Mah. Two-Hours Exam December, 7. JE/JKL..7 Problem resar;wih(linearalgebra): Given he inhomogeneous linear sysem of equaions lign:=x-*x+3*x3=a^+*a-3; lign x x 3 x3 = a a 3 lign:=x+*x-*x3=a^+3; lign x x x3 = a 3 where a is an arbirary real number. The coefficien marix A and he righ-hand side b is A,b:=GeneraeMarix([lign,lign],[x,x,x3]); Quesion a:=: The linear sysem of equaions is hen lign; x x 3 x3 = 8 lign; A, b The augmened marix T = [A b] is T:=GeneraeMarix([lign,lign],[x,x,x3],augmened=rue); 3 8 T 4 ha has he reduced row echelon form rap('t'):=reducedrowechelonform(t); 6 rap T Wih Maple we ge x:=linearsolve(t,free=); 3 a, a 3 a 3 x x x3 = 4 The compleely reduced linear sysem of equaions is hen x x 3 = 6 x x 3 = if we se x 3 = we ge he complee soluion in sandard parameric form (x, x, x 3 ) = 6,,,,,.
6 3 x 3 From his we read he complee soluion in sandard parameric form (x, x, x 3 ) = 6,,,,,. 3 Quesion a:='a': x:=<-7,7,-; x 7 7 x = 7, 7, is a soluion o he sysem of equaions Ax = b. resar;wih(linearalgebra): Given he symmeric marix A:=<<,,,- <,,-, <,-,, <-,,,; The characerisic polynomial for A in compleely facorized form is P(lambda):=facor(Deerminan(A-lambda*IdeniyMarix(4))); P A.x=b; 4 = a a 3 a 3 From his we see ha Ax = b a 3 = and a a 3 = 4 a = 9 and a a = a = 3 and a a =. By inserion we see ha only a = 3 fulfills he quadraic equaion a a =. Thus, (x, x, x 3 ) = 7, 7, is a soluion o he sysem of equaions a = 3. Problem Quesion A From his we read ha he characerisic polynomial for A has he double roos and which means ha A has he eigenvalues and boh wih he algebraic mulipliciy. Quesion g:=(x,y)-evalb(x[]<y[]):
ev:=sor(eigenvecors(a,oupu=lis),g); ev,,,,,,, Maple gives orhogonal basis vecors for boh E and E. Since he wo eigenvecor spaces E and E are orhogonal, because A is symmeric, we ge an orhonormal basis for 4 equipped wih he ordinary scalar produc by norming he four basis vecors shown. q:=/sqr()*<,,,: q:=/sqr()*<,,,: q3:=/sqr()*<-,,,: q4:=/sqr()*<,-,,: If Q:=<q q q3 q4; and Lambda:=DiagonalMarix([-,-,,]); Q hen Q is orhogonal and Q A Q = A = Q Q = Q Q T. Check using Maple Transpose(Q).Q; A=Q.Lambda.Transpose(Q); =
Quesion 3 We consider he linear map f : 4 4, ha has e F e = [ e f (e ) e f(e ) e f (e 3 ) e f(e 4 ) ] = A. efe:=a; efe Furhermore we consider he -dimensional subspace U = {u 4 f(u) = u} in 4. I is seen ha U = E. v:=</,/,/,/; v Since ef('v')=efe.v; ef v = hen f(v ) = v. From his i follows ha v U. Also, i is seen ha v = (,,, ) + (,,, ), where (,,, ) and (,,, ) according o quesion are wo basis vecors for U. From his i follows ha v U. v = (,,, ) (,,, ) = (,,, ) is also a uni vecor in U. Since he wo vecors v and v are orhogonal, hen he se (v, v ) is an orhonormal basis for U. Of course, we could also have used "Gram-Schmid" wih e.g. u = (,,, ) U: w = u (u v ) v v =
w w. Problem 3 resar;wih(linearalgebra): assume(,real): inerface(showassumed=): In he vecor space (R,C) a 4-dimensional subspace U is given by is basis: a = (cos, sin, e cos, e i ). A linear map f : U U is given by he expression f:=x-diff(x,,)-*diff(x,)+*x; f x x x x Quesion The images of he wo las basis vecors are f(exp()*cos()); f(exp((+i)*)); Since f (e cos ) = for all and f (e i ) = for all, hen he wo basis vecors e cos and e i belongs o he kernel for f. Quesion The images of he wo firs basis vecors f(cos()); cos f(sin()); sin sin cos From his and quesion we read ha afa:=<<,,, <-,,, <,,, <,,,; afa (Concerning e.g. he second column: f (sin()) = - cos() + sin() ) Quesion 3 The coordinae vecor for cos wih respec o he basis a is a_cos:=<,,,;
a_cos f (x()) = cos a f (x()) = a cos a F a a x() = a cos. x:=linearsolve(afa,a_cos,free=c); x c 3 I.e. all soluions o he equaion are x() = cos sin c e cos c e i, R, c, c. f(cos()-*sin()+c*exp()*cos()+c*exp((+i)*)); cos as expeced. Exercise 4 resar;wih(linearalgebra):wih(plos): x x = 4 Quesion x x, R. c 4 A:=<<4, <-,-; A 4 g:=(x,y)-evalb(x[]<y[]): sor(eigenvecors(a,oupu=lis),g);,,,,, From his we read ha all eigenvalues for he sysem marix A are and - wih he corresponding eigenvecor spaces E = span and E =. E and E are linearly inddependen, since he corresponding eigenvecors are differen. The complee soluion o he sysem of differenial uquaions is hen
Or wrien ou in Maple x:=-c*+c**exp(-): x:=-c*+c*exp(-): 'x[]()'=x(); x = c x x = c 'x[]()'=x(); Quesion c e wih corresponding augmened marix T:=<<, <, <,; ha has he reduced row echelon form rap('t'):=reducedrowechelonform(t); rap T, R, c, c R. x = c From he figure we read ha (x, x ) = (,). c e c e Therefore, we have for he deerminaion of he consans c og c he linear syem of equaions x x = c c = T c c =. From his we read ha c:=; and c:=-; The soluion wished for is hen x = e, R. x Or wrien ou in Maple 'x[]()'=x(); 'x[]()'=x(); c c x = x = 4 e e where R. Since e ends owards for ending owards infiniy, hen x ends owards and x owards, as ends owards infiniy.
P:=plo(x(),=..,hickness=3,color=blue): P:=plo(x(),=..,hickness=3,color=red): punker:=poinplo([[,],[,]],symbol=solidcircle, symbolsize=): display(p,p,punker,scaling=consrained,view=..); 8 6 4 Quesion 3 Or wrien ou in Maple 'x[]()'=x(); 4 6 8 Oher non-consan soluions, ha fulfills he limi values wished for, can be found by fixing c = and changing c. E.g. c:=;c:=; The soluion wished for is hen x = e, R. x 'x[]()'=x(); c c x = x = e e
where R. P:=plo(x(),=..,hickness=3,color=blue): P:=plo(x(),=..,hickness=3,color=red): display(p,p,scaling=consrained,view=..); 8 6 4 4 6 8