Ataraxy

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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 \[ A= \begin{pmatrix}1 & 1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 5 & 0 & -\dfrac{1}{3} & -\dfrac{11}{3} & \dfrac{13}{2} & 1 \\ -\dfrac{4}{5} & -2 & -\dfrac{9}{4} & 1 & 4 & -4 \\ -\dfrac{4}{3} & -3 & -3 & 0 & -3 & -5 \\ -2 & -\dfrac{17}{4} & 2 & -2 & 2 & -3 \\ -\dfrac{14}{3} & 4 & \dfrac{18}{5} & 5 & 3 & 1\end{pmatrix}\]

Donner les mineurs d'ordre $ (2, 3)$ et $ (5, 5)$ : $ \widehat{A}_{2, 3}=$ $ \widehat{A}_{5, 5}=$
Expliquer pourquoi $ \det(A)=\det \begin{pmatrix}1 & 1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 0 & -5 & \dfrac{14}{3} & \dfrac{19}{3} & -\dfrac{71}{6} & -20 \\ 0 & -\dfrac{6}{5} & -\dfrac{61}{20} & -\dfrac{3}{5} & \dfrac{104}{15} & -\dfrac{16}{25} \\ 0 & -\dfrac{5}{3} & -\dfrac{13}{3} & -\dfrac{8}{3} & \dfrac{17}{9} & \dfrac{3}{5} \\ 0 & -\dfrac{9}{4} & 0 & -6 & \dfrac{28}{3} & \dfrac{27}{5} \\ 0 & \dfrac{26}{3} & -\dfrac{16}{15} & -\dfrac{13}{3} & \dfrac{181}{9} & \dfrac{103}{5}\end{pmatrix} $
Calculer $ \det(A)$ .
Pourquoi la matrice $ A $ est inversible.
Donner $ B=A^{-1}$ l'inverse de la matrice $ A $ , en ne détaillant que le calcul de $ A^{-1}_{2, 3}$ .
Donner $ B^{-1}$ l'inverse de la matrice $ B$ . Justifier.
Résoudre le système suivant. $ \left\{\begin{array}{*{7}{cr}} &x&+&y &-&z &-&2t &+&\dfrac{11}{3}u &+&\dfrac{21}{5}v &=&0\\ &5x&&&-&\dfrac{1}{3}z &-&\dfrac{11}{3}t &+&\dfrac{13}{2}u &+&v &=&\dfrac{10}{9}\\ &-\dfrac{4}{5}x&-&2y &-&\dfrac{9}{4}z &+&t &+&4u &-&4v &=&0\\ &-\dfrac{4}{3}x&-&3y &-&3z &&&-&3u &-&5v &=&8\\ &-2x&-&\dfrac{17}{4}y &+&2z &-&2t &+&2u &-&3v &=&\dfrac{41}{8}\\ &-\dfrac{14}{3}x&+&4y &+&\dfrac{18}{5}z &+&5t &+&3u &+&v &=&1\\ \end{array} \right. $
Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{3.676256580672E+22}{1.4976437202132E+23}x&-&\dfrac{2.206452332736E+22}{5.6161639507997E+23}y &+&\dfrac{1.7421253868736E+23}{8.4242459261995E+23}z &-&\dfrac{1.4912510639259E+27}{4.0436380445758E+27}t &-&\dfrac{3.594332470464E+22}{4.2121229630998E+23}u &-&\dfrac{1.3646713298112E+23}{6.7393967409596E+23}v &=&5\\ &\dfrac{1.893948018624E+22}{2.1060614815499E+23}x&+&\dfrac{1.7099392791802E+24}{4.2121229630998E+24}y &-&\dfrac{3.991799320128E+22}{8.4242459261995E+22}z &+&\dfrac{1.0242327058361E+28}{1.5163642667159E+28}t &-&\dfrac{1.0620490219275E+29}{1.0235458800332E+30}u &+&\dfrac{5.432761202112E+22}{1.4040409876999E+23}v &=&9\\ &-\dfrac{2.980638575424E+22}{1.6848491852399E+23}x&+&\dfrac{7.33859082432E+21}{2.1060614815499E+23}y &-&\dfrac{6.584211033408E+22}{1.0109095111439E+24}z &-&\dfrac{7.5637596833856E+23}{5.0545475557197E+24}t &+&\dfrac{4.347975329856E+22}{4.2121229630998E+23}u &+&\dfrac{4.602749562432E+22}{5.0545475557197E+24}v &=&-8\\ &-\dfrac{7.3370001859776E+23}{3.3696983704798E+24}x&-&\dfrac{9.833942987712E+22}{2.8080819753998E+23}y &+&\dfrac{1.5918319965829E+28}{3.0706376400997E+28}z &-&\dfrac{1.3218740492287E+33}{1.9652080896638E+33}t &-&\dfrac{2.6301479033741E+29}{3.0706376400997E+30}u &-&\dfrac{3.6926392848858E+28}{1.3101387264425E+29}v &=&3\\ &\dfrac{1.4413779648576E+23}{3.3696983704798E+24}x&+&\dfrac{5.647423685184E+22}{1.8720546502666E+24}y &+&\dfrac{2.745591432768E+22}{2.8080819753998E+23}z &-&\dfrac{2.4285759640128E+23}{5.6161639507997E+24}t &+&\dfrac{4.423754564928E+22}{2.1060614815499E+24}u &+&\dfrac{3.153964868928E+22}{1.1232327901599E+24}v &=&0\\ &\dfrac{1.0331677873728E+23}{1.1232327901599E+24}x&-&\dfrac{9.1691026439616E+23}{3.3696983704798E+24}y &+&\dfrac{2.1186202992192E+23}{1.0109095111439E+24}z &-&\dfrac{5.1252200542162E+28}{1.3101387264425E+29}t &+&\dfrac{5.294354125248E+22}{5.0545475557197E+24}u &-&\dfrac{4.0532628101184E+23}{2.0218190222879E+24}v &=&6\\ \end{array} \right. \)

Cliquer ici pour afficher la solution

Exercice

$ \widehat{A}_{2, 3}=\begin{pmatrix}1 & 1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ -\dfrac{4}{5} & -2 & 1 & 4 & -4 \\ -\dfrac{4}{3} & -3 & 0 & -3 & -5 \\ -2 & -\dfrac{17}{4} & -2 & 2 & -3 \\ -\dfrac{14}{3} & 4 & 5 & 3 & 1\end{pmatrix}$ $ \widehat{A}_{5, 5}=\begin{pmatrix}1 & 1 & -1 & -2 & \dfrac{21}{5} \\ 5 & 0 & -\dfrac{1}{3} & -\dfrac{11}{3} & 1 \\ -\dfrac{4}{5} & -2 & -\dfrac{9}{4} & 1 & -4 \\ -\dfrac{4}{3} & -3 & -3 & 0 & -5 \\ -\dfrac{14}{3} & 4 & \dfrac{18}{5} & 5 & 1\end{pmatrix}$
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(5\right)L_1$ , $ L_{3}\leftarrow L_{3}-\left(-\dfrac{4}{5}\right)L_1$ , $ L_{4}\leftarrow L_{4}-\left(-\dfrac{4}{3}\right)L_1$ , $ L_{5}\leftarrow L_{5}-\left(-2\right)L_1$ et $ L_{6}\leftarrow L_{6}-\left(-\dfrac{14}{3}\right)L_1$
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 & 1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 0 & -5 & \dfrac{14}{3} & \dfrac{19}{3} & -\dfrac{71}{6} & -20 \\ 0 & -\dfrac{6}{5} & -\dfrac{61}{20} & -\dfrac{3}{5} & \dfrac{104}{15} & -\dfrac{16}{25} \\ 0 & -\dfrac{5}{3} & -\dfrac{13}{3} & -\dfrac{8}{3} & \dfrac{17}{9} & \dfrac{3}{5} \\ 0 & -\dfrac{9}{4} & 0 & -6 & \dfrac{28}{3} & \dfrac{27}{5} \\ 0 & \dfrac{26}{3} & -\dfrac{16}{15} & -\dfrac{13}{3} & \dfrac{181}{9} & \dfrac{103}{5}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-5 & \dfrac{14}{3} & \dfrac{19}{3} & -\dfrac{71}{6} & -20 \\ -\dfrac{6}{5} & -\dfrac{61}{20} & -\dfrac{3}{5} & \dfrac{104}{15} & -\dfrac{16}{25} \\ -\dfrac{5}{3} & -\dfrac{13}{3} & -\dfrac{8}{3} & \dfrac{17}{9} & \dfrac{3}{5} \\ -\dfrac{9}{4} & 0 & -6 & \dfrac{28}{3} & \dfrac{27}{5} \\ \dfrac{26}{3} & -\dfrac{16}{15} & -\dfrac{13}{3} & \dfrac{181}{9} & \dfrac{103}{5}\end{pmatrix}\\ &=&\dfrac{9.3602732513328E+20}{113374080000000000} \end{eqnarray*}
On observe que le déterminant de $ A$ est non nul. D'après le cours, cela signifie que la matrice est inversible.
