Can you please help me with number 12 and number 15. 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 1 =2.2.2 parts of the complex-valued solutions wie" and (vit + v,jel OLEN 1 = -1. -1.2.2 to find four independent real-valued solutions of x' = Ax. GO- ALL- 34. The characteristic equation of the coefficient matrix A of the system x: A=-1. -1.2.2 12 -25 33 10 90 32 x 1=LI LI i -27 -3 () = 0' - 4 +13) =0. -1 26 Therefore, A has the repeated complex conjugate pair 3 2 + 3/ of eigenvalues. First show that the complex vee- tors -48 -138 V = [-4 3+37 0 - ]. 1 =2.2.3.3.3 v1 = [3 -10 +91 -i o]' The characteristic equation of the coefficient matrix A of the system form a length 2 chain [v1. V;) associated with the cigen- value A = 2 + 3/. Then calculate (as in Problem 33) four independent real-valued solutions of x' = Ax. 35. Find the position functions , (f) and x; (f) of the railway cars of Fig. 5.6.1 if the physical parameters are given by (2) = (23 - 62 + 25)' = 0. and the initial conditions are Therefore. A has the repeated complex conjugate pair 3 + 4/ of eigenvalues. First show that the complex vec- (0) -x(0) =0. .(0) = 1;(0) = 1. tors How far do the cars travel before stopping? v = [ 1 i o o]" and v; = [9 0 1 6 ]' 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. v=] associated with the eigen- that, before stopping. the cars travel twice as far as those value 2. = 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 a homogeneous linear system x' = Ax (1) can be used to construct a square matrix X = (r) that satisfies the matrix differen- rial equation X' = AX (1')734 Answers to Selected Problems 20. Eigenvalue A = 2 with multiplicity 4 and defect 3: all) =( + + er + at + fart + lar'jed. 21. Eigenvalue & = I with multiplicity 4 and defect 2: n(1) = (-2 + 1 - 20,1). not) = (+el); 21. Eigenvalue & = I with multiplicity 4 and defect 2: 13. "(1) = give" + (eng + civile" with m = [ 1 -1 2]. -5-4-3-2-1012 ") = [4 0 9].= [0 2 1 ] 24. x(1) = give" + (eavg + cifiled withw, = [$ 3 -3]. 7. Eigenvalues A = 2, 2, 9 with three linearly independent 1=[4 0 -1].) = [2 -1 0] cigenvectors; () - ge' + gr .x(!)=ge +ge". 25. x(1) = [ avita(vir + ")+ c (fvr' + var+ ")led with =[-1 0 -1]. - [-4 -1 0],and 8. Eigenvalues A = 7. 13, 13 with three linearly independent =[1 0 0] eigenvectors; x, (1) = 2ce - ce . 1,(1) m -3ce" +gjel. 26. x(1) = [ av + atvs+ ")+ a(lvr + was+ ")le" with 9. Eigenvalues A = 5, 5, 9 with three linearly independent v =[0 2 2].", = [2 1 -3 ]. and eigenvectors; (1) = ge* + 7ge* + 3r". .,(1) = 2ge". [1 0 0] n() = 2ce* + ge" 27. "(1 ) = lov, + cq (v,1 + vi) + civile" with 10. Eigenvalues A = 3. 3, 7 with three linearly independent " =[ -5 3 8] . ", = [ 1 0 0 ] . ", = [ 1 1 0] igenvectors; (1) = Scie - Jeje" + 2cje". 28. x(0 ) = [ av, + elvis+ w.)+ a(/vr'+ +.)] with V1 = [119 -289 0 ] . V3 = [-17 34 17 ] , and 11. Triple eigenvalue & = -1 of defect 2: x() =(-20+ 5 - 2rifle". 29. x(1) = lan + galvis + ville"+ long + alvis + ville" with 1= [1 -3 -1 -2 ]. = [0 1 0 0 ]. 12. Triple eigenvalue & = -I of defect 2: =[0 -1 10]. . =[0 0 2 1 ] 30. "(1 ) = lev + c(v/+ willed+ lev, + alve + ville". with 1 = [0 1 -1 -3] . > = [0 0 1 2 ]. =[-10 0 0] . 0 0 3 5] 13. Triple eigenvalue A = -1 of defect 2: 31. x(1) =[av + (vi+vi) +g(w/+ vit v) +able with n() = (2+ + 2gile". 1 = [42 7 -21 -42 ]'. 14. Triple eigenvalue & = -I of defect 2: = [ 34 22 -10 -27 ] . . = [-1 0 0 0]. W=[0 1 30] n() =e"(-250 - 50 - 25cat - 5cy - Hey!). 32. x(1) = (give + savile" + (ev, + cave te,valet with n(!) =e"(-50 + 40) - Sept + 4ejl - lar) V =[8 0 -3 1 0] . > =[1 0 0 0 3 ]. 15. Triple eigenvalue & = 1 of defect I; V) = [3 -2 -1 0 0 ]. W. = [2 -2 0 -3 0 ]. = [1 -1 0 0 3]' 16. Triple eigenvalue A = 1 of defect I; 33. x, (1) = [cos4r sin4 0 0]e. x;(1) = [ - sin4r cos4r 0 0 ]fe. x()=(-2n - 21). x() =(-2 +261) x,(1) = [reosir f sinar coser sin4: ]" ". 