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Unformatted text preview: v
0 −1 −1 0
1
0
(12 pts) Solution: First, we ﬁnd orthogonal vectors 1 , 2 , and 3 , and, at the end of
uu
u
the process, normalize each of them to 1 , 2 , and 3 , respectively. q q
q 0
1
1 0 1+0+1+0 0 0 1
uv We have 1 = 1 , 2 = 2 − 1 ··2 1 = u
vu
v
u
u u1 −1 − 1+0+1+0 −1 = 0 , and
1
1
0 2
1
0
1 −1 2+0+0+0 0 0+0+0+0 0 −1 1
uv
2
uv 3 = 3 − 1 ··3 1 − 2 ··3 2 = u
v
u
u
u u1
u u2 0 − 1+0+1+0 −1 − 0+0+0+1 0 = 1 .
0
0
1
0
√
√
√
√
1
u=
u
+
Since  1  = + 0 + 1 + 0 = 2,  2  1, and  3  = 1 + 1 + 1 0 = 3, we have
u 1
0
1
0
0 −1 u
2
u
3
u
1
1
q √
1 =  1  = √2 q
q
1
u
u
u −1 , 2 =  2  = 0 , and 3 =  3  = 3 1 .
0
1
0
Answer: An orthonormal basis under question is {1 , 2 , 3 }, where vectors i are as above.
qqq
q 2 MATH 441.001
Instr. K. Ciesielski
Spring 2011 NAME (print): Review for TEST # 4
Solve the followin...
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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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