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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 \[ A= \begin{pmatrix}1 & -\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ \dfrac{6}{5} & \dfrac{1}{2} & 3 & \dfrac{7}{5} & 0 & 4 \\ -\dfrac{5}{2} & \dfrac{3}{2} & 1 & -5 & \dfrac{14}{5} & \dfrac{9}{2} \\ 3 & 4 & 0 & -1 & -\dfrac{2}{3} & 0 \\ -\dfrac{9}{2} & 4 & 3 & -2 & -\dfrac{17}{3} & -5 \\ 3 & -\dfrac{8}{5} & -4 & -\dfrac{9}{4} & \dfrac{14}{5} & 2\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (3, 1)\) et \( (5, 4)\) : \( \widehat{A}_{3, 1}=\) \( \widehat{A}_{5, 4}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & -\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ 0 & \dfrac{21}{10} & 6 & \dfrac{31}{5} & -\dfrac{18}{5} & \dfrac{16}{25} \\ 0 & -\dfrac{11}{6} & -\dfrac{21}{4} & -15 & \dfrac{103}{10} & \dfrac{23}{2} \\ 0 & 8 & \dfrac{15}{2} & 11 & -\dfrac{29}{3} & -\dfrac{42}{5} \\ 0 & -2 & -\dfrac{33}{4} & -20 & \dfrac{47}{6} & \dfrac{38}{5} \\ 0 & \dfrac{12}{5} & \dfrac{7}{2} & \dfrac{39}{4} & -\dfrac{31}{5} & -\dfrac{32}{5}\end{pmatrix} \)
  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 \( A^{-1}_{1, 3}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{4}{3}y &-&\dfrac{5}{2}z &-&4t &+&3u &+&\dfrac{14}{5}v &=&-4\\ &\dfrac{6}{5}x&+&\dfrac{1}{2}y &+&3z &+&\dfrac{7}{5}t &&&+&4v &=&0\\ &-\dfrac{5}{2}x&+&\dfrac{3}{2}y &+&z &-&5t &+&\dfrac{14}{5}u &+&\dfrac{9}{2}v &=&4\\ &3x&+&4y &&&-&t &-&\dfrac{2}{3}u &&&=&9\\ &-\dfrac{9}{2}x&+&4y &+&3z &-&2t &-&\dfrac{17}{3}u &-&5v &=&-\dfrac{71}{6}\\ &3x&-&\dfrac{8}{5}y &-&4z &-&\dfrac{9}{4}t &+&\dfrac{14}{5}u &+&2v &=&-4\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{3.192468E+24}{3.92406288E+24}x&+&\dfrac{2.572908E+24}{5.88609432E+25}y &-&\dfrac{7.145124E+24}{1.96203144E+25}z &+&\dfrac{3.52394205E+28}{1.47152358E+29}t &-&\dfrac{2.148684E+24}{2.354437728E+25}u &-&\dfrac{1.2462444E+25}{1.96203144E+25}v &=&-3\\ &-\dfrac{5.5452E+22}{7.84812576E+22}x&-&\dfrac{2.8596E+22}{2.94304716E+23}y &+&\dfrac{3.159492E+24}{9.8101572E+24}z &+&\dfrac{1.0524444E+25}{1.177218864E+26}t &-&\dfrac{6.4788E+22}{1.177218864E+25}u &+&\dfrac{4.366668E+24}{9.8101572E+24}v &=&9\\ &\dfrac{1.5698868E+25}{1.177218864E+25}x&-&\dfrac{6.912876E+24}{1.765828296E+26}y &-&\dfrac{2.4906108E+25}{5.88609432E+25}z &+&\dfrac{8.691141996E+36}{3.1784909328E+37}t &-&\dfrac{8.67247533E+35}{3.973113666E+36}u &-&\dfrac{8.1365028E+25}{5.88609432E+25}v &=&8\\ &-\dfrac{2.34384E+23}{2.94304716E+23}x&-&\dfrac{2.51976E+23}{2.20728537E+25}y &+&\dfrac{1.728312E+24}{7.3576179E+24}z &-&\dfrac{5.74365024E+28}{5.297484888E+29}t &-&\dfrac{7.73808E+23}{2.20728537E+25}u &+&\dfrac{7.675764E+24}{1.47152358E+25}v &=&-6\\ &\dfrac{4.836372E+24}{7.84812576E+24}x&-&\dfrac{8.691684E+24}{2.354437728E+25}y &-&\dfrac{2.304804E+24}{3.92406288E+25}z &+&\dfrac{2.084559936E+34}{7.5342007296E+34}t &-&\dfrac{9.2084412E+30}{2.354437728E+31}u &-&\dfrac{7.613004E+24}{7.84812576E+24}v &=&-\dfrac{65}{7}\\ &-\dfrac{5.1631356E+25}{5.88609432E+25}x&+&\dfrac{9.972996E+24}{3.531656592E+25}y &+&\dfrac{7.16052E+23}{2.354437728E+24}z &-&\dfrac{2.65030146E+28}{1.0594969776E+29}t &+&\dfrac{1.4412372E+25}{7.063313184E+25}u &+&\dfrac{2.328756E+24}{2.354437728E+24}v &=&-5\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{3, 1}=\begin{pmatrix}-\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ \dfrac{1}{2} & 3 & \dfrac{7}{5} & 0 & 4 \\ 4 & 0 & -1 & -\dfrac{2}{3} & 0 \\ 4 & 3 & -2 & -\dfrac{17}{3} & -5 \\ -\dfrac{8}{5} & -4 & -\dfrac{9}{4} & \dfrac{14}{5} & 2\end{pmatrix}\) \( \widehat{A}_{5, 4}=\begin{pmatrix}1 & -\dfrac{4}{3} & -\dfrac{5}{2} & 3 & \dfrac{14}{5} \\ \dfrac{6}{5} & \dfrac{1}{2} & 3 & 0 & 4 \\ -\dfrac{5}{2} & \dfrac{3}{2} & 1 & \dfrac{14}{5} & \dfrac{9}{2} \\ 3 & 4 & 0 & -\dfrac{2}{3} & 0 \\ 3 & -\dfrac{8}{5} & -4 & \dfrac{14}{5} & 2\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{6}{5}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{5}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(3\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-\dfrac{9}{2}\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(3\right)L_1\)
  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{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ 0 & \dfrac{21}{10} & 6 & \dfrac{31}{5} & -\dfrac{18}{5} & \dfrac{16}{25} \\ 0 & -\dfrac{11}{6} & -\dfrac{21}{4} & -15 & \dfrac{103}{10} & \dfrac{23}{2} \\ 0 & 8 & \dfrac{15}{2} & 11 & -\dfrac{29}{3} & -\dfrac{42}{5} \\ 0 & -2 & -\dfrac{33}{4} & -20 & \dfrac{47}{6} & \dfrac{38}{5} \\ 0 & \dfrac{12}{5} & \dfrac{7}{2} & \dfrac{39}{4} & -\dfrac{31}{5} & -\dfrac{32}{5}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{21}{10} & 6 & \dfrac{31}{5} & -\dfrac{18}{5} & \dfrac{16}{25} \\ -\dfrac{11}{6} & -\dfrac{21}{4} & -15 & \dfrac{103}{10} & \dfrac{23}{2} \\ 8 & \dfrac{15}{2} & 11 & -\dfrac{29}{3} & -\dfrac{42}{5} \\ -2 & -\dfrac{33}{4} & -20 & \dfrac{47}{6} & \dfrac{38}{5} \\ \dfrac{12}{5} & \dfrac{7}{2} & \dfrac{39}{4} & -\dfrac{31}{5} & -\dfrac{32}{5}\end{pmatrix}\\ &=&\dfrac{1.96203144E+22}{1.2E+19} \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{1.96203144E+22}{1.2E+19}\right)^{-1}{}^tCo\begin{pmatrix}1 & -\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ \dfrac{6}{5} & \dfrac{1}{2} & 3 & \dfrac{7}{5} & 0 & 4 \\ -\dfrac{5}{2} & \dfrac{3}{2} & 1 & -5 & \dfrac{14}{5} & \dfrac{9}{2} \\ 3 & 4 & 0 & -1 & -\dfrac{2}{3} & 0 \\ -\dfrac{9}{2} & 4 & 3 & -2 & -\dfrac{17}{3} & -5 \\ 3 & -\dfrac{8}{5} & -4 & -\dfrac{9}{4} & \dfrac{14}{5} & 2\end{pmatrix} =\dfrac{1.2E+19}{1.96203144E+22}\begin{pmatrix}\dfrac{6.2637225871939E+46}{4.708875456E+43} & \dfrac{5.0481263882275E+46}{7.063313184E+44} & -\dfrac{1.4018957930699E+47}{2.354437728E+44} & \dfrac{6.9140850948381E+50}{1.765828296E+48} & -\dfrac{4.215785562625E+46}{2.8253252736E+44} & -\dfrac{2.4451706947239E+47}{2.354437728E+44} \\ -\dfrac{1.0879856741088E+45}{9.417750912E+41} & -\dfrac{5.610625105824E+44}{3.531656592E+42} & \dfrac{6.1990226384285E+46}{1.177218864E+44} & \dfrac{2.0649290016519E+47}{1.4126626368E+45} & -\dfrac{1.2711609293472E+45}{1.4126626368E+44} & \dfrac{8.5675399040419E+46}{1.177218864E+44} \\ \dfrac{3.080167258841E+47}{1.4126626368E+44} & -\dfrac{1.3563280052821E+47}{2.1189939552E+45} & -\dfrac{4.8866566944036E+47}{7.063313184E+44} & \dfrac{1.7052293845656E+59}{3.81418911936E+56} & -\dfrac{1.7015669260084E+58}{4.7677363992E+55} & -\dfrac{1.5964074305248E+48}{7.063313184E+44} \\ -\dfrac{4.5986877703296E+45}{3.531656592E+42} & -\dfrac{4.9438483412544E+45}{2.648742444E+44} & \dfrac{3.3910024821293E+46}{8.82914148E+43} & -\dfrac{1.1269222351244E+51}{6.3569818656E+48} & -\dfrac{1.5182356245235E+46}{2.648742444E+44} & \dfrac{1.506009029402E+47}{1.765828296E+44} \\ \dfrac{9.4891139195357E+46}{9.417750912E+43} & -\dfrac{1.7053357274545E+47}{2.8253252736E+44} & -\dfrac{4.5220979110378E+46}{4.708875456E+44} & \dfrac{4.0899721329964E+56}{9.04104087552E+53} & -\dfrac{1.8067251147791E+53}{2.8253252736E+50} & -\dfrac{1.4936953200846E+47}{9.417750912E+43} \\ -\dfrac{1.0130234376183E+48}{7.063313184E+44} & \dfrac{1.9567331702994E+47}{4.2379879104E+44} & \dfrac{1.4049165366749E+46}{2.8253252736E+43} & -\dfrac{5.1999747899979E+50}{1.27139637312E+48} & \dfrac{2.8277526988976E+47}{8.4759758208E+44} & \dfrac{4.5690924880886E+46}{2.8253252736E+43}\end{pmatrix} =\begin{pmatrix}\dfrac{3.192468E+24}{3.92406288E+24} & \dfrac{2.572908E+24}{5.88609432E+25} & -\dfrac{7.145124E+24}{1.96203144E+25} & \dfrac{3.52394205E+28}{1.47152358E+29} & -\dfrac{2.148684E+24}{2.354437728E+25} & -\dfrac{1.2462444E+25}{1.96203144E+25} \\ -\dfrac{5.5452E+22}{7.84812576E+22} & -\dfrac{2.8596E+22}{2.94304716E+23} & \dfrac{3.159492E+24}{9.8101572E+24} & \dfrac{1.0524444E+25}{1.177218864E+26} & -\dfrac{6.4788E+22}{1.177218864E+25} & \dfrac{4.366668E+24}{9.8101572E+24} \\ \dfrac{1.5698868E+25}{1.177218864E+25} & -\dfrac{6.912876E+24}{1.765828296E+26} & -\dfrac{2.4906108E+25}{5.88609432E+25} & \dfrac{8.691141996E+36}{3.1784909328E+37} & -\dfrac{8.67247533E+35}{3.973113666E+36} & -\dfrac{8.1365028E+25}{5.88609432E+25} \\ -\dfrac{2.34384E+23}{2.94304716E+23} & -\dfrac{2.51976E+23}{2.20728537E+25} & \dfrac{1.728312E+24}{7.3576179E+24} & -\dfrac{5.74365024E+28}{5.297484888E+29} & -\dfrac{7.73808E+23}{2.20728537E+25} & \dfrac{7.675764E+24}{1.47152358E+25} \\ \dfrac{4.836372E+24}{7.84812576E+24} & -\dfrac{8.691684E+24}{2.354437728E+25} & -\dfrac{2.304804E+24}{3.92406288E+25} & \dfrac{2.084559936E+34}{7.5342007296E+34} & -\dfrac{9.2084412E+30}{2.354437728E+31} & -\dfrac{7.613004E+24}{7.84812576E+24} \\ -\dfrac{5.1631356E+25}{5.88609432E+25} & \dfrac{9.972996E+24}{3.531656592E+25} & \dfrac{7.16052E+23}{2.354437728E+24} & -\dfrac{2.65030146E+28}{1.0594969776E+29} & \dfrac{1.4412372E+25}{7.063313184E+25} & \dfrac{2.328756E+24}{2.354437728E+24}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{1, 3}=B_{1, 3}= \left(\dfrac{1.96203144E+22}{1.2E+19}\right)^{-1}Co(A)_{3, 1}= \left(\dfrac{1.96203144E+22}{1.2E+19}\right)^{-1}\times(-1)^{3+1}\det\begin{pmatrix}-\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ \dfrac{1}{2} & 3 & \dfrac{7}{5} & 0 & 4 \\ 4 & 0 & -1 & -\dfrac{2}{3} & 0 \\ 4 & 3 & -2 & -\dfrac{17}{3} & -5 \\ -\dfrac{8}{5} & -4 & -\dfrac{9}{4} & \dfrac{14}{5} & 2\end{pmatrix}=-\dfrac{7.145124E+24}{1.96203144E+25}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ \dfrac{6}{5} & \dfrac{1}{2} & 3 & \dfrac{7}{5} & 0 & 4 \\ -\dfrac{5}{2} & \dfrac{3}{2} & 1 & -5 & \dfrac{14}{5} & \dfrac{9}{2} \\ 3 & 4 & 0 & -1 & -\dfrac{2}{3} & 0 \\ -\dfrac{9}{2} & 4 & 3 & -2 & -\dfrac{17}{3} & -5 \\ 3 & -\dfrac{8}{5} & -4 & -\dfrac{9}{4} & \dfrac{14}{5} & 2\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{4}{3}y &-&\dfrac{5}{2}z &-&4t &+&3u &+&\dfrac{14}{5}v &=&-4\\ &\dfrac{6}{5}x&+&\dfrac{1}{2}y &+&3z &+&\dfrac{7}{5}t &&&+&4v &=&0\\ &-\dfrac{5}{2}x&+&\dfrac{3}{2}y &+&z &-&5t &+&\dfrac{14}{5}u &+&\dfrac{9}{2}v &=&4\\ &3x&+&4y &&&-&t &-&\dfrac{2}{3}u &&&=&9\\ &-\dfrac{9}{2}x&+&4y &+&3z &-&2t &-&\dfrac{17}{3}u &-&5v &=&-\dfrac{71}{6}\\ &3x&-&\dfrac{8}{5}y &-&4z &-&\dfrac{9}{4}t &+&\dfrac{14}{5}u &+&2v &=&-4\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-4 \\ 0 \\ 4 \\ 9 \\ -\dfrac{71}{6} \\ -4\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{3.192468E+24}{3.92406288E+24} & \dfrac{2.572908E+24}{5.88609432E+25} & -\dfrac{7.145124E+24}{1.96203144E+25} & \dfrac{3.52394205E+28}{1.47152358E+29} & -\dfrac{2.148684E+24}{2.354437728E+25} & -\dfrac{1.2462444E+25}{1.96203144E+25} \\ -\dfrac{5.5452E+22}{7.84812576E+22} & -\dfrac{2.8596E+22}{2.94304716E+23} & \dfrac{3.159492E+24}{9.8101572E+24} & \dfrac{1.0524444E+25}{1.177218864E+26} & -\dfrac{6.4788E+22}{1.177218864E+25} & \dfrac{4.366668E+24}{9.8101572E+24} \\ \dfrac{1.5698868E+25}{1.177218864E+25} & -\dfrac{6.912876E+24}{1.765828296E+26} & -\dfrac{2.4906108E+25}{5.88609432E+25} & \dfrac{8.691141996E+36}{3.1784909328E+37} & -\dfrac{8.67247533E+35}{3.973113666E+36} & -\dfrac{8.1365028E+25}{5.88609432E+25} \\ -\dfrac{2.34384E+23}{2.94304716E+23} & -\dfrac{2.51976E+23}{2.20728537E+25} & \dfrac{1.728312E+24}{7.3576179E+24} & -\dfrac{5.74365024E+28}{5.297484888E+29} & -\dfrac{7.73808E+23}{2.20728537E+25} & \dfrac{7.675764E+24}{1.47152358E+25} \\ \dfrac{4.836372E+24}{7.84812576E+24} & -\dfrac{8.691684E+24}{2.354437728E+25} & -\dfrac{2.304804E+24}{3.92406288E+25} & \dfrac{2.084559936E+34}{7.5342007296E+34} & -\dfrac{9.2084412E+30}{2.354437728E+31} & -\dfrac{7.613004E+24}{7.84812576E+24} \\ -\dfrac{5.1631356E+25}{5.88609432E+25} & \dfrac{9.972996E+24}{3.531656592E+25} & \dfrac{7.16052E+23}{2.354437728E+24} & -\dfrac{2.65030146E+28}{1.0594969776E+29} & \dfrac{1.4412372E+25}{7.063313184E+25} & \dfrac{2.328756E+24}{2.354437728E+24}\end{pmatrix}\times\begin{pmatrix}-4 \\ 0 \\ 4 \\ 9 \\ -\dfrac{71}{6} \\ -4\end{pmatrix}=\begin{pmatrix}\dfrac{3.3442844124267E+130}{3.1401729422092E+130} \\ \dfrac{2.0120798672286E+125}{6.2803458844183E+124} \\ \dfrac{1.0959915634481E+152}{3.0904012010751E+151} \\ \dfrac{3.3035836106464E+129}{2.2355332762407E+129} \\ \dfrac{2.1347290382597E+143}{2.5724296742577E+142} \\ -\dfrac{5.709436976458E+130}{1.4650790879171E+130}\end{pmatrix}\) . Ainsi \( x=\dfrac{3.3442844124267E+130}{3.1401729422092E+130}\) , \( y=\dfrac{2.0120798672286E+125}{6.2803458844183E+124}\) , \( z=\dfrac{1.0959915634481E+152}{3.0904012010751E+151}\) , \( t=\dfrac{3.3035836106464E+129}{2.2355332762407E+129}\) , \( u=\dfrac{2.1347290382597E+143}{2.5724296742577E+142}\) et \( v=\dfrac{5.709436976458E+130}{1.4650790879171E+130}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{3.192468E+24}{3.92406288E+24}x&+&\dfrac{2.572908E+24}{5.88609432E+25}y &-&\dfrac{7.145124E+24}{1.96203144E+25}z &+&\dfrac{3.52394205E+28}{1.47152358E+29}t &-&\dfrac{2.148684E+24}{2.354437728E+25}u &-&\dfrac{1.2462444E+25}{1.96203144E+25}v &=&-3\\ &-\dfrac{5.5452E+22}{7.84812576E+22}x&-&\dfrac{2.8596E+22}{2.94304716E+23}y &+&\dfrac{3.159492E+24}{9.8101572E+24}z &+&\dfrac{1.0524444E+25}{1.177218864E+26}t &-&\dfrac{6.4788E+22}{1.177218864E+25}u &+&\dfrac{4.366668E+24}{9.8101572E+24}v &=&9\\ &\dfrac{1.5698868E+25}{1.177218864E+25}x&-&\dfrac{6.912876E+24}{1.765828296E+26}y &-&\dfrac{2.4906108E+25}{5.88609432E+25}z &+&\dfrac{8.691141996E+36}{3.1784909328E+37}t &-&\dfrac{8.67247533E+35}{3.973113666E+36}u &-&\dfrac{8.1365028E+25}{5.88609432E+25}v &=&8\\ &-\dfrac{2.34384E+23}{2.94304716E+23}x&-&\dfrac{2.51976E+23}{2.20728537E+25}y &+&\dfrac{1.728312E+24}{7.3576179E+24}z &-&\dfrac{5.74365024E+28}{5.297484888E+29}t &-&\dfrac{7.73808E+23}{2.20728537E+25}u &+&\dfrac{7.675764E+24}{1.47152358E+25}v &=&-6\\ &\dfrac{4.836372E+24}{7.84812576E+24}x&-&\dfrac{8.691684E+24}{2.354437728E+25}y &-&\dfrac{2.304804E+24}{3.92406288E+25}z &+&\dfrac{2.084559936E+34}{7.5342007296E+34}t &-&\dfrac{9.2084412E+30}{2.354437728E+31}u &-&\dfrac{7.613004E+24}{7.84812576E+24}v &=&-\dfrac{65}{7}\\ &-\dfrac{5.1631356E+25}{5.88609432E+25}x&+&\dfrac{9.972996E+24}{3.531656592E+25}y &+&\dfrac{7.16052E+23}{2.354437728E+24}z &-&\dfrac{2.65030146E+28}{1.0594969776E+29}t &+&\dfrac{1.4412372E+25}{7.063313184E+25}u &+&\dfrac{2.328756E+24}{2.354437728E+24}v &=&-5\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-3 \\ 9 \\ 8 \\ -6 \\ -\dfrac{65}{7} \\ -5\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -\dfrac{4}{3} & -\dfrac{5}{2} & -4 & 3 & \dfrac{14}{5} \\ \dfrac{6}{5} & \dfrac{1}{2} & 3 & \dfrac{7}{5} & 0 & 4 \\ -\dfrac{5}{2} & \dfrac{3}{2} & 1 & -5 & \dfrac{14}{5} & \dfrac{9}{2} \\ 3 & 4 & 0 & -1 & -\dfrac{2}{3} & 0 \\ -\dfrac{9}{2} & 4 & 3 & -2 & -\dfrac{17}{3} & -5 \\ 3 & -\dfrac{8}{5} & -4 & -\dfrac{9}{4} & \dfrac{14}{5} & 2\end{pmatrix}\times\begin{pmatrix}-3 \\ 9 \\ 8 \\ -6 \\ -\dfrac{65}{7} \\ -5\end{pmatrix}=\begin{pmatrix}-\dfrac{370}{7} \\ -\dfrac{7}{2} \\ \dfrac{21}{2} \\ \dfrac{823}{21} \\ \dfrac{6851}{42} \\ -\dfrac{779}{10}\end{pmatrix}\) . Ainsi \( x=\dfrac{370}{7}\) , \( y=\dfrac{7}{2}\) , \( z=\dfrac{21}{2}\) , \( t=\dfrac{823}{21}\) , \( u=\dfrac{6851}{42}\) et \( v=\dfrac{779}{10}\)