This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: D20) = 0 This implies 4 C + 8 D = 36 8 C + 26 D = 112 . 1 A T Ax = A t b is given by ± 4 8 8 26 ²± C D ² = ± 36 112 ² . 6. Problem Set 4.5, p. 217, # 12 b = ( b 1 , . . . , b m ) and a = (1 , . . . , 1). Observe that there is a change in notation and b, a denote a row vectors. Thus b T , a T will denote column vectors. (a) a T a ˆ x = a T implies that m ˆ x = ∑ m i =1 b i and ˆ x = 1 m ∑ m i =1 b i . (b) e = ba ˆ x = ( b 1ˆ x, . . . , b mˆ x ). Then k e k 2 2 = ∑ m i =1 ( b iˆ x ) 2 and k e k 2 = { ∑ m i =1 ( b iˆ x ) 2 } 1 / 2 . (c) b = (1 , 2 , 6), a = (1 , 1 , 1), ˆ b = 3 and p = (3 , 3 , 3). Then for e = bp = (2 ,1 , 3) we have pe T = 0 and P = a T a k a T k 2 2 = 1 3 1 1 1 1 1 1 1 1 1 . 2...
View
Full Document
 Fall '11
 AnargyrosPapageorgiou
 Linear Algebra, Pallavolo Modena, Sisley Volley Treviso

Click to edit the document details