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p b 0 01 0 1 1 p2 13 13 0 2 3 0 its lenght is

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Unformatted text preview: 3 and a A −1 1 −1 0 −1 1 ￿ ￿ 10 1 −1 1 0 1 = 1 −1 A(AT A)−1 AT = 1 AAT = 1 3 3 3 0 0 00 0 1 −1 10 1 1/3 −1/3 0 −1/3 2/3 1/3 0 1 /3 −1/3 1/3 0 1 /3 2/3 0 − 1/3 and P = I −P ⊥ = 1/3 . Answer: P ⊥ = 0 0 00 0 01 0 −1/3 1/3 0 1 /3 1/3 −1/3 0 2 /3 (d) The projection is equal to 2 2 2/3 1/3 0 1 /3 3p 2p 1/3 0 p2 2/3 0 − 1/3 . = P￿ = b 0 01 0 1 1 p2 1/3 −1/3 0 2 /3 0 ￿ ￿ Its lenght is ||P￿ || = (2p2 )2 + (p2 )2 + 1 + (p2 )2 = 6(p2 )2 + 1. It is the smallest, when b ￿ 6p4 +1 is the smallest, that is, when p = 0. In this case the vector has length 6(0)2 + 1 = 1. 2 Ex. 1. Find the equation of the line in the plane that fits the best (in the least square approximation sense) the data points (−3, −2), (0, 0), (1, 0), (2, 2). ￿ ￿ C (12 pts) Solution: The solution is of the form y = C + Dt, where x = ˆ constitute D −2 1 −3 0 1 0 . Then, and ￿ = b the solution of the system AT Ax = AT￿ , where A = ˆ b 1 0 1 2 1 2￿ ￿ ￿ ￿ ￿ ￿￿ ￿ ￿ ￿ m ￿ i ti 40 b 0 AT A = ...
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This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.

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