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Exercice

L'exercice suivant est automatiquement et aléatoirement généré par ataraXy.
Si vous regénérez la page (F5) les valeurs seront changées.
La correction se trouve en bas de page.

Exercice


On considère la matrice \[ A= \begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ -\dfrac{4}{5} & 0 & -\dfrac{2}{3} & -\dfrac{25}{4} & -\dfrac{3}{2} & \dfrac{19}{5} \\ \dfrac{8}{3} & 2 & -2 & \dfrac{5}{4} & \dfrac{1}{2} & -5 \\ -\dfrac{24}{5} & -2 & -2 & \dfrac{11}{5} & 1 & -\dfrac{9}{5} \\ -3 & -\dfrac{3}{4} & -\dfrac{9}{4} & -3 & -\dfrac{21}{5} & -2 \\ -5 & \dfrac{3}{2} & -2 & -1 & -2 & \dfrac{19}{5}\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (2, 2)\) et \( (4, 6)\) : \( \widehat{A}_{2, 2}=\) \( \widehat{A}_{4, 6}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ 0 & -\dfrac{12}{5} & -\dfrac{194}{75} & -\dfrac{209}{20} & \dfrac{53}{10} & \dfrac{23}{5} \\ 0 & 10 & \dfrac{22}{5} & \dfrac{61}{4} & -\dfrac{133}{6} & -\dfrac{23}{3} \\ 0 & -\dfrac{82}{5} & -\dfrac{338}{25} & -23 & \dfrac{209}{5} & 3 \\ 0 & -\dfrac{39}{4} & -\dfrac{189}{20} & -\dfrac{75}{4} & \dfrac{213}{10} & 1 \\ 0 & -\dfrac{27}{2} & -14 & -\dfrac{109}{4} & \dfrac{81}{2} & \dfrac{44}{5}\end{pmatrix} \)
  3. Calculer \( \det(A)\) .
  4. Pourquoi la matrice \( A \) est inversible.
  5. Donner \( B=A^{-1}\) l'inverse de la matrice \( A \) , en ne détaillant que le calcul de \( A^{-1}_{5, 5}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&-&3y &-&\dfrac{12}{5}z &-&\dfrac{21}{4}t &+&\dfrac{17}{2}u &+&v &=&-2\\ &-\dfrac{4}{5}x&&&-&\dfrac{2}{3}z &-&\dfrac{25}{4}t &-&\dfrac{3}{2}u &+&\dfrac{19}{5}v &=&7\\ &\dfrac{8}{3}x&+&2y &-&2z &+&\dfrac{5}{4}t &+&\dfrac{1}{2}u &-&5v &=&-8\\ &-\dfrac{24}{5}x&-&2y &-&2z &+&\dfrac{11}{5}t &+&u &-&\dfrac{9}{5}v &=&-4\\ &-3x&-&\dfrac{3}{4}y &-&\dfrac{9}{4}z &-&3t &-&\dfrac{21}{5}u &-&2v &=&7\\ &-5x&+&\dfrac{3}{2}y &-&2z &-&t &-&2u &+&\dfrac{19}{5}v &=&-\dfrac{5}{6}\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{2.6781158664806E+74}{1.9398664651235E+74}x&-&\dfrac{8.795403421655E+78}{2.1554071834706E+78}y &-&\dfrac{1.1743777878221E+71}{8.0827769380147E+70}z &-&\dfrac{2.1256765032038E+72}{6.4662215504118E+71}t &+&\dfrac{5.133899708928E+66}{2.0206942345037E+66}u &+&\dfrac{9.6118953170688E+75}{6.0620827035111E+75}v &=&-\dfrac{2}{3}\\ &-\dfrac{5.846475362256E+75}{3.7888016896944E+75}x&+&\dfrac{2.2950168607872E+77}{5.0517355862592E+76}y &+&\dfrac{8.540637959124E+79}{4.736002112118E+79}z &+&\dfrac{1.893265783799E+74}{5.4558744331599E+73}t &-&\dfrac{4.910797435392E+66}{1.6839118620864E+66}u &-&\dfrac{7.83938141298E+74}{4.736002112118E+74}v &=&1\\ &-\dfrac{4.3521712065792E+74}{2.0206942345037E+74}x&+&\dfrac{1.6203155760415E+71}{2.5864886201647E+70}y &+&\dfrac{1.810011378528E+75}{8.419559310432E+74}z &+&\dfrac{1.3168018119168E+67}{2.6942589793382E+66}t &-&\dfrac{1.5875684522496E+67}{4.0413884690074E+66}u &-&\dfrac{2.1189127343616E+67}{8.0827769380147E+66}v &=&3\\ &\dfrac{1.403422184448E+66}{1.5155206758778E+66}x&-&\dfrac{1.4800933935936E+74}{5.0517355862592E+73}y &-&\dfrac{7.910668243872E+74}{7.5776033793888E+74}z &-&\dfrac{3.3255316806912E+69}{1.5155206758778E+69}t &+&\dfrac{2.597746733568E+66}{1.5155206758778E+66}u &+&\dfrac{2.226483783072E+70}{1.8944008448472E+70}v &=&-2\\ &-\dfrac{2.301260917248E+66}{3.0310413517555E+66}x&+&\dfrac{5.2532220930048E+71}{2.1554071834706E+71}y &+&\dfrac{1.8600062265216E+75}{2.0206942345037E+75}z &+&\dfrac{1.5767259710976E+67}{8.0827769380147E+66}t &-&\dfrac{2.87190939474E+73}{1.7760007920443E+73}u &-&\dfrac{1.284266866176E+66}{1.3471294896691E+66}v &=&-4\\ &\dfrac{2.295852553728E+66}{2.0206942345037E+66}x&-&\dfrac{1.8071372616192E+74}{5.3885179586765E+73}y &-&\dfrac{2.7641193272525E+71}{2.1554071834706E+71}z &-&\dfrac{4.3221523815936E+67}{1.6165553876029E+67}t &+&\dfrac{1.2285570267648E+67}{6.0620827035111E+66}u &+&\dfrac{9.636206694912E+66}{6.7356474483456E+66}v &=&\dfrac{13}{5}\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{2, 2}=\begin{pmatrix}1 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ \dfrac{8}{3} & -2 & \dfrac{5}{4} & \dfrac{1}{2} & -5 \\ -\dfrac{24}{5} & -2 & \dfrac{11}{5} & 1 & -\dfrac{9}{5} \\ -3 & -\dfrac{9}{4} & -3 & -\dfrac{21}{5} & -2 \\ -5 & -2 & -1 & -2 & \dfrac{19}{5}\end{pmatrix}\) \( \widehat{A}_{4, 6}=\begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} \\ -\dfrac{4}{5} & 0 & -\dfrac{2}{3} & -\dfrac{25}{4} & -\dfrac{3}{2} \\ \dfrac{8}{3} & 2 & -2 & \dfrac{5}{4} & \dfrac{1}{2} \\ -3 & -\dfrac{3}{4} & -\dfrac{9}{4} & -3 & -\dfrac{21}{5} \\ -5 & \dfrac{3}{2} & -2 & -1 & -2\end{pmatrix}\)
  2. On sait, d'après le cours, que l'on ne modifie la valeur du déterminant d'une matrice lorsqu'on ajoute à une ligne une combinaison linéaire des autres. On est donc partie de la matrice \( A\) et on a fait : \( L_{2}\leftarrow L_{2}-\left(-\dfrac{4}{5}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(\dfrac{8}{3}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(-\dfrac{24}{5}\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-3\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-5\right)L_1\)
  3. En développant par rapport à la première colonne, en se servant de la précédente remarque on a \begin{eqnarray*} \det(A) &=&\det\begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ 0 & -\dfrac{12}{5} & -\dfrac{194}{75} & -\dfrac{209}{20} & \dfrac{53}{10} & \dfrac{23}{5} \\ 0 & 10 & \dfrac{22}{5} & \dfrac{61}{4} & -\dfrac{133}{6} & -\dfrac{23}{3} \\ 0 & -\dfrac{82}{5} & -\dfrac{338}{25} & -23 & \dfrac{209}{5} & 3 \\ 0 & -\dfrac{39}{4} & -\dfrac{189}{20} & -\dfrac{75}{4} & \dfrac{213}{10} & 1 \\ 0 & -\dfrac{27}{2} & -14 & -\dfrac{109}{4} & \dfrac{81}{2} & \dfrac{44}{5}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-\dfrac{12}{5} & -\dfrac{194}{75} & -\dfrac{209}{20} & \dfrac{53}{10} & \dfrac{23}{5} \\ 10 & \dfrac{22}{5} & \dfrac{61}{4} & -\dfrac{133}{6} & -\dfrac{23}{3} \\ -\dfrac{82}{5} & -\dfrac{338}{25} & -23 & \dfrac{209}{5} & 3 \\ -\dfrac{39}{4} & -\dfrac{189}{20} & -\dfrac{75}{4} & \dfrac{213}{10} & 1 \\ -\dfrac{27}{2} & -14 & -\dfrac{109}{4} & \dfrac{81}{2} & \dfrac{44}{5}\end{pmatrix}\\ &=&-\dfrac{1.3471294896691E+63}{1.119744E+60} \end{eqnarray*}
