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 \[ A= \begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ \dfrac{17}{4} & 0 & 0 & -\dfrac{17}{2} & -\dfrac{21}{4} & -\dfrac{17}{5} \\ \dfrac{7}{4} & 5 & \dfrac{15}{4} & -\dfrac{14}{3} & 5 & 0 \\ 3 & 4 & 2 & 0 & -\dfrac{23}{2} & \dfrac{3}{2} \\ -\dfrac{22}{3} & -4 & -1 & -\dfrac{15}{2} & \dfrac{24}{5} & -\dfrac{13}{2} \\ \dfrac{13}{5} & 2 & -\dfrac{14}{3} & \dfrac{5}{2} & 2 & 4\end{pmatrix}\]

Donner les mineurs d'ordre $ (3, 3)$ et $ (4, 5)$ : $ \widehat{A}_{3, 3}=$ $ \widehat{A}_{4, 5}=$
Expliquer pourquoi $ \det(A)=\det \begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ 0 & -\dfrac{425}{8} & \dfrac{51}{4} & -\dfrac{493}{8} & -\dfrac{21}{4} & \dfrac{1479}{40} \\ 0 & -\dfrac{135}{8} & 9 & -\dfrac{637}{24} & 5 & \dfrac{133}{8} \\ 0 & -\dfrac{67}{2} & 11 & -\dfrac{75}{2} & -\dfrac{23}{2} & 30 \\ 0 & \dfrac{263}{3} & -23 & \dfrac{505}{6} & \dfrac{24}{5} & -\dfrac{457}{6} \\ 0 & -\dfrac{61}{2} & \dfrac{47}{15} & -30 & 2 & \dfrac{287}{10}\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}_{6, 2}$ .
Donner $ B^{-1}$ l'inverse de la matrice $ B$ . Justifier.
Résoudre le système suivant. $ \left\{\begin{array}{*{7}{cr}} &x&+&\dfrac{25}{2}y &-&3z &+&\dfrac{25}{2}t &&&-&\dfrac{19}{2}v &=&5\\ &\dfrac{17}{4}x&&&&&-&\dfrac{17}{2}t &-&\dfrac{21}{4}u &-&\dfrac{17}{5}v &=&7\\ &\dfrac{7}{4}x&+&5y &+&\dfrac{15}{4}z &-&\dfrac{14}{3}t &+&5u &&&=&2\\ &3x&+&4y &+&2z &&&-&\dfrac{23}{2}u &+&\dfrac{3}{2}v &=&4\\ &-\dfrac{22}{3}x&-&4y &-&z &-&\dfrac{15}{2}t &+&\dfrac{24}{5}u &-&\dfrac{13}{2}v &=&-3\\ &\dfrac{13}{5}x&+&2y &-&\dfrac{14}{3}z &+&\dfrac{5}{2}t &+&2u &+&4v &=&4\\ \end{array} \right. $
Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{1.0298539508844E+52}{1.1516651092845E+54}x&+&\dfrac{4.1059071755891E+53}{3.3167955147393E+54}y &-&\dfrac{7.8367264752439E+47}{6.3981394960249E+49}z &-&\dfrac{2.0254932204852E+53}{1.6379237109824E+54}t &-&\dfrac{1.238357383853E+55}{9.4344405752584E+55}u &-&\dfrac{4.1421301625056E+51}{1.023702319364E+53}v &=&-5\\ &\dfrac{5.4486606161347E+62}{2.445406997107E+64}x&-&\dfrac{4.7011258696002E+54}{9.9503865442179E+55}y &+&\dfrac{1.6153114629495E+51}{1.9194418488075E+52}z &+&\dfrac{2.9431100415513E+58}{2.9482626797683E+59}t &+&\dfrac{2.8542908376772E+60}{4.3473902170791E+61}u &+&\dfrac{3.0950397811207E+60}{3.7737762301034E+61}v &=&2\\ &-\dfrac{3.8918895559306E+54}{3.4549953278534E+56}x&-&\dfrac{2.2356637740695E+51}{7.6777673952298E+52}y &+&\dfrac{5.3278982872713E+51}{1.023702319364E+53}z &-&\dfrac{1.0134337105311E+57}{4.7172202876292E+58}t &-&\dfrac{4.374073413578E+51}{6.1422139161839E+52}u &-&\dfrac{1.5016738162805E+57}{9.4344405752584E+57}v &=&-2\\ &\dfrac{3.9731798037784E+51}{1.9194418488075E+53}x&-&\dfrac{4.0663658222505E+50}{1.535553479046E+52}y &-&\dfrac{2.5581369447572E+54}{5.7583255464224E+55}z &-&\dfrac{2.4369659034259E+56}{5.8965253595365E+57}t &-&\dfrac{1.2118747851266E+51}{2.047404638728E+52}u &-&\dfrac{2.2657006934426E+56}{4.1930847001149E+57}v &=&-5\\ &\dfrac{4.0718393263952E+50}{1.7274976639267E+53}x&+&\dfrac{5.9982913385438E+47}{5.7583255464224E+50}y &+&\dfrac{7.6075733875013E+48}{1.9194418488075E+50}z &-&\dfrac{9.056135725299E+58}{1.1462845298939E+60}t &-&\dfrac{3.2935494353199E+58}{1.5283793731919E+60}u &+&\dfrac{8.3877130770398E+47}{7.6777673952298E+50}v &=&5\\ &-\dfrac{1.9551790839246E+54}{4.4223940196524E+55}x&-&\dfrac{1.4882453526291E+55}{1.9900773088436E+56}y &+&\dfrac{3.3160235629122E+49}{9.5972092440373E+50}z &+&\dfrac{2.7806510924002E+53}{3.9310169063577E+54}t &+&\dfrac{2.5929706384801E+54}{1.5095104920413E+56}u &+&\dfrac{1.017406640455E+51}{1.2284427832368E+52}v &=&2\\ \end{array} \right. \)

Cliquer ici pour afficher la solution

Exercice

$ \widehat{A}_{3, 3}=\begin{pmatrix}1 & \dfrac{25}{2} & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ \dfrac{17}{4} & 0 & -\dfrac{17}{2} & -\dfrac{21}{4} & -\dfrac{17}{5} \\ 3 & 4 & 0 & -\dfrac{23}{2} & \dfrac{3}{2} \\ -\dfrac{22}{3} & -4 & -\dfrac{15}{2} & \dfrac{24}{5} & -\dfrac{13}{2} \\ \dfrac{13}{5} & 2 & \dfrac{5}{2} & 2 & 4\end{pmatrix}$ $ \widehat{A}_{4, 5}=\begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & -\dfrac{19}{2} \\ \dfrac{17}{4} & 0 & 0 & -\dfrac{17}{2} & -\dfrac{17}{5} \\ \dfrac{7}{4} & 5 & \dfrac{15}{4} & -\dfrac{14}{3} & 0 \\ -\dfrac{22}{3} & -4 & -1 & -\dfrac{15}{2} & -\dfrac{13}{2} \\ \dfrac{13}{5} & 2 & -\dfrac{14}{3} & \dfrac{5}{2} & 4\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(\dfrac{17}{4}\right)L_1$ , $ L_{3}\leftarrow L_{3}-\left(\dfrac{7}{4}\right)L_1$ , $ L_{4}\leftarrow L_{4}-\left(3\right)L_1$ , $ L_{5}\leftarrow L_{5}-\left(-\dfrac{22}{3}\right)L_1$ et $ L_{6}\leftarrow L_{6}-\left(\dfrac{13}{5}\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 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ 0 & -\dfrac{425}{8} & \dfrac{51}{4} & -\dfrac{493}{8} & -\dfrac{21}{4} & \dfrac{1479}{40} \\ 0 & -\dfrac{135}{8} & 9 & -\dfrac{637}{24} & 5 & \dfrac{133}{8} \\ 0 & -\dfrac{67}{2} & 11 & -\dfrac{75}{2} & -\dfrac{23}{2} & 30 \\ 0 & \dfrac{263}{3} & -23 & \dfrac{505}{6} & \dfrac{24}{5} & -\dfrac{457}{6} \\ 0 & -\dfrac{61}{2} & \dfrac{47}{15} & -30 & 2 & \dfrac{287}{10}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-\dfrac{425}{8} & \dfrac{51}{4} & -\dfrac{493}{8} & -\dfrac{21}{4} & \dfrac{1479}{40} \\ -\dfrac{135}{8} & 9 & -\dfrac{637}{24} & 5 & \dfrac{133}{8} \\ -\dfrac{67}{2} & 11 & -\dfrac{75}{2} & -\dfrac{23}{2} & 30 \\ \dfrac{263}{3} & -23 & \dfrac{505}{6} & \dfrac{24}{5} & -\dfrac{457}{6} \\ -\dfrac{61}{2} & \dfrac{47}{15} & -30 & 2 & \dfrac{287}{10}\end{pmatrix}\\ &=&\dfrac{1.3329457283385E+47}{1.6543163447904E+41} \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{1.3329457283385E+47}{1.6543163447904E+41}\right)^{-1}{}^tCo\begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ \dfrac{17}{4} & 0 & 0 & -\dfrac{17}{2} & -\dfrac{21}{4} & -\dfrac{17}{5} \\ \dfrac{7}{4} & 5 & \dfrac{15}{4} & -\dfrac{14}{3} & 5 & 0 \\ 3 & 4 & 2 & 0 & -\dfrac{23}{2} & \dfrac{3}{2} \\ -\dfrac{22}{3} & -4 & -1 & -\dfrac{15}{2} & \dfrac{24}{5} & -\dfrac{13}{2} \\ \dfrac{13}{5} & 2 & -\dfrac{14}{3} & \dfrac{5}{2} & 2 & 4\end{pmatrix} =\dfrac{1.6543163447904E+41}{1.3329457283385E+47}\begin{pmatrix}\dfrac{1.3727394246439E+99}{1.9052184140141E+95} & \dfrac{5.472951430656E+100}{5.4870290323606E+95} & -\dfrac{1.0445931079334E+95}{1.0584546744523E+91} & -\dfrac{2.6998725360244E+100}{2.7096439665978E+95} & -\dfrac{1.6506631849634E+102}{1.5607549247604E+97} & -\dfrac{5.5212347063339E+98}{1.6935274791236E+94} \\ \dfrac{7.2627688934431E+109}{4.0454767649788E+105} & -\dfrac{6.2663456462652E+101}{1.6461087097082E+97} & \dfrac{2.1531225144748E+98}{3.1753640233568E+93} & \dfrac{3.923005957916E+105}{4.8773591398761E+100} & \dfrac{3.8046147795176E+107}{7.1919586932957E+102} & \dfrac{4.1255200552826E+107}{6.2430196990414E+102} \\ -\dfrac{5.187677558743E+101}{5.7156552420423E+97} & -\dfrac{2.9800184776471E+98}{1.2701456093427E+94} & \dfrac{7.1017992630404E+98}{1.6935274791236E+94} & -\dfrac{1.3508521354067E+104}{7.8037746238018E+99} & -\dfrac{5.8304024720679E+98}{1.0161164874742E+94} & -\dfrac{2.0016496987689E+104}{1.5607549247604E+99} \\ \dfrac{5.2960330473673E+98}{3.1753640233568E+94} & -\dfrac{5.4202449526305E+97}{2.5402912186855E+93} & -\dfrac{3.4098577130191E+101}{9.5260920700705E+96} & -\dfrac{3.2483432910781E+103}{9.7547182797522E+98} & -\dfrac{1.6153633181157E+98}{3.3870549582473E+93} & -\dfrac{3.0200560610179E+103}{6.9366885544904E+98} \\ \dfrac{5.4275408365992E+97}{2.8578276210212E+94} & \dfrac{7.9953968170418E+94}{9.5260920700705E+91} & \dfrac{1.0140482449892E+96}{3.1753640233568E+91} & -\dfrac{1.2071337430291E+106}{1.8963172335838E+101} & -\dfrac{4.3901226508814E+105}{2.5284229781118E+101} & \dfrac{1.1180366316569E+95}{1.2701456093427E+92} \\ -\dfrac{2.6061476080542E+101}{7.3160387098141E+96} & -\dfrac{1.9837502855066E+102}{3.2922174194164E+97} & \dfrac{4.4200794432537E+96}{1.5876820116784E+92} & \dfrac{3.7064569956147E+100}{6.5031455198348E+95} & \dfrac{3.4562891362693E+101}{2.4972078796166E+97} & \dfrac{1.3561478353778E+98}{2.0322329749484E+93}\end{pmatrix} =\begin{pmatrix}\dfrac{1.0298539508844E+52}{1.1516651092845E+54} & \dfrac{4.1059071755891E+53}{3.3167955147393E+54} & -\dfrac{7.8367264752439E+47}{6.3981394960249E+49} & -\dfrac{2.0254932204852E+53}{1.6379237109824E+54} & -\dfrac{1.238357383853E+55}{9.4344405752584E+55} & -\dfrac{4.1421301625056E+51}{1.023702319364E+53} \\ \dfrac{5.4486606161347E+62}{2.445406997107E+64} & -\dfrac{4.7011258696002E+54}{9.9503865442179E+55} & \dfrac{1.6153114629495E+51}{1.9194418488075E+52} & \dfrac{2.9431100415513E+58}{2.9482626797683E+59} & \dfrac{2.8542908376772E+60}{4.3473902170791E+61} & \dfrac{3.0950397811207E+60}{3.7737762301034E+61} \\ -\dfrac{3.8918895559306E+54}{3.4549953278534E+56} & -\dfrac{2.2356637740695E+51}{7.6777673952298E+52} & \dfrac{5.3278982872713E+51}{1.023702319364E+53} & -\dfrac{1.0134337105311E+57}{4.7172202876292E+58} & -\dfrac{4.374073413578E+51}{6.1422139161839E+52} & -\dfrac{1.5016738162805E+57}{9.4344405752584E+57} \\ \dfrac{3.9731798037784E+51}{1.9194418488075E+53} & -\dfrac{4.0663658222505E+50}{1.535553479046E+52} & -\dfrac{2.5581369447572E+54}{5.7583255464224E+55} & -\dfrac{2.4369659034259E+56}{5.8965253595365E+57} & -\dfrac{1.2118747851266E+51}{2.047404638728E+52} & -\dfrac{2.2657006934426E+56}{4.1930847001149E+57} \\ \dfrac{4.0718393263952E+50}{1.7274976639267E+53} & \dfrac{5.9982913385438E+47}{5.7583255464224E+50} & \dfrac{7.6075733875013E+48}{1.9194418488075E+50} & -\dfrac{9.056135725299E+58}{1.1462845298939E+60} & -\dfrac{3.2935494353199E+58}{1.5283793731919E+60} & \dfrac{8.3877130770398E+47}{7.6777673952298E+50} \\ -\dfrac{1.9551790839246E+54}{4.4223940196524E+55} & -\dfrac{1.4882453526291E+55}{1.9900773088436E+56} & \dfrac{3.3160235629122E+49}{9.5972092440373E+50} & \dfrac{2.7806510924002E+53}{3.9310169063577E+54} & \dfrac{2.5929706384801E+54}{1.5095104920413E+56} & \dfrac{1.017406640455E+51}{1.2284427832368E+52}\end{pmatrix}\) . Précisément on a calculé $ A^{-1}_{6, 2}=B_{6, 2}= \left(\dfrac{1.3329457283385E+47}{1.6543163447904E+41}\right)^{-1}Co(A)_{2, 6}= \left(\dfrac{1.3329457283385E+47}{1.6543163447904E+41}\right)^{-1}\times(-1)^{2+6}\det\begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 \\ \dfrac{7}{4} & 5 & \dfrac{15}{4} & -\dfrac{14}{3} & 5 \\ 3 & 4 & 2 & 0 & -\dfrac{23}{2} \\ -\dfrac{22}{3} & -4 & -1 & -\dfrac{15}{2} & \dfrac{24}{5} \\ \dfrac{13}{5} & 2 & -\dfrac{14}{3} & \dfrac{5}{2} & 2\end{pmatrix}=-\dfrac{1.4882453526291E+55}{1.9900773088436E+56}$ .
