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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...
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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 MichiganDearborn.
 Spring '09
 eveet
 Math, Linear Algebra, Algebra

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