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. Then we ﬁnd that the eigenvalues are λ1 = 1 and
2/5 1/5
λ2 = −1/5, with corresponding eigenvectors 2. Let A = 2
1 X1 = and −1
1 X2 = . Then if P = [X1 X2 ], P is non–singular and
P −1 AP = 1
0
0 −1/5 and 1
0
0 −1/5 A=P P −1 . Hence
An = P →P = 10
00 1
3
1
3 P −1 P −1 2 −1
1
1 =
= 1
0
0 (−1/5)n 1
3 10
00 20
10 11
−1 2 11
−1 2 22
11 = 2/3 2/3
1/3 1/3 . ˙
3. The given system of diﬀerential equations is equivalent to X = AX ,
where
x
3 −2
.
and X =
A=
y
5 −4
21
is a nonsingular matrix of eigenvectors corre51
sponding to eigenvalues λ1 = −2 and λ2 = 1. Then The matrix P = P −1 AP = −2 0
01 . The substitution X = P Y , where Y = [x1 , y1 ]t , gives
˙
Y= −2 0
01
66 Y, ...
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This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.
 Fall '10
 Dreibelbis

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