Can you please help me with number 31. I have provided the math chapter it came from and the answer from the back of the book: Thank you so much and Best wishes!

5.7 Matrix Exponentials and Linear Systems 407 x 1 = 2,2,2 parts of the complex-valued solutions view and ( vit + vill x: A= -1. -1.2.2 to find four independent real-valued solutions of X' = Ax. HOW - OWENL 34. The characteristic equation of the coefficient matrix A of the system OOON -1 x Am-1.-1.2.2 -25 -9 33 10 90 32 _ x A= l. 1. 1. 1 is () = 01 -41 + 13) = 0. bCEMENT -1 Therefore, A has the repeated complex conjugate pair 2 + 3/ of eigenvalues. First show that the complex vee- tors -WOW .48 -138 v = [-4 3+37 0 - ]". 1 = 2.2.3.3.3 13 = [3 -10 +9/ -i o]' U. The characteristic equation of the coefficient matrix A of the system form a length 2 chain (v1. V;] associated with the eigen- value A = 2 + 3/. Then calculate (as in Problem 33) four independent real-valued solutions of x' = Ax. 35. Find the position functions x, (f) and x; (f) of the railway cars of Fig. 5.6.1 if the physical parameters are given by *(2) = (23 - 61 +25)' = 0. and the initial conditions are Therefore. A has the repeated complex conjugate pair 3 1 4/ of eigenvalues. First show that the complex vec- *(0) = x;(0) =0. .;(0) = x;(0) = 16- tors How far do the cars travel before stopping? v = [1 / o o]" and v3 = [9 0 1 / ]' 36. Repeat Problem 35 under the assumption that car I is shielded from air resistance by car 2, so now c = 0. Show form a length 2 chain (v1. v2] associated with the eigen- that, before stopping. the cars travel twice as far as those value A = 3 - 4/. Then calculate the real and imaginary of Problem 35. 5.7 Matrix Exponentials and Linear Systems The solution vectors of an a x n homogeneous linear system X' = Ax (1) can be used to construct a square matrix X = $(r) that satisfies the matrix differen- rial equation Page 2 / 15 (1')Open with Chapter 5 Linear Systems of Ations 394 CASE 1: A1 = 5. The eigenvector equation (A-Al)v = 0. where v = [a b ep is (A - SD) = (-8 8 8][:] - [8]- Each of the first two equations, 4a + 4b = 0 and -6a - 6b = 0. yields b = -a Then the third equation reduces to 2a - 2c = 0, so that c = a. The choice a = | then yields the eigenvector associated with the eigenvalue A, = 5. CASE 2: 22 = 3. Now the cigenvector equation is (A - 30)0- -8 -4 8][21-[8] so the nonzero vector v = [a b c ] is an eigenvector if and only if 60 + 46 = 0; (6) that is. b = - ga. The fact that Eq. (6) does not involve c means that c is arbitrary, subject to the condition v # 0. If c = 1. then we may choose a == b = 0: this gives the eigenvector v2 = [0 0 1 ] associated with 22 = 3. If c = 0. then we must choose a to be nonzero. For instance. if a = 2 (to avoid fractions). then b = -3. so VJ = [2 -3 0] is a second linearly independent eigenvector associated with the multiplicity 2 eigen- value 22 = 3. Thus we have found a complete set v1. V2. vy of three eigenvectors associated with the eigenvalues 5, 3. 3. The corresponding general solution of Eq. (5) is x(1) = civje" + cavze" + cavse" (7) with scalar component functions given by * (1 ) = get x(1) = -get - 3cge". x3(1) = ge"+ cze. Page 3 / 15 +5.6 Multiple Eigenvalue Solutions 395 Remark: Our choice in Example 1 of the two cigenvectors v2 = [0 0 1] and v) = [2 -3 0 ]' associated with the repeated eigenvalue Ag = 3 bears comment. The fact that b = -ta for any eigenvector associated with A? = 3 means that any such cigenvector can be written as and thus is a linear combination of vy and vy. Therefore, given a and c not both zero, we could choose v rather than v, as our third eigenvector, and the new general solution x(1 ) = civic" + cavie" + cave" would be equivalent to the one in Eq. (7). Thus we need not worry about making the "right" choice of independent eigenvectors associated with a multiple