60 30 24 0 5 2 0 19 2 4 | 0 1 0 1 − 1 3 0 0 − 1 12 1 ] R 2 ↔ R 3 [ 60 30 24 0 19 2 4 0 5 2 | 0 1 0 0 − 1 12 1 1 − 1 3 0 ] R 3 → R 3 − 10 19 ⋅ R 2 [ 60 30 24 0 19 2 4 0 0 − 2 19 | 0 1 0 0 − 1 12 1 1 − 11 38 − 10 19 ] R 3 → − 19 2 ⋅ R 3 [ 60 30 24 0 19 2 4 0 0 1 | 0 1 0 0 − 1 12 1 − 19 2 11 4 5 ]

Bundle: Webassign Printed Access Card For Gustafson/hughes' College Algebra, Single-term + Student Solutions Manual For Gustafson/hughes? College Algebra, 12th

12th Edition

ISBN: 9781337605304

Author: R. David Gustafson; Jeff Hughes

Publisher: Cengage Learning

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Chapter 6.4, Problem 38E

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How many each should be made to use all available labor hours.

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Chapter 6.4, Problem 37E

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Chapter 6 Solutions

Bundle: Webassign Printed Access Card For Gustafson/hughes' College Algebra, Single-term + Student Solutions Manual For Gustafson/hughes? College Algebra, 12th

Ch. 6.1 - Self Check Use the graphing method to solve...Ch. 6.1 - Prob. 2SC Ch. 6.1 - Prob. 3SC Ch. 6.1 - Prob. 4SC Ch. 6.1 - Self Check 5 Use the elimination method to solve...Ch. 6.1 - Prob. 6SC Ch. 6.1 - Prob. 7SC Ch. 6.1 - Prob. 8SC Ch. 6.1 - Fill in the blanks. Any set of several equations...Ch. 6.1 - Prob. 2E

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How many each should be made to use all available labor hours.

Solution Summary: The author explains how a clothing manufacturer makes coats, shirts, and slacks using all available labor hours.

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To find: How many each should be made to use all available labor hours.

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Chapter 6 Solutions

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How many each should be made to use all available labor hours.