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Unformatted text preview: nd
x0 = 3v1 + 4v2 . 1
2k 6. xk = 6(2k )v1 + 2 v2 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 . But v1 , v2 are eigenvectors corresponding to
1
respective eigenvalues 2, 2 so To compute c1 , c2 we apply row reduction to
the augmented matrix x1 = Ax0 = 3Av1 + 4Av2
−1
1
+2
= 6v1 + 2v2 = 6
1
1 [ v1 v2 x0 ] = . Consequently, 1
5
∼
0
1 1 −1
11 03
,
1 −2 for then c1 = 3, c2 = −2 and
x0 = 3v1 − 2v2 . x1 = 4
8 But v1 , v2 are eigenvectors corresponding to
1
respective eigenvalues 2, 2 so set . . 1
2k xk = 3(2k )v1 − 2
003 10.0 points When A is a 2 × 2 matrix with eigenvalues
1
2, 2 and corresponding eigenvectors
1
,
v1 =
1 −1
,
v2 =
1 determine the solution {xk } of the diﬀerence
equation
xk+1 = Axk , x0 = 5
.
1 v2 for k ≥ 0. Then x0 is the given initial value
and
Axk = 3 2k Av1 − 2
= 3 2k+1 v1 − 2 1
2k Av2 1
2k+1 v2 = xk+1 , so {xk } solve the diﬀerence equation.
004 10.0 points cruz (fmc326)...
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This document was uploaded on 03/16/2014 for the course M 340L at University of Texas.
 Spring '08
 PAVLOVIC
 Matrices

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