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Unformatted text preview: Solutions to quiz # 9 (November 4) 1. We write the system as Ax = b, where 13
A = 1 −1 ,
11 x
y x= and 5
b = 1.
0 Computing
AT A = 11
3 −1 13
1
33
1 −1 =
1
3 11
11 and AT b = 1
3 1
−1 5
1
6
1=
,
1
14
0 we get the system x
3
3 y
1
3  6 −→
3
11  14 12
11
−→
11  14
08 2
1
−→
8
0 from which x = 1 and y = 1 is the least squares solution.
Answer: The least squares solution is x = 1 and y = 1.
2. We have
CT C =
Answer: C T C = 4
1 a·a
b·a 4
a·b
=
1
b·b 1
.
9 1 1
.
9 01
,
11 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.
 Fall '08
 Conger
 Linear Algebra, Algebra

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