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Unformatted text preview: Solutions to quiz # 11 (November 18)
1. To ﬁnd the eigenvalues, we solve the equation
0 = det 2−λ
−3 1
= (2 − λ)(6 − λ) + 3 = λ2 − 8λ + 15,
6−λ and hence √ 82 − 4 · 15
= 4 ± 1.
2
Thus there are two eigenvalues λ = 3 and λ = 5.
The eigenvectors with eigenvalue 3 are nonzero solutions to the system
λ= 2−3
−3 1
6−3 8± 0
−1
−→
0
−3 1
3 0
1 −1
−→
0
00 0
,
0 1
for t = 0.
1
The eigenvectors with eigenvalue 5 are nonzero solutions to the system so the eigenvectors are t 2−5
−3 1
6−5 so the eigenvectors are t 0
−3
−→
0
−3
1
3 1
1 0
1 −1/3
−→
0
0
0 for t = 0. Answer: The eigenvalues are λ = 3 with the eigenvectors t
the eigenvectors t 1
3 0
,
0 1
1 for t = 0 and λ = 5 with for t = 0. 2. We have Av = 2v and hence
A2 v = A (Av) = A (2v ) = 2Av = 4v
and
A−1 (Av) = A−1 (2v ) =⇒ A−1 A v = 2A−1 v =⇒ A−1 v =
Therefore
A2 − A−1 1
v.
2 7/2
7
1
v = A2 v − A−1 v = 4v − v = v = 7 .
2
2
21/2 7/2
Answer: Bv = 7 .
21/2
1 ...
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 Fall '08
 Conger
 Linear Algebra, Algebra, Eigenvectors, Vectors

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