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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 \[ A= \begin{pmatrix}1 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 & -2 \\ \dfrac{1}{3} & 5 & 1 & -4 & 1 & 4 \\ \dfrac{21}{5} & -\dfrac{9}{2} & \dfrac{20}{3} & 3 & 0 & 4 \\ 4 & 3 & \dfrac{19}{5} & 1 & -5 & \dfrac{15}{4} \\ -2 & -3 & -3 & -4 & -3 & -4 \\ -\dfrac{19}{5} & 0 & -\dfrac{21}{5} & -\dfrac{9}{4} & 2 & -2\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (3, 6)\) et \( (2, 4)\) : \( \widehat{A}_{3, 6}=\) \( \widehat{A}_{2, 4}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 & -2 \\ 0 & \dfrac{47}{9} & -\dfrac{1}{3} & -\dfrac{73}{15} & 0 & \dfrac{14}{3} \\ 0 & -\dfrac{17}{10} & -\dfrac{152}{15} & -\dfrac{198}{25} & -\dfrac{63}{5} & \dfrac{62}{5} \\ 0 & \dfrac{17}{3} & -\dfrac{61}{5} & -\dfrac{47}{5} & -17 & \dfrac{47}{4} \\ 0 & -\dfrac{13}{3} & 5 & \dfrac{6}{5} & 3 & -8 \\ 0 & -\dfrac{38}{15} & 11 & \dfrac{763}{100} & \dfrac{67}{5} & -\dfrac{48}{5}\end{pmatrix} \)
  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 \( A^{-1}_{3, 5}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{2}{3}y &+&4z &+&\dfrac{13}{5}t &+&3u &-&2v &=&2\\ &\dfrac{1}{3}x&+&5y &+&z &-&4t &+&u &+&4v &=&0\\ &\dfrac{21}{5}x&-&\dfrac{9}{2}y &+&\dfrac{20}{3}z &+&3t &&&+&4v &=&6\\ &4x&+&3y &+&\dfrac{19}{5}z &+&t &-&5u &+&\dfrac{15}{4}v &=&6\\ &-2x&-&3y &-&3z &-&4t &-&3u &-&4v &=&1\\ &-\dfrac{19}{5}x&&&-&\dfrac{21}{5}z &-&\dfrac{9}{4}t &+&2u &-&2v &=&-3\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{89748000000000000}{251319943125000000}x&+&\dfrac{1.098947655E+20}{6.031678635E+20}y &-&\dfrac{8539303500000000000}{2.51319943125E+19}z &-&\dfrac{1.61165295E+19}{1.8848995734375E+19}t &+&\dfrac{2.27095245E+19}{6.031678635E+20}u &-&\dfrac{1.02824235E+19}{6282998578125000000}v &=&-\dfrac{67}{4}\\ &\dfrac{585670500000000000}{7539598293750000000}x&+&\dfrac{1.37172285E+19}{5.0263988625E+20}y &-&\dfrac{1.678143015E+20}{1.50791965875E+21}z &+&\dfrac{2.52311355E+19}{2.8273493601562E+20}t &-&\dfrac{2.213881065E+20}{4.52375897625E+21}u &+&\dfrac{2.20830705E+19}{1.1309397440625E+21}v &=&-6\\ &\dfrac{2644204500000000000}{8042238180000000000}x&-&\dfrac{2161606500000000000}{2.010559545E+20}y &+&\dfrac{5.74987005E+19}{3.0158393175E+20}z &+&\dfrac{1.007954685E+20}{2.261879488125E+20}t &+&\dfrac{6.21205605E+19}{9.0475179525E+20}u &+&\dfrac{3.298948155E+20}{4.52375897625E+20}v &=&-5\\ &\dfrac{276088500000000000}{1.0052797725E+19}x&-&\dfrac{3423829500000000000}{1.8848995734375E+19}y &+&\dfrac{3236314500000000000}{2.261879488125E+20}z &+&\dfrac{3.0609819E+19}{1.6964096160937E+20}t &-&\dfrac{2.033904735E+20}{1.357127692875E+21}u &+&\dfrac{3.744606105E+20}{1.357127692875E+21}v &=&8\\ &-\dfrac{509692500000000000}{8042238180000000000}x&+&\dfrac{8.26598115405E+26}{7.328489541525E+27}y &-&\dfrac{1.719665235E+20}{2.261879488125E+21}z &-&\dfrac{1.2559703027063E+26}{3.4352294725898E+26}t &-&\dfrac{5.441546269575E+29}{1.0992734312287E+31}u &-&\dfrac{1.2356367221475E+26}{2.7481835780719E+26}v &=&-1\\ &-\dfrac{133488000000000000}{1256599715625000000}x&-&\dfrac{810688500000000000}{1.256599715625E+20}y &+&\dfrac{4.3280163E+19}{2.8273493601562E+20}z &+&\dfrac{3.301644298725E+25}{2.7481835780719E+26}t &-&\dfrac{6.6383693623125E+26}{6.8704589451797E+27}u &+&\dfrac{1.213723855125E+23}{3.8169216362109E+23}v &=&2\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{3, 6}=\begin{pmatrix}1 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 \\ \dfrac{1}{3} & 5 & 1 & -4 & 1 \\ 4 & 3 & \dfrac{19}{5} & 1 & -5 \\ -2 & -3 & -3 & -4 & -3 \\ -\dfrac{19}{5} & 0 & -\dfrac{21}{5} & -\dfrac{9}{4} & 2\end{pmatrix}\) \( \widehat{A}_{2, 4}=\begin{pmatrix}1 & -\dfrac{2}{3} & 4 & 3 & -2 \\ \dfrac{21}{5} & -\dfrac{9}{2} & \dfrac{20}{3} & 0 & 4 \\ 4 & 3 & \dfrac{19}{5} & -5 & \dfrac{15}{4} \\ -2 & -3 & -3 & -3 & -4 \\ -\dfrac{19}{5} & 0 & -\dfrac{21}{5} & 2 & -2\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{1}{3}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(\dfrac{21}{5}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(4\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-2\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-\dfrac{19}{5}\right)L_1\)
  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 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 & -2 \\ 0 & \dfrac{47}{9} & -\dfrac{1}{3} & -\dfrac{73}{15} & 0 & \dfrac{14}{3} \\ 0 & -\dfrac{17}{10} & -\dfrac{152}{15} & -\dfrac{198}{25} & -\dfrac{63}{5} & \dfrac{62}{5} \\ 0 & \dfrac{17}{3} & -\dfrac{61}{5} & -\dfrac{47}{5} & -17 & \dfrac{47}{4} \\ 0 & -\dfrac{13}{3} & 5 & \dfrac{6}{5} & 3 & -8 \\ 0 & -\dfrac{38}{15} & 11 & \dfrac{763}{100} & \dfrac{67}{5} & -\dfrac{48}{5}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{47}{9} & -\dfrac{1}{3} & -\dfrac{73}{15} & 0 & \dfrac{14}{3} \\ -\dfrac{17}{10} & -\dfrac{152}{15} & -\dfrac{198}{25} & -\dfrac{63}{5} & \dfrac{62}{5} \\ \dfrac{17}{3} & -\dfrac{61}{5} & -\dfrac{47}{5} & -17 & \dfrac{47}{4} \\ -\dfrac{13}{3} & 5 & \dfrac{6}{5} & 3 & -8 \\ -\dfrac{38}{15} & 11 & \dfrac{763}{100} & \dfrac{67}{5} & -\dfrac{48}{5}\end{pmatrix}\\ &=&\dfrac{251319943125000000}{40500000000000} \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{251319943125000000}{40500000000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 & -2 \\ \dfrac{1}{3} & 5 & 1 & -4 & 1 & 4 \\ \dfrac{21}{5} & -\dfrac{9}{2} & \dfrac{20}{3} & 3 & 0 & 4 \\ 4 & 3 & \dfrac{19}{5} & 1 & -5 & \dfrac{15}{4} \\ -2 & -3 & -3 & -4 & -3 & -4 \\ -\dfrac{19}{5} & 0 & -\dfrac{21}{5} & -\dfrac{9}{4} & 2 & -2\end{pmatrix} =\dfrac{40500000000000}{251319943125000000}\begin{pmatrix}-\dfrac{2.2555462255583E+34}{1.0178457696562E+31} & \dfrac{2.7618746215195E+37}{2.442829847175E+34} & -\dfrac{2.1460972699471E+36}{1.0178457696562E+33} & -\dfrac{4.0504052773124E+36}{7.6338432724219E+32} & \dfrac{5.7073564057358E+36}{2.442829847175E+34} & -\dfrac{2.5841780892072E+36}{2.5446144241406E+32} \\ \dfrac{1.4719067674999E+35}{3.0535373089687E+32} & \dfrac{3.4474130864526E+36}{2.0356915393125E+34} & -\dfrac{4.2175080708542E+37}{6.1070746179375E+34} & \dfrac{6.3410875388392E+36}{1.1450764908633E+34} & -\dfrac{5.5639246334131E+37}{1.8321223853813E+35} & \dfrac{5.5499160220854E+36}{4.5803059634531E+34} \\ \dfrac{6.6454132455087E+35}{3.2571064629E+32} & -\dfrac{5.4325482263863E+35}{8.14276615725E+33} & \dfrac{1.4450570139421E+37}{1.2214149235875E+34} & \dfrac{2.5331911410678E+37}{9.1606119269063E+33} & \dfrac{1.5612135731753E+37}{3.6642447707625E+34} & \dfrac{8.2909146268692E+37}{1.8321223853813E+34} \\ \dfrac{6.9386546117467E+34}{4.071383078625E+32} & -\dfrac{8.604766352097E+35}{7.6338432724219E+32} & \dfrac{8.1335037607461E+35}{9.1606119269063E+33} & \dfrac{7.6928579701465E+36}{6.8704589451797E+33} & -\dfrac{5.1116082232187E+37}{5.4963671561437E+34} & \dfrac{9.4109419333413E+37}{5.4963671561437E+34} \\ -\dfrac{1.2809589011124E+35}{3.2571064629E+32} & \dfrac{2.0774059135082E+44}{2.9680382643176E+41} & -\dfrac{4.3218616905424E+37}{9.1606119269063E+34} & -\dfrac{3.1565038504282E+43}{1.3912679363989E+40} & -\dfrac{1.3675690989816E+47}{4.4520573964764E+44} & -\dfrac{3.1054015073327E+43}{1.1130143491191E+40} \\ -\dfrac{3.354819656787E+34}{5.0892288482812E+31} & -\dfrac{2.0374218771209E+35}{5.0892288482812E+33} & \dfrac{1.0877168103601E+37}{1.1450764908633E+34} & \dfrac{8.2976905737455E+42}{1.1130143491191E+40} & -\dfrac{1.6683546105791E+44}{2.7825358727978E+41} & \dfrac{3.0503301023947E+40}{1.5458532626654E+37}\end{pmatrix} =\begin{pmatrix}-\dfrac{89748000000000000}{251319943125000000} & \dfrac{1.098947655E+20}{6.031678635E+20} & -\dfrac{8539303500000000000}{2.51319943125E+19} & -\dfrac{1.61165295E+19}{1.8848995734375E+19} & \dfrac{2.27095245E+19}{6.031678635E+20} & -\dfrac{1.02824235E+19}{6282998578125000000} \\ \dfrac{585670500000000000}{7539598293750000000} & \dfrac{1.37172285E+19}{5.0263988625E+20} & -\dfrac{1.678143015E+20}{1.50791965875E+21} & \dfrac{2.52311355E+19}{2.8273493601562E+20} & -\dfrac{2.213881065E+20}{4.52375897625E+21} & \dfrac{2.20830705E+19}{1.1309397440625E+21} \\ \dfrac{2644204500000000000}{8042238180000000000} & -\dfrac{2161606500000000000}{2.010559545E+20} & \dfrac{5.74987005E+19}{3.0158393175E+20} & \dfrac{1.007954685E+20}{2.261879488125E+20} & \dfrac{6.21205605E+19}{9.0475179525E+20} & \dfrac{3.298948155E+20}{4.52375897625E+20} \\ \dfrac{276088500000000000}{1.0052797725E+19} & -\dfrac{3423829500000000000}{1.8848995734375E+19} & \dfrac{3236314500000000000}{2.261879488125E+20} & \dfrac{3.0609819E+19}{1.6964096160937E+20} & -\dfrac{2.033904735E+20}{1.357127692875E+21} & \dfrac{3.744606105E+20}{1.357127692875E+21} \\ -\dfrac{509692500000000000}{8042238180000000000} & \dfrac{8.26598115405E+26}{7.328489541525E+27} & -\dfrac{1.719665235E+20}{2.261879488125E+21} & -\dfrac{1.2559703027063E+26}{3.4352294725898E+26} & -\dfrac{5.441546269575E+29}{1.0992734312287E+31} & -\dfrac{1.2356367221475E+26}{2.7481835780719E+26} \\ -\dfrac{133488000000000000}{1256599715625000000} & -\dfrac{810688500000000000}{1.256599715625E+20} & \dfrac{4.3280163E+19}{2.8273493601562E+20} & \dfrac{3.301644298725E+25}{2.7481835780719E+26} & -\dfrac{6.6383693623125E+26}{6.8704589451797E+27} & \dfrac{1.213723855125E+23}{3.8169216362109E+23}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{3, 5}=B_{3, 5}= \left(\dfrac{251319943125000000}{40500000000000}\right)^{-1}Co(A)_{5, 3}= \left(\dfrac{251319943125000000}{40500000000000}\right)^{-1}\times(-1)^{5+3}\det\begin{pmatrix}1 & -\dfrac{2}{3} & \dfrac{13}{5} & 3 & -2 \\ \dfrac{1}{3} & 5 & -4 & 1 & 4 \\ \dfrac{21}{5} & -\dfrac{9}{2} & 3 & 0 & 4 \\ 4 & 3 & 1 & -5 & \dfrac{15}{4} \\ -\dfrac{19}{5} & 0 & -\dfrac{9}{4} & 2 & -2\end{pmatrix}=\dfrac{6.21205605E+19}{9.0475179525E+20}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 & -2 \\ \dfrac{1}{3} & 5 & 1 & -4 & 1 & 4 \\ \dfrac{21}{5} & -\dfrac{9}{2} & \dfrac{20}{3} & 3 & 0 & 4 \\ 4 & 3 & \dfrac{19}{5} & 1 & -5 & \dfrac{15}{4} \\ -2 & -3 & -3 & -4 & -3 & -4 \\ -\dfrac{19}{5} & 0 & -\dfrac{21}{5} & -\dfrac{9}{4} & 2 & -2\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{2}{3}y &+&4z &+&\dfrac{13}{5}t &+&3u &-&2v &=&2\\ &\dfrac{1}{3}x&+&5y &+&z &-&4t &+&u &+&4v &=&0\\ &\dfrac{21}{5}x&-&\dfrac{9}{2}y &+&\dfrac{20}{3}z &+&3t &&&+&4v &=&6\\ &4x&+&3y &+&\dfrac{19}{5}z &+&t &-&5u &+&\dfrac{15}{4}v &=&6\\ &-2x&-&3y &-&3z &-&4t &-&3u &-&4v &=&1\\ &-\dfrac{19}{5}x&&&-&\dfrac{21}{5}z &-&\dfrac{9}{4}t &+&2u &-&2v &=&-3\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}2 \\ 0 \\ 6 \\ 6 \\ 1 \\ -3\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{89748000000000000}{251319943125000000} & \dfrac{1.098947655E+20}{6.031678635E+20} & -\dfrac{8539303500000000000}{2.51319943125E+19} & -\dfrac{1.61165295E+19}{1.8848995734375E+19} & \dfrac{2.27095245E+19}{6.031678635E+20} & -\dfrac{1.02824235E+19}{6282998578125000000} \\ \dfrac{585670500000000000}{7539598293750000000} & \dfrac{1.37172285E+19}{5.0263988625E+20} & -\dfrac{1.678143015E+20}{1.50791965875E+21} & \dfrac{2.52311355E+19}{2.8273493601562E+20} & -\dfrac{2.213881065E+20}{4.52375897625E+21} & \dfrac{2.20830705E+19}{1.1309397440625E+21} \\ \dfrac{2644204500000000000}{8042238180000000000} & -\dfrac{2161606500000000000}{2.010559545E+20} & \dfrac{5.74987005E+19}{3.0158393175E+20} & \dfrac{1.007954685E+20}{2.261879488125E+20} & \dfrac{6.21205605E+19}{9.0475179525E+20} & \dfrac{3.298948155E+20}{4.52375897625E+20} \\ \dfrac{276088500000000000}{1.0052797725E+19} & -\dfrac{3423829500000000000}{1.8848995734375E+19} & \dfrac{3236314500000000000}{2.261879488125E+20} & \dfrac{3.0609819E+19}{1.6964096160937E+20} & -\dfrac{2.033904735E+20}{1.357127692875E+21} & \dfrac{3.744606105E+20}{1.357127692875E+21} \\ -\dfrac{509692500000000000}{8042238180000000000} & \dfrac{8.26598115405E+26}{7.328489541525E+27} & -\dfrac{1.719665235E+20}{2.261879488125E+21} & -\dfrac{1.2559703027063E+26}{3.4352294725898E+26} & -\dfrac{5.441546269575E+29}{1.0992734312287E+31} & -\dfrac{1.2356367221475E+26}{2.7481835780719E+26} \\ -\dfrac{133488000000000000}{1256599715625000000} & -\dfrac{810688500000000000}{1.256599715625E+20} & \dfrac{4.3280163E+19}{2.8273493601562E+20} & \dfrac{3.301644298725E+25}{2.7481835780719E+26} & -\dfrac{6.6383693623125E+26}{6.8704589451797E+27} & \dfrac{1.213723855125E+23}{3.8169216362109E+23}\end{pmatrix}\times\begin{pmatrix}2 \\ 0 \\ 6 \\ 6 \\ 1 \\ -3\end{pmatrix}=\begin{pmatrix}-\dfrac{1.3245634832694E+96}{4.511773380565E+95} \\ -\dfrac{1.3888781957346E+102}{1.6445413972159E+103} \\ \dfrac{5.2904606207067E+101}{2.2453471876655E+101} \\ \dfrac{1.7460998744404E+101}{7.1044188359729E+101} \\ -\dfrac{2.7887447058859E+124}{1.8877897253041E+124} \\ \dfrac{9.634050755139E+115}{2.5604786855796E+116}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.3245634832694E+96}{4.511773380565E+95}\) , \( y=\dfrac{1.3888781957346E+102}{1.6445413972159E+103}\) , \( z=\dfrac{5.2904606207067E+101}{2.2453471876655E+101}\) , \( t=\dfrac{1.7460998744404E+101}{7.1044188359729E+101}\) , \( u=\dfrac{2.7887447058859E+124}{1.8877897253041E+124}\) et \( v=\dfrac{9.634050755139E+115}{2.5604786855796E+116}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{89748000000000000}{251319943125000000}x&+&\dfrac{1.098947655E+20}{6.031678635E+20}y &-&\dfrac{8539303500000000000}{2.51319943125E+19}z &-&\dfrac{1.61165295E+19}{1.8848995734375E+19}t &+&\dfrac{2.27095245E+19}{6.031678635E+20}u &-&\dfrac{1.02824235E+19}{6282998578125000000}v &=&-\dfrac{67}{4}\\ &\dfrac{585670500000000000}{7539598293750000000}x&+&\dfrac{1.37172285E+19}{5.0263988625E+20}y &-&\dfrac{1.678143015E+20}{1.50791965875E+21}z &+&\dfrac{2.52311355E+19}{2.8273493601562E+20}t &-&\dfrac{2.213881065E+20}{4.52375897625E+21}u &+&\dfrac{2.20830705E+19}{1.1309397440625E+21}v &=&-6\\ &\dfrac{2644204500000000000}{8042238180000000000}x&-&\dfrac{2161606500000000000}{2.010559545E+20}y &+&\dfrac{5.74987005E+19}{3.0158393175E+20}z &+&\dfrac{1.007954685E+20}{2.261879488125E+20}t &+&\dfrac{6.21205605E+19}{9.0475179525E+20}u &+&\dfrac{3.298948155E+20}{4.52375897625E+20}v &=&-5\\ &\dfrac{276088500000000000}{1.0052797725E+19}x&-&\dfrac{3423829500000000000}{1.8848995734375E+19}y &+&\dfrac{3236314500000000000}{2.261879488125E+20}z &+&\dfrac{3.0609819E+19}{1.6964096160937E+20}t &-&\dfrac{2.033904735E+20}{1.357127692875E+21}u &+&\dfrac{3.744606105E+20}{1.357127692875E+21}v &=&8\\ &-\dfrac{509692500000000000}{8042238180000000000}x&+&\dfrac{8.26598115405E+26}{7.328489541525E+27}y &-&\dfrac{1.719665235E+20}{2.261879488125E+21}z &-&\dfrac{1.2559703027063E+26}{3.4352294725898E+26}t &-&\dfrac{5.441546269575E+29}{1.0992734312287E+31}u &-&\dfrac{1.2356367221475E+26}{2.7481835780719E+26}v &=&-1\\ &-\dfrac{133488000000000000}{1256599715625000000}x&-&\dfrac{810688500000000000}{1.256599715625E+20}y &+&\dfrac{4.3280163E+19}{2.8273493601562E+20}z &+&\dfrac{3.301644298725E+25}{2.7481835780719E+26}t &-&\dfrac{6.6383693623125E+26}{6.8704589451797E+27}u &+&\dfrac{1.213723855125E+23}{3.8169216362109E+23}v &=&2\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-\dfrac{67}{4} \\ -6 \\ -5 \\ 8 \\ -1 \\ 2\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -\dfrac{2}{3} & 4 & \dfrac{13}{5} & 3 & -2 \\ \dfrac{1}{3} & 5 & 1 & -4 & 1 & 4 \\ \dfrac{21}{5} & -\dfrac{9}{2} & \dfrac{20}{3} & 3 & 0 & 4 \\ 4 & 3 & \dfrac{19}{5} & 1 & -5 & \dfrac{15}{4} \\ -2 & -3 & -3 & -4 & -3 & -4 \\ -\dfrac{19}{5} & 0 & -\dfrac{21}{5} & -\dfrac{9}{4} & 2 & -2\end{pmatrix}\times\begin{pmatrix}-\dfrac{67}{4} \\ -6 \\ -5 \\ 8 \\ -1 \\ 2\end{pmatrix}=\begin{pmatrix}-\dfrac{379}{20} \\ -\dfrac{787}{12} \\ -\dfrac{2681}{60} \\ -\dfrac{167}{2} \\ \dfrac{59}{2} \\ \dfrac{1213}{20}\end{pmatrix}\) . Ainsi \( x=\dfrac{379}{20}\) , \( y=\dfrac{787}{12}\) , \( z=\dfrac{2681}{60}\) , \( t=\dfrac{167}{2}\) , \( u=\dfrac{59}{2}\) et \( v=\dfrac{1213}{20}\)