D'après le cours \( B=A^{-1}=\left(\dfrac{9.3602732513328E+20}{113374080000000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & 1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 5 & 0 & -\dfrac{1}{3} & -\dfrac{11}{3} & \dfrac{13}{2} & 1 \\ -\dfrac{4}{5} & -2 & -\dfrac{9}{4} & 1 & 4 & -4 \\ -\dfrac{4}{3} & -3 & -3 & 0 & -3 & -5 \\ -2 & -\dfrac{17}{4} & 2 & -2 & 2 & -3 \\ -\dfrac{14}{3} & 4 & \dfrac{18}{5} & 5 & 3 & 1\end{pmatrix} =\dfrac{113374080000000000}{9.3602732513328E+20}\begin{pmatrix}-\dfrac{3.44107661371E+43}{1.6979397894695E+40} & -\dfrac{2.065299675045E+43}{6.3672742105108E+40} & \dfrac{1.6306769659221E+44}{9.5509113157662E+40} & -\dfrac{1.3958517444687E+48}{4.5844374315678E+44} & -\dfrac{3.3643934079681E+43}{4.7754556578831E+40} & -\dfrac{1.2773696545293E+44}{7.6407290526129E+40} \\ \dfrac{1.7727870978141E+43}{2.3877278289415E+40} & \dfrac{1.6005498896313E+45}{4.7754556578831E+41} & -\dfrac{3.7364332400883E+43}{9.5509113157662E+39} & \dfrac{9.5870979995776E+48}{1.7191640368379E+45} & -\dfrac{9.9410690515523E+49}{1.1604357248656E+47} & \dfrac{5.0852129361008E+43}{1.5918185526277E+40} \\ -\dfrac{2.7899591529432E+43}{1.9101822631532E+40} & \dfrac{6.8691215395359E+42}{2.3877278289415E+40} & -\dfrac{6.1630014417139E+43}{1.1461093578919E+41} & -\dfrac{7.0798857443904E+44}{5.7305467894597E+41} & \dfrac{4.0698237177506E+43}{4.7754556578831E+40} & \dfrac{4.3082993611816E+43}{5.7305467894597E+41} \\ -\dfrac{6.867632658583E+44}{3.8203645263065E+41} & -\dfrac{9.2048393503012E+43}{3.1836371052554E+40} & \dfrac{1.489998245823E+49}{3.4813071745968E+45} & -\dfrac{1.2373102304626E+54}{2.2280365917419E+50} & -\dfrac{2.4618903067002E+50}{3.4813071745968E+47} & -\dfrac{3.4564112725137E+49}{1.485357727828E+46} \\ \dfrac{1.3491691609517E+44}{3.8203645263065E+41} & \dfrac{5.2861428859371E+43}{2.1224247368369E+41} & \dfrac{2.5699486047227E+43}{3.1836371052554E+40} & -\dfrac{2.2732134634779E+44}{6.3672742105108E+41} & \dfrac{4.1407551524557E+43}{2.3877278289415E+41} & \dfrac{2.952197299827E+43}{1.2734548421022E+41} \\ \dfrac{9.6707328042843E+43}{1.2734548421022E+41} & -\dfrac{8.5825306216999E+44}{3.8203645263065E+41} & \dfrac{1.9830864916512E+44}{1.1461093578919E+41} & -\dfrac{4.7973460180675E+49}{1.485357727828E+46} & \dfrac{4.9556601301642E+43}{5.7305467894597E+41} & -\dfrac{3.7939647462173E+44}{2.2922187157839E+41}\end{pmatrix} =\begin{pmatrix}-\dfrac{3.676256580672E+22}{1.4976437202132E+23} & -\dfrac{2.206452332736E+22}{5.6161639507997E+23} & \dfrac{1.7421253868736E+23}{8.4242459261995E+23} & -\dfrac{1.4912510639259E+27}{4.0436380445758E+27} & -\dfrac{3.594332470464E+22}{4.2121229630998E+23} & -\dfrac{1.3646713298112E+23}{6.7393967409596E+23} \\ \dfrac{1.893948018624E+22}{2.1060614815499E+23} & \dfrac{1.7099392791802E+24}{4.2121229630998E+24} & -\dfrac{3.991799320128E+22}{8.4242459261995E+22} & \dfrac{1.0242327058361E+28}{1.5163642667159E+28} & -\dfrac{1.0620490219275E+29}{1.0235458800332E+30} & \dfrac{5.432761202112E+22}{1.4040409876999E+23} \\ -\dfrac{2.980638575424E+22}{1.6848491852399E+23} & \dfrac{7.33859082432E+21}{2.1060614815499E+23} & -\dfrac{6.584211033408E+22}{1.0109095111439E+24} & -\dfrac{7.5637596833856E+23}{5.0545475557197E+24} & \dfrac{4.347975329856E+22}{4.2121229630998E+23} & \dfrac{4.602749562432E+22}{5.0545475557197E+24} \\ -\dfrac{7.3370001859776E+23}{3.3696983704798E+24} & -\dfrac{9.833942987712E+22}{2.8080819753998E+23} & \dfrac{1.5918319965829E+28}{3.0706376400997E+28} & -\dfrac{1.3218740492287E+33}{1.9652080896638E+33} & -\dfrac{2.6301479033741E+29}{3.0706376400997E+30} & -\dfrac{3.6926392848858E+28}{1.3101387264425E+29} \\ \dfrac{1.4413779648576E+23}{3.3696983704798E+24} & \dfrac{5.647423685184E+22}{1.8720546502666E+24} & \dfrac{2.745591432768E+22}{2.8080819753998E+23} & -\dfrac{2.4285759640128E+23}{5.6161639507997E+24} & \dfrac{4.423754564928E+22}{2.1060614815499E+24} & \dfrac{3.153964868928E+22}{1.1232327901599E+24} \\ \dfrac{1.0331677873728E+23}{1.1232327901599E+24} & -\dfrac{9.1691026439616E+23}{3.3696983704798E+24} & \dfrac{2.1186202992192E+23}{1.0109095111439E+24} & -\dfrac{5.1252200542162E+28}{1.3101387264425E+29} & \dfrac{5.294354125248E+22}{5.0545475557197E+24} & -\dfrac{4.0532628101184E+23}{2.0218190222879E+24}\end{pmatrix}\) . Précisément on a calculé $ A^{-1}_{2, 3}=B_{2, 3}= \left(\dfrac{9.3602732513328E+20}{113374080000000000}\right)^{-1}Co(A)_{3, 2}= \left(\dfrac{9.3602732513328E+20}{113374080000000000}\right)^{-1}\times(-1)^{3+2}\det\begin{pmatrix}1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 5 & -\dfrac{1}{3} & -\dfrac{11}{3} & \dfrac{13}{2} & 1 \\ -\dfrac{4}{3} & -3 & 0 & -3 & -5 \\ -2 & 2 & -2 & 2 & -3 \\ -\dfrac{14}{3} & \dfrac{18}{5} & 5 & 3 & 1\end{pmatrix}=-\dfrac{3.991799320128E+22}{8.4242459261995E+22}$ .
On observe que $ B^{-1}=\left(A^{-1}\right)^{-1}=A$ . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 5 & 0 & -\dfrac{1}{3} & -\dfrac{11}{3} & \dfrac{13}{2} & 1 \\ -\dfrac{4}{5} & -2 & -\dfrac{9}{4} & 1 & 4 & -4 \\ -\dfrac{4}{3} & -3 & -3 & 0 & -3 & -5 \\ -2 & -\dfrac{17}{4} & 2 & -2 & 2 & -3 \\ -\dfrac{14}{3} & 4 & \dfrac{18}{5} & 5 & 3 & 1\end{pmatrix}\]
Le système $ \left\{\begin{array}{*{7}{cr}} &x&+&y &-&z &-&2t &+&\dfrac{11}{3}u &+&\dfrac{21}{5}v &=&0\\ &5x&&&-&\dfrac{1}{3}z &-&\dfrac{11}{3}t &+&\dfrac{13}{2}u &+&v &=&\dfrac{10}{9}\\ &-\dfrac{4}{5}x&-&2y &-&\dfrac{9}{4}z &+&t &+&4u &-&4v &=&0\\ &-\dfrac{4}{3}x&-&3y &-&3z &&&-&3u &-&5v &=&8\\ &-2x&-&\dfrac{17}{4}y &+&2z &-&2t &+&2u &-&3v &=&\dfrac{41}{8}\\ &-\dfrac{14}{3}x&+&4y &+&\dfrac{18}{5}z &+&5t &+&3u &+&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}0 \\ \dfrac{10}{9} \\ 0 \\ 8 \\ \dfrac{41}{8} \\ 1\end{pmatrix}$ de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{3.676256580672E+22}{1.4976437202132E+23} & -\dfrac{2.206452332736E+22}{5.6161639507997E+23} & \dfrac{1.7421253868736E+23}{8.4242459261995E+23} & -\dfrac{1.4912510639259E+27}{4.0436380445758E+27} & -\dfrac{3.594332470464E+22}{4.2121229630998E+23} & -\dfrac{1.3646713298112E+23}{6.7393967409596E+23} \\ \dfrac{1.893948018624E+22}{2.1060614815499E+23} & \dfrac{1.7099392791802E+24}{4.2121229630998E+24} & -\dfrac{3.991799320128E+22}{8.4242459261995E+22} & \dfrac{1.0242327058361E+28}{1.5163642667159E+28} & -\dfrac{1.0620490219275E+29}{1.0235458800332E+30} & \dfrac{5.432761202112E+22}{1.4040409876999E+23} \\ -\dfrac{2.980638575424E+22}{1.6848491852399E+23} & \dfrac{7.33859082432E+21}{2.1060614815499E+23} & -\dfrac{6.584211033408E+22}{1.0109095111439E+24} & -\dfrac{7.5637596833856E+23}{5.0545475557197E+24} & \dfrac{4.347975329856E+22}{4.2121229630998E+23} & \dfrac{4.602749562432E+22}{5.0545475557197E+24} \\ -\dfrac{7.3370001859776E+23}{3.3696983704798E+24} & -\dfrac{9.833942987712E+22}{2.8080819753998E+23} & \dfrac{1.5918319965829E+28}{3.0706376400997E+28} & -\dfrac{1.3218740492287E+33}{1.9652080896638E+33} & -\dfrac{2.6301479033741E+29}{3.0706376400997E+30} & -\dfrac{3.6926392848858E+28}{1.3101387264425E+29} \\ \dfrac{1.4413779648576E+23}{3.3696983704798E+24} & \dfrac{5.647423685184E+22}{1.8720546502666E+24} & \dfrac{2.745591432768E+22}{2.8080819753998E+23} & -\dfrac{2.4285759640128E+23}{5.6161639507997E+24} & \dfrac{4.423754564928E+22}{2.1060614815499E+24} & \dfrac{3.153964868928E+22}{1.1232327901599E+24} \\ \dfrac{1.0331677873728E+23}{1.1232327901599E+24} & -\dfrac{9.1691026439616E+23}{3.3696983704798E+24} & \dfrac{2.1186202992192E+23}{1.0109095111439E+24} & -\dfrac{5.1252200542162E+28}{1.3101387264425E+29} & \dfrac{5.294354125248E+22}{5.0545475557197E+24} & -\dfrac{4.0532628101184E+23}{2.0218190222879E+24}\end{pmatrix}\times\begin{pmatrix}0 \\ \dfrac{10}{9} \\ 0 \\ 8 \\ \dfrac{41}{8} \\ 1\end{pmatrix}=\begin{pmatrix}-\dfrac{1.6866565682028E+101}{4.6415882534639E+100} \\ \dfrac{3.7735360484624E+108}{6.6088238999516E+107} \\ -\dfrac{1.0121925947094E+98}{1.6318083703584E+98} \\ -\dfrac{1.0375562178498E+119}{1.5984412363499E+118} \\ -\dfrac{3.1641005047438E+98}{1.7907362088981E+99} \\ -\dfrac{1.1624961013194E+105}{3.2483691233042E+104}\end{pmatrix}\) . Ainsi $ x=\dfrac{1.6866565682028E+101}{4.6415882534639E+100}$ , $ y=\dfrac{3.7735360484624E+108}{6.6088238999516E+107}$ , $ z=\dfrac{1.0121925947094E+98}{1.6318083703584E+98}$ , $ t=\dfrac{1.0375562178498E+119}{1.5984412363499E+118}$ , $ u=\dfrac{3.1641005047438E+98}{1.7907362088981E+99}$ et $ v=\dfrac{1.1624961013194E+105}{3.2483691233042E+104}$
Le système \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{3.676256580672E+22}{1.4976437202132E+23}x&-&\dfrac{2.206452332736E+22}{5.6161639507997E+23}y &+&\dfrac{1.7421253868736E+23}{8.4242459261995E+23}z &-&\dfrac{1.4912510639259E+27}{4.0436380445758E+27}t &-&\dfrac{3.594332470464E+22}{4.2121229630998E+23}u &-&\dfrac{1.3646713298112E+23}{6.7393967409596E+23}v &=&5\\ &\dfrac{1.893948018624E+22}{2.1060614815499E+23}x&+&\dfrac{1.7099392791802E+24}{4.2121229630998E+24}y &-&\dfrac{3.991799320128E+22}{8.4242459261995E+22}z &+&\dfrac{1.0242327058361E+28}{1.5163642667159E+28}t &-&\dfrac{1.0620490219275E+29}{1.0235458800332E+30}u &+&\dfrac{5.432761202112E+22}{1.4040409876999E+23}v &=&9\\ &-\dfrac{2.980638575424E+22}{1.6848491852399E+23}x&+&\dfrac{7.33859082432E+21}{2.1060614815499E+23}y &-&\dfrac{6.584211033408E+22}{1.0109095111439E+24}z &-&\dfrac{7.5637596833856E+23}{5.0545475557197E+24}t &+&\dfrac{4.347975329856E+22}{4.2121229630998E+23}u &+&\dfrac{4.602749562432E+22}{5.0545475557197E+24}v &=&-8\\ &-\dfrac{7.3370001859776E+23}{3.3696983704798E+24}x&-&\dfrac{9.833942987712E+22}{2.8080819753998E+23}y &+&\dfrac{1.5918319965829E+28}{3.0706376400997E+28}z &-&\dfrac{1.3218740492287E+33}{1.9652080896638E+33}t &-&\dfrac{2.6301479033741E+29}{3.0706376400997E+30}u &-&\dfrac{3.6926392848858E+28}{1.3101387264425E+29}v &=&3\\ &\dfrac{1.4413779648576E+23}{3.3696983704798E+24}x&+&\dfrac{5.647423685184E+22}{1.8720546502666E+24}y &+&\dfrac{2.745591432768E+22}{2.8080819753998E+23}z &-&\dfrac{2.4285759640128E+23}{5.6161639507997E+24}t &+&\dfrac{4.423754564928E+22}{2.1060614815499E+24}u &+&\dfrac{3.153964868928E+22}{1.1232327901599E+24}v &=&0\\ &\dfrac{1.0331677873728E+23}{1.1232327901599E+24}x&-&\dfrac{9.1691026439616E+23}{3.3696983704798E+24}y &+&\dfrac{2.1186202992192E+23}{1.0109095111439E+24}z &-&\dfrac{5.1252200542162E+28}{1.3101387264425E+29}t &+&\dfrac{5.294354125248E+22}{5.0545475557197E+24}u &-&\dfrac{4.0532628101184E+23}{2.0218190222879E+24}v &=&6\\ \end{array} \right.\) 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}5 \\ 9 \\ -8 \\ 3 \\ 0 \\ 6\end{pmatrix}$ de sorte que la solution est $ X=B^{-1}b=Ab=\begin{pmatrix}1 & 1 & -1 & -2 & \dfrac{11}{3} & \dfrac{21}{5} \\ 5 & 0 & -\dfrac{1}{3} & -\dfrac{11}{3} & \dfrac{13}{2} & 1 \\ -\dfrac{4}{5} & -2 & -\dfrac{9}{4} & 1 & 4 & -4 \\ -\dfrac{4}{3} & -3 & -3 & 0 & -3 & -5 \\ -2 & -\dfrac{17}{4} & 2 & -2 & 2 & -3 \\ -\dfrac{14}{3} & 4 & \dfrac{18}{5} & 5 & 3 & 1\end{pmatrix}\times\begin{pmatrix}5 \\ 9 \\ -8 \\ 3 \\ 0 \\ 6\end{pmatrix}=\begin{pmatrix}\dfrac{206}{5} \\ \dfrac{68}{3} \\ -25 \\ -\dfrac{119}{3} \\ -\dfrac{353}{4} \\ \dfrac{73}{15}\end{pmatrix}$ . Ainsi $ x=\dfrac{206}{5}$ , $ y=\dfrac{68}{3}$ , $ z=25$ , $ t=\dfrac{119}{3}$ , $ u=\dfrac{353}{4}$ et $ v=\dfrac{73}{15}$