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Unformatted text preview: = 2( C2 D4)(2)+2( CD2)(1)+0+2( C + D )(1)+2( C +2 D )(2) = 20 D +20 so we need to solve 5 C = 5, 10 D =10 so C = 1 and D =1. The line of best ﬁt is 1t . 26 ) We are trying to solve C + D = 0, C + E = 1, CD = 3 and CE = 4, that is 1 1 1 1 11 11 = 1 3 4 . So A T A = 1 1 1 1 1 01 0 11 1 1 1 1 11 11 = 4 0 0 0 2 0 0 0 2 . Now A T ~ b = 833 So solving A T A~x = 833 gives ~x = 23 / 23 / 2 . The best plane is 2(3 / 2) x(3 / 2) y which at (0 , 0) gives 2, the average of 0, 1, 3 and 4....
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 '05
 HUI
 Linear Algebra, Algebra, Derivative, Normal Distribution, Standard Deviation, Dot Product, normal equations, shortest distance

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