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Unformatted text preview: xk } be a solution of the diﬀerence equation
0
P −1 ,
d2 d1
0 002 xk+1 = Axk , then 2 , so + e−t 2(e −3t −t correct A=P 1
1 v2 = 2e−t + e−3t = 4. e −3t e−3t − e−t = tA −t +e −3t tA i.e., λ1 = −3 and λ2 = −1. Corresponding
eigenvectors are 10.0 points Using the fact that
1
1
ex = 1 + x + x2 + . . . + xn + . . . ,
2!
n! 1. etA = 1 4
1−λ = λ − λ + 3 = (λ + 3)(λ + 1) = 0 , Compute x1 .
1. x1 = 8
4 2. x1 = 4
correct
8 3. x1 = 7
3 4. x1 = 3
7 x0 = −1
.
7 cruz (fmc326) – HW10 – gilbert – (56540) 2 5. x1 = 3
8 1. xk = 3(2k )v1 − 1
2k v2 6. x1 = 4
7 2. xk = 3(2k )v1 + 1
2k v2 Explanation:
Since v1 , v2 are eigenvectors corresponding
to distinct eigenvalues of A, they form an
eigenbasis for R2 . Thus
x0 = c1 v1 + c2 v2 . v2 x0 ] = 1
1 1
−1
∼
0
7 −1
1 1
2k v2 correct 4. xk = 6(2k )v1 − 2 1
2k v2 5. xk = 6(2k )v1 − To compute c1 , c2 we apply row reduction to
the augmented matrix
[ v1 3. xk = 3(2k )v1 − 2 03
,
14 for then c1 = 3, c2 = 4 a...
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 Spring '08
 PAVLOVIC
 Matrices

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