eigenvalue. Any choice will do; we generally make the simplest one we can. Defective Eigenvalues The following example shows that-unfortunately-not all multiple eigenvalues are complete. Example 2 The matrix A = [: 4] (8) has characteristic equation IA - All = 3 7-2 = (1 -2)(7 -2) +9 = 12 - 81 + 16 = (2 - 4)' = 0. Thus A has the single eigenvalue 21 = 4 of multiplicity 2. The eigenvector equation (A-40)= [3 3]-[8] then amounts to the equivalent scalar equations -3a - 3b =0, 3a + 3b = 0. Hence b = -a if v = [a b ]' is to be an eigenvector of A. Therefore any eigen- vector associated with 1. a s nonzem multiple of v = [ 1 -1 ] . Thus the Page 4 / 15 4 independent cigenvector, and henceChapter 5 Linear System pen 26 An cigenvalue A of multiplicity & > I is called defective il it is not complete, If A has only p I. The Case of Multiplicity k = 2 Let us begin with the case & = 2. and suppose that we have found (as in Example 2) that there is only a single eigenvector v1 associated with the defective eigenvalue 2. Then at this point we have found only the single solution *I(1) = VIed (10) of x' = Ax. By analogy with the case of a repeated characteristic root for a single linear differential equation (Section 2.3). we might hope to find a second solution of the form x2 (1) = (vit)ed = vite". (11) When we substitute x = vafell in x' = Ax, we get the equation vzell + Avgled = Avare". But because the coefficients of both el and red must balance, it follows that v; = 0, and hence that x2(1) = 0. This means that-contrary to our hope-the system x' = Ax does not have a nontrivial solution of the form assumed in (1 1). Instead of simply giving up on the idea behind Eq. (1 1). let us extend it slightly and replace vat with vit + v2. Thus we explore the possibility of a second solution of the form x2 (1) = (vit + vz)e" = vite" + vzed (12) where v, and va are nonzero constant vectors. When we substitute x = vite" +vie in x' = Ax, we get the equation vie" + avite" + avge" = Avite" + Avie". (13) We equate coefficients of el and tell here, and thereby obtain the two equations (A - AD)V) = 0 (14) and (A - AI)V2 = VI (15) that the vectors v, and vy must satisfy in order for (12) to give a solution of x' = Ax. Page 5 / 15Note that Eq. (14) merely confirms that vi is an eigenvector of A associated with the eigenvalue A. Then Eq. (15) says that the vector vy satisfies the equation (A - 21) v2 = (A - AD)[(A - ADV2] = (A - ADV, = 0. It follows that, in order to solve simultaneously the two equations in (14) and (15). it suffices to find a solution va of the single equation (A - Al)-vy = 0 such that the resulting vector v1 = (A - Al)v; is nonzero. It turns out that this is always possible if the defective eigenvalue ) of A is of multiplicity 2. Consequently, the procedure described in the following algorithm always succeeds in finding two independent solutions associated with such an eigenvalue. ALGORITHM Defective Multiplicity 2 Eigenvalues 1. First find a nonzero solution vz of the equation (A - AD) v2 = 0 (16) such that (A - AD)V1 = VI (17) is nonzero, and therefore is an eigenvector v, associated with A. 2. Then form the two independent solutions *1 (1 ) = Viel (18) and x2 (1 ) = (vit + vz)ed (19) of x' = Ax corresponding to 1. Example 3 Find a general solution of the system (20) Solution In Example 2 we found that the coefficient matrix A in Eq. (20) has the defective cigenvalue > = 4 of multiplicity 2. We therefore begin by calculating (A-40) = [3 3][-3 3 ]-[8 8] Hence Eq. (16) is [8 8 ] .