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MAth 22 - 8 For-3 0-l A 3 23 2 0 0 wehave —3——}t 0 ~1...

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Unformatted text preview: 8. For -3 0 -l A: 3 23 2 0 0 wehave —3——}t 0 ~1 A-—AI= 3 2~A 3 We can compute the characteristic polynomial pot) = —- det(A — M) by expanding along the sec— ond column to get -3 _ A —-1 = (A m 2)(i2 +3)» +2) = o. — 2m +1)(?~ + 2). Hence the eigenvalues are A1 = ——2, A2 = —1, and A3 = 2. For M = —2 we have ~1 —1 A—MI=A+21= 3 0&0 3 2 The nullspace is generated by the vector v1 (—1, 0,1)T. For A2 = —l we have ll —2 0 ~l A—AZI=A+I: 3 3 3 2 0 l The nullspace is generated by the vector v2 (1,1, —Z)T. For A3 : 2 we have ~5 0 4‘ A—Agle—M: 3 0 3 l ,4, n r: i \ L U ”A l \ ,1 ll The nullspace is generated by the vector v3 = (0, ], 0)T. Thus wehave three exponential solutions: —1 371(1) = eAltVl = 6‘2’ 0 l 1 mo = elz’Vz = e” -—2 0 y3(t) = eA3’v3 = ezt l 0 Since the three eigenvalues are distinct, these solu— tions are linearly independent and form a fundamen— tal set of solutions. The general solution is YO) = C1Y1(t) + CZYZU) + C3Y3(t)‘ 14. The solution has the form Y“) = CiY1(t)+ C2Y2(_t) + 033%). where y1, y2, and y3 are the fundamental set of solu— tions found in Exercise 9.4.8. Hence we must have 1 "1 = y (0) 2 = C1Y1(0) + C2Y2(0) + C3Y3(0) =QW+QW+QW C1 =[V1,V2,V3] C2 , C3 where v], V2, and V3 are the eigenvectors of A found in Exercise 9.4.8. To solve the system we form the augmented matrix —1 l 0 1 [V1, V2, V3, y(0)] : 0 l l —-l 1 —2 0 2 This is reduced to the row echelon form 100- 010—3. 0012 Backsolving, we find that C; = —4, C2 = —3, and C3 = 2. Hence the solution is —1 1 0 y(t) = —4e*2‘ 0 — 3e“ 1 +2e2t 1 1 —2 0 46‘” ~ Seq = 9—3?’ + 262‘ . \«46‘2’ + 66"} W ‘ 'z‘atmumwv nth at t."y‘r'.x.=5.'i'1':l«rzf'7frfi".\47r.,"v‘i".#"."nv. E i ...
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