hw6sol

# hw6sol - {ii 14 The matrix 5 4 A184 has characteristic...

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Unformatted text preview: {ii 14. The matrix 5 4 A184), has characteristic polynomial 1700 = A2 — 2A — 3. Note that MA) = A2 + 2A — 31 4;; act 30%? 33> (3 3). 2. Thematrix 1 6 ' A=(—3 8) has the following eigenvalue-eigenvector pairs. 1 A1: 2 —-> and 1.2 = 5 '—> Thus, the general solution is YO) = Ciezt + C285! v -—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} = 516/ \ ~— /"'\ \._/ 8, The system in Exercise 2 had the general solution y(t) = Clez’ + Czes’ . Thus, if y(0) = (1, ——2)T, then ('32) = 613+ 6211);“ (i 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’ is a solution. For M = 4, —7 14 A --41 —— < 0 0) and vz = (2, 1)T is an eigenvector. Thus, yza) = e‘“ 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) i 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) :e4it I) . . —-1 , 1 :(cos4t+zsm4t) 1 +1 0 : cos4t — Sin4t . 1 . . 4 —1 +zcos4t 0 +zsmt 1 g” % é h g _ —- cos 4t — sin 4t T cos 4! + t. (cos 4t — Sin 4t) ‘ sin 4t Therefore, _ — cos 4t — sin 4t Yi(l‘)~( COS,” ) and __ cos 4t —— sin 4t yz (I) _ ( sin 4t ) form a fundamental set of real solutions. we 21> has one eigenvalue, A = —2. However, the nullspace of 1 1 A + 21 : (#1 1) is generated by a single eigenvector, v1 = (1, 1)T, with corresponding solution ‘ 1 mt) = e‘z’ (1) . To ﬁnd another solution, we need to ﬁnd a vec« tor vz which satisﬁes (A + 21)vz = v1. Choose 30. The matrix w = (1, 0)T, which is independent of v; and note that _ —1 1) 1 _ ~1 _ 1/ (Otto—«1- Thus, choose V2 = —w = (—1,0)T. Our second solution is y2(t) = e‘z’m + M) q r /_-1\ [/1\ "i __ —Ai *3 lio)+‘ll)i- Thus, the general solution can be written __ —2z 1 —21 [ *1 1 y(t)——C1€ <1 +C2e L 0 +t l -2, 1 —1 =6 [(C1+C2t)(1)+C2<0)]. 42. Thematrix ~ ——8 ——10 A~(s 7) has characteristic polynomial p()t) = A2+A ——6 with eigenvalues A1 = ——3 and A2 = 2. The nullspace of —5 —10 A+3I‘(5 10) 34. The matrix viz, he a has one eigenvalue, A = 3. However, the nullspace of 2 1 A—3I=(~4 _2) is generated by a single eigenvector, v1 = (1, ——2)T, with corresponding solution \\ i» v ‘* 3'10) = eat t / \t \ \ 1 To ﬁnd another solution, we needxtg ﬁnd/a vec— tor vz which satisﬁes (A - 31 )vz = v1} Choose that 22><a>=<34>=2w Thus, choose v2 = (1/2)w = (1/2, 0)T. Our sec— L 0nd solution is 3’20) = €3t(V2 + “’0 = [(162) +t (32)} Thus, the general solution can be written (I) — C e3’ 1 y " l _2 3: 1/2 1 + C26 O + l ‘2 1 1 2 2 e3’ [(C1 + Cy) (_2) + C2( 6 . is generated by the single eigenvector, v1 = (——2, 1)T, with corresponding solution no) = {3‘ (“12). The nullspace of —10 —-10 A—-21—_(5 5) is generated by the single eigenvector, V2 2 (1, ——l)T, with corresponding solution 3’20) = 62’ ‘ Thus, the general solution can be written 3’“) = Che—3t + C2ezr ' w = (1,0)7‘, which is independent of v1 and note E g 50. From Exercise 42, the general solution is y(t) = Cleﬂ3r + C2e2‘ . Because y(0) = (3, 1)T, (?)=Cl (32)”2 (31)- Reduce the augmented matrix —21310—4 1—11601—5' Thus, C1 = —4 and C2 = —5 and the particular solution is y(t) : —4e‘3‘ -— 562: _ 863’ — 5e2’ # —4e‘3’ + 562’ ‘ ...
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