= = .. and therefore is satisfied by any choice of v2. In principle, it could happen that (A - 41)v2 is nonzero (as desired) for some choices of vy though not for others. IfChapter 5 Linear Systems of Differential Equations 398 we try v2 = [1 we find that (A - 41)v2 [3 3]-[3]= is nonzero, and therefore is an eigenvector associated with ) = 4. (It is -3 time the eigenvector found in Example 2.) Therefore the two solutions of Eq. (20) gives by Eqs. (18) and (19) are x, (1) = ve"= [ 3]. x2(1) = (vit + vale" = -31+ 1]. The resulting general solution x(1) = CIXI (1)+ czxz(1) has scalar component functions *(1) = (-3cat + c2 - 3elle#. x2(1) = (3car + 3e;)e". With cz = 0 these solution equations reduce to the equations x, (1) = -3cie". x2(1) = 3cje". which parametrization, = -x2 in the x1x2-plane. The point (x) (1). x2(1)) then recedes along this line away from the origin as / -+ +00, to the FIGURE 5.6.1. Direction field northwest if c, > 0 and to the southeast if c, (fvir? + var + vs)el = X'. Therefore, X3(/) in (24) does, indeed, define a solution of x' = Ax. Consequently, in order to "handle" a multiplicity 3 cigenvalue 2. it suffices to find a length 3 chain (v1. V2. }} of generalized cigenvalues associated with A. Looking at Eq. (23), we see that we need only find a solution v, of (A - AD)'V) = 0 such that the vectors V2 = (A - AD)v) and VI = (A - Al)v2 are both nonzero (although, as we will see, this is not always possible). Example 4 Find three linearly independent solutions of the system (25) Solution The characteristic equation of the coefficient matrix in Eq. (25) is IA - X1 = =1 .[-7-2 . (-3- A)]+ (-A)[(-2)(-3 ->) +5] -(2+ 1) =0. Page 8 / 15 - +400 Chapter 5 Linear Systems ations Open with and thus The eigenvector co (A - Al)v = 0 for an eigenvector v = [ a b c] is (A + DV - - -3 -3] [:]-[8] The third row a + c = 0 gives c = -a. then the first row a + b + 2c = 0 give b = a. Thus, to within a constant multiple. the eigenvalue A = -1 has only the single associated eigenvector v = [ a -a ]' with a # 0. and so the defec of 1 = -1 is 2. To apply the method described here for triple eigenvalues, we first calculate and (+w-[4 4 ][1 7 3]-(8::] Thus any nonzero vector vy will be a solution of the equation (A + D)'v) = 0 Beginning with vy = [ 1 0 0 ] . for instance, we calculate V2 = (A + 1)V) = Note that v1 is the previously found eigenvector v with a = -2; this agreement serves as a check of the accuracy of our matrix computations. Thus we have found a length 3 chain (v1. v2. vy] of generalized eigenvectors associated with the triple eigenvalue A = -1. Substitution in (24) now yields the linearly independent solutions xi(!) = Ve NN -2r + 1 x2(1 ) = (vit+ vz)e"'= -21 - 5 21 + 1 -13+1+1 x ,(1 ) = (qvist vrtv)e= -13 - 51 of the system x' = Ax.5.6 Multiple Eigenvalue Solutions 401 The General Case A fundamental theorem of linear algebra states that every n x n matrix A has n linearly independent generalized eigenvectors. These n generalized eigenvectors may be arranged in chains, with the sum of the lengths of the chains associated with a given eigenvalue A equal to the multiplicity of A. But the structure of these chains depends on the defect of 2, and can be quite complicated. For instance. a multiplicity 4 eigenvalue can correspond to . Four length I chains (defect 0): Two length I chains and a length 2 chain (defect 1); Two length 2 chains (defect 2); . A length I chain and a length 3 chain (defect 2): or . A length 4 chain (defect 3). Note that, in each of these cases, the length of the longest chain is at most d + 1. where d is the defect of the eigenvalue. Consequently, once we have found all the ordinary cigenvectors associated with a multiple eigenvalue A. and therefore know the defect d of A, we can begin with the equation (A - AD)+u=0 (26) to start building the chains of generalized eigenvectors associated with A. ALGORITHM Chains of Generalized Eigenvectors Begin with a nonzero solution u, of Eq. (26) and successively multiply by the matrix A - AI until the zero vector is obtained. If (A - AD)u) = uz # 0. (A - AD)UI-1 = Ut # 0, but (A - AD)u) = 0, then the vectors ( v1. V2, ... , VA) = (us. UA-1. . .. . U2, uj) (listed in reverse order of their appearance) form a length & chain of generalized cigenvectors based on the (ordinary) eigenvector VI. Each length & chain (v1. V2. .... VA] of generalized eigenvectors (with v, an ordinary eigenvector associated with A) determines a set of & independent solutions of x' = Ax corresponding to the eigenvalue A: *1 (1) = Vieh x2 (1) = (vil + vz)ed. x, (1) = (fvir] + var + vy)el. (27) *(1) = VIN-1 (K - 1)! 