  4. On observe que le déterminant de \( A\) est non nul. D'après le cours, cela signifie que la matrice est inversible.
  5. D'après le cours \( B=A^{-1}=\left(-\dfrac{1.3471294896691E+63}{1.119744E+60}\right)^{-1}{}^tCo\begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ -\dfrac{4}{5} & 0 & -\dfrac{2}{3} & -\dfrac{25}{4} & -\dfrac{3}{2} & \dfrac{19}{5} \\ \dfrac{8}{3} & 2 & -2 & \dfrac{5}{4} & \dfrac{1}{2} & -5 \\ -\dfrac{24}{5} & -2 & -2 & \dfrac{11}{5} & 1 & -\dfrac{9}{5} \\ -3 & -\dfrac{3}{4} & -\dfrac{9}{4} & -3 & -\dfrac{21}{5} & -2 \\ -5 & \dfrac{3}{2} & -2 & -1 & -2 & \dfrac{19}{5}\end{pmatrix} =-\dfrac{1.119744E+60}{1.3471294896691E+63}\begin{pmatrix}-\dfrac{3.6077688604868E+137}{2.1721538351233E+134} & \dfrac{1.1848547322848E+142}{2.4135042612481E+138} & \dfrac{1.5820389499875E+134}{9.0506409796804E+130} & \dfrac{2.8635615029626E+135}{7.2405127837443E+131} & -\dfrac{6.9160276949006E+129}{2.2626602449201E+126} & -\dfrac{1.2948467633236E+139}{6.7879807347603E+135} \\ \dfrac{7.875959371119E+138}{4.2424879592252E+135} & -\dfrac{3.0916848924543E+140}{5.6566506123002E+136} & -\dfrac{1.1505345255323E+143}{5.3031099490315E+139} & -\dfrac{2.5504741691372E+137}{6.1091826612843E+133} & \dfrac{6.6154800430081E+129}{1.8855502041001E+126} & \dfrac{1.0560661882189E+138}{5.3031099490315E+134} \\ \dfrac{5.8629381764717E+137}{2.2626602449201E+134} & -\dfrac{2.1827748950557E+134}{2.8962051134977E+130} & -\dfrac{2.4383197046517E+138}{9.4277510205004E+134} & -\dfrac{1.7739025528829E+130}{3.0168803265601E+126} & \dfrac{2.1386602788938E+130}{4.5253204898402E+126} & \dfrac{2.8544498304939E+130}{9.0506409796804E+126} \\ -\dfrac{1.8905914111258E+129}{1.6969951836901E+126} & \dfrac{1.9938774579744E+137}{5.6566506123002E+133} & \dfrac{1.0656694474309E+138}{8.4849759184503E+134} & \dfrac{4.479921795888E+132}{1.6969951836901E+129} & -\dfrac{3.4995012314811E+129}{1.6969951836901E+126} & -\dfrac{2.9993619624464E+133}{2.1212439796126E+130} \\ \dfrac{3.1000964450478E+129}{3.3939903673801E+126} & -\dfrac{7.0767703972681E+134}{2.4135042612481E+131} & -\dfrac{2.5056692387154E+138}{2.2626602449201E+135} & -\dfrac{2.1240540527928E+130}{9.0506409796804E+126} & \dfrac{3.8688338373121E+136}{1.9886662308868E+133} & \dfrac{1.7300737680306E+129}{1.5084401632801E+126} \\ -\dfrac{3.0928106790592E+129}{2.2626602449201E+126} & \dfrac{2.4344478970071E+137}{6.0337606531202E+133} & \dfrac{3.7236266587062E+134}{2.4135042612481E+131} & \dfrac{5.8224989320884E+130}{1.8101281959361E+127} & -\dfrac{1.6550254004951E+130}{6.7879807347603E+126} & -\dfrac{1.2981218207263E+130}{7.5422008164003E+126}\end{pmatrix} =\begin{pmatrix}\dfrac{2.6781158664806E+74}{1.9398664651235E+74} & -\dfrac{8.795403421655E+78}{2.1554071834706E+78} & -\dfrac{1.1743777878221E+71}{8.0827769380147E+70} & -\dfrac{2.1256765032038E+72}{6.4662215504118E+71} & \dfrac{5.133899708928E+66}{2.0206942345037E+66} & \dfrac{9.6118953170688E+75}{6.0620827035111E+75} \\ -\dfrac{5.846475362256E+75}{3.7888016896944E+75} & \dfrac{2.2950168607872E+77}{5.0517355862592E+76} & \dfrac{8.540637959124E+79}{4.736002112118E+79} & \dfrac{1.893265783799E+74}{5.4558744331599E+73} & -\dfrac{4.910797435392E+66}{1.6839118620864E+66} & -\dfrac{7.83938141298E+74}{4.736002112118E+74} \\ -\dfrac{4.3521712065792E+74}{2.0206942345037E+74} & \dfrac{1.6203155760415E+71}{2.5864886201647E+70} & \dfrac{1.810011378528E+75}{8.419559310432E+74} & \dfrac{1.3168018119168E+67}{2.6942589793382E+66} & -\dfrac{1.5875684522496E+67}{4.0413884690074E+66} & -\dfrac{2.1189127343616E+67}{8.0827769380147E+66} \\ \dfrac{1.403422184448E+66}{1.5155206758778E+66} & -\dfrac{1.4800933935936E+74}{5.0517355862592E+73} & -\dfrac{7.910668243872E+74}{7.5776033793888E+74} & -\dfrac{3.3255316806912E+69}{1.5155206758778E+69} & \dfrac{2.597746733568E+66}{1.5155206758778E+66} & \dfrac{2.226483783072E+70}{1.8944008448472E+70} \\ -\dfrac{2.301260917248E+66}{3.0310413517555E+66} & \dfrac{5.2532220930048E+71}{2.1554071834706E+71} & \dfrac{1.8600062265216E+75}{2.0206942345037E+75} & \dfrac{1.5767259710976E+67}{8.0827769380147E+66} & -\dfrac{2.87190939474E+73}{1.7760007920443E+73} & -\dfrac{1.284266866176E+66}{1.3471294896691E+66} \\ \dfrac{2.295852553728E+66}{2.0206942345037E+66} & -\dfrac{1.8071372616192E+74}{5.3885179586765E+73} & -\dfrac{2.7641193272525E+71}{2.1554071834706E+71} & -\dfrac{4.3221523815936E+67}{1.6165553876029E+67} & \dfrac{1.2285570267648E+67}{6.0620827035111E+66} & \dfrac{9.636206694912E+66}{6.7356474483456E+66}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{5, 5}=B_{5, 5}= \left(-\dfrac{1.3471294896691E+63}{1.119744E+60}\right)^{-1}Co(A)_{5, 5}= \left(-\dfrac{1.3471294896691E+63}{1.119744E+60}\right)^{-1}\times(-1)^{5+5}\det\begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & 1 \\ -\dfrac{4}{5} & 0 & -\dfrac{2}{3} & -\dfrac{25}{4} & \dfrac{19}{5} \\ \dfrac{8}{3} & 2 & -2 & \dfrac{5}{4} & -5 \\ -\dfrac{24}{5} & -2 & -2 & \dfrac{11}{5} & -\dfrac{9}{5} \\ -5 & \dfrac{3}{2} & -2 & -1 & \dfrac{19}{5}\end{pmatrix}=-\dfrac{2.87190939474E+73}{1.7760007920443E+73}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ -\dfrac{4}{5} & 0 & -\dfrac{2}{3} & -\dfrac{25}{4} & -\dfrac{3}{2} & \dfrac{19}{5} \\ \dfrac{8}{3} & 2 & -2 & \dfrac{5}{4} & \dfrac{1}{2} & -5 \\ -\dfrac{24}{5} & -2 & -2 & \dfrac{11}{5} & 1 & -\dfrac{9}{5} \\ -3 & -\dfrac{3}{4} & -\dfrac{9}{4} & -3 & -\dfrac{21}{5} & -2 \\ -5 & \dfrac{3}{2} & -2 & -1 & -2 & \dfrac{19}{5}\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&-&3y &-&\dfrac{12}{5}z &-&\dfrac{21}{4}t &+&\dfrac{17}{2}u &+&v &=&-2\\ &-\dfrac{4}{5}x&&&-&\dfrac{2}{3}z &-&\dfrac{25}{4}t &-&\dfrac{3}{2}u &+&\dfrac{19}{5}v &=&7\\ &\dfrac{8}{3}x&+&2y &-&2z &+&\dfrac{5}{4}t &+&\dfrac{1}{2}u &-&5v &=&-8\\ &-\dfrac{24}{5}x&-&2y &-&2z &+&\dfrac{11}{5}t &+&u &-&\dfrac{9}{5}v &=&-4\\ &-3x&-&\dfrac{3}{4}y &-&\dfrac{9}{4}z &-&3t &-&\dfrac{21}{5}u &-&2v &=&7\\ &-5x&+&\dfrac{3}{2}y &-&2z &-&t &-&2u &+&\dfrac{19}{5}v &=&-\dfrac{5}{6}\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-2 \\ 7 \\ -8 \\ -4 \\ 7 \\ -\dfrac{5}{6}\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{2.6781158664806E+74}{1.9398664651235E+74} & -\dfrac{8.795403421655E+78}{2.1554071834706E+78} & -\dfrac{1.1743777878221E+71}{8.0827769380147E+70} & -\dfrac{2.1256765032038E+72}{6.4662215504118E+71} & \dfrac{5.133899708928E+66}{2.0206942345037E+66} & \dfrac{9.6118953170688E+75}{6.0620827035111E+75} \\ -\dfrac{5.846475362256E+75}{3.7888016896944E+75} & \dfrac{2.2950168607872E+77}{5.0517355862592E+76} & \dfrac{8.540637959124E+79}{4.736002112118E+79} & \dfrac{1.893265783799E+74}{5.4558744331599E+73} & -\dfrac{4.910797435392E+66}{1.6839118620864E+66} & -\dfrac{7.83938141298E+74}{4.736002112118E+74} \\ -\dfrac{4.3521712065792E+74}{2.0206942345037E+74} & \dfrac{1.6203155760415E+71}{2.5864886201647E+70} & \dfrac{1.810011378528E+75}{8.419559310432E+74} & \dfrac{1.3168018119168E+67}{2.6942589793382E+66} & -\dfrac{1.5875684522496E+67}{4.0413884690074E+66} & -\dfrac{2.1189127343616E+67}{8.0827769380147E+66} \\ \dfrac{1.403422184448E+66}{1.5155206758778E+66} & -\dfrac{1.4800933935936E+74}{5.0517355862592E+73} & -\dfrac{7.910668243872E+74}{7.5776033793888E+74} & -\dfrac{3.3255316806912E+69}{1.5155206758778E+69} & \dfrac{2.597746733568E+66}{1.5155206758778E+66} & \dfrac{2.226483783072E+70}{1.8944008448472E+70} \\ -\dfrac{2.301260917248E+66}{3.0310413517555E+66} & \dfrac{5.2532220930048E+71}{2.1554071834706E+71} & \dfrac{1.8600062265216E+75}{2.0206942345037E+75} & \dfrac{1.5767259710976E+67}{8.0827769380147E+66} & -\dfrac{2.87190939474E+73}{1.7760007920443E+73} & -\dfrac{1.284266866176E+66}{1.3471294896691E+66} \\ \dfrac{2.295852553728E+66}{2.0206942345037E+66} & -\dfrac{1.8071372616192E+74}{5.3885179586765E+73} & -\dfrac{2.7641193272525E+71}{2.1554071834706E+71} & -\dfrac{4.3221523815936E+67}{1.6165553876029E+67} & \dfrac{1.2285570267648E+67}{6.0620827035111E+66} & \dfrac{9.636206694912E+66}{6.7356474483456E+66}\end{pmatrix}\times\begin{pmatrix}-2 \\ 7 \\ -8 \\ -4 \\ 7 \\ -\dfrac{5}{6}\end{pmatrix}=\begin{pmatrix}\dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF}\end{pmatrix}\) . Ainsi \( x=\dfrac{NAN}{INF}\) , \( y=\dfrac{NAN}{INF}\) , \( z=\dfrac{NAN}{INF}\) , \( t=\dfrac{NAN}{INF}\) , \( u=\dfrac{NAN}{INF}\) et \( v=\dfrac{NAN}{INF}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{2.6781158664806E+74}{1.9398664651235E+74}x&-&\dfrac{8.795403421655E+78}{2.1554071834706E+78}y &-&\dfrac{1.1743777878221E+71}{8.0827769380147E+70}z &-&\dfrac{2.1256765032038E+72}{6.4662215504118E+71}t &+&\dfrac{5.133899708928E+66}{2.0206942345037E+66}u &+&\dfrac{9.6118953170688E+75}{6.0620827035111E+75}v &=&-\dfrac{2}{3}\\ &-\dfrac{5.846475362256E+75}{3.7888016896944E+75}x&+&\dfrac{2.2950168607872E+77}{5.0517355862592E+76}y &+&\dfrac{8.540637959124E+79}{4.736002112118E+79}z &+&\dfrac{1.893265783799E+74}{5.4558744331599E+73}t &-&\dfrac{4.910797435392E+66}{1.6839118620864E+66}u &-&\dfrac{7.83938141298E+74}{4.736002112118E+74}v &=&1\\ &-\dfrac{4.3521712065792E+74}{2.0206942345037E+74}x&+&\dfrac{1.6203155760415E+71}{2.5864886201647E+70}y &+&\dfrac{1.810011378528E+75}{8.419559310432E+74}z &+&\dfrac{1.3168018119168E+67}{2.6942589793382E+66}t &-&\dfrac{1.5875684522496E+67}{4.0413884690074E+66}u &-&\dfrac{2.1189127343616E+67}{8.0827769380147E+66}v &=&3\\ &\dfrac{1.403422184448E+66}{1.5155206758778E+66}x&-&\dfrac{1.4800933935936E+74}{5.0517355862592E+73}y &-&\dfrac{7.910668243872E+74}{7.5776033793888E+74}z &-&\dfrac{3.3255316806912E+69}{1.5155206758778E+69}t &+&\dfrac{2.597746733568E+66}{1.5155206758778E+66}u &+&\dfrac{2.226483783072E+70}{1.8944008448472E+70}v &=&-2\\ &-\dfrac{2.301260917248E+66}{3.0310413517555E+66}x&+&\dfrac{5.2532220930048E+71}{2.1554071834706E+71}y &+&\dfrac{1.8600062265216E+75}{2.0206942345037E+75}z &+&\dfrac{1.5767259710976E+67}{8.0827769380147E+66}t &-&\dfrac{2.87190939474E+73}{1.7760007920443E+73}u &-&\dfrac{1.284266866176E+66}{1.3471294896691E+66}v &=&-4\\ &\dfrac{2.295852553728E+66}{2.0206942345037E+66}x&-&\dfrac{1.8071372616192E+74}{5.3885179586765E+73}y &-&\dfrac{2.7641193272525E+71}{2.1554071834706E+71}z &-&\dfrac{4.3221523815936E+67}{1.6165553876029E+67}t &+&\dfrac{1.2285570267648E+67}{6.0620827035111E+66}u &+&\dfrac{9.636206694912E+66}{6.7356474483456E+66}v &=&\dfrac{13}{5}\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-\dfrac{2}{3} \\ 1 \\ 3 \\ -2 \\ -4 \\ \dfrac{13}{5}\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -3 & -\dfrac{12}{5} & -\dfrac{21}{4} & \dfrac{17}{2} & 1 \\ -\dfrac{4}{5} & 0 & -\dfrac{2}{3} & -\dfrac{25}{4} & -\dfrac{3}{2} & \dfrac{19}{5} \\ \dfrac{8}{3} & 2 & -2 & \dfrac{5}{4} & \dfrac{1}{2} & -5 \\ -\dfrac{24}{5} & -2 & -2 & \dfrac{11}{5} & 1 & -\dfrac{9}{5} \\ -3 & -\dfrac{3}{4} & -\dfrac{9}{4} & -3 & -\dfrac{21}{5} & -2 \\ -5 & \dfrac{3}{2} & -2 & -1 & -2 & \dfrac{19}{5}\end{pmatrix}\times\begin{pmatrix}-\dfrac{2}{3} \\ 1 \\ 3 \\ -2 \\ -4 \\ \dfrac{13}{5}\end{pmatrix}=\begin{pmatrix}-\dfrac{953}{30} \\ \dfrac{4037}{150} \\ -\dfrac{419}{18} \\ -\dfrac{447}{25} \\ \dfrac{121}{10} \\ \dfrac{2807}{150}\end{pmatrix}\) . Ainsi \( x=\dfrac{953}{30}\) , \( y=\dfrac{4037}{150}\) , \( z=\dfrac{419}{18}\) , \( t=\dfrac{447}{25}\) , \( u=\dfrac{121}{10}\) et \( v=\dfrac{2807}{150}\)