Explanation Given: The table shows the sales (in billions of dollars) for Wal-Mart stores from 2006 through 2013. Year 2006 2007 2008 2009 Sales (in billions) 348.7 378.8 405.6 408.2 Year 2010 2011 2012 2013 Sales (in billions) 421.8 447.0 469.2 476.2 Calculation: Consider the points ( 2006 , 348.7 ) , ( 2007 , 378.8 ) , ( 2008 , 405.6 ) , ( 2009 , 408.2 ) and ( 2010 , 421.8 ) . As the x -values are large, so use the translation z = x − 2000 , Then the points becomes, ( 6 , 348.7 ) , ( 7 , 378.8 ) , ( 8 , 405.6 ) , ( 9 , 408.2 ) and ( 10 , 421.8 ) The quadratic polynomial function be, p ( z ) = a 0 + a 1 z + a 2 z 2 + a 3 z 3 Substituting z = 6 , 7 , 8 , 9 and 10 into p ( z ) and equating the results to the respective y -values produces the system of linear equations in the variables a 0 , a 1 , a 2 , a 3 and a 4 as shown below. p ( 6 ) = a 0 + a 1 ( 6 ) + a 2 ( 6 ) 2 + a 3 ( 6 ) 3 + a 4 ( 6 ) 4 = a 0 + 6 a 1 + 36 a 2 + 216 a 3 + 1296 a 4 = 348.7 p ( 7 ) = a 0 + a 1 ( 7 ) + a 2 ( 7 ) 2 + a 3 ( 7 ) 3 + a 4 ( 7 ) 4 = a 0 + 7 a 1 + 49 a 2 + 343 a 3 + 2401 a 4 = 378.8 p ( 8 ) = a 0 + a 1 ( 8 ) + a 2 ( 8 ) 2 + a 3 ( 8 ) 3 + a 4 ( 8 ) 4 = a 0 + 8 a 1 + 64 a 2 + 512 a 3 + 4096 a 4 = 405.6 p ( 9 ) = a 0 + a 1 ( 9 ) + a 2 ( 9 ) 2 + a 3 ( 9 ) 3 + a 4 ( 9 ) 4 = a 0 + 9 a 1 + 81 a 2 + 729 a 3 + 6561 a 4 = 408.2 p ( 10 ) = a 0 + a 1 ( 10 ) + a 2 ( 10 ) 2 + a 3 ( 10 ) 3 + a 4 ( 10 ) 4 = a 0 + 10 a 1 + 100 a 2 + 1000 a 3 + 10000 a 4 = 421.8 The augmented matrix is, [ 1 6 36 216 1296 348.7 1 7 49 343 2401 378.8 1 8 64 512 4096 405.6 1 9 81 729 6561 408.2 1 10 100 1000 10000 421.8 ] [ 1 6 36 216 1296 3487 10 1 7 49 343 2401 1894 5 1 8 64 512 4096 2028 5 1 9 81 729 6561 2041 5 1 10 100 1000 10000 2109 5 ] Solving the system by using Gauss elimination with back substitution: − R 1 + R 2 → R 2 , − R 1 + R 3 → R 3 , − R 1 + R 4 → R 4 , − R 1 + R 5 → R 5 [ 1 6 36 216 1296 3487 10 0 1 13 127 1105 301 10 0 2 28 296 2800 569 10 0 3 45 513 5265 119 2 0 4 64 786 8704 731 10 ] R 1 − 6 R 2 → R 1 , R 3 − 2 R 2 → R 3 , R 4 − 3 R 2 → R 4 , R 5 − 4 R 2 → R 5 [ 1 0 − 42 − 546 − 5334 1681 10 0 1 13 127 1105 301 10 0 0 2 42 590 − 33 10 0 0 6 132 1950 − 154 To determine b) To calculate: The solution of the system and also determine does the solution produce reasonable model for determining sales after 2010.

Bundle: Elementary Linear Algebra, Loose-leaf Version, 8th + WebAssign Printed Access Card for Larson's Elementary Linear Algebra, 8th Edition, Single-Term

8th Edition

ISBN: 9781337604925

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 1.3, Problem 20E

To determine

a)

To calculate:

The system of equations to fit the data for the years 2006, 2007, 2008, 2009 and 2010 to a quadratic model.

To determine

b)

To calculate:

The solution of the system and also determine does the solution produce reasonable model for determining sales after 2010.

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Chapter 1.3, Problem 19E

Chapter 1.3, Problem 21E

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Bundle: Elementary Linear Algebra, Loose-leaf Version, 8th + WebAssign Printed Access Card for Larson's Elementary Linear Algebra, 8th Edition, Single-Term

Ch. 1.1 - Linear Equations. In Exercises 1-6, determine...Ch. 1.1 - Linear Equations. In Exercises 1-6, determine...Ch. 1.1 - Linear Equations. In Exercises 1-6, determine...Ch. 1.1 - Linear Equations. In Exercises 1-6, determine...Ch. 1.1 - Linear Equations. In Exercises 1-6, determine...Ch. 1.1 - Linear Equations. In Exercises 1-6, determine...Ch. 1.1 - Parametric Representation. In Exercises 7-10, find...Ch. 1.1 - Parametric Representation. In Exercises 7-10, find...Ch. 1.1 - Parametric Representation. In Exercises 7-10, find...Ch. 1.1 - Parametric Representation. In Exercises 7-10, find...

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a) To calculate: The system of equations to fit the data for the years 2006, 2007, 2008, 2009 and 2010 to a quadratic model.

Solution Summary: The author calculates the system of equations to fit the data for the years 2006, 2007, 2008, 2009, and 2010 to a quadratic model.