2! Page 10 / 15 +5.6 Multiple Eigenvalue Solutions 403 where M and K are mass and stiffness matrices (as in Eqs. (2) and (3) of Section 5.5), and R = -(etc) -("+ c3)] is the resistance matrix. Unfortunately, because of the presence of the term involv- ing x'. the methods of Section 5.5 cannot be used. Instead, we write (28) as a first-order system in the four unknown functions x (1). x(1). x() = x;(1). and x4(!) = x;(1). If my = my = I we get X' = Ax (30) where now x = [x x2 xy x4 ] and (31) -k - (e + (2 ) Example 6 With m, = my = c = 1 and A = c1 = cy = 2, the system in Eq. (30) is - 3 (32) It is not too tedious to calculate manually-although a computer algebra system such as Maple, Mathematica, or MATLAB is useful here-the characteristic equa- ion 24 + 623 + 1212+ 82 = 2(2 + 2)3 = 0 of the coefficient matrix A in Eq. (32). Thus A has the distinct eigenvalue 20 = 0 and the triple eigenvalue A1 = -2. CASE 1: 20 = 0. The eigenvalue equation (A - Al)v = 0 for the eigenvector v= [a b c d] is The first two rows give c = d = 0, then the last two rows yield a = b. Thus vo = [ 1 10 0 ] is an eigenvector associated with do = 0.Open with inear Systems o Ions 404 CASE 2: 21 = -2. The eigenvalue equation (A - Al)v = 0 is (A + 20- 2 8 7 1708]-08] The third and fourth scalar equations here are the differences of the first and second equations, and therefore are redundant. Hence v is determined by the first two equations, 2a + c =0 and 26 + d = 0. We can choose a and b independently. then solve for c and d. Thereby we obtain no cigenvectors associated with the triple eigenvalue 2, = -2. The choice a = 1. b = 0 yields c = -2, d = 0 and thereby the eigenvector u = [1 0 -2 0]. The choice a = 0. b = 1 yields c =0. d = -2 and thereby the eigenvector uz = [0 1 0 -2 ]. Because 2, = -2 has defect 1. we need a generalized eigenvector of rank 2 and hence a nonzero solution vy of the equation 2 (A + 20)*v2 = 00 0 V2 = 0. 0 0 0 0 Obviously. V2 = [0 0 1 -1 ] is such a vector, and we find that (A + 20302 - 8 6 - 170 8]-[-41 --. is nonzero, and therefore is an eigenvector associated with A, = -2. Then (v1. val is the length 2 chain we need. The eigenvector , just found is neither of the two cigenvectors u, and u; found previously. but we observe that v1 == uj - U2. For a length I chain w; to complete the picture we can choose any linear combination of up and up that is independent of v1. For instance, we could choose either wi = uj or w| = up. However, we will see momentarily that the particular choice WI=uj + up = [1 1 -2 -2 ] yields a solution of the are of physical interest. Page 13 /5.6 Multiple Eigenvalue Solutions 405 Finally, the chains (vol. (wj]. and (v1. V2) yield the four independent solutions XI (!) = Voe"' = [1 1 0 0]. x2(1) = We-# = [1 1 -2 -2]e-# x)(1) = Ve = [1 -1 -2 2]e-3. (33) x. (!) = (vit + Vz)e-2 =[1 -1 -21 +1 21 - 1 ] e-2 of the system x' = Ax in (32). The four scalar components of the general solution x (1) = cixi (1) + czx=(1) + cyx(1) + cox.