plugin-hw07 - Assignment 7 Math 417 — Winter 2009 Due...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Assignment 7 Math 417 — Winter 2009 Due March 13 § 3.4#10. The question is whether we can express ~x =  - 5 1 3   as a linear combination c 1 ~v 1 + c 2 ~v 2 of ~v 1 =  - 1 1   and ~v 2 =  - 2 1   . By inspection we see that, if there is to be such an expression, the coefficient c 1 must be 3 in order to make the last component right (3 = c 1 · 1 + c 2 · 0), and similarly c 2 must be 1 in order to make the middle component right. These values for the two coefficients happen to make the first component right also (- 5 = 3 · (- 1)+1 · (- 2)), so ~x is in the span of ~v 1 ,~v 2 , and its coordinates are the coefficients we found, 3 and 1. Its coordinate vector is [ ~x ] B = 3 1 . § 3.4#12. The question is whether we can express ~x =   1- 2- 2   as a linear combination c 1 ~v 1 + c 2 ~v 2 of ~v 1 =   8 4- 1   and ~v 2 =   5 2- 1   . In contrast to the preceding problem, there seems to be no way to guess the coefficients c i by inspection (or even to guess whether such coefficients exist). So we compute instead. We need 8 c 1 + 5 c 2 = 1 to make the first component of c 1 ~v 1 + c 2 ~v 2 match the first component of ~x . We need 4 c 1 + 2 c 2 =- 2 to make the second components match. And we need- c 1- c 2 =- 2 to make the third components match. So ~x will be in the span of ~v 1 and ~v 2 if and only if this system of three equations has a solution for the two unknowns c 1 and c 2 . The augmented matrix of the system is   8 5 1 4 2- 2- 1- 1- 2   . To reduce it to rref, I’ll first interchange the last row with the first (indulging my dislike for fractions); while I’m at it, I’ll also reverse the sign of the top (formerly bottom) row.   1 1 2 4 2- 2 8 5 1   →   1 1 2- 2- 10- 3- 15   →   1 1 2 0 1 5 0 1 5   →   1 0- 3 0 1 5 0 0   . 1 So there is a solution, namely c 1 =- 3 and c 2 = 5, which means that ~x is in the span of ~v 1 and ~v 2 . Its coordinate vector with respect to this basis is therefore [ ~x ] B =- 3 5 . § 3.4#22. (a) To use the formula B = S- 1 AS , recall that S is the matrix whose columns are the vectors ~v i of the basis. In this problem, therefore, S = 1- 2 2 1 . We need its inverse S- 1 as well, so we compute it by row reduction: 1- 2 1 0 2 1 0 1 → 1- 2 1 5- 2 1 → 1- 2 1 1- 2 / 5 1 / 5 → 1 0 1 / 5 2 / 5 0 1- 2 / 5 1 / 5 . So B = S- 1 AS = 1 / 5 2 / 5- 2 / 5 1 / 5- 3 4 4 3 1- 2 2 1 = 5- 5 . (b) The commutative diagram method (as in the book’s Examples 3 and 4) requires us to take an arbitrary vector expressed as a linear combination c 1 ~v 1 + c 2 ~v 2 , apply the linear transformation (represented by) A to it, and express the result again as a linear combination of ~v 1 and ~v 2 . There are two ways to do this. The more obvious but less efficient is to use the given ~v 1 and ~v 2 to compute that c 1 ~v 1 + c 2 ~v 2 = c 1- 2 c 2 2 c 1 + c 2 , and applying A yields- 3 4 4 3 c 1- 2...
View Full Document

This note was uploaded on 10/13/2009 for the course MATH 417 taught by Professor Eveet during the Spring '09 term at University of Michigan-Dearborn.

Page1 / 8

plugin-hw07 - Assignment 7 Math 417 — Winter 2009 Due...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online