Explanation Given: Let p ( x ) = a 0 + a 1 x + a 2 x 2 be a vector in P 2 , q ( x ) = 1 − x + 2 x 2 and r ( x ) = 2 − 3 x + x 2 and the spanning set is { 1 , x + 1 , ( x + 1 ) 2 } Approach: The vector v is a linear combination of the vectors v 1 , v 2 , ... v m provided that there exist scalars c 1 , c 2 , ... c m such that v = c 1 v 1 , c 2 v 2 , ... + c m v m Calculation: Setting p ( x ) = b 0 + b 1 ( x + 1 ) + b 2 ( x + 1 ) 2 yields, p ( x ) = b 0 + b 1 ( x + 1 ) + b 2 ( x + 1 ) 2 a 0 + a 1 x + a 2 x 2 = b 0 + b 1 ( x + 1 ) + b 2 ( x + 1 ) 2 a 0 + a 1 x + a 2 x 2 = b 0 + b 1 x + b 1 + b 2 ( x 2 + 2 x + 1 ) a 0 + a 1 x + a 2 x 2 = b 0 + b 1 x + b 1 + b 2 x 2 + 2 b 2 x + b 2 Equating the coefficients for system of equations: b 0 + b 1 + b 2 = a 0 ( 1 ) b 1 + 2 b 2 = a 1 ( 2 ) b 2 = a 2 ( 3 ) Substitute the value b 2 = a 2 in equation ( 2 ) , b 1 + 2 a 2 = a 1 b 1 = a 1 − 2 a 2 Substitute the values in equation ( 1 ) , b 0 + a 1 − 2 a 2 + a 2 = a 0 b 0 = a 0 − a 1 + 2 a 2 − a 2 Now, q ( x ) = 1 − x + 2 x 2 Setting q ( x ) = c 0 ( 1 ) + c 1 ( x + 1 ) + c 2 ( x + 1 ) 2 yields, 1 − x + 2 x 2 = c 0 ( 1 ) + c 1 ( x + 1 ) + c 2 ( x + 1 ) 2 1 − x + 2 x 2 = c 0 + c 1 x + c 1 + c 2 ( x 2 + 2 x + 1 ) 1 − x + 2 x 2 = c 0 + c 1 x + c 1 + c 2 x 2 + 2 c 2 x + c 2 Equating the coefficients for system of equations: c 0 + c 1 + c 2 = 1 ( 1 ) c 1 + 2 c 2 = − 1 ( 2 ) c 2 = 2 ( 3 ) Substitute the value c 2 = 2 in equation ( 2 ) ,

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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In mathematics, problems can be solved by the arithmetic operators like addition, subtraction, multiplication, and division. In algebra, we represent the numbers by any letter called a variable. If these variables are of degree 1, then it is studied unde…

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Chapter 5.3, Problem 32E

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The value of b0,b1 and b2 in terms of a0,a1 and a2, also represent q(x)=1−x+2x2 and r(x)=2−3x+x2 in terms of the spanning set {1,x+1,(x+1)2}.

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Chapter 5.3, Problem 31E

Chapter 5.3, Problem 33E

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Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - Prob. 4E Ch. 5.2 - Prob. 5E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - Prob. 7E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...

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The value of b 0 , b 1 and b 2 in terms of a 0 , a 1 and a 2 , also represent q ( x ) = 1 − x + 2 x 2 and r ( x ) = 2 − 3 x + x 2 in terms of the spanning set { 1 , x + 1 , ( x + 1 ) 2 } .

Solution Summary: The author explains that the value of b_0, q(x)=1-x+2x

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To find:

The value of b0,b1 and b2 in terms of a0,a1 and a2, also represent q(x)=1−x+2x2 and r(x)=2−3x+x2 in terms of the spanning set {1,x+1,(x+1)2}.

Want to see the full answer?

Chapter 5 Solutions

The value of b 0 , b 1 and b 2 in terms of a 0 , a 1 and a 2 , also represent q ( x ) = 1 − x + 2 x 2 and r ( x ) = 2 − 3 x + x 2 in terms of the spanning set { 1 , x + 1 , ( x + 1 ) 2 } .

Solution Summary: The author explains that the value of b_0, q(x)=1-x+2x

Concept explainers

Videos

To find: The value of b0,b1 and b2 in terms of a0,a1 and a2, also represent q(x)=1−x+2x2 and r(x)=2−3x+x2 in terms of the spanning set {1,x+1,(x+1)2}.

Want to see the full answer?

Chapter 5 Solutions

To find:

The value of b0,b1 and b2 in terms of a0,a1 and a2, also represent q(x)=1−x+2x2 and r(x)=2−3x+x2 in terms of the spanning set {1,x+1,(x+1)2}.