MAT4375 HW#2, 1999

MAT4375 HW#2, 1999 - MAT4375 HW#2, 1999

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MAT4375 HW#2, 1999 http://aix1.uottawa.ca/~mat4375/Solutions/hw2.html 1 of 3 10/2/2007 8:25 PM -3 0 3 0 1 -1 -3 1 2 X-X 1 = 9 -3/2 15/2 -3/2 1 -1/2 15/2 -1/2 7 S = MAT4375 HW#2, 1999 © A.R. Dabrowski, 1999 Due to html limitations, ``X- bar'' will be noted X . From the third edition of Johnson and Wichern Problem 3.4 Problem 3.6 Problem 4.4 Problem 4.14 (#4.19 in the fourth edition) Problem 4.20 (#4.27 in the fourth edition) Problems 4.21 and 4.22 (#4.28 and 4.29 in the fourth edition) Problem 3.4 For visual simplicity we work in millions of dollars rather than dollars. (a) Here p 1 = (x 1 ·1/ || 1 || ) 1/ || 1 || , which equals x 1 ·1 2.10121. (b) e 1 = y 1 -x i 1 (1.3967, .3843, -.3183, -.3757, -.4556, -.6314) . Note that (s 11 ) 1/2 = || y 1 -x 1 || /6 1/2 . (c) By construction p 1 and e 1 are orthogonal. We have that || e 1 || 1.717 and || p 1 || 5.147. We can sketch the relationship among y 1 , e 1 and p 1 as a right-angle triangle with hypotenuse y 1 , short side e 1 and long side p
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MAT4375 HW#2, 1999 - MAT4375 HW#2, 1999

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