253Solutions1

Linear Algebra with Applications (3rd Edition)

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ma253 ,Spr ing 2007 — Problem Set 1 Solutions 1. Problems from the textbook: a. Section 1 . 1 , problems 5 , 7 , 9 ,* 20 28 . Problems 5, 7, and 9 should be straightforward; the answers are in the back of the book. Let’s do the other two. 20. The total demand for the product of A is 1000 (from consumers) plus 0.1b (from B). So we must have a = 1000 + 0.1b . Similarly, b = 780 + 0.2b .Th isleads to the system a - 0.1b = 1000 - 0.2a + b = 780 The unique solution is a = 1100 , b = 1000 . 28. Thermal equilibrium requires that T 1 = T 2 + 200 + 0 + 0 4 T 2 = T 1 + T 3 + 200 + 0 4 T 3 = T 2 + 400 + 0 + 0 4 Rewriting as a linear system and solving gives ( T 1 ,T 2 3 )=( 75, 100, 125 ) . b. Section 1 . 2 , problems 5 , 11 , 13 38 44 61 . Again, 5, 11, and 13 should be routine; the answers are in the back of the book. You should be able to do 5 by hand, but getting 11 right by hand is challenging. 38. This is in the same genre as problem 20 in section 1.1, but is a little closer to what would happen in real economic modeling. For (a), the vectors are ~ v 1 = 0 0.1 0.2 ~ v 2 = 0.2 0 0.5 ~ v 3 = 0.3 0.4 0 ~ b = 320 90 150 . For (b), notice ±rst x j is the output of I j and the i -th entry a ij in ~ v j is the per-unit demand of I j for the goods made by I i .S o x j a is the actual demand of I j for Never Give an Iguana Viagra – headline, Reuters, January 25, 2007
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the goods made by I i . That’s the i -th entry in x j ~ v j ,sothevec tor x j ~ v j gives the demands made by I j for the goods of each of the other industries. Now if we add all of those up, x 1 ~ v 1 + ··· + x n ~ v n + ~ b we get a vector whose i -th entry represents the total demand for the goods pro- duced by I i . So this could be called the total demand vector. Finally, the equation x 1 ~ v 1 + + x n ~ v n + ~ b = ~ x says that the total demand should equal the total production. 44. Kyle’s predicament boils down to the following system of equations: x 1 + x 2 + x 3 = 24 3x 1 + 2x 2 + 1 2 x 3 = 24 Two equations and three unknowns, so we expect one degree of freedom. Solving this (and taking x 3 = 2a to avoid denominators) gives x 1 x 2 x 3 = 3a - 24 - 5a + 48 2a . All three numbers need to be positive, so a 8 (from the value of x 1 )and a 9.6 (from the value of x 2 ). Since a needs to be a whole number, either a = 8 or a = 9 , which leads to two possible solutions: x 1 x 2 x 3 = 0 8 16 or 3 3 18 . Since Kate loves lillies, the second choice is better: Kyle buys 3 lillies, 3 roses, and 18 daisies. 61. This is very similar to the previous one, except that the handling of the inequal- ities is much messier and there are many more solutions.
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This homework help was uploaded on 01/23/2008 for the course MATH 253 taught by Professor Ghitza during the Spring '07 term at Colby.

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253Solutions1 - ma253, Spring 2007 - Problem Set 1...

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