17. Triple eigenvalue & = 1 of defect I: (() = [-Isin4r rcoser - siner cosAre x() = (20 + cle. x;(1) = (-3+ +6cle. sin 3r x (1) = -9( + cthe 34. x, (1) = 3cos 3/ - 3 sin 3r 18. Triple eigenvalue A = 1 of defect 1: 0 sin 3r 10)=(-0 -2+6). ()=(+gr). - cos 3r *()= 3 sin 3r + 3 cos 3r 0 19. Double eigemalues A = -I and & = 1, each with defect I; cos 3r 3 cos 3r + 1 sin Jr *()= (3r - 10) cos 3r - (3r + 9) sin 3r sin 3r f sin 3rChapter 5 Linear Systems of Differential Equations 394 CASE 1: 21 = 5. The eigenvector equation (A-Al)v = 0. where v = [a b ep is (A - 50)- - -8 -3 8][2]- [8] Each of the first two equations, 4a + 46 = 0 and -60 - 6b = 0. yields b = -. Then the third equation reduces to 2a - 2c = 0. so that c = a. The choice a = | then yields the cigenvector associated with the eigenvalue A, = 5. CASE 2: 2g = 3. Now the cigenvector equation is (A - 30)0 - -8 -3 8][8]-[8] so the nonzero vector v = [a b c ] is an cigenvector 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 V) = [2 -3 0] is a second linearly independent eigenvector associated with the multiplicity 2 eigen- value A? = 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 x() = get x(1)=-ge - 3cge". x(1) = ge"+ eze".5.6 Multiple Eigenvalue Solutions 395 Remark: Our choice in Example I of the two cigenvectors v2 = [0 0 1 ]" and v) = [2 -3 0] associated with the repeated eigenvalue &2 = 3 bears comment. The fact that b = -ga for any eigenvector associated with A? = 3 means that any such eigenvector can be written as and thus is a linear combination of v; and vy. Therefore. given a and c not both zero, we could choose v rather than vy as our third eigenvector, and the new general solution x (1 ) = give" + cave"+ cave" would be equivalent to the one in Eq. (7). Thus we need not worry about making the "right" choice of independent cigenvectors 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 = [1 4] (8) has characteristic equation IA - All = 3 7- 2 = (1 -2)(7-2) +9 = X' - 81 + 16 = (. -4)' =0. Thus A has the single eigenvalue 21 = 4 of multiplicity 2. The eigenvector equation (-40=[3 3]-[8] then amounts to the equivalent scalar equations -3a - 3b =0, 3a + 3b =0. Hence b = -a ifv = [a b ] is to be an eigenvector of A. Therefore any eigen- vector associated with 21 = 4 is a nonzero multiple of v = [ 1 -1 ] . Thus the multiplicity 2 eigenvalue 21 = 4 has only one independent cigenvector, and hence is incomplete. Page 4 / 15An eigenvalue & of multiplicity & > I is called defective if 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 vi associated with the defective cigenvalue A. Then at this point we have found only the single solution *1(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 x(1) = (vr)e = view. (11) When we substitute x = vifell in x' = Ax, we get the equation view + Avare" = Avate". But because the coefficients of both cy and red must balance. it follows that vy = 0, and hence that x;(/) = 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 vit with vit + 12. Thus we explore the possibility of a second solution of the form * = (1) = ( vit + vz)e" = vile"+ view (12) where v, and v, are nonzero constant vectors. When we substitute x = vice" +vie in x' = Ax, we get the equation vie" + avite" + avge" = Avite" + Avge". (13) We equate coefficients of e" and fell here. and thereby obtain the two equations (A - AD)V) = 0 (14) and (A - AD)V? = VI (15) that the vectors v) and va must satisfy in order for (12) to give a solution of x' = Ax.5.6 Multiple Eigenvalue Solutions 397 Note that Eq. (14) merely confirms that V, is an eigenvector of A associated with the eigenvalue A. Then Eq. (15) says that the vector va 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)-v; = 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 vy of the equation (A - ADV2 =0 (16) such that (A - AD)Y] = VI (17) is nonzero, and therefore is an eigenvector v1 associated with A. 2. Then form the two independent solutions XI (1) = View (18) and x; (1) = (vir+ valet (19) of x' = Ax corresponding to A. 