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"Heading 1" -1 263 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 257 "" 0 "" {TEXT 326 22 "Laboratoire 10 - Bases " }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 52 "Par Claude St-Hilaire, clau de.sthilaire@videotron.ca" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg):" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 346 51 "Pr incipales commandes utilis\351es dans ce laboratoire" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "vector, augment, gausselim, backsub, basis, Gra mSchmidt." }}{PARA 0 "" 0 "" {TEXT -1 69 "Note : S\351lectionner un mo t et utiliser l'aide pour plus d'information" }}}}{SECT 1 {PARA 263 " " 0 "" {TEXT -1 5 "Base " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "D\351 finition: Un ensemble de vecteurs \{v1,v2,..vn\}forme une base d'un en semble de vecteurs V(espace vectoriel) si et seulement si :" }}{PARA 0 "" 0 "" {TEXT -1 43 "a) \{v1,v2,..vn\}est lin\351airement ind\351pen dant" }}{PARA 0 "" 0 "" {TEXT -1 34 "b) \{v1,v2,..vn\}est g\351n\351ra teur de V" }}{PARA 0 "" 0 "" {TEXT -1 75 "Remarque : Une base de V, es t un ensemble g\351n\351rateur de dimension minimale." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "v1:=vector([3,4,-5,1]);v2:=vector( [8,-9,4,12]);v3:=vector([11,0,6,-3]);v4:=vector([5, -18, 2, 27]);v5:=v ector([34, 6, -9,12]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 330 10 "Exempl e 1)" }{TEXT -1 16 " \{v1,v2,v3\}forme" }{TEXT 327 1 "-" }{TEXT 328 11 "il une base" }{TEXT 329 1 " " }{TEXT -1 60 "de B = l'ensemble des \+ combinaisons lin\351aires de v1,v2,v3,v4?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 331 2 "a)" }{TEXT -1 131 " \{v1,v2,v3\}est-il lin\351airement in d\351pendant? R\351solvons x1v1+x2v2+x3v3 = (0,0,0,0) dont la matrice \+ augment\351e est [v1,v2,v3,(0,0,0,0)]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "A:=augment(v1,v2,v3,vector([0,0,0,0]));gausselim(A); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "[x1,x2,x3]=backsub(gaus selim(A));" }}}{PARA 0 "" 0 "" {TEXT -1 40 "\{v1,v2,v3\} est lin\351ai rement ind\351pendant." }}{EXCHG {PARA 0 "" 0 "" {TEXT 332 2 "b)" } {TEXT -1 94 " \{v1,v2,v3\} est-il g\351n\351rateur de B ? B = \{a*v1 + b*v2 + c*v3 + d*v4 | a,b,c,d sont des r\351els\}." }}{PARA 0 "" 0 "" {TEXT -1 85 "Un vecteur quelconque v = a*v1 + b*v2 + c*v3 + d*v4 de B \+ est-il g\351n\351r\351 par v1,v2,v3? " }}{PARA 0 "" 0 "" {TEXT -1 106 "Il faut r\351soudre v = xv1 + yv2 + zv3 dont la matrice augment\351e \+ est [v1,v2,v3 | a*v1 + b*v2 + c*v3 + d*v4] " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "A1:=augment(v1,v2,v3,a*v1+b*v2+c*v3+d*v4);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "gausselim(A1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "On a 3 variables, 3 pivots donc \{v1,v2,v 3\} est g\351n\351rateur de B." }}{PARA 0 "" 0 "" {TEXT -1 27 "De a) e t b) \{v1,v2,v3\}forme" }{TEXT 342 9 " une base" }{TEXT 343 1 " " } {TEXT -1 60 "de B = l'ensemble des combinaisons lin\351aires de v1,v2, v3,v4 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 340 11 "Exemple 2) " }{TEXT -1 20 "\{v2,v3,v4,v5\} forme" }{TEXT 337 1 "-" }{TEXT 338 11 "il une \+ base" }{TEXT 339 1 " " }{TEXT -1 8 "de R^4 ?" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 150 "a) \{v2,v3,v4,v5\}est-il lin\351airement ind\351pendan t? R\351solvons x1v2 + x2v3 + x3v4 + x4v5 = (0,0,0,0) dont la matrice \+ augment\351e est [v2,v3,v4,v5,(0,0,0,0)]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "A2:=augment(v2,v3,v4,v5,vector([0,0,0,0]));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "gausselim(A2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "On a une variable libre x3 donc une infi nit\351 de solutions. \{v2,v3,v4,v5\}n'est pas lin\351airement ind\351 pendant donc donc \{v2,v3,v4,v5\} " }{TEXT 364 18 "n'est pas une base " }{TEXT 365 1 " " }{TEXT -1 6 "de R^4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 336 11 "Exemple 3) \+ " }{TEXT -1 20 "\{v1,v2,v3,v5\} forme" }{TEXT 333 1 "-" }{TEXT 334 11 "il une base" }{TEXT 335 1 " " }{TEXT -1 8 "de R^4 ?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "a) \{v1,v2,v3,v5\}est-il lin\351airement ind\351pendant? R\351solvons x1v1 + x2v2 + x3v3 + x4v5 = (0,0,0,0) do nt la matrice augment\351e est [v1,v2,v3,v5,(0,0,0,0)]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "A:=augment(v1,v2,v3,v5,vector([0,0, 0,0]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "gausselim(A);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "\{v1,v2,v3,v5\} est lin\351airem ent ind\351pendant car on a une solution unique x1 = x2 = x3 = 4 = 0. \+ " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 208 "\{v1,v2,v3,v5\} est-il g\351 n\351rateur de R^4? Un vecteur quelconque de R^4, v = (a,b,c,d), est-i l g\351n\351r\351 par v1,v2,v3,v5? Il faut r\351soudre v = xv1 + yv2 + zv3 + wv5 dont la matrice augment\351e est [v1,v2,v3,v5 | v] " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "M:=augment(v1,v2,v3,v5,vecto r([a,b,c,d]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "gausselim (%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "On a 4 variables, 4 pivot s donc une solution unique pour tout vecteur (a,b,c,d). " }}{PARA 0 " " 0 "" {TEXT -1 95 "\{v1,v2,v3,v5\}est donc lin\351airement ind\351pen dant et g\351n\351rateur de R^4 donc forme une base de R^4." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 261 12 " Exemple 4) " }{TEXT -1 20 "\{v1,v2,v3,v4\} forme" }{TEXT 258 1 "-" }{TEXT 259 11 "il une base" }{TEXT 260 1 " " }{TEXT -1 8 "de R^4 ?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Puisque \{v1,v2,v3\}est " }{TEXT 318 9 " une base" }{TEXT 319 1 " " }{TEXT -1 184 "de B = l'en semble des combinaisons lin\351aires de v1,v2,v3,v4 alors v4 s'exprime comme combinaison lin\351aire de v1, v2, v3 et donc \{v1,v2,v3,v4\}es t lin\351airement d\351pendant et \{v1,v2,v3,v4\}" }{TEXT 320 21 "ne f orme pas une base" }{TEXT 321 1 " " }{TEXT -1 22 "de R^4. V\351rifions -le :" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "a) \{v1,v2,v3,v4\}est-i l lin\351airement ind\351pendant? R\351solvons x1v1+x2v2+x3v3+x4v4 = ( 0,0,0,0) dont la matrice augment\351e est [v1,v2,v3,v4,(0,0,0,0)]" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "A:=augment(v1,v2,v3,v4,vecto r([0,0,0,0]));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "gausselim (A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "\{v1,v2,v3,v4\} est lin\351airement d\351pendant car on a une variable libre, donc une infinit\351 de solutions \340 x 1v1 + x2v2 + x3v3 + x4v4 = (0,0,0,0) donc \{v1,v2,v3,v4\}n'est pas une base de R^4." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 257 81 "Un ensemble de \+ n vecteurs de R^n, lin\351airement ind\351pendant, forme une base de R ^n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 341 7 "Crit\350re" }{TEXT -1 178 " : \{b1,b2,...bn\}forme une base de R^n si et seulement si la matrice \+ A(nxn) = [b1,b2,...,bn], \351chelonn\351e donne une matrice H qui a n \+ pivots (un sur chaque ligne et chaque colonne)" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 256 11 "Explication" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 534 "Si un ensemble \{v1,v2,..vn\} de n vecteurs de R^n est lin\351airemen t ind\351pendant alors la matrice nxn \351chelonn\351e de [v1,v2..vn] \+ poss\350dera n pivots (un pivot sur chaque ligne et chaque colonne). \+ \{v1,v2,..vn\} sera g\351n\351rateur de R^n car pour un vecteur quelco nque b de R^n, la matrice augment\351e [v1,v2..vn | b] donnera une ma trice \351chelonn\351e avec n pivots pour n variables. On aura donc un e solution unique qui exprimera le vecteur quelconque b comme combinai son lin\351aire de \{v1,v2,..vn\} et par cons\351quent \{v1,v2,..vn\}e st g\351n\351rateur de R^n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Une base de R^n comprend exactement n vecteurs, lin\351airement ind\351pe ndants" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 21 "Avantage d'une base : " }{TEXT -1 62 "\nDans une base d'un ensemble de vecteurs V(espace vec toriel), " }{TEXT 263 99 "tout vecteur de V s'y exprime d'une fa\347on unique comme combinaison lin\351aire des vecteurs de la base" }{TEXT -1 262 " alors qu'avec des vecteurs lin\351airement d\351pendants, ce \+ n'est pas le cas. Certains vecteurs vont s'exprimer d'une infinit\351 \+ de fa\347ons comme combinaison lin\351aire de ces vecteurs, ou encore , d'autres ne s'exprimeront pas comme combinaison lin\351aire de ces v ecteurs." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 263 "" 0 "" {TEXT 264 13 "Base ordonn\351e" }{TEXT -1 12 " de ve cteurs" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "D\351finition : Une bas e ordonn\351e est une base pour laquelle, on tient compte de l'ordre d es vecteurs de la base. " }}{PARA 0 "" 0 "" {TEXT -1 57 "On \351crit l es vecteurs de la base ordonn\351e entour\351s de < >" }}{PARA 0 "" 0 "" {TEXT -1 42 "Soit la base ordonn\351e B = " }} {PARA 261 "" 0 "" {TEXT -1 38 "Si v = x1b1 + x2b2 + ...+ xnbn alors \+ " }{TEXT 265 62 "on \351crira v = (x1,x2,..xn) dans la base B ou vB = \+ (x1,x2,..xn)" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Ex emple : Soit \{b1, b2, b3\}, une base de R^3." }}{PARA 0 "" 0 "" {TEXT -1 52 " Si v = 20b1 + 30b2 + 40b3 alors on a, par exemple :" }} {PARA 0 "" 0 "" {TEXT -1 48 "v = (20,30,40) dans la base ordonn\351e < b1,b2,b3> " }}{PARA 0 "" 0 "" {TEXT -1 48 "v = (40,30,20) dans la bas e ordonn\351e " }}{PARA 0 "" 0 "" {TEXT -1 48 "v = (30,20,40 ) dans la base ordonn\351e " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 266 107 "Note: On peut ne pas indiquer la base dans le cas de la base canonique E = : On \351crira v ou vE" }{TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 109 "Remarque: ei est le vecteur dont tou tes les composantes sont nulles sauf \340 la position i o\371 on a la \+ valeur 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 263 "" 0 "" {TEXT 267 42 "Exprimer un vecteur dans une base ordo