MATH 286 Sections D1 & X1: Assignment 6Solutions for the graded problems.Section 5.1, #28 [2 points] To verify thatx1=16−13= eλ1tv1, x2= e3t23−2= eλ2tv2, x3= e−4t−121= eλ3tv3are solutions tox0=1 2 16 −1 2−1 −2 −1| {z }=Ax,we just need to check that viis an eigenvector of A with eigenvalue λifor each i = 1, 2, 3.That is, we must show that(A − λiI)vi= 0 (i = 1, 2, 3).We leave this calculation to the reader. Next, to verify linear independence, we calculatethe WronskianW(t) = det[ x1(t) x2(t) x3(t)] = det1 2e3t−1e−4t6 3e3t2e−4t−13 −2e3t1e−4t= e3te−4tdet1 2 −16 3 2−13 −2 1= −84e−t.Since W(t) 6= 0 for all t, we see that these solutions are linearly independent. The generalsolution is then given by a general linear combinationx(t) =3∑i=1cixi(t).Section 5.2, #38 [2 points] Given A =1 0 0 02 0 0 00 3 3 00 0 4 4, the eigenvalues λ are given bydet(A − λ I) = 0 ⇐⇒ det1 − λ 0 0 02 2 − λ 0 00 3 3 − λ 00 0 4 4 − λ= 0⇐⇒ (1 − λ)(2 − λ)(3 − λ)(4 − λ) = 0.1So the eigenvalues are λ = 1, 2, 3, 4. The corresponding solutions to the ODE x0= Axarex1(t) = etv1, x2(t) = e2tv2, x3(t) = e3tv3, x4(t) = e4tv4,where vλ=aλbλcλdλ6= 0 is an eigenvextor associated to the eigenvalue λ. I.e., vλsolvesthe equation(A − λI)vλ= 0 (λ = 1, 2, 3, 4).For λ = 1, this gives0 0 0 02 1 0 00 3 2 00 0 4 3a1b1c1d1=0000⇐⇒0 = 02a1+ b1= 03b1+ 2c1= 04c1+ 3d1= 0.Setting a1= 1, we get b1= −2, c1= 3, d1= −4. Sox1(t) = eta1b1c1d1=et−2et3et−4etFor λ = 2, 3, 4, we proceed similarly to getx2(t) =0e2t−3e2t6e2t, x3(t) =00e3t−4e3t, x4(t) =000e4t.Since v1, v2, v3, v4are linearly independent, the general solution is thenx(t) = c1x1(t) + c2x2(t) + c3x3(t) + c4x4(t).Section 5.2, #24 [2 points] The given system of equations, written in matrix form, isx0=2 1 −1−4 −3 −14 4 2| {z }Ax.To compute the eigenvalues of A, we solve the characteristic equation0 = det(A − λI) = det2 − λ 1 −1−4 −3 − λ −14 4 2 − λ= −(λ − 1)(λ2+ 4) =⇒ λ = 1, ±2i.2Now we compute eigenvectors. First consider λ = 1 and an associated eigenvectorv1= [a, b, c]T. Then(A − I)v1= 0 ⇐⇒1 1 −1−4 −4 −14 4 1abc=000=⇒ v1=abc=1−10is an eigenvector andx1(t) = et1−10is a solution. Next consider λ = 2i and an associated eigenvector v = [a, b, c]T. Then(A − I)v = 0 ⇐⇒2 − 2i 1 −1−4 −3 − 2i −14 4 2 − 2iabc=000=⇒ v =abc=−1 − i2−2is a solution giving a complex eigenvector. This gives a (complex) solutionx(t) = e2it−1 − i2−2= (cos( 2t) + i sin(2t))−1 − i2−2=− cos(2t) − i cos(2t) − i sin(2t) + sin(2t)2 cos(2t) + 2i sin(2t)−2 cos(2t) − 2i sin(2t).Taking real and imaginary parts of x, we get two real-valued solutionsx2(t) =− cos(2t) + sin(2t)2 cos(2t)−2 cos(2t), x3(t) =− cos(2t) − sin(2t)2 sin(2t)−2 sin(2t)Thus the general solution is of the form x(t) = c1x1(t) + c2x2(t) + c3x3(t). Equivalently:x(t) =c1et+ c2(− cos(2t) + sin(2t)) + c3(− cos(2t) − sin(2t))−c1et+ 2c2cos(2t) + 2c3sin(2t)−2c2cos(2t) − 2c3sin(2t).Section 5.4, #14 [2 points] For this system,A =0 0 1−5 −1 −54 1 −2=⇒ det(A − λI) = −(λ − 1)3.So A has a multiplicity 3 eigenvalue λ = −1. To find the defect, we solve for eigenvectorsv = [a, b, c]T:(A + I)v = 0 ⇐⇒1 0 1−5 0 −54 1 −1abc=000⇐⇒ v =a−5a−a.3Since the only free variable is a, we get just one eigenvector (up to scaling) given byv = [1, −5, −1]. Therefore the defect of λ is 2 and we need to build a length 3 chain{v1, v2, v3} of generalized eigenvectors based at an eigenvector v1. In this case, thesethree non-zero vectors must satisfy the equations v1= (A + I)v2and v2= (A + I)v3.Taking our base eigenvector to be v1= [5, −25, −5] (so that our solution matches theback of the texbook!), we can then solve for v2, v3:(A + I)v2= v1⇐⇒1 0 1−5 0 −54 1 −1abc|{z}v2=5−25−5=⇒ a + c = 5 & 4a − (5 − a) + b = −5.Setting the free variable a = 1, we get v2= [1, −5, 4]T. Now we solve for v3= [a, b, c]T:(A + I)v3= v2⇐⇒1 0 1−5 0 −54 1 −1abc|{z}v3=1−5−4=⇒ a + c = 1 & 4a − (1 − a ) + b = −4Setting the free variable a = 1, we get v3= [1, −8, 0]T. The general solution then isx(t) =3∑i=1cixi(t), where x1(t) = e−tv1, x2(t) = e−t(tv1+ v2), and x3(t) = e−t(t22v1+ tv2+ v3).Section 5.4, #24 [2 points] We havex0=28 50 10015 33 60−15 −30 −57xFor the eigenvalue λ1= 2, we look for an eigenvector:v1=abc=⇒ (A + 2I)v1= 0 ⇐⇒30 50 10015 35 60−15 −30 −55abc=000=⇒ v1=53−3is an eigenvector. Next, we consider λ2= 3, which has multiplicity 2. An eigenvectorw = [a, b, c]Tin this case must satisfy(A − 3I)w =25 50 10015 30 60−15 −30 −60abc=000⇐⇒ a + 2b + 4c = 0 ⇐⇒ w =ab14(−a − 2b).4Since a, b ∈ R can take any value, we get two linearly independent eigenvectors. Forexamplew1=40−1(setting a = 4, b = 0) w2=2−10(setting a = 2, b = −1)Then the general solution isx(t) = c1e−2tv1+ c2e3tw1+