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 eigenvalue ) = 4 of multiplicity 2. We therefore begin by calculating (A-40)3 - [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 va though not for others. IfChapter 5 Linear Systems of Differential Equations 398 we wy vz = [ 1 0 ]' we find that (A - 40)2 - [3 3]-[-3]-~ is nonzero, and therefore is an eigenvector associated with ) = 4. (It is -3 times the eigenvector found in Example 2.) Therefore the two solutions of Eq. (20) gives by Eqs. (18) and (19) are x (1) = ve"=[3]. x(1) = (vil +vzle" = -35+17. The resulting general solution x (1) = CIxi (1) + cax=(1) has scalar component functions x(1) = (-3cat + cz -3cile". x(1) = (3cyr + 3ci)e". With cz = 0 these solution equations reduce to the equations x (1) = -3cje". x(1) = 3cje". which parametrization) = -x2 in the xx2-plane. The point (x (1). x2(1)) then recedes along this line away from the origin as / - +co. to the FIGURE 5.6.1. Direction field northwest if c, > 0 and to the southeast if c, (recalling the convention that the Oth power of a square matrix is the identity matrix). Thus a rank I generalized eigenvector is an ordinary eigenvector. The vector vz in (16) is a rank 2 generalized eigenvector (and not an ordinary eigenvector). The multiplicity 2 method described earlier boils down to finding a pair ( v. v;] of generalized eigenvectors, one of rank I and one of rank 2. such that (A - AI)] = V1. Higher multiplicity methods involve longer "chains" of generalized eigenvec- tors. A length & chain of generalized eigenvectors based on the eigenvector v, is a set (v1. V2. .... Vi] of k generalized eigenvectors such that (A - AI)VI = VI-1. (A - AI)VA-1 = VA-2. (22) (A - AI)VE = VI.5.6 Multiple Eigenvalue Solutions 399 Because , is an ordinary eigenvector. (A - Al)v, = 0. Therefore, it follows from (22) that (A - 21)'V1 =0. (23) If (v1. V2. va] is a length 3 chain of generalized eigenvectors associated with the multiple eigenvalue & of the matrix A, then it is easy to verify that three linearly independent solutions of x' = Ax are given by XI (!) = vie" x1 (1) = (vil + vile". (24) x] (1) = (;virtualtv)e". For instance, the equations in (22) give Avj = V1 + AVJ. AV; = VI + AVE. AVI = AVI. SO AX) = [;AVI + Avar + AVI] = (vit + vale" + > (jvir? + var+ vs)" Therefore, Xy(1) in (24) does, indeed, define a solution of x' = Ax. Consequently. in order to "handle" a multiplicity 3 cigenvalue A, it suffices to find a length 3 chain (v1. V2. v,] of generalized eigenvalues associated with A. Looking at Eq. (23), we see that we need only find a solution vy of (A - 10)'V) =0 such that the vectors V2 = (A - AD)v, and VI = (A - Al)v; are both nonzero (although, as we will see, this is not always possible). Example 4 Find three linearly independent solutions of the system x (25) Solution The characteristic equation of the coefficient matrix in Eq. (25) is IA - All = = 1.1-7-2 .(-3-A)]+ (-A)[(-2)(-3-2) +51 = -3 -323 - 34 - 1 = -(. +1)' =0,400 Chapter 5 Linear Systems of Differential Equations and thus A has the eigenvalue & = -1 of multiplicity 3. The cigenvector equas (A - Al)v = 0 for an eigenvector v = [a b c] is (A + 1)V - 4 -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 is 2. To apply the method described here for triple eigenvalues, we first calculate (A + 1)? = and Thus any nonzero vector v) will be a solution of the equation (A + D'v, = 0 Beginning with vy = [ 1 0 0 ] . for instance, we calculate Note that v, 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. v;] of generalized eigenvectors associated with the triple eigenvalue A = -1. Substitution in (24) now yields the linearly independent solutions -2r + 1 x (1 ) = (vit+ vz)e-'= -21 - 5 21 + 1 x 3(1 ) = (fvs'+ with)e'= -13 -51 of the Page 9 / 15 +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 1 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 eigenvectors associated with a multiple cigenvalue A. and therefore know the defect d of A, we can begin with the equation (A - AD) +0=0 (26) to start building the chains of generalized eigenvectors associated with A. ALGORITHM Chains of Generalized Eigenvectors Begin with a nonzero solution uj of Eq. (26) and successively multiply by the matrix A - 21 until the zero vector is obtained. If (A - AD)uj = 13 # 0. (A - AD)-1 = 1) # 0, but (A - AD)up = 0, then the vectors (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 2: *1 (1 ) = VIP". x3 (1) = (vil + Valet, x,(1) = ()vir + var+ vy)eh. (27) x(1) = (4 - 1)!