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are155-homework-5

# are155-homework-5 - University of California Davis...

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University of California, Davis Department of Agricultural and Resource Economics “We are what we repeatedly do. Excellence then is not an act, but a habit.” Aristotle Copyright c 2010 by Quirino Paris. ARE 155 Winter 2010 Prof. Quirino Paris HOMEWORK #5 Due Tuesday, February 9 1. On the pivot method. Solve the following system of equations using the pivot method. You must also produce the inverse of the basis. Hence, your initial tableau must include an identity matrix. 3 x 1 + 4 x 2 + 2 x 3 = 18 x 1 x 3 = 20 2 x 2 + 3 x 3 = 12 (1) The emphasis of this exercise is on the inverse of the basis and on the three roles it plays. I have insisted that a basis is for a solution of a system of equations what a sca ff olding is for a building. To illustrate this notion let us rewrite the above system in a compact way that will allow the use of a short-hand notation: 3 4 2 1 0 1 0 2 3 x 1 x 2 x 3 = 18 20 12 (2) The two systems (1) and (2) are equivalent because we know how to multiply out vectors and matrices. The second way of writing the system is convenient because we can see that the square matrix in front of the vector of unknowns [ x 1 , x 2 , x 3 ] is a basis for this system, while the vector [ x 1 , x 2 , x 3 ] will be a solution once the system is solved. Therefore, it should be clear now that without a basis we cannot find a solution (just as without a sca ff olding we cannot have a building). From now on, (promise?), we will not confuse bases and solutions. When we wish to emphasize the conceptual framework rather than the actual numerical structure, the second system can be written in short hand notation as B x = b (3) where B is the square matrix, x = [ x 1 , x 2 , x 3 ] and b = [18 , 20 , 12] , in this case. To approach the meaning of the inverse of a basis, we first consider a simple equation in one variable that can easily be solved numerically and can be represented in a symbolic fashion for the purpose of generalizing the discussion 9 x = 45 or Bx = b (4) x = 9 1 45 or x = B 1 b 1

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where B = 9 and b = 45 . If 9 is considered the basis for our “system” of one equation in one unknown, 9 1 is the inverse (or reciprocal) of the basis, that is, if B = 9 , then B 1 = 9 1 . The solution to the “system”, then, is obtained by taking the inverse of the basis and multiplying such an inverse by the RHS (in this case, 45). In symbolic notation, x = B 1 b = 9 1 45 . The extension to a non trivial system of equations such as given in (2) and (3) is straightforward. We apply to system (2) and (3) the same line of reasoning that was developed to solve “system” (4). The solution of system (2) and (3) will be obtained by computing the inverse of the basis and multiplying such an inverse by the RHS vector b . That is x = B 1 b (5) x 1 x 2 x 3 = 3 4 2 1 0 1 0 2 3 1 18 20 12 The conclusion is that for solving a system of equations, the inverse matrix must be computed either explicitly or implicitly; we might as well compute it explicitly.
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