(!) are described by the equations x(1 ) = ate" ( c+ + car). x(1) = ate=(c-c - car). (34) x(1) = e-#(-202 - 2c3 + ca - 2cat). x(1) = e-=(-202 + 20) - cs + 2cat). Recall that x, (f) and x2(f) are the position functions of the two masses, whereas x3(1) = x;(1) and x4(1) = x;(1) are their respective velocity functions. For instance, suppose that x, (0) = x:(0) = 0 and that x; (0) = x;(0) = 10- Then the equations *(0 )=+ + cy = 0, x2(0) = 01 + C - C 0 . x;(0) = - 202 - 2cy + ca = to. x;(0) = - 202 + 2cy - c4 = to are readily solved for ci = ;up. cz = - vo. and cy = c4 = 0. so x(1) = x(1) = (1 -e-2). x;(1) = x;(1) = the-*. In this case the two railway cars continue in the same direction with equal but ex- ponentially damped velocities, approaching the displacements x1 = x2 = to as 1+ +00. It is of interest to interpret physically the individual generalized eigenvector solutions given in (33). The degenerate (Ag = 0) solution x1 (1) = [1 1 0 0] describes the two masses at rest with position functions x (f) = 1 and x;(!) = 1. The solution x2(1) = [1 1 -2 -2]e-2 corresponding to the carefully chosen eigenvector wj describes damped motions x (p) = and maff) = e of the two masses, with equal velocities in the sa Page 14 / 15 the solution x, (f) resulting from the length 2 damped he two masses moving in oppositeto Selected Problems 20. Eigenvalue A = 2 with multiplicity 4 and defect 3; 21. Eigenvalue A = I with multiplicity 4 and defect 2: n(1) = (-2 + c - 21), x;(1) = (c +elle. a,(t ) = ( + + elle' . x.(1) = (+er+ tere ALONSO-NW 22. Eigenvalue A = I with multiplicity 4 and defect 2: x(1) = -(2c - c + 2elle'. .(1) = lc+cle. 23. x(1) = give" + (cav + civile" with v, = [ 1 -1 2]". -5-4-3-2-101 "> =[ 4 0 9 ]'. ", = [0 2 1 ] 24. x(1) = give" + (avg + civiled with v, = [5 3 -3 ]. 7. Eigenvalues A = 2, 2, 9 with three linearly independent 1 =[4 0 -1]'. ) = [2 -1 0] eigenvectors; a, (!) = ce) + gra. x(!)=ge" +gr". 25. x(1) = [ av + c(vr+ ")+c(;vr' + vr+v.)le" with = [-1 0 -1] . = [-4 -1 0] . and 8. Eigenvalues A = 7. 13, 13 with three linearly independent " =[1 0 0] eigenvectors; x,(1) = 2get - gel. x,(1) =-3ce"+ cell. 26. x(1 ) = [ avita(vis+ vi)+ c (fvir + v+ ville" with 9. Eigenvalues A = 5, 5, 9 with three linearly independent v = [0 2 2] . v, = [2 1 -3 ]". and eigenvectors; x,(!) = ge* + 7ce* + 3eye". x.(1) =2ge*. =[1 0 0] x(1) = 2ce* + ge" 27. x(1 ) = lev, + ca(vit + vi)+ cavsle" with 10. Eigenvalues A = 3. 3, 7 with three linearly independent 1 = [-5 3 8] . = [1 0 0 ] . ", = [1 1 0] eigenvectors; x(1) = Scie - Jeje" + 2eye". 28. x (1 ) = [ avi + c(v+ v.)+c(tvs' + v+v)] with x(1) =2ge" + ge".x,(1) =( V1 = [ 119 -289 0 ]7 . W; = [-17 34 17 ] . and 11. Triple eigenvalue a = -I of defect 2; " =[10 0] 29. x(t) = lev, + ca(vit + ville" + lev, + calves + valled with (1) = (+ elle V = [1 -3 -1 -2 ] . , =[0 1 0 0 ]. 12. Triple eigenvalue & = -I of defect 2: " =[0 -1 1 0 ]. . = [0 0 2 1 ] 30. x(1 ) = [avg + c (vis+ ville"'+ lev, + c(v, + v.)led. with x()=e"(a+ gr+ for'). =[0 1 -1 -3]. > =[0 0 1 2]. n(!) =e"'(a + cl) " =[-10 0 0]. . =0 0 3 5] 13. Triple eigenvalue A = -1 of defect 2; 31. x (1) = [avi + c(v/+ v.) + q (; vr' + vi+ vy)+ cule with x(1) = (20+g+201)e". V1 = [ 42 7 -21 -42 ]'. 14. Triple eigenvalue & = -1 of defect 2; V2 = [34 22 -10 -27 ]". V, =[-1 0 0 0 ]. . =[0 1 3 0 ] x:(1) =e"(-25c) -5eg - 25egl - 5cy - #of!). 32. x(1 ) = (civi+ cavile# + (ov, + cave+ cavsle" with x(1) =e"(-50 + 40 - 5egl + 4ejt - jet?) VI = [ 8 0 -3 1 0 ] .=[1 0 0 0 3 ]. 15. Triple eigenvalue A = 1 of defect I; V) = [3 -2 -1 0 0]. I(1) = (36 + 6 - 3of)e. V. = [2 -2 0 -3 0 ]. , = [ 1 -1 0 0 3] 16. Triple eigenvalue A = 1 of defect 1; 33. x, (1) = [ cos 4/ sin4 0 0 ]'e". x(1) = (36 + 30 + cy) x,(1) = [-sin4r cos4r 0 0 ]e". x(1) =(-20 - 201). n(1) =(-20 + 2gr) x)(1) = [ cos4r / sin4r cos4r sin 4: ] e". 17. Triple eigenvalue A = 1 of defect I; (1) =[-Isin4r rcos4r - sin4r cos4: ]' e n (1) = (2g + cale, x(1) = (-30+ +6cle'. sin 3r x,(1) =-9(6 + cyl)e 34. x, (1) = 3 cos 3/ - 3 sin 3r 0 18. Triple eigenvalue A = 1 of defect 1: sin 3 "() =(-6-20+6). - cos 3r x;()= 3 sin 3r + 3 cos 3r 19. Double eigenvalues A = -I and A = 1, each with defect I; - cos 3r x(1)=gr. 3 cos 3r + / sin 3r (31 - 10) cos 3r - (3r + 9) sin 3r sin 3r / sin 3r Page 15 / 15