Math 18 - Qi 14 The matrix 5 4 A 4 has characteristic...

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Unformatted text preview: Qi 14. The matrix 5 4 A_(_, 4) has characteristic polynomial 1700 = A2 — 2A — 3. Note that MA) = A2 + 2A — 31 4;; act -E4)+(’03 33> (3 3). 92 2. Thematrix 1 6 ' A=(—3 8) has the following eigenvalue-eigenvector pairs. 1 A1: 2 —-> (?) and 1.2 = 5 '—> (1) Thus, the general solution is YO) = C16” (i) + C285! G). -—1 1 A = ( 1 4) has the following eigenvalue—eigenvector pairs. A1=0~><D and A22—2——>(_11> Thus, the general solution is 6. The matrix ——1 1 r \..n to rege " l 1/ Y3} = C1( \ ~— /"'\ \._/ 8, The system in Exercise 2 had the general solution y(t) = Clez’ (i) + Czes’ (1). Thus, if y(0) = (1, ——2)T, then ('32) = 613+ 02(1) = (E i) (5:) The augmented matrix reduces 211_>103 11—2 01—5' Thus, C1 = 3 and C2 = —5, giving particular solu— tion 2 1 _ Zr _ 5t y(t) — 3e (1) Se <1>' 22. If ~3 14 A = ( o 4), then —3 ~ A 14 p(}\.) = det< O 4 _ A.) .-: (——3 ~ A)(4 - A). Thus, A; = -—3 and A2 = 4 are eigenvalues. For 1 A1 = —3, ) 0 14 A+31=(O 7) and v1 = (l, 0)T is an eigenvector. Thus, y1(t) = e‘s’ ((1)) is a solution. For M = 4, —7 14 A --41 —— < 0 0) and v2 = (2, 1)T is an eigenvector. Thus, yza) = e‘“ (f) is a solution. Because y](0) = (1, 0)T and y2(0) = (2, l)T are independent, the solutions y] (t) and y2 (t) are independent for all t and form a fundamental set of solutions. 16. If g —4 —8 A—(4 4) then the characteristic polynomial of A is p()») _ 2 . A +4 and the eigenvalues are A] = 41‘ and Ag 2 —4i . Trusting that /__A /. A \ xii—’47: “‘4’ *3 (l) E 4 4—41‘ is singular, examination of the second row shows that (~1 + i, 1)T generates the nullspace of A — (4i )1 . Thus, we have a complex solution which we must break into real and imaginary parts. 1(1) :64” <—11+l) . . —-1 , 1 :(cos4t+zsm4t) 1 +1 0 : cos4t (~11) — Sin4t ((1)) . 1 . . 4 —1 +zcos4t 0 +zsmt 1 g” % E E h g E ...
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