nn\351e" }{TEXT -1 1 " " }{TEXT 268 5 "de R^" }{TEXT -1 1 "n" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 173 "Exprimer un vecteur v dans la bas e ordonn\351e B = , revient \340 r\351soudre v = x1b1 \+ + x2b2 + ..xnbn dont la matrice augment\351e: [b1,b2,b3,...bn | v] et \+ alors vB = (" }{TEXT 350 11 "x1,x2,..xn)" }}{PARA 0 "" 0 "" {TEXT -1 96 "De plus, v = x1b1 + x2b2 + ..xnbn = [b1,b2,..,bn]vB o\371 vB est \+ le vecteur colonne (x1,x2,..xn). " }}{PARA 0 "" 0 "" {TEXT -1 32 "Posa nt MB = [b1,b2,..,bn] alors" }{TEXT 270 10 " v = MBvB " }{TEXT -1 3 " et " }{TEXT 269 13 "vB = MB^(-1)v" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "Remarque : la matrice carr\351e MB est inversible car ses colonne s forment une base et donc sont lin\351airement ind\351pendantes." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 314 11 " Exemple 5)" }{TEXT -1 97 " Exprimer v = [3,7,-9] dans l a base o\371 b1 = [2,10,-43], b2 = [9,0,-15], b3 = [0,23,5] " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "b1:=vector([2,10,-43]); b2:=vector([9,0,-15]);b3:=vector([0,23,5]);v:=vector([3,7,-9]);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 344 19 "Premi\350re solution :" }{TEXT -1 154 " Exprimons v dans cette base , c'est-\340-dire r\351 solvons le syst\350me d'\351quations x*b1 + y*b2 + z*b3 = v qui a comm e matrice augment\351e [b1,b2,b3 | v]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "M2:=augment(b1,b2,b3,v);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "vB=backsub(gausselim(M2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Donc v = " }{XPPEDIT 18 0 "381/2887;" "6#*&\"$\"Q\"\"\" \"%()G!\"\"" }{TEXT -1 5 "b1 + " }{XPPEDIT 18 0 "2633/8661;" "6#*&\"%L E\"\"\"\"%h')!\"\"" }{TEXT -1 5 "b2 + " }{XPPEDIT 18 0 "713/2887;" "6# *&\"$8(\"\"\"\"%()G!\"\"" }{TEXT -1 2 "b3" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 345 20 "Deuxi\350me solution : " }{TEXT -1 38 "Avec vB = MB^(-1) v o\371 MB = [b1,b2,b3]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "MB:=augment(b1,b2,b3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 " vB:=multiply(inverse(MB),v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 352 41 "Bases orthogonales et bases orthonormales" }}{EXCHG {PARA 0 "" 0 "" {TEXT 353 64 "Une base \+ est orthogonale si ses vecteurs sont orthogonaux 2 \340 2." }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 354 11 "Exemple 6) " }{TEXT 355 8 "La base " }{TEXT 356 34 " est-elle orthogonale ?" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "dotprod(b1,b2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Non, puisque b1 et b2 ne sont pas orthogo naux. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "\300 partir d'une base \+ B de V, la commande GramSchmidt donne une base orthogonale de V" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Ortho:=GramSchmidt([b1,b2,b3 ]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "V\351rifions que Ortho est bien une base orthogonale." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "dotprod(Ortho[1],Ortho[2]);dotprod(Ortho[1],Ortho[3]);dotprod(Orth o[2],Ortho[3]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 357 80 "Une base est \+ orthonormale si elle est orthogonale et ses vecteurs sont unitaires" } {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "Trouvons une bas e orthonormale \340 partir de la base orthogonale Ortho en rendant ses vecteurs unitaires" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "u1:= evalm(Ortho[1]/norm(Ortho[1],2));u2:=normalize(Ortho[2]);u3:=normalize (Ortho[3]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 358 11 "Exemple 7) " } {TEXT -1 66 "V\351rifier que la base de R^3 suivante est or thonormale :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "u1 := vect or([1/14*sqrt(14), -1/7*sqrt(14), 3/14*sqrt(14)]);\nu2 := vector([3/10 *sqrt(10), 0, -1/10*sqrt(10)]);\nu3 := vector([1/35*sqrt(35), 1/7*sqrt (35), 3/35*sqrt(35)]); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "u1,u2,u3 sont orthogonaux 2 \340 2 :" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "dotprod(u1,u2);dotprod(u1,u3) ;dotprod(u2,u3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "u1,u2,u3 sont unitaires :" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "norm(u1,2);norm(u2, 2);norm(u3,2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 " donc est une base orthonormale" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 271 66 "Exprimer simultan\351me nt plusieurs vecteurs dans une base ordonn\351e B" }}{EXCHG {PARA 0 " " 0 "" {TEXT -1 237 "Pour exprimer plusieurs vecteurs v1,v2,...vk dans la base B = , on r\351soud simultan\351ment les syst \350mes d'\351quations v1 = MB(v1B), v2 = MB(v2B), v3 = MB(v3B), ..., \+ vk = MB(vkB), o\371 viB exprime le vecteur vi dans la base B. " } {TEXT 298 37 "Pour trouver v1B , v2B , ... , vkB\n" }{TEXT -1 48 "a) On utilise alors le sch\351ma avec Gauss-Jordan:" }}{PARA 261 "" 0 " " {TEXT 292 30 " [ MB | v1, v2, v3, ...,vk] " }{TEXT 360 1 "~" } {TEXT 361 32 " [ I | v1B , v2B , ... , vkB ]" }{TEXT 274 9 " ou bien " }}{PARA 0 "" 0 "" {TEXT -1 2 "b)" }{TEXT 362 27 "MB^(-1)*[v1, v2, v 3, ...,vk" }{TEXT 363 2 "] " }{TEXT -1 35 "= [v1B , v2B , ... , vkB] ou bien" }}{PARA 0 "" 0 "" {TEXT -1 12 "c) linsolve(" }{TEXT 359 2 "M B" }{TEXT -1 1 "," }{TEXT 297 21 "[v1, v2, v3, ...,vk])" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 11 "Explication" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "a) Pour trouver v1B , v2B , ... , vkB on peut appliquer k fois le sch\351ma " }{TEXT 284 24 "[ MB | v] ~ [ I | vB ]. " } {TEXT 285 11 "On aurait :" }}{PARA 0 "" 0 "" {TEXT 276 12 "[ MB | v1 ] " }{TEXT 275 1 "~" }{TEXT 277 13 " [ I | v1B], " }{TEXT -1 1 " " } {TEXT 279 12 "[ MB | v2 ] " }{TEXT 278 1 "~" }{TEXT 280 15 " [ I | v2B ]...." }{TEXT -1 1 " " }{TEXT 282 12 "[ MB | vk ] " }{TEXT 281 1 "~" } {TEXT 283 13 " [ I | vkB] " }{TEXT 286 3 "ou " }}{PARA 0 "" 0 "" {TEXT 300 66 "On peut l'appliquer une seule fois en r\351solvant simul tan\351ment : [" }{TEXT 288 23 " MB | v1 v2 ,v3 ..vk ] " }{TEXT 287 1 "~" }{TEXT 289 32 " [ I | v1B , v2B , ... , vkB] " }{TEXT 299 1 ": " }}{PARA 0 "" 0 "" {TEXT -1 96 "Pour la r\351solution simultan\351e d e syst\350me d'\351quations, voir le laboratoire 8 M\351thode Gauss-Jo rdan" }}{PARA 0 "" 0 "" {TEXT -1 35 " b) On a : MB^(-1)*[v1 v2 ,v3 ..v k]" }{TEXT 273 3 " = " }{TEXT -1 73 " [MB^(-1)* v1 , MB^(-1)* v2 , MB( -1) *vk] = [v1B , v2B , ... , vkB] car" }}{PARA 0 "" 0 "" {TEXT -1 17 " pour chaque i, " }{TEXT 272 16 "viB = MB^(-1)*vi" }{TEXT -1 22 " puisque vi = MB(viB) " }}{PARA 0 "" 0 "" {TEXT -1 7 "donc : " }{TEXT 290 24 "MB^(-1)*[v1 v2 ,v3 ..vk]" }{TEXT 291 1 " " }{TEXT -1 28 "= [v1 B , v2B , ... , vkB] " }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 315 11 " Exe mple 8)" }{TEXT -1 132 " Exprimer les vecteurs v1 = [1,2,3], v2 = [45, 9,-111], v3 = [-44,67,0], v4 = [85,1,67] dans la base B = d onn\351e plus haut." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Premi\350r e solution : avec a) Gauss-Jordan" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "v1:=vector([1,2,3]);v2:=vector([45,9,-111]);v3:=vecto r([-44,67,0]);v4:=vector([85,1,67]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "M:=augment(b1,b2,b3,v1,v2,v3,v4);#pour une solution s imultan\351e." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "GJ:=gaussj ord(M);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "La quatri\350me colonn e donne les composantes du vecteur v1 dans la base MB. v1B = " } {XPPEDIT 18 0 "-292/2887;" "6#,$*&\"$#H\"\"\"\"%()G!\"\"F(" }{TEXT -1 5 "b1 + " }{XPPEDIT 18 0 "1157/8661;" "6#*&\"%d6\"\"\"\"%h')!\"\"" } {TEXT -1 4 "b2 +" }{XPPEDIT 18 0 "378/2887;" "6#*&\"$y$\"\"\"\"%()G!\" \"" }{TEXT -1 2 "b3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 80 "La cinqui\350me colonne donne les composantes du vecteu r v2 dans la base MB. v2B = " }{XPPEDIT 18 0 "2619/2887;" "6#*&\"%>E\" \"\"\"%()G!\"\"" }{TEXT -1 5 "b1 + " }{XPPEDIT 18 0 "13853/2887;" "6#* &\"&`Q\"\"\"\"\"%()G!\"\"" }{TEXT -1 3 "b2 " }{XPPEDIT 18 0 "-9/2887; " "6#,$*&\"\"*\"\"\"\"%()G!\"\"F(" }{TEXT -1 2 "b3" }}{PARA 0 "" 0 "" {TEXT -1 4 "etc " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Deuxi\350me s olution : avec b) " }{TEXT 347 24 "MB^(-1)*[v1 v2 ,v3 ..vk]" }{TEXT 348 1 " " }{TEXT -1 28 "= [v1B , v2B , ... , vkB] " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "MB:=augment(b1,b2,b3);MatriceVecteurs:=au gment(v1,v2,v3,v4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "mult iply(inverse(MB),MatriceVecteurs);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Troisi\350me solution : avec c) " }{TEXT 349 28 "linsolve(A,B) \+ :linsolve(A,B)" }{TEXT -1 135 " qui r\351soud Ax = bi pour chaque colo nne bi de B, c'est-\340-dire exprime chaque colonne de B comme combina ison lin\351aire des colonnes de A." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "linsolve(MB,MatriceVecteurs);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "Remarque: Pour r\351soudre " }{TEXT 293 1 " " } {TEXT 294 1 "[" }{TEXT 296 1 " " }{TEXT 295 179 "MB | v1 v2 ,v3 ..vk ] on ne doit pas utiliser gausselim avec backsub car Maple donnerait la solution de la derni\350re colonne vk comme combinaison lin\351aire d es colonnes pr\351c\351dentes." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 262 "" 0 "" {TEXT 301 85 "Extraire une base de l'ensemble des combinaisons lin\351aires de vecteurs v1,v2,v3,...v n" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "Notons " } {TEXT 305 16 "L\{v1,v2,v3,...vn" }{TEXT -1 3 "\}, " }{TEXT 307 65 "l'e nsemble des combinaisons lin\351aires des vecteurs v1,v2,v3,...vn" } {TEXT 308 1 " " }}{PARA 0 "" 0 "" {TEXT -1 15 "c'est-\340-dire : " } {TEXT 309 16 "L\{v1,v2,v3,...vn" }{TEXT -1 51 "\} = \{x1v1+x2v2+...xnv n | x1,x2,..xn sont des r\351els\}" }}{PARA 0 "" 0 "" {TEXT 302 42 "Po ur extraire une base de L\{v1,v2,v3,...vn" }{TEXT -1 3 "\}: " }}{PARA 0 "" 0 "" {TEXT 316 2 "a)" }{TEXT -1 25 " On construit la matrice " } {TEXT 312 3 "A =" }{TEXT -1 2 " [" }{TEXT 303 14 "v1,v2,v3,...vn" } {TEXT -1 26 "] et on l'\351chelonne. Soit " }{TEXT 313 17 "H = [h1,h2, ...hn]" }{TEXT -1 102 ", la matrice A \351chelonn\351e. On choisit al ors les colonnes de A qui donnent un pivot dans la matrice H." }} {PARA 0 "" 0 "" {TEXT -1 80 "Pour plus d'explications sur cet algorith me, voir l'annexe A-2, ExtraireBase.mws" }}{PARA 0 "" 0 "" {TEXT 317 2 "b)" }{TEXT -1 36 " Avec Maple, on utilise la commande " }{TEXT 322 6 "basis(" }{TEXT 306 15 "\{v1,v2,v3,...vn" }{TEXT 324 2 "\})" }{TEXT -1 4 "; ou" }{TEXT 323 6 " basis" }{TEXT -1 1 "(" }{TEXT 311 15 "[v1,v 2,v3,...vn" }{TEXT 325 2 "])" }{TEXT -1 23 " pour une base ordonn\351e " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 240 "v1:=vector([2,-3,4,-6,12 ,34]);v2:=vector([12,-2,5,-5,9,18]);v3:=vector([-34, 19, -30, 40, -78, -206]);v4:=vector([6, -9, 12, -18, 36, 102]);\nv5:=vector([9,-8,32,6,3 4,10]);v6:=vector([7,-7,3,0,6,0]);v7:=vector([-7, -18, 232, 175, 109, \+ -500]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 351 11 "Exemple 9) " }{TEXT -1 46 "Extraire une base de L\{v1,v2,v3,v4,v5,v6,v6\} :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "A:=augment(v1,v2,v3,v4,v5,v6,v7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "H:=gausselim(%);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "h1,h2,h5,h6 ont des pivots donc est une base de L\{" }{TEXT 304 20 "v1,v2,v3,v4,v5,v6,v7" }{TEXT -1 2 "\}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "basis( \{v1,v2,v3,v4,v5,v6,v7\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "basis([v1,v2,v3,v4,v5,v6,v7]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 310 9 "Exercices" }}{EXCHG {PARA 0 "" 0 "" {TEXT 370 5 "No 1)" }{TEXT -1 80 " Soit les v ecteurs v1= (5,-7,12) et v2 = (3,4,9), v3 = (2,0,5) et v4 = (-1,26,3) " }}{PARA 0 "" 0 "" {TEXT 387 2 "a)" }{TEXT -1 48 " Montrer que forment une base de R^3 " }}{PARA 0 "" 0 "" {TEXT 388 2 "b)" } {TEXT -1 36 " Exprimer v4 dans la base " }}{PARA 0 "" 0 "" {TEXT 389 2 "c)" }{TEXT -1 36 " Exprimer v4 dans la base " } }}{SECT 1 {PARA 3 "" 0 "" {TEXT 395 1 " " }{TEXT 394 31 "Espace de tra vail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 379 5 "No 2)" } {TEXT -1 1 " " }{TEXT 381 2 "a)" }{TEXT -1 77 " \300 partir de la r \351ponse b) du No 1), dire si \{v1,v2,v4\} forme une base de R^3" }} {PARA 0 "" 0 "" {TEXT 382 3 "b) " }{TEXT -1 40 "Est-ce que \{v1,v2\}fo rme une base de B = " }{TEXT 380 63 "l'ensemble des combinaisons lin \351aires des vecteurs v1, v2, v4 ?" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 397 1 " " }{TEXT 396 31 "Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT 371 5 "No 3)" }{TEXT -1 105 " Soit vB = ( 5,-7 2) o\371 B = <(4,3,2),(2,5,9),(10,-11,-17)>. Trouver v (dans la b ase canonique E = )" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 399 1 " \+ " }{TEXT 398 31 "Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 372 5 "No 4)" }{TEXT -1 68 " Exprimer simultan\351ment \+ b5,b6,b7 dans la base B = " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 186 "b1:=vector([-1,12,3,-7]);b2:=vector([5,0,-7,11]); b3:=vector([34,-1,8,-12]);b4:=vector([7,-8,0,-6]);b5:=vector([-13,5,7, 9]);b6:=vector([67, -39, 3, 49]);b7:=vector([148, -102, -18, -14]);" } }}{SECT 1 {PARA 3 "" 0 "" {TEXT 401 1 " " }{TEXT 400 31 "Espace de tra vail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 366 5 "No 5)" } {TEXT -1 197 " \340 partir du crit\350re : \{b1,b2,...bn\}forme une ba se de R^n si et seulement si la matrice A(nxn) = [b1,b2,...,bn] \351ch elonn\351e donne une matrice H qui a n pivots (un sur chaque ligne et \+ chaque colonne)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 383 2 "a)" }{TEXT -1 49 " Est-ce que \{ b1,b2,b3,b5\} forme une base de R^4?" }}{PARA 0 "" 0 "" {TEXT 384 2 "b)" }{TEXT -1 48 " Est-ce que \{b4,b5,b6,b7\} forme \+ une base de R^4?" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 403 1 " " }{TEXT 402 31 "Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 367 5 "No 6)" }{TEXT -1 1 " " }{TEXT 385 2 "a)" }{TEXT -1 69 " E xprimer, si possible, b7 comme combinaison lin\351aire de b4,b5 et b6 \+ " }}{PARA 0 "" 0 "" {TEXT -1 5 " " }{TEXT 386 2 "b)" }{TEXT -1 69 " Exprimer, si possible, b5 comme combinaison lin\351aire de b2,b3 et \+ b4 " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 405 1 " " }{TEXT 404 31 "Espace \+ de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 393 5 "No \+ 7)" }{TEXT -1 68 " Montrer que la base B = est une base ort hogonale de R^3." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "u1:=vector([2,1 0,-1]);u2:=vector([9,0,18]);u3:=vector([36,-9,-18]);u4:=vector([3,7,-9 ]);" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 407 1 " " }{TEXT 406 31 "Espace \+ de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 368 5 "No \+ 8)" }{TEXT -1 44 " Rendre orthonormale la base B = " }}} {SECT 1 {PARA 3 "" 0 "" {TEXT 409 1 " " }{TEXT 408 31 "Espace de trava il de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 373 5 "No 9)" } {TEXT 374 1 " " }{TEXT 390 2 "a)" }{TEXT -1 18 " Trouver || u4||." }} {PARA 0 "" 0 "" {TEXT 391 2 "b)" }{TEXT -1 80 " Exprimer u4 dans la ba se orthogonale B = , i.e trouver u4B = (x,y,z) " }}{PARA 0 " " 0 "" {TEXT -1 32 "A-t-on |u4| = sqrt(x^2+y^2+z^2)?" }}{PARA 0 "" 0 " " {TEXT 392 2 "c)" }{TEXT -1 95 " Exprimer u4 dans la base orthonormal e B' obtenue au num\351ro 8), i.e trouver u4B' = (x1,y1,z1) " }} {PARA 0 "" 0 "" {TEXT -1 35 "A-t-on |u4| = sqrt(x1^2+y1^2+z1^2)?" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT 411 1 " " }{TEXT 410 31 "Espace de trava il de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 369 7 "No 10) " } {TEXT -1 48 "Des 5 vecteurs u1,u2,u3,u4,u5, donn\351s plus bas :" }} {PARA 0 "" 0 "" {TEXT 377 2 "a)" }{TEXT -1 41 " Extraire une base B de L\{u1,u2,u3,u4,u5\}" }}{PARA 0 "" 0 "" {TEXT 378 2 "b)" }{TEXT -1 121 " Exprimer simultan\351ment les vecteurs qui ne sont pas dans la b ase B comme combinaison lin\351aire des vecteurs de la base B " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "u1:=vector([21.45,-33.25,11 .65]);u2:=vector([32.175,-49.875,17.475]);u3:=vector([12.75,89.45,-18. 35]);u4:=vector([-110.5125, -1739.9375, 382.2875]);u5:=vector([-8.95,6 9.15,45.45]);" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 413 1 " " }{TEXT 412 31 "Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT 375 6 "No 11)" }{TEXT -1 13 " Extraire de " }{TEXT 376 1 " " } {TEXT -1 58 "L\{u1,u2,u3,u4,u5\}une autre base que celle obtenue au No 10" }{MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 415 1 " " } {TEXT 414 31 "Espace de travail de l'\351tudiant" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "2 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }