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
\[ A= \begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 & -\dfrac{8}{3} \\ 1 & -\dfrac{23}{5} & -\dfrac{1}{4} & -4 & \dfrac{21}{5} & 1 \\ -\dfrac{11}{2} & -\dfrac{19}{2} & -4 & 5 & 4 & -2 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{11}{5} & -\dfrac{12}{5} & \dfrac{14}{3} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & -\dfrac{7}{4} & \dfrac{9}{2} & \dfrac{1}{5} \\ -\dfrac{18}{5} & -3 & \dfrac{1}{4} & 0 & -2 & 2\end{pmatrix}\]
- Donner les mineurs d'ordre \( (3, 4)\) et \( (2, 6)\) :
\( \widehat{A}_{3, 4}=\)
\( \widehat{A}_{2, 6}=\)
- Expliquer pourquoi
\( \det(A)=\det
\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 & -\dfrac{8}{3} \\ 0 & -\dfrac{109}{15} & \dfrac{37}{12} & -2 & \dfrac{6}{5} & \dfrac{11}{3} \\ 0 & \dfrac{31}{6} & -\dfrac{67}{3} & -6 & \dfrac{41}{2} & -\dfrac{50}{3} \\ 0 & -11 & \dfrac{58}{5} & \dfrac{19}{5} & -\dfrac{57}{5} & \dfrac{38}{3} \\ 0 & -\dfrac{193}{12} & \dfrac{215}{12} & \dfrac{33}{4} & -\dfrac{21}{2} & \dfrac{203}{15} \\ 0 & \dfrac{33}{5} & -\dfrac{47}{4} & -\dfrac{36}{5} & \dfrac{44}{5} & -\dfrac{38}{5}\end{pmatrix}
\)
- 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 \( A^{-1}_{6, 6}\) .
- Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&x&+&\dfrac{8}{3}y &-&\dfrac{10}{3}z &-&2t &+&3u &-&\dfrac{8}{3}v &=&7\\
&x&-&\dfrac{23}{5}y &-&\dfrac{1}{4}z &-&4t &+&\dfrac{21}{5}u &+&v &=&4\\
&-\dfrac{11}{2}x&-&\dfrac{19}{2}y &-&4z &+&5t &+&4u &-&2v &=&-3\\
&3x&-&3y &+&\dfrac{8}{5}z &-&\dfrac{11}{5}t &-&\dfrac{12}{5}u &+&\dfrac{14}{3}v &=&-\dfrac{50}{9}\\
&5x&-&\dfrac{11}{4}y &+&\dfrac{5}{4}z &-&\dfrac{7}{4}t &+&\dfrac{9}{2}u &+&\dfrac{1}{5}v &=&9\\
&-\dfrac{18}{5}x&-&3y &+&\dfrac{1}{4}z &&&-&2u &+&2v &=&\dfrac{1}{2}\\
\end{array}
\right.
\)
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&-\dfrac{1.3597428460865E+46}{1.2093062068408E+46}x&+&\dfrac{2.1259359222689E+46}{1.4511674482089E+46}y &+&\dfrac{6.8701473976709E+52}{2.612101406776E+53}z &+&\dfrac{8.1735001749156E+51}{1.8574943337074E+52}t &-&\dfrac{4.5112446894293E+45}{2.4186124136815E+45}u &-&\dfrac{1.37570668227E+55}{4.8976901377051E+54}v &=&5\\
&\dfrac{1.9211183472721E+49}{7.5581637927548E+48}x&-&\dfrac{2.7137329788867E+46}{8.0620413789384E+45}y &-&\dfrac{9.8341724604779E+49}{1.8139593102611E+50}z &-&\dfrac{3.3371998357178E+46}{4.837224827363E+46}t &+&\dfrac{1.4900951288346E+47}{3.6279186205223E+46}u &+&\dfrac{3.7411608089903E+51}{6.5302535169401E+50}v &=&8\\
&-\dfrac{5.1474284828099E+44}{5.0387758618365E+44}x&+&\dfrac{1.6623938678833E+45}{1.511632758551E+45}y &+&\dfrac{5.1315910852461E+45}{9.0697965513057E+46}z &-&\dfrac{1.6824786031411E+43}{8.0620413789384E+44}t &-&\dfrac{1.0505429310962E+46}{9.0697965513057E+45}u &-&\dfrac{6.1342629157793E+46}{3.6279186205223E+46}v &=&-7\\
&\dfrac{4.1969507119519E+49}{2.0155103447346E+49}x&-&\dfrac{2.1720031581385E+47}{7.2558372410446E+46}y &-&\dfrac{2.4964949777464E+46}{7.2558372410446E+46}z &-&\dfrac{1.6760605781017E+45}{3.2248165515754E+45}t &+&\dfrac{7.177820465288E+45}{2.0155103447346E+45}u &+&\dfrac{5.0002829132898E+53}{1.0448405627104E+53}v &=&1\\
&\dfrac{2.9417230918902E+50}{8.0620413789384E+49}x&-&\dfrac{3.8912015629276E+46}{8.0620413789384E+45}y &-&\dfrac{3.289645914109E+49}{4.5348982756529E+49}z &-&\dfrac{5.0768043472016E+46}{4.837224827363E+46}t &+&\dfrac{2.2162436469789E+47}{3.6279186205223E+46}u &+&\dfrac{1.6438674220523E+55}{1.959076055082E+54}v &=&\dfrac{8}{3}\\
&\dfrac{3.5894156815024E+45}{6.4496331031507E+44}x&-&\dfrac{7.1360296402172E+46}{9.6744496547261E+45}y &-&\dfrac{6.4834826257318E+45}{6.0465310342038E+45}z &-&\dfrac{6.2386573042631E+45}{4.837224827363E+45}t &+&\dfrac{2.7382311838664E+46}{3.0232655171019E+45}u &+&\dfrac{1.2228350875551E+47}{9.6744496547261E+45}v &=&1\\
\end{array}
\right.
\)
Cliquer ici pour afficher la solution
Exercice
-
\( \widehat{A}_{3, 4}=\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & 3 & -\dfrac{8}{3} \\ 1 & -\dfrac{23}{5} & -\dfrac{1}{4} & \dfrac{21}{5} & 1 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{12}{5} & \dfrac{14}{3} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & \dfrac{9}{2} & \dfrac{1}{5} \\ -\dfrac{18}{5} & -3 & \dfrac{1}{4} & -2 & 2\end{pmatrix}\)
\( \widehat{A}_{2, 6}=\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 \\ -\dfrac{11}{2} & -\dfrac{19}{2} & -4 & 5 & 4 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{11}{5} & -\dfrac{12}{5} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & -\dfrac{7}{4} & \dfrac{9}{2} \\ -\dfrac{18}{5} & -3 & \dfrac{1}{4} & 0 & -2\end{pmatrix}\)
- 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 : \( L_{2}\leftarrow L_{2}-\left(1\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{11}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(3\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(5\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-\dfrac{18}{5}\right)L_1\)
- 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{8}{3} & -\dfrac{10}{3} & -2 & 3 & -\dfrac{8}{3} \\ 0 & -\dfrac{109}{15} & \dfrac{37}{12} & -2 & \dfrac{6}{5} & \dfrac{11}{3} \\ 0 & \dfrac{31}{6} & -\dfrac{67}{3} & -6 & \dfrac{41}{2} & -\dfrac{50}{3} \\ 0 & -11 & \dfrac{58}{5} & \dfrac{19}{5} & -\dfrac{57}{5} & \dfrac{38}{3} \\ 0 & -\dfrac{193}{12} & \dfrac{215}{12} & \dfrac{33}{4} & -\dfrac{21}{2} & \dfrac{203}{15} \\ 0 & \dfrac{33}{5} & -\dfrac{47}{4} & -\dfrac{36}{5} & \dfrac{44}{5} & -\dfrac{38}{5}\end{pmatrix}\\
&=&1\times\det\begin{pmatrix}-\dfrac{109}{15} & \dfrac{37}{12} & -2 & \dfrac{6}{5} & \dfrac{11}{3} \\ \dfrac{31}{6} & -\dfrac{67}{3} & -6 & \dfrac{41}{2} & -\dfrac{50}{3} \\ -11 & \dfrac{58}{5} & \dfrac{19}{5} & -\dfrac{57}{5} & \dfrac{38}{3} \\ -\dfrac{193}{12} & \dfrac{215}{12} & \dfrac{33}{4} & -\dfrac{21}{2} & \dfrac{203}{15} \\ \dfrac{33}{5} & -\dfrac{47}{4} & -\dfrac{36}{5} & \dfrac{44}{5} & -\dfrac{38}{5}\end{pmatrix}\\
&=&\dfrac{4.0310206894692E+42}{7.8364164096E+39}
\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{4.0310206894692E+42}{7.8364164096E+39}\right)^{-1}{}^tCo\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 & -\dfrac{8}{3} \\ 1 & -\dfrac{23}{5} & -\dfrac{1}{4} & -4 & \dfrac{21}{5} & 1 \\ -\dfrac{11}{2} & -\dfrac{19}{2} & -4 & 5 & 4 & -2 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{11}{5} & -\dfrac{12}{5} & \dfrac{14}{3} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & -\dfrac{7}{4} & \dfrac{9}{2} & \dfrac{1}{5} \\ -\dfrac{18}{5} & -3 & \dfrac{1}{4} & 0 & -2 & 2\end{pmatrix}
=\dfrac{7.8364164096E+39}{4.0310206894692E+42}\begin{pmatrix}-\dfrac{5.4811515449325E+88}{9.4766270035181E+85} & \dfrac{8.5696916871517E+88}{1.1371952404222E+86} & \dfrac{2.7693706299714E+95}{2.0469514327599E+93} & \dfrac{3.2947548310465E+94}{1.4556099077404E+92} & -\dfrac{1.8184920678348E+88}{1.8953254007036E+85} & -\dfrac{5.5455020988713E+97}{3.8380339364248E+94} \\ \dfrac{7.7440678047728E+91}{5.9228918771988E+88} & -\dfrac{1.0939113783587E+89}{6.3177513356787E+85} & -\dfrac{3.9641752651995E+92}{1.4214940505277E+90} & -\dfrac{1.3452321582672E+89}{3.7906508014072E+86} & \dfrac{6.0066042936095E+89}{2.8429881010554E+86} & \dfrac{1.5080696623671E+94}{5.1173785818998E+90} \\ -\dfrac{2.074939071177E+87}{3.9485945847992E+84} & \dfrac{6.7011440754844E+87}{1.1845783754398E+85} & \dfrac{2.0685549834523E+88}{7.1074702526385E+86} & -\dfrac{6.7821060588511E+85}{6.3177513356787E+84} & -\dfrac{4.2347602904244E+88}{7.1074702526385E+85} & -\dfrac{2.472734072815E+89}{2.8429881010554E+86} \\ \dfrac{1.6917995152561E+92}{1.5794378339197E+89} & -\dfrac{8.7553896680489E+89}{5.6859762021108E+86} & -\dfrac{1.0063422906452E+89}{5.6859762021108E+86} & -\dfrac{6.7562348671315E+87}{2.5271005342715E+85} & \dfrac{2.8933942800871E+88}{1.5794378339197E+85} & \dfrac{2.015624387667E+96}{8.1878057310396E+92} \\ \dfrac{1.1858146646099E+93}{6.3177513356787E+89} & -\dfrac{1.5685514007056E+89}{6.3177513356787E+85} & -\dfrac{1.3260630740801E+92}{3.5537351263193E+89} & -\dfrac{2.0464703359957E+89}{3.7906508014072E+86} & \dfrac{8.9337239938768E+89}{2.8429881010554E+86} & \dfrac{6.6264635890373E+97}{1.5352135745699E+94} \\ \dfrac{1.4469008875241E+88}{5.054201068543E+84} & -\dfrac{2.8765483120381E+89}{7.5813016028145E+85} & -\dfrac{2.6135052604139E+88}{4.738313501759E+85} & -\dfrac{2.5148156667993E+88}{3.7906508014072E+85} & \dfrac{1.1037866554715E+89}{2.3691567508795E+85} & \dfrac{4.9292735377437E+89}{7.5813016028145E+85}\end{pmatrix}
=\begin{pmatrix}-\dfrac{1.3597428460865E+46}{1.2093062068408E+46} & \dfrac{2.1259359222689E+46}{1.4511674482089E+46} & \dfrac{6.8701473976709E+52}{2.612101406776E+53} & \dfrac{8.1735001749156E+51}{1.8574943337074E+52} & -\dfrac{4.5112446894293E+45}{2.4186124136815E+45} & -\dfrac{1.37570668227E+55}{4.8976901377051E+54} \\ \dfrac{1.9211183472721E+49}{7.5581637927548E+48} & -\dfrac{2.7137329788867E+46}{8.0620413789384E+45} & -\dfrac{9.8341724604779E+49}{1.8139593102611E+50} & -\dfrac{3.3371998357178E+46}{4.837224827363E+46} & \dfrac{1.4900951288346E+47}{3.6279186205223E+46} & \dfrac{3.7411608089903E+51}{6.5302535169401E+50} \\ -\dfrac{5.1474284828099E+44}{5.0387758618365E+44} & \dfrac{1.6623938678833E+45}{1.511632758551E+45} & \dfrac{5.1315910852461E+45}{9.0697965513057E+46} & -\dfrac{1.6824786031411E+43}{8.0620413789384E+44} & -\dfrac{1.0505429310962E+46}{9.0697965513057E+45} & -\dfrac{6.1342629157793E+46}{3.6279186205223E+46} \\ \dfrac{4.1969507119519E+49}{2.0155103447346E+49} & -\dfrac{2.1720031581385E+47}{7.2558372410446E+46} & -\dfrac{2.4964949777464E+46}{7.2558372410446E+46} & -\dfrac{1.6760605781017E+45}{3.2248165515754E+45} & \dfrac{7.177820465288E+45}{2.0155103447346E+45} & \dfrac{5.0002829132898E+53}{1.0448405627104E+53} \\ \dfrac{2.9417230918902E+50}{8.0620413789384E+49} & -\dfrac{3.8912015629276E+46}{8.0620413789384E+45} & -\dfrac{3.289645914109E+49}{4.5348982756529E+49} & -\dfrac{5.0768043472016E+46}{4.837224827363E+46} & \dfrac{2.2162436469789E+47}{3.6279186205223E+46} & \dfrac{1.6438674220523E+55}{1.959076055082E+54} \\ \dfrac{3.5894156815024E+45}{6.4496331031507E+44} & -\dfrac{7.1360296402172E+46}{9.6744496547261E+45} & -\dfrac{6.4834826257318E+45}{6.0465310342038E+45} & -\dfrac{6.2386573042631E+45}{4.837224827363E+45} & \dfrac{2.7382311838664E+46}{3.0232655171019E+45} & \dfrac{1.2228350875551E+47}{9.6744496547261E+45}\end{pmatrix}\) .
Précisément on a calculé \( A^{-1}_{6, 6}=B_{6, 6}=
\left(\dfrac{4.0310206894692E+42}{7.8364164096E+39}\right)^{-1}Co(A)_{6, 6}=
\left(\dfrac{4.0310206894692E+42}{7.8364164096E+39}\right)^{-1}\times(-1)^{6+6}\det\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 \\ 1 & -\dfrac{23}{5} & -\dfrac{1}{4} & -4 & \dfrac{21}{5} \\ -\dfrac{11}{2} & -\dfrac{19}{2} & -4 & 5 & 4 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{11}{5} & -\dfrac{12}{5} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & -\dfrac{7}{4} & \dfrac{9}{2}\end{pmatrix}=\dfrac{1.2228350875551E+47}{9.6744496547261E+45}\) .
- On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul,
\[ B^{-1}=\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 & -\dfrac{8}{3} \\ 1 & -\dfrac{23}{5} & -\dfrac{1}{4} & -4 & \dfrac{21}{5} & 1 \\ -\dfrac{11}{2} & -\dfrac{19}{2} & -4 & 5 & 4 & -2 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{11}{5} & -\dfrac{12}{5} & \dfrac{14}{3} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & -\dfrac{7}{4} & \dfrac{9}{2} & \dfrac{1}{5} \\ -\dfrac{18}{5} & -3 & \dfrac{1}{4} & 0 & -2 & 2\end{pmatrix}\]
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&x&+&\dfrac{8}{3}y &-&\dfrac{10}{3}z &-&2t &+&3u &-&\dfrac{8}{3}v &=&7\\
&x&-&\dfrac{23}{5}y &-&\dfrac{1}{4}z &-&4t &+&\dfrac{21}{5}u &+&v &=&4\\
&-\dfrac{11}{2}x&-&\dfrac{19}{2}y &-&4z &+&5t &+&4u &-&2v &=&-3\\
&3x&-&3y &+&\dfrac{8}{5}z &-&\dfrac{11}{5}t &-&\dfrac{12}{5}u &+&\dfrac{14}{3}v &=&-\dfrac{50}{9}\\
&5x&-&\dfrac{11}{4}y &+&\dfrac{5}{4}z &-&\dfrac{7}{4}t &+&\dfrac{9}{2}u &+&\dfrac{1}{5}v &=&9\\
&-\dfrac{18}{5}x&-&3y &+&\dfrac{1}{4}z &&&-&2u &+&2v &=&\dfrac{1}{2}\\
\end{array}
\right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}7 \\ 4 \\ -3 \\ -\dfrac{50}{9} \\ 9 \\ \dfrac{1}{2}\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{1.3597428460865E+46}{1.2093062068408E+46} & \dfrac{2.1259359222689E+46}{1.4511674482089E+46} & \dfrac{6.8701473976709E+52}{2.612101406776E+53} & \dfrac{8.1735001749156E+51}{1.8574943337074E+52} & -\dfrac{4.5112446894293E+45}{2.4186124136815E+45} & -\dfrac{1.37570668227E+55}{4.8976901377051E+54} \\ \dfrac{1.9211183472721E+49}{7.5581637927548E+48} & -\dfrac{2.7137329788867E+46}{8.0620413789384E+45} & -\dfrac{9.8341724604779E+49}{1.8139593102611E+50} & -\dfrac{3.3371998357178E+46}{4.837224827363E+46} & \dfrac{1.4900951288346E+47}{3.6279186205223E+46} & \dfrac{3.7411608089903E+51}{6.5302535169401E+50} \\ -\dfrac{5.1474284828099E+44}{5.0387758618365E+44} & \dfrac{1.6623938678833E+45}{1.511632758551E+45} & \dfrac{5.1315910852461E+45}{9.0697965513057E+46} & -\dfrac{1.6824786031411E+43}{8.0620413789384E+44} & -\dfrac{1.0505429310962E+46}{9.0697965513057E+45} & -\dfrac{6.1342629157793E+46}{3.6279186205223E+46} \\ \dfrac{4.1969507119519E+49}{2.0155103447346E+49} & -\dfrac{2.1720031581385E+47}{7.2558372410446E+46} & -\dfrac{2.4964949777464E+46}{7.2558372410446E+46} & -\dfrac{1.6760605781017E+45}{3.2248165515754E+45} & \dfrac{7.177820465288E+45}{2.0155103447346E+45} & \dfrac{5.0002829132898E+53}{1.0448405627104E+53} \\ \dfrac{2.9417230918902E+50}{8.0620413789384E+49} & -\dfrac{3.8912015629276E+46}{8.0620413789384E+45} & -\dfrac{3.289645914109E+49}{4.5348982756529E+49} & -\dfrac{5.0768043472016E+46}{4.837224827363E+46} & \dfrac{2.2162436469789E+47}{3.6279186205223E+46} & \dfrac{1.6438674220523E+55}{1.959076055082E+54} \\ \dfrac{3.5894156815024E+45}{6.4496331031507E+44} & -\dfrac{7.1360296402172E+46}{9.6744496547261E+45} & -\dfrac{6.4834826257318E+45}{6.0465310342038E+45} & -\dfrac{6.2386573042631E+45}{4.837224827363E+45} & \dfrac{2.7382311838664E+46}{3.0232655171019E+45} & \dfrac{1.2228350875551E+47}{9.6744496547261E+45}\end{pmatrix}\times\begin{pmatrix}7 \\ 4 \\ -3 \\ -\dfrac{50}{9} \\ 9 \\ \dfrac{1}{2}\end{pmatrix}=\begin{pmatrix}-\dfrac{4.2548429752827E+300}{1.81552163408E+299} \\ \dfrac{1.1313079586792E+292}{2.2800544570232E+290} \\ -\dfrac{4.6431744500711E+276}{3.2986898973137E+275} \\ \dfrac{5.3137639046024E+289}{1.2970976466621E+288} \\ \dfrac{1.3391740148327E+296}{1.8240435656186E+294} \\ \dfrac{1.0344984293992E+277}{9.6081307160156E+274}\end{pmatrix}\) . Ainsi \( x=\dfrac{4.2548429752827E+300}{1.81552163408E+299}\) , \( y=\dfrac{1.1313079586792E+292}{2.2800544570232E+290}\) , \( z=\dfrac{4.6431744500711E+276}{3.2986898973137E+275}\) , \( t=\dfrac{5.3137639046024E+289}{1.2970976466621E+288}\) , \( u=\dfrac{1.3391740148327E+296}{1.8240435656186E+294}\) et \( v=\dfrac{1.0344984293992E+277}{9.6081307160156E+274}\)
