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

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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 0 1 and v 2 = - 2 1 0 . 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 0 - 2 - 10 0 - 3 - 15 1 1 2 0 1 5 0 1 5 1 0 - 3 0 1 5 0 0 0 . 1
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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 0 0 5 - 2 1 1 - 2 1 0 0 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 0 0 - 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 c 2 2 c 1 + c 2 = 5 c 1 + 10 c 2 10 c 1 - 5 c 2 .
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