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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.


On considère la matrice $$ A= \begin{pmatrix}1 & \dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ -1 & \dfrac{6}{5} & -1 & \dfrac{6}{5} & 3 & -4 \\ \dfrac{23}{4} & 5 & \dfrac{15}{4} & 1 & 0 & -2 \\ \dfrac{7}{5} & 0 & -\dfrac{3}{2} & -4 & -3 & 2 \\ -1 & \dfrac{7}{5} & -\dfrac{23}{2} & \dfrac{10}{3} & -2 & -\dfrac{17}{3} \\ -3 & \dfrac{5}{3} & -\dfrac{5}{4} & -\dfrac{11}{4} & 1 & 1\end{pmatrix}$$
  1. Donner les mineurs d'ordre $ (2, 5)$ et $ (4, 2)$ : $ \widehat{A}_{2, 5}=$ $ \widehat{A}_{4, 2}=$
  2. Expliquer pourquoi $ \det(A)=\det \begin{pmatrix}1 & \dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ 0 & \dfrac{17}{10} & \dfrac{14}{5} & \dfrac{19}{5} & \dfrac{21}{2} & -\dfrac{9}{2} \\ 0 & \dfrac{17}{8} & -\dfrac{181}{10} & -\dfrac{279}{20} & -\dfrac{345}{8} & \dfrac{7}{8} \\ 0 & -\dfrac{7}{10} & -\dfrac{341}{50} & -\dfrac{191}{25} & -\dfrac{27}{2} & \dfrac{27}{10} \\ 0 & \dfrac{19}{10} & -\dfrac{77}{10} & \dfrac{89}{15} & \dfrac{11}{2} & -\dfrac{37}{6} \\ 0 & \dfrac{19}{6} & \dfrac{203}{20} & \dfrac{101}{20} & \dfrac{47}{2} & -\dfrac{1}{2}\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, 4}$ .
  6. Donner $ B^{-1}$ l'inverse de la matrice $ B$ . Justifier.
  7. Résoudre le système suivant. $ \left\{\begin{array}{*{7}{cr}} &x&+&\dfrac{1}{2}y &+&\dfrac{19}{5}z &+&\dfrac{13}{5}t &+&\dfrac{15}{2}u &-&\dfrac{1}{2}v &=&-8\\ &-x&+&\dfrac{6}{5}y &-&z &+&\dfrac{6}{5}t &+&3u &-&4v &=&1\\ &\dfrac{23}{4}x&+&5y &+&\dfrac{15}{4}z &+&t &&&-&2v &=&6\\ &\dfrac{7}{5}x&&&-&\dfrac{3}{2}z &-&4t &-&3u &+&2v &=&5\\ &-x&+&\dfrac{7}{5}y &-&\dfrac{23}{2}z &+&\dfrac{10}{3}t &-&2u &-&\dfrac{17}{3}v &=&-6\\ &-3x&+&\dfrac{5}{3}y &-&\dfrac{5}{4}z &-&\dfrac{11}{4}t &+&u &+&v &=&1\\ \end{array} \right. $
  8. Résoudre le système suivant. $ \left\{\begin{array}{*{7}{cr}} &\dfrac{3.7047232E+21}{3.0679162176E+22}x&-&\dfrac{8.335808E+20}{6.1358324352E+22}y &+&\dfrac{1.781184E+20}{6.1358324352E+21}z &+&\dfrac{1.0505897984E+27}{4.90866594816E+27}t &+&\dfrac{2.3615424E+21}{6.1358324352E+22}u &-&\dfrac{2.7968448E+21}{1.917447636E+22}v &=&2\\ &-\dfrac{2.0613952E+21}{1.22716648704E+23}x&-&\dfrac{1.84263872E+22}{1.22716648704E+23}y &+&\dfrac{1.11056064E+22}{7.669790544E+22}z &-&\dfrac{1.416416512E+27}{9.81733189632E+27}t &+&\dfrac{5.83960896E+22}{1.22716648704E+24}u &+&\dfrac{1.83145152E+22}{7.669790544E+22}v &=&3\\ &-\dfrac{2.30309696E+22}{2.3009371632E+23}x&+&\dfrac{8.23739072E+22}{9.2037486528E+23}y &+&\dfrac{9.98464E+19}{3.834895272E+21}z &-&\dfrac{1.84632512E+22}{1.84074973056E+23}t &-&\dfrac{3.9674304E+21}{3.834895272E+22}u &-&\dfrac{9.824192E+20}{3.834895272E+22}v &=&-\dfrac{75}{4}\\ &\dfrac{1.313024E+20}{1.1504685816E+22}x&-&\dfrac{2.218854976E+25}{9.2037486528E+25}y &+&\dfrac{1.3294912E+21}{3.834895272E+22}z &-&\dfrac{4.69827776E+22}{1.84074973056E+23}t &+&\dfrac{5.2449984E+21}{7.669790544E+22}u &+&\dfrac{1.3355584E+21}{1.5339581088E+23}v &=&7\\ &\dfrac{3.19046336E+22}{1.84074973056E+23}x&+&\dfrac{4.70711488E+22}{1.84074973056E+24}y &-&\dfrac{1.310160096E+30}{3.4514057448E+31}z &+&\dfrac{4.282092416E+30}{3.68149946112E+31}t &+&\dfrac{1.53776832E+22}{6.1358324352E+23}u &+&\dfrac{8.5976512E+21}{3.834895272E+23}v &=&9\\ &\dfrac{9.0720832E+21}{7.36299892224E+22}x&-&\dfrac{2.16278147072E+29}{5.890399137792E+29}y &+&\dfrac{8.8992448E+21}{7.669790544E+23}z &-&\dfrac{7.1921895424E+30}{1.1780798275584E+32}t &+&\dfrac{1.713861504E+27}{2.45433297408E+28}u &+&\dfrac{1.02723904E+22}{7.669790544E+22}v &=&\dfrac{37}{3}\\ \end{array} \right. $
Cliquer ici pour afficher la solution
  1. $ \widehat{A}_{2, 5}=\begin{pmatrix}1 & \dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & -\dfrac{1}{2} \\ \dfrac{23}{4} & 5 & \dfrac{15}{4} & 1 & -2 \\ \dfrac{7}{5} & 0 & -\dfrac{3}{2} & -4 & 2 \\ -1 & \dfrac{7}{5} & -\dfrac{23}{2} & \dfrac{10}{3} & -\dfrac{17}{3} \\ -3 & \dfrac{5}{3} & -\dfrac{5}{4} & -\dfrac{11}{4} & 1\end{pmatrix}$ $ \widehat{A}_{4, 2}=\begin{pmatrix}1 & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ -1 & -1 & \dfrac{6}{5} & 3 & -4 \\ \dfrac{23}{4} & \dfrac{15}{4} & 1 & 0 & -2 \\ -1 & -\dfrac{23}{2} & \dfrac{10}{3} & -2 & -\dfrac{17}{3} \\ -3 & -\dfrac{5}{4} & -\dfrac{11}{4} & 1 & 1\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(-1\right)L_1$ , $ L_{3}\leftarrow L_{3}-\left(\dfrac{23}{4}\right)L_1$ , $ L_{4}\leftarrow L_{4}-\left(\dfrac{7}{5}\right)L_1$ , $ L_{5}\leftarrow L_{5}-\left(-1\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{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ 0 & \dfrac{17}{10} & \dfrac{14}{5} & \dfrac{19}{5} & \dfrac{21}{2} & -\dfrac{9}{2} \\ 0 & \dfrac{17}{8} & -\dfrac{181}{10} & -\dfrac{279}{20} & -\dfrac{345}{8} & \dfrac{7}{8} \\ 0 & -\dfrac{7}{10} & -\dfrac{341}{50} & -\dfrac{191}{25} & -\dfrac{27}{2} & \dfrac{27}{10} \\ 0 & \dfrac{19}{10} & -\dfrac{77}{10} & \dfrac{89}{15} & \dfrac{11}{2} & -\dfrac{37}{6} \\ 0 & \dfrac{19}{6} & \dfrac{203}{20} & \dfrac{101}{20} & \dfrac{47}{2} & -\dfrac{1}{2}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{17}{10} & \dfrac{14}{5} & \dfrac{19}{5} & \dfrac{21}{2} & -\dfrac{9}{2} \\ \dfrac{17}{8} & -\dfrac{181}{10} & -\dfrac{279}{20} & -\dfrac{345}{8} & \dfrac{7}{8} \\ -\dfrac{7}{10} & -\dfrac{341}{50} & -\dfrac{191}{25} & -\dfrac{27}{2} & \dfrac{27}{10} \\ \dfrac{19}{10} & -\dfrac{77}{10} & \dfrac{89}{15} & \dfrac{11}{2} & -\dfrac{37}{6} \\ \dfrac{19}{6} & \dfrac{203}{20} & \dfrac{101}{20} & \dfrac{47}{2} & -\dfrac{1}{2}\end{pmatrix}\\ &=&\dfrac{1.5339581088E+20}{6400000000000000} \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{1.5339581088E+20}{6400000000000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & \dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ -1 & \dfrac{6}{5} & -1 & \dfrac{6}{5} & 3 & -4 \\ \dfrac{23}{4} & 5 & \dfrac{15}{4} & 1 & 0 & -2 \\ \dfrac{7}{5} & 0 & -\dfrac{3}{2} & -4 & -3 & 2 \\ -1 & \dfrac{7}{5} & -\dfrac{23}{2} & \dfrac{10}{3} & -2 & -\dfrac{17}{3} \\ -3 & \dfrac{5}{3} & -\dfrac{5}{4} & -\dfrac{11}{4} & 1 & 1\end{pmatrix} =\dfrac{6400000000000000}{1.5339581088E+20}\begin{pmatrix}\dfrac{5.6828901934995E+41}{1.963466379264E+38} & -\dfrac{1.2786780275E+41}{3.926932758528E+38} & \dfrac{2.7322616400648E+40}{3.926932758528E+37} & \dfrac{1.6115607402782E+47}{3.1415462068224E+43} & \dfrac{3.622507113755E+41}{3.926932758528E+38} & -\dfrac{4.2902427600151E+41}{1.22716648704E+38} \\ -\dfrac{3.1620938824814E+41}{7.853865517056E+38} & -\dfrac{2.8265306061329E+42}{7.853865517056E+38} & \dfrac{1.7035534990421E+42}{4.90866594816E+38} & -\dfrac{2.1727235940206E+47}{6.2830924136448E+43} & \dfrac{8.9577155164131E+42}{7.853865517056E+39} & \dfrac{2.8093699099781E+42}{4.90866594816E+38} \\ -\dfrac{3.5328542571446E+42}{1.472599784448E+39} & \dfrac{1.2635812290298E+43}{5.890399137792E+39} & \dfrac{1.5316019491449E+40}{2.45433297408E+37} & -\dfrac{2.8321853893051E+42}{1.1780798275584E+39} & -\dfrac{6.0858720331796E+41}{2.45433297408E+38} & -\dfrac{1.5069898980808E+41}{2.45433297408E+38} \\ \dfrac{2.014123811849E+40}{7.36299892224E+37} & -\dfrac{3.4036305826864E+45}{5.890399137792E+41} & \dfrac{2.0393838068182E+41}{2.45433297408E+38} & -\dfrac{7.2069612673467E+42}{1.1780798275584E+39} & \dfrac{8.045607826323E+41}{4.90866594816E+38} & \dfrac{2.048690637456E+41}{9.81733189632E+38} \\ \dfrac{4.8940371419013E+42}{1.1780798275584E+39} & \dfrac{7.2205170392291E+42}{1.1780798275584E+40} & -\dfrac{2.0097307030854E+50}{2.208899676672E+47} & \dfrac{6.5685503841542E+50}{2.3561596551168E+47} & \dfrac{2.3588721839198E+42}{3.926932758528E+39} & \dfrac{1.3188436774874E+42}{2.45433297408E+39} \\ \dfrac{1.3916195588348E+42}{4.7123193102336E+38} & -\dfrac{3.3176161745733E+49}{3.7698554481869E+45} & \dfrac{1.3651068723156E+42}{4.90866594816E+39} & -\dfrac{1.1032517468591E+51}{7.5397108963738E+47} & \dfrac{2.628991751421E+47}{1.5707731034112E+44} & \dfrac{1.5757416550839E+42}{4.90866594816E+38}\end{pmatrix} =\begin{pmatrix}\dfrac{3.7047232E+21}{3.0679162176E+22} & -\dfrac{8.335808E+20}{6.1358324352E+22} & \dfrac{1.781184E+20}{6.1358324352E+21} & \dfrac{1.0505897984E+27}{4.90866594816E+27} & \dfrac{2.3615424E+21}{6.1358324352E+22} & -\dfrac{2.7968448E+21}{1.917447636E+22} \\ -\dfrac{2.0613952E+21}{1.22716648704E+23} & -\dfrac{1.84263872E+22}{1.22716648704E+23} & \dfrac{1.11056064E+22}{7.669790544E+22} & -\dfrac{1.416416512E+27}{9.81733189632E+27} & \dfrac{5.83960896E+22}{1.22716648704E+24} & \dfrac{1.83145152E+22}{7.669790544E+22} \\ -\dfrac{2.30309696E+22}{2.3009371632E+23} & \dfrac{8.23739072E+22}{9.2037486528E+23} & \dfrac{9.98464E+19}{3.834895272E+21} & -\dfrac{1.84632512E+22}{1.84074973056E+23} & -\dfrac{3.9674304E+21}{3.834895272E+22} & -\dfrac{9.824192E+20}{3.834895272E+22} \\ \dfrac{1.313024E+20}{1.1504685816E+22} & -\dfrac{2.218854976E+25}{9.2037486528E+25} & \dfrac{1.3294912E+21}{3.834895272E+22} & -\dfrac{4.69827776E+22}{1.84074973056E+23} & \dfrac{5.2449984E+21}{7.669790544E+22} & \dfrac{1.3355584E+21}{1.5339581088E+23} \\ \dfrac{3.19046336E+22}{1.84074973056E+23} & \dfrac{4.70711488E+22}{1.84074973056E+24} & -\dfrac{1.310160096E+30}{3.4514057448E+31} & \dfrac{4.282092416E+30}{3.68149946112E+31} & \dfrac{1.53776832E+22}{6.1358324352E+23} & \dfrac{8.5976512E+21}{3.834895272E+23} \\ \dfrac{9.0720832E+21}{7.36299892224E+22} & -\dfrac{2.16278147072E+29}{5.890399137792E+29} & \dfrac{8.8992448E+21}{7.669790544E+23} & -\dfrac{7.1921895424E+30}{1.1780798275584E+32} & \dfrac{1.713861504E+27}{2.45433297408E+28} & \dfrac{1.02723904E+22}{7.669790544E+22}\end{pmatrix}$ . Précisément on a calculé $ A^{-1}_{1, 4}=B_{1, 4}= \left(\dfrac{1.5339581088E+20}{6400000000000000}\right)^{-1}Co(A)_{4, 1}= \left(\dfrac{1.5339581088E+20}{6400000000000000}\right)^{-1}\times(-1)^{4+1}\det\begin{pmatrix}\dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ \dfrac{6}{5} & -1 & \dfrac{6}{5} & 3 & -4 \\ 5 & \dfrac{15}{4} & 1 & 0 & -2 \\ \dfrac{7}{5} & -\dfrac{23}{2} & \dfrac{10}{3} & -2 & -\dfrac{17}{3} \\ \dfrac{5}{3} & -\dfrac{5}{4} & -\dfrac{11}{4} & 1 & 1\end{pmatrix}=\dfrac{1.0505897984E+27}{4.90866594816E+27}$ .
  6. On observe que $ B^{-1}=\left(A^{-1}\right)^{-1}=A$ . Trivialement, et sans calcul, $$ B^{-1}=\begin{pmatrix}1 & \dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ -1 & \dfrac{6}{5} & -1 & \dfrac{6}{5} & 3 & -4 \\ \dfrac{23}{4} & 5 & \dfrac{15}{4} & 1 & 0 & -2 \\ \dfrac{7}{5} & 0 & -\dfrac{3}{2} & -4 & -3 & 2 \\ -1 & \dfrac{7}{5} & -\dfrac{23}{2} & \dfrac{10}{3} & -2 & -\dfrac{17}{3} \\ -3 & \dfrac{5}{3} & -\dfrac{5}{4} & -\dfrac{11}{4} & 1 & 1\end{pmatrix}$$