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&-\dfrac{1.3597428460865E+46}{1.2093062068408E+46}x&+&\dfrac{2.1259359222689E+46}{1.4511674482089E+46}y &+&\dfrac{6.8701473976709E+52}{2.612101406776E+53}z &+&\dfrac{8.1735001749156E+51}{1.8574943337074E+52}t &-&\dfrac{4.5112446894293E+45}{2.4186124136815E+45}u &-&\dfrac{1.37570668227E+55}{4.8976901377051E+54}v &=&5\\
&\dfrac{1.9211183472721E+49}{7.5581637927548E+48}x&-&\dfrac{2.7137329788867E+46}{8.0620413789384E+45}y &-&\dfrac{9.8341724604779E+49}{1.8139593102611E+50}z &-&\dfrac{3.3371998357178E+46}{4.837224827363E+46}t &+&\dfrac{1.4900951288346E+47}{3.6279186205223E+46}u &+&\dfrac{3.7411608089903E+51}{6.5302535169401E+50}v &=&8\\
&-\dfrac{5.1474284828099E+44}{5.0387758618365E+44}x&+&\dfrac{1.6623938678833E+45}{1.511632758551E+45}y &+&\dfrac{5.1315910852461E+45}{9.0697965513057E+46}z &-&\dfrac{1.6824786031411E+43}{8.0620413789384E+44}t &-&\dfrac{1.0505429310962E+46}{9.0697965513057E+45}u &-&\dfrac{6.1342629157793E+46}{3.6279186205223E+46}v &=&-7\\
&\dfrac{4.1969507119519E+49}{2.0155103447346E+49}x&-&\dfrac{2.1720031581385E+47}{7.2558372410446E+46}y &-&\dfrac{2.4964949777464E+46}{7.2558372410446E+46}z &-&\dfrac{1.6760605781017E+45}{3.2248165515754E+45}t &+&\dfrac{7.177820465288E+45}{2.0155103447346E+45}u &+&\dfrac{5.0002829132898E+53}{1.0448405627104E+53}v &=&1\\
&\dfrac{2.9417230918902E+50}{8.0620413789384E+49}x&-&\dfrac{3.8912015629276E+46}{8.0620413789384E+45}y &-&\dfrac{3.289645914109E+49}{4.5348982756529E+49}z &-&\dfrac{5.0768043472016E+46}{4.837224827363E+46}t &+&\dfrac{2.2162436469789E+47}{3.6279186205223E+46}u &+&\dfrac{1.6438674220523E+55}{1.959076055082E+54}v &=&\dfrac{8}{3}\\
&\dfrac{3.5894156815024E+45}{6.4496331031507E+44}x&-&\dfrac{7.1360296402172E+46}{9.6744496547261E+45}y &-&\dfrac{6.4834826257318E+45}{6.0465310342038E+45}z &-&\dfrac{6.2386573042631E+45}{4.837224827363E+45}t &+&\dfrac{2.7382311838664E+46}{3.0232655171019E+45}u &+&\dfrac{1.2228350875551E+47}{9.6744496547261E+45}v &=&1\\
\end{array}
\right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}5 \\ 8 \\ -7 \\ 1 \\ \dfrac{8}{3} \\ 1\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & \dfrac{8}{3} & -\dfrac{10}{3} & -2 & 3 & -\dfrac{8}{3} \\ 1 & -\dfrac{23}{5} & -\dfrac{1}{4} & -4 & \dfrac{21}{5} & 1 \\ -\dfrac{11}{2} & -\dfrac{19}{2} & -4 & 5 & 4 & -2 \\ 3 & -3 & \dfrac{8}{5} & -\dfrac{11}{5} & -\dfrac{12}{5} & \dfrac{14}{3} \\ 5 & -\dfrac{11}{4} & \dfrac{5}{4} & -\dfrac{7}{4} & \dfrac{9}{2} & \dfrac{1}{5} \\ -\dfrac{18}{5} & -3 & \dfrac{1}{4} & 0 & -2 & 2\end{pmatrix}\times\begin{pmatrix}5 \\ 8 \\ -7 \\ 1 \\ \dfrac{8}{3} \\ 1\end{pmatrix}=\begin{pmatrix}53 \\ -\dfrac{437}{20} \\ -\dfrac{371}{6} \\ -\dfrac{362}{15} \\ \dfrac{47}{10} \\ -\dfrac{565}{12}\end{pmatrix}\) . Ainsi \( x=53\) , \( y=\dfrac{437}{20}\) , \( z=\dfrac{371}{6}\) , \( t=\dfrac{362}{15}\) , \( u=\dfrac{47}{10}\) et \( v=\dfrac{565}{12}\)