On observe que $ B^{-1}=\left(A^{-1}\right)^{-1}=A$ . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ \dfrac{17}{4} & 0 & 0 & -\dfrac{17}{2} & -\dfrac{21}{4} & -\dfrac{17}{5} \\ \dfrac{7}{4} & 5 & \dfrac{15}{4} & -\dfrac{14}{3} & 5 & 0 \\ 3 & 4 & 2 & 0 & -\dfrac{23}{2} & \dfrac{3}{2} \\ -\dfrac{22}{3} & -4 & -1 & -\dfrac{15}{2} & \dfrac{24}{5} & -\dfrac{13}{2} \\ \dfrac{13}{5} & 2 & -\dfrac{14}{3} & \dfrac{5}{2} & 2 & 4\end{pmatrix}\]
Le système $ \left\{\begin{array}{*{7}{cr}} &x&+&\dfrac{25}{2}y &-&3z &+&\dfrac{25}{2}t &&&-&\dfrac{19}{2}v &=&5\\ &\dfrac{17}{4}x&&&&&-&\dfrac{17}{2}t &-&\dfrac{21}{4}u &-&\dfrac{17}{5}v &=&7\\ &\dfrac{7}{4}x&+&5y &+&\dfrac{15}{4}z &-&\dfrac{14}{3}t &+&5u &&&=&2\\ &3x&+&4y &+&2z &&&-&\dfrac{23}{2}u &+&\dfrac{3}{2}v &=&4\\ &-\dfrac{22}{3}x&-&4y &-&z &-&\dfrac{15}{2}t &+&\dfrac{24}{5}u &-&\dfrac{13}{2}v &=&-3\\ &\dfrac{13}{5}x&+&2y &-&\dfrac{14}{3}z &+&\dfrac{5}{2}t &+&2u &+&4v &=&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}5 \\ 7 \\ 2 \\ 4 \\ -3 \\ 4\end{pmatrix}$ de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{1.0298539508844E+52}{1.1516651092845E+54} & \dfrac{4.1059071755891E+53}{3.3167955147393E+54} & -\dfrac{7.8367264752439E+47}{6.3981394960249E+49} & -\dfrac{2.0254932204852E+53}{1.6379237109824E+54} & -\dfrac{1.238357383853E+55}{9.4344405752584E+55} & -\dfrac{4.1421301625056E+51}{1.023702319364E+53} \\ \dfrac{5.4486606161347E+62}{2.445406997107E+64} & -\dfrac{4.7011258696002E+54}{9.9503865442179E+55} & \dfrac{1.6153114629495E+51}{1.9194418488075E+52} & \dfrac{2.9431100415513E+58}{2.9482626797683E+59} & \dfrac{2.8542908376772E+60}{4.3473902170791E+61} & \dfrac{3.0950397811207E+60}{3.7737762301034E+61} \\ -\dfrac{3.8918895559306E+54}{3.4549953278534E+56} & -\dfrac{2.2356637740695E+51}{7.6777673952298E+52} & \dfrac{5.3278982872713E+51}{1.023702319364E+53} & -\dfrac{1.0134337105311E+57}{4.7172202876292E+58} & -\dfrac{4.374073413578E+51}{6.1422139161839E+52} & -\dfrac{1.5016738162805E+57}{9.4344405752584E+57} \\ \dfrac{3.9731798037784E+51}{1.9194418488075E+53} & -\dfrac{4.0663658222505E+50}{1.535553479046E+52} & -\dfrac{2.5581369447572E+54}{5.7583255464224E+55} & -\dfrac{2.4369659034259E+56}{5.8965253595365E+57} & -\dfrac{1.2118747851266E+51}{2.047404638728E+52} & -\dfrac{2.2657006934426E+56}{4.1930847001149E+57} \\ \dfrac{4.0718393263952E+50}{1.7274976639267E+53} & \dfrac{5.9982913385438E+47}{5.7583255464224E+50} & \dfrac{7.6075733875013E+48}{1.9194418488075E+50} & -\dfrac{9.056135725299E+58}{1.1462845298939E+60} & -\dfrac{3.2935494353199E+58}{1.5283793731919E+60} & \dfrac{8.3877130770398E+47}{7.6777673952298E+50} \\ -\dfrac{1.9551790839246E+54}{4.4223940196524E+55} & -\dfrac{1.4882453526291E+55}{1.9900773088436E+56} & \dfrac{3.3160235629122E+49}{9.5972092440373E+50} & \dfrac{2.7806510924002E+53}{3.9310169063577E+54} & \dfrac{2.5929706384801E+54}{1.5095104920413E+56} & \dfrac{1.017406640455E+51}{1.2284427832368E+52}\end{pmatrix}\times\begin{pmatrix}5 \\ 7 \\ 2 \\ 4 \\ -3 \\ 4\end{pmatrix}=\begin{pmatrix}\dfrac{NAN}{INF} \\ \dfrac{INF}{INF} \\ -\dfrac{INF}{INF} \\ -\dfrac{INF}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF}\end{pmatrix}\) . Ainsi $ x=\dfrac{NAN}{INF}$ , $ y=\dfrac{INF}{INF}$ , $ z=\dfrac{INF}{INF}$ , $ t=\dfrac{INF}{INF}$ , $ u=\dfrac{NAN}{INF}$ et $ v=\dfrac{NAN}{INF}$
Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{1.0298539508844E+52}{1.1516651092845E+54}x&+&\dfrac{4.1059071755891E+53}{3.3167955147393E+54}y &-&\dfrac{7.8367264752439E+47}{6.3981394960249E+49}z &-&\dfrac{2.0254932204852E+53}{1.6379237109824E+54}t &-&\dfrac{1.238357383853E+55}{9.4344405752584E+55}u &-&\dfrac{4.1421301625056E+51}{1.023702319364E+53}v &=&-5\\ &\dfrac{5.4486606161347E+62}{2.445406997107E+64}x&-&\dfrac{4.7011258696002E+54}{9.9503865442179E+55}y &+&\dfrac{1.6153114629495E+51}{1.9194418488075E+52}z &+&\dfrac{2.9431100415513E+58}{2.9482626797683E+59}t &+&\dfrac{2.8542908376772E+60}{4.3473902170791E+61}u &+&\dfrac{3.0950397811207E+60}{3.7737762301034E+61}v &=&2\\ &-\dfrac{3.8918895559306E+54}{3.4549953278534E+56}x&-&\dfrac{2.2356637740695E+51}{7.6777673952298E+52}y &+&\dfrac{5.3278982872713E+51}{1.023702319364E+53}z &-&\dfrac{1.0134337105311E+57}{4.7172202876292E+58}t &-&\dfrac{4.374073413578E+51}{6.1422139161839E+52}u &-&\dfrac{1.5016738162805E+57}{9.4344405752584E+57}v &=&-2\\ &\dfrac{3.9731798037784E+51}{1.9194418488075E+53}x&-&\dfrac{4.0663658222505E+50}{1.535553479046E+52}y &-&\dfrac{2.5581369447572E+54}{5.7583255464224E+55}z &-&\dfrac{2.4369659034259E+56}{5.8965253595365E+57}t &-&\dfrac{1.2118747851266E+51}{2.047404638728E+52}u &-&\dfrac{2.2657006934426E+56}{4.1930847001149E+57}v &=&-5\\ &\dfrac{4.0718393263952E+50}{1.7274976639267E+53}x&+&\dfrac{5.9982913385438E+47}{5.7583255464224E+50}y &+&\dfrac{7.6075733875013E+48}{1.9194418488075E+50}z &-&\dfrac{9.056135725299E+58}{1.1462845298939E+60}t &-&\dfrac{3.2935494353199E+58}{1.5283793731919E+60}u &+&\dfrac{8.3877130770398E+47}{7.6777673952298E+50}v &=&5\\ &-\dfrac{1.9551790839246E+54}{4.4223940196524E+55}x&-&\dfrac{1.4882453526291E+55}{1.9900773088436E+56}y &+&\dfrac{3.3160235629122E+49}{9.5972092440373E+50}z &+&\dfrac{2.7806510924002E+53}{3.9310169063577E+54}t &+&\dfrac{2.5929706384801E+54}{1.5095104920413E+56}u &+&\dfrac{1.017406640455E+51}{1.2284427832368E+52}v &=&2\\ \end{array} \right.\) 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}-5 \\ 2 \\ -2 \\ -5 \\ 5 \\ 2\end{pmatrix}$ de sorte que la solution est $ X=B^{-1}b=Ab=\begin{pmatrix}1 & \dfrac{25}{2} & -3 & \dfrac{25}{2} & 0 & -\dfrac{19}{2} \\ \dfrac{17}{4} & 0 & 0 & -\dfrac{17}{2} & -\dfrac{21}{4} & -\dfrac{17}{5} \\ \dfrac{7}{4} & 5 & \dfrac{15}{4} & -\dfrac{14}{3} & 5 & 0 \\ 3 & 4 & 2 & 0 & -\dfrac{23}{2} & \dfrac{3}{2} \\ -\dfrac{22}{3} & -4 & -1 & -\dfrac{15}{2} & \dfrac{24}{5} & -\dfrac{13}{2} \\ \dfrac{13}{5} & 2 & -\dfrac{14}{3} & \dfrac{5}{2} & 2 & 4\end{pmatrix}\times\begin{pmatrix}-5 \\ 2 \\ -2 \\ -5 \\ 5 \\ 2\end{pmatrix}=\begin{pmatrix}-\dfrac{111}{2} \\ -\dfrac{59}{5} \\ \dfrac{505}{12} \\ -\dfrac{131}{2} \\ \dfrac{475}{6} \\ \dfrac{35}{6}\end{pmatrix}$ . Ainsi $ x=\dfrac{111}{2}$ , $ y=\dfrac{59}{5}$ , $ z=\dfrac{505}{12}$ , $ t=\dfrac{131}{2}$ , $ u=\dfrac{475}{6}$ et $ v=\dfrac{35}{6}$