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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.
 Spring '08
 STAFF
 Math, Linear Algebra, Algebra

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