253Solutions2

Linear Algebra with Applications (3rd Edition)

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ma253 ,Fa l l 2007 — Problem Set 2 Solutions 1. Problems from the textbook: a. Section 1 . 1 , problem * 46 The equations are x 1 + x 2 + x 3 = 1000 .2x 1 + .5x 2 + 2x 3 = 1000. Row-reducing and solving leads to x 1 x 2 x 3 = 5a - 5000 3 - 6a + 800 3 a , where a is arbitrary. Unfortunately for the meter-maids, whenever a is a whole number, the expressions for x 1 and x 2 are not whole numbers! Since we can’t have fractions of coins, no one will ever get the Ferrari. Unless they ±gure this out and sue, of course. b. Section 1 . 2 , problem 50 ,* 53 . 50. Write the polynomial f ( t ) as f ( t )= a + bt + ct 2 + dt 3 . The ±rst condition says that f ( 0 3 ; plugging in, we get a = 3 .Thesecondg ives f ( 1 a + b + c + d = 2 , f ( 2 a + 2b + 4c + 8d = 0 , and the ±nal one says Z 2 0 f ( t ) dt = 2a + 2b + 8 3 c + 4d = 4. That gives us a system of equations in the four unknown a , b , c , d . Setting up the row-reduction, we see that the system is inconsistent: there is no solution, hence there is no such polynomial! The reason is this: for a cubic polynomial Simpson’s Rule gives the exact value of the integral, and the information that goes into Simpson’s rule is exactly the values f ( 0 ) , f ( 1 ) and f ( 2 ) . Using that, we can see that for the given values the integral must be equal to 11/3 . Hence, asking that the integral be 4 is asking for something that is impossible. 53. The (only) hard part of this is extracting a system of equations from the given data. Let’s start by letting x 1 be the number of students who start out as liberals and x 2 be the number of students who start out as conservatives. So we know that x 1 + The ±rst piece of luggage out of the chute doesn’t belong to anyone, ever.
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x 2 = 260 . At the end, 30% of the liberals, i.e. .3x 1 have become conservatives; on the other hand, .4x 2 are now liberals. So how many conservatives are there at the end? Well, x 2 - .4x 2 + .3x 1 . But the Fnal number of conservatives is equal to the initial number of liberals, so this expression is equal to x 1 . So the second equation is x 1 = x 2 - .4x 2 + .3x 1 ,wh ichshakesoutto .7x 1 - .6x 2 = 0 . That gives us the other equation. Sovling, we see that x 1 = 120 and x 2 = 140 . So we started with 120 liberals and 140 conservatives and ended with 140 liberals and 120 conservatives.
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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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253Solutions2 - ma253, Fall 2007 - Problem Set 2 Solutions...

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