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solutions2HW5

# solutions2HW5 - — —2 5.6 Determine a closed—form...

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Unformatted text preview: — —2 5.6 Determine a closed—form expression for eA , where A = l: 3 6] . Solution: IA — MIZMDL + 7) = 0 . Thus, we have eigenvalues I. : 0, i 7. f0») 2 e1, gm=a11 +ao fm=am no) : e“ = gm) =ug f(—7) : e” = g(—7) = Jot] + a0 a 3:8"? ' 7 (10:1 Therefore, # 77 e":1 e A+I 7 _ 0.3573 43.2355 " 43.4232 0.l436 —4 2 5 5.7 IfA: 1 —1 —1 ,ﬁnd A‘“. 71 2 2 Solution: EA w M|=7LJ +3?»2 —7L- 3 =0 .Thue, we have eigenvalues l =73, — l, 1. f(x)=x‘“, g(?\.)=u27\.2+o:]l+un f(l)=g(7\) 11—3) = (4)” = g(—3) = 9&2 — 3a. + “a f(—1)=1=g(#1)=0€2 “11"‘0‘0 f(1)=1:g(])=U—2 +€1| er0 a; =7381 0:1 =0 an =—7380 Therefore, A10 = 7381.42 -73801 88573 0 788572 = 49524 1 29524 29524 0 —29523 5.8 Compute em for the following two matrices: —14 —2 #14 #60 30 210 a) —7 F3 —8 b) 40 —20 440 1 1 2 l l "-20 [0 70 Solution: 0) Using 2’“ = L_i{(sI — Afl} L_l:fnverse Laplace Transform — l 4 —2 —] 4 A: -7 —3 78 I] 2 11 52—85—17 —2 —2(7s+13) s3+652+lls+6 52 +3s+2 :3 +632 +lls+6 —1 *7: +11 3 .2913 + 7) 5[_A =._._.__._._.,._.___.._ +— — {( )} 33+652+lls+6 32+3s+2 s3+632+lls+6 115+19 2 32+173+28 9+5;2 +lls+6 52+35+2 53+652+lls+6 Sta—3’ —3e'2" —4e" 2am —Ze" 89'3" —2e'2' —69" e" = K‘ {(51 — A)"} = 524‘ — 32"" — 2e“ 22‘” ﬂ e“ 524' 7 22'” u 3e“ —3! —7e +3!” + 42” —29A2' +22" 49‘3" + 29‘” +62" 5) —60 30 210 B = 40 —20 “I40 —20 10 70 “L5 6 3.5 l 4125 —3.5 To convert Jordan form, use modal matrix M = 4 -4 0 and M "i = l —05 —3.5 —l 2 1 -—l 0.75 4.5 0 0 0 J=M"AM= o —10 o 0 0 0 New, compute e" l 0 0 air _ 0 9—10! 0 0 0 I Now, ﬁnd e‘“ = Inﬁe‘l'jlafl e)“ = MeJlM—l 5+6e"°’ 3—324“ 21—212'1‘“ = 4—424” —1+Ze"°' “14+l4e'w' ﬁ2+2e#lm l_e-IOI 8—72-WI‘ .9 Find a" for the given matrices: _4 0 (1) b a w ) —8 7 _ c ' a) 0 7 ) —m a 4 3 0 -9 -1 Solution: —4 0 O a) A x 0 —7 l eigenvalues: A. = —4, —4, — 4 (m = 3) 0 —9 —1 complete Jordan form: Rank[(A — 9.01] = o (= n — m) Index = 2 l 0 0 Solutionof(A—m2x=0 are x]: 0,x2=1,x3= 0 0 0 1 Only one of these satisﬁes (A — m2): at O [NDEPENDENTLY (i.e., gives an independent result) 0 0 0 (14—le3: 1 (A—ml = o 3 3 o I l 0 The other eigenvector is 0 , (Aim 0 = 0 0 U 0 Modal Matrix: 1 {l O 1 0 0 M: 010 :henM": 010 0 3 1 0 73 1 —4 0 0 J=M“AM: 0 —4 1 0 D —4 Now, compute 2”" ed" 0 0 E.” _ 0 6—4: (8—4! 0 0 e“: Now, ﬁnd e"if = MeJ'M’] ed: = M61: M71 2"“ 0 0 = 0 Bah—3:64: re“h 0 —9Ie"“ 5""l +3re_‘" a to . :| eigenvalues: A = a i 1m b) A-[ ~(D0‘. Use the e’“ =(1lA-l-(101 method. (eh :alk+ao) e(“+""“‘)' : aim + jcu) + an (em—"W 2 ohm — jm) + an 8“)”ij = emﬂm" -0L](C! +jm) +u](a —j<u) _ m 8}“)! _e—;wr em solvmg, a; =e ——- I =——sinwr 2,10) co then I a0 = e‘“”"’}’ ——sinmt(u + jun) u) _ _ G. . _ . = ea'{cosmr + 15m 0347—2” smcot - Je‘“ 5mm! 0) 0'. . = em C030}: ——Sm0)! (n 50, 0c . e“’[coswz — —sm not] 0 eAI=0l|A+0LDI= m (X . 0 em[cosmr — —smcot] (I) 01'. + —e°‘sinmr emsinmt [garcoswt ea'sinmt] m = C” ' CU _eut sin an em coswr E! Slﬂmf 8 C050}! »3 7 , _ 1 7 c) A: 3 elgenvalues: kl =7], l2 =—4 elgenvectors: x1 = , x2 = -4 l 4 50, M: 1 7 7| _ -% Z J: _1 O I 4” )4 —K’ 0 74 eAszeJ'M_l= I 7 9—! Eu 7% 7/; 4 0 e Z 34 ‘— _%e—.‘ +%e-4r _}e—{“%€—4r _%e~t +%e—4f %e-I _%e-4f 0 0 O 5.10 Findaclosed-form expression for AR, k21,where A: 9 23 30 ‘ -7 —18 —23.5 Solution: 1 Eigenvalues: 0, —1 solving, ...
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