  7. Le système $ \left\{\begin{array}{*{7}{cr}} &x&+&\dfrac{1}{2}y &+&\dfrac{19}{5}z &+&\dfrac{13}{5}t &+&\dfrac{15}{2}u &-&\dfrac{1}{2}v &=&-8\\ &-x&+&\dfrac{6}{5}y &-&z &+&\dfrac{6}{5}t &+&3u &-&4v &=&1\\ &\dfrac{23}{4}x&+&5y &+&\dfrac{15}{4}z &+&t &&&-&2v &=&6\\ &\dfrac{7}{5}x&&&-&\dfrac{3}{2}z &-&4t &-&3u &+&2v &=&5\\ &-x&+&\dfrac{7}{5}y &-&\dfrac{23}{2}z &+&\dfrac{10}{3}t &-&2u &-&\dfrac{17}{3}v &=&-6\\ &-3x&+&\dfrac{5}{3}y &-&\dfrac{5}{4}z &-&\dfrac{11}{4}t &+&u &+&v &=&1\\ \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}-8 \\ 1 \\ 6 \\ 5 \\ -6 \\ 1\end{pmatrix}$ de sorte que la solution est $ X=A^{-1}a=\begin{pmatrix}\dfrac{3.7047232E+21}{3.0679162176E+22} & -\dfrac{8.335808E+20}{6.1358324352E+22} & \dfrac{1.781184E+20}{6.1358324352E+21} & \dfrac{1.0505897984E+27}{4.90866594816E+27} & \dfrac{2.3615424E+21}{6.1358324352E+22} & -\dfrac{2.7968448E+21}{1.917447636E+22} \\ -\dfrac{2.0613952E+21}{1.22716648704E+23} & -\dfrac{1.84263872E+22}{1.22716648704E+23} & \dfrac{1.11056064E+22}{7.669790544E+22} & -\dfrac{1.416416512E+27}{9.81733189632E+27} & \dfrac{5.83960896E+22}{1.22716648704E+24} & \dfrac{1.83145152E+22}{7.669790544E+22} \\ -\dfrac{2.30309696E+22}{2.3009371632E+23} & \dfrac{8.23739072E+22}{9.2037486528E+23} & \dfrac{9.98464E+19}{3.834895272E+21} & -\dfrac{1.84632512E+22}{1.84074973056E+23} & -\dfrac{3.9674304E+21}{3.834895272E+22} & -\dfrac{9.824192E+20}{3.834895272E+22} \\ \dfrac{1.313024E+20}{1.1504685816E+22} & -\dfrac{2.218854976E+25}{9.2037486528E+25} & \dfrac{1.3294912E+21}{3.834895272E+22} & -\dfrac{4.69827776E+22}{1.84074973056E+23} & \dfrac{5.2449984E+21}{7.669790544E+22} & \dfrac{1.3355584E+21}{1.5339581088E+23} \\ \dfrac{3.19046336E+22}{1.84074973056E+23} & \dfrac{4.70711488E+22}{1.84074973056E+24} & -\dfrac{1.310160096E+30}{3.4514057448E+31} & \dfrac{4.282092416E+30}{3.68149946112E+31} & \dfrac{1.53776832E+22}{6.1358324352E+23} & \dfrac{8.5976512E+21}{3.834895272E+23} \\ \dfrac{9.0720832E+21}{7.36299892224E+22} & -\dfrac{2.16278147072E+29}{5.890399137792E+29} & \dfrac{8.8992448E+21}{7.669790544E+23} & -\dfrac{7.1921895424E+30}{1.1780798275584E+32} & \dfrac{1.713861504E+27}{2.45433297408E+28} & \dfrac{1.02723904E+22}{7.669790544E+22}\end{pmatrix}\times\begin{pmatrix}-8 \\ 1 \\ 6 \\ 5 \\ -6 \\ 1\end{pmatrix}=\begin{pmatrix}-\dfrac{7.4786714378889E+138}{6.6703858754336E+139} \\ \dfrac{9.060609432743E+142}{1.0672617400694E+144} \\ \dfrac{2.5064403795477E+137}{2.1984914384364E+137} \\ -\dfrac{1.5848185950293E+140}{8.7939657537455E+139} \\ -\dfrac{1.1499971802736E+158}{1.0130648548315E+158} \\ -\dfrac{1.3820976280978E+160}{7.3769131473596E+159}\end{pmatrix}$ . Ainsi $ x=\dfrac{7.4786714378889E+138}{6.6703858754336E+139}$ , $ y=\dfrac{9.060609432743E+142}{1.0672617400694E+144}$ , $ z=\dfrac{2.5064403795477E+137}{2.1984914384364E+137}$ , $ t=\dfrac{1.5848185950293E+140}{8.7939657537455E+139}$ , $ u=\dfrac{1.1499971802736E+158}{1.0130648548315E+158}$ et $ v=\dfrac{1.3820976280978E+160}{7.3769131473596E+159}$
  8. Le système $ \left\{\begin{array}{*{7}{cr}} &\dfrac{3.7047232E+21}{3.0679162176E+22}x&-&\dfrac{8.335808E+20}{6.1358324352E+22}y &+&\dfrac{1.781184E+20}{6.1358324352E+21}z &+&\dfrac{1.0505897984E+27}{4.90866594816E+27}t &+&\dfrac{2.3615424E+21}{6.1358324352E+22}u &-&\dfrac{2.7968448E+21}{1.917447636E+22}v &=&2\\ &-\dfrac{2.0613952E+21}{1.22716648704E+23}x&-&\dfrac{1.84263872E+22}{1.22716648704E+23}y &+&\dfrac{1.11056064E+22}{7.669790544E+22}z &-&\dfrac{1.416416512E+27}{9.81733189632E+27}t &+&\dfrac{5.83960896E+22}{1.22716648704E+24}u &+&\dfrac{1.83145152E+22}{7.669790544E+22}v &=&3\\ &-\dfrac{2.30309696E+22}{2.3009371632E+23}x&+&\dfrac{8.23739072E+22}{9.2037486528E+23}y &+&\dfrac{9.98464E+19}{3.834895272E+21}z &-&\dfrac{1.84632512E+22}{1.84074973056E+23}t &-&\dfrac{3.9674304E+21}{3.834895272E+22}u &-&\dfrac{9.824192E+20}{3.834895272E+22}v &=&-\dfrac{75}{4}\\ &\dfrac{1.313024E+20}{1.1504685816E+22}x&-&\dfrac{2.218854976E+25}{9.2037486528E+25}y &+&\dfrac{1.3294912E+21}{3.834895272E+22}z &-&\dfrac{4.69827776E+22}{1.84074973056E+23}t &+&\dfrac{5.2449984E+21}{7.669790544E+22}u &+&\dfrac{1.3355584E+21}{1.5339581088E+23}v &=&7\\ &\dfrac{3.19046336E+22}{1.84074973056E+23}x&+&\dfrac{4.70711488E+22}{1.84074973056E+24}y &-&\dfrac{1.310160096E+30}{3.4514057448E+31}z &+&\dfrac{4.282092416E+30}{3.68149946112E+31}t &+&\dfrac{1.53776832E+22}{6.1358324352E+23}u &+&\dfrac{8.5976512E+21}{3.834895272E+23}v &=&9\\ &\dfrac{9.0720832E+21}{7.36299892224E+22}x&-&\dfrac{2.16278147072E+29}{5.890399137792E+29}y &+&\dfrac{8.8992448E+21}{7.669790544E+23}z &-&\dfrac{7.1921895424E+30}{1.1780798275584E+32}t &+&\dfrac{1.713861504E+27}{2.45433297408E+28}u &+&\dfrac{1.02723904E+22}{7.669790544E+22}v &=&\dfrac{37}{3}\\ \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}2 \\ 3 \\ -\dfrac{75}{4} \\ 7 \\ 9 \\ \dfrac{37}{3}\end{pmatrix}$ de sorte que la solution est $ X=B^{-1}b=Ab=\begin{pmatrix}1 & \dfrac{1}{2} & \dfrac{19}{5} & \dfrac{13}{5} & \dfrac{15}{2} & -\dfrac{1}{2} \\ -1 & \dfrac{6}{5} & -1 & \dfrac{6}{5} & 3 & -4 \\ \dfrac{23}{4} & 5 & \dfrac{15}{4} & 1 & 0 & -2 \\ \dfrac{7}{5} & 0 & -\dfrac{3}{2} & -4 & -3 & 2 \\ -1 & \dfrac{7}{5} & -\dfrac{23}{2} & \dfrac{10}{3} & -2 & -\dfrac{17}{3} \\ -3 & \dfrac{5}{3} & -\dfrac{5}{4} & -\dfrac{11}{4} & 1 & 1\end{pmatrix}\times\begin{pmatrix}2 \\ 3 \\ -\dfrac{75}{4} \\ 7 \\ 9 \\ \dfrac{37}{3}\end{pmatrix}=\begin{pmatrix}\dfrac{707}{60} \\ \dfrac{77}{12} \\ -\dfrac{2951}{48} \\ \dfrac{71}{120} \\ \dfrac{55177}{360} \\ \dfrac{1177}{48}\end{pmatrix}$ . Ainsi $ x=\dfrac{707}{60}$ , $ y=\dfrac{77}{12}$ , $ z=\dfrac{2951}{48}$ , $ t=\dfrac{71}{120}$ , $ u=\dfrac{55177}{360}$ et $ v=\dfrac{1177}{48}$