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) -(etc)] 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(). x() = x(1). and x(1) = x;(1). Ifm = m; = ] we get X' = Ax (30) where now x = [x x2 x3 x4 ] and 0 A = - (c+ c ) (31) - (e + c). Example 6 With my = my = e = 1 and * = c) = cy = 2, the system in Eq. (30) is - NNOO NNOO (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- tion 24 +623 + 1213 + 82 = 2(2 + 2)' =0 of the coefficient matrix A in Eq. (32). Thus A has the distinct eigenvalue do = 0 and the triple cigenvalue 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 No = [1 1 0 0] is an eigenvector associated with do = 0.Chapter 5 Linear Systems of Differential Equations 404 CASE 2: 21 = -2. The eigenvalue equation (A - AD)v = 0 is 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 two cigenvectors associated with the triple eigenvalue 2, = -2. The choice a = 1, b = 0 yields c = -2, d = 0 and thereby the eigenvector U1 = [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 cigenvector of rank 2 and hence a nonzero solution vy of the equation NN (A + 21) v2 = 00 0 0 V2 = 0. 0 0 0 0 Obviously. V2=[0 0 1-1 ] is such a vector, and we find that (A + 20 2 = ) 9 3 8 170 81-[-31 --- is nonzero, and therefore is an eigenvector associated with A, = -2. Then (v. v,] is the length 2 chain we need. The eigenvector v, just found is neither of the two cigenvectors up and u; found previously, but we observe that v1 = Uj - uz. For a length I chain w, to complete the picture we can choose any linear combination of up and u, that is independent of v1. For instance, we could choose either w, = up or w| = up However, we will see momentarily that the particular choice W=u+ = [1 1 -2 -2] yields a solution of the system that is of physical interest.5.6 Multiple Eigenvalue Solutions 405 Finally, the chains ( vol. [wil. and (v1. Wal yield the four independent solutions XI (!) = voe" = [1 1 0 0]. x(!) = we # = [1 1 -2 -2]e-". x(1)=ve * = [1 -1 -2 2]e-2. (33) x (1) = (vit + Vile-2 = [1 -1 -21 +1 20 - 1 ]e-2 of the system x' = Ax in (32). The four scalar components of the general solution x (1 ) = ex, (1)+ ex(1)+ ex(1)+ cx(1) are described by the equations x(1)=ate( c - - car). (34) x(1) =e-#(-2c2 - 2c) + ca - 2car). x(1) =e-"(-202 + 20) - co + 2car). Recall that x, (f) and x; (r) are the position functions of the two masses, whereas x() = x(1) and x (1) = x,(r) are their respective velocity functions. For instance. suppose that x (0) = x:(0) = 0 and that x; (0) = x;(0) = up. Then the equations x(0 )= + + c = 0, x2(0) =0 + 0 - c = 0. x;(0) = - 202 - 20) + c = 10. x;(0) = - 2+2cy - c = 10 are readily solved for c = job. cz = -fro. and cy = 04 = 0, so x (1) = x(1) = (1-e-=). 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 x, = x] = to as 1- +00. It is of interest to interpret physically the individual generalized eigenvector solutions given in (33). The degenerate (Ag = 0) solution *(1) = [1 1 0 0] describes the two masses at rest with position functions . (f) = 1 and x;(!) = 1. The solution x2(1) = [1 1 -2 -2]e-2 corresponding to the carefully chosen eigenvector w, describes damped motions x (r) = e" and x2(1) = e" of the two masses, with equal velocities in the same direction. Finally, the solutions x;(f) and x,(f) resulting from the length 2 chain (v1. vz) both describe damped motion with the two masses moving in opposite directions