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
\[ A= \begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ \dfrac{25}{4} & -\dfrac{13}{3} & \dfrac{9}{4} & \dfrac{2}{3} & \dfrac{13}{5} & -\dfrac{3}{5} \\ \dfrac{7}{2} & 1 & 4 & \dfrac{23}{3} & -5 & 5 \\ -4 & \dfrac{14}{3} & 4 & 3 & \dfrac{21}{5} & \dfrac{21}{4} \\ 3 & -2 & -\dfrac{9}{2} & -4 & 0 & -3 \\ 3 & 5 & -\dfrac{20}{3} & \dfrac{23}{5} & -\dfrac{19}{3} & 1\end{pmatrix}\]
- Donner les mineurs d'ordre \( (2, 4)\) et \( (5, 1)\) :
\( \widehat{A}_{2, 4}=\)
\( \widehat{A}_{5, 1}=\)
- Expliquer pourquoi
\( \det(A)=\det
\begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ 0 & \dfrac{173}{12} & \dfrac{151}{6} & 34 & \dfrac{138}{5} & -\dfrac{131}{10} \\ 0 & \dfrac{23}{2} & \dfrac{101}{6} & \dfrac{79}{3} & 9 & -2 \\ 0 & -\dfrac{22}{3} & -\dfrac{32}{3} & -\dfrac{55}{3} & -\dfrac{59}{5} & \dfrac{53}{4} \\ 0 & 7 & \dfrac{13}{2} & 12 & 12 & -9 \\ 0 & 14 & \dfrac{13}{3} & \dfrac{103}{5} & \dfrac{17}{3} & -5\end{pmatrix}
\)
- 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 \( A^{-1}_{5, 6}\) .
- Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&x&-&3y &-&\dfrac{11}{3}z &-&\dfrac{16}{3}t &-&4u &+&2v &=&-2\\
&\dfrac{25}{4}x&-&\dfrac{13}{3}y &+&\dfrac{9}{4}z &+&\dfrac{2}{3}t &+&\dfrac{13}{5}u &-&\dfrac{3}{5}v &=&3\\
&\dfrac{7}{2}x&+&y &+&4z &+&\dfrac{23}{3}t &-&5u &+&5v &=&-2\\
&-4x&+&\dfrac{14}{3}y &+&4z &+&3t &+&\dfrac{21}{5}u &+&\dfrac{21}{4}v &=&-7\\
&3x&-&2y &-&\dfrac{9}{2}z &-&4t &&&-&3v &=&7\\
&3x&+&5y &-&\dfrac{20}{3}z &+&\dfrac{23}{5}t &-&\dfrac{19}{3}u &+&v &=&\dfrac{43}{6}\\
\end{array}
\right.
\)
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&-\dfrac{5.9908095522196E+54}{3.3127560662745E+55}x&-&\dfrac{5.107817026881E+48}{1.8404200368192E+49}y &+&\dfrac{3.1553345248322E+56}{6.9878448272979E+56}z &+&\dfrac{4.932335347757E+48}{2.7606300552288E+49}t &+&\dfrac{3.4751495789134E+60}{3.7734362067408E+60}u &-&\dfrac{8.7500859626173E+48}{3.6808400736384E+49}v &=&-5\\
&-\dfrac{1.2989613616971E+55}{3.6808400736384E+55}x&-&\dfrac{2.0382066322751E+48}{3.067366728032E+48}y &+&\dfrac{3.1022345847054E+53}{4.313484461295E+53}z &+&\dfrac{2.7602160746478E+47}{9.2021001840959E+47}t &+&\dfrac{2.981757545359E+60}{2.0124993102618E+60}u &-&\dfrac{5.1381037553921E+46}{1.2269466912128E+47}v &=&0\\
&-\dfrac{2.437737897505E+48}{9.2021001840959E+48}x&-&\dfrac{7.8249591992351E+47}{1.533683364016E+48}y &+&\dfrac{2.1529596455425E+58}{3.4939224136489E+58}z &+&\dfrac{6.5332663902508E+47}{4.601050092048E+48}t &+&\dfrac{1.1955500248903E+63}{1.1320308620223E+63}u &-&\dfrac{4.4420654299538E+47}{1.0224555760107E+48}v &=&7\\
&\dfrac{1.4704761335194E+48}{8.1796446080853E+48}x&+&\dfrac{7.8075430494792E+48}{1.4723360294554E+49}y &-&\dfrac{3.3947610577073E+51}{5.8893441178214E+51}z &-&\dfrac{5.487439525177E+47}{2.4538933824256E+48}t &-&\dfrac{2.5469037015488E+52}{2.1201638824157E+52}u &+&\dfrac{8.7123825970563E+55}{2.1201638824157E+56}v &=&-7\\
&\dfrac{5.789881444928E+48}{7.3616801472768E+49}x&+&\dfrac{4.43883289152E+48}{1.4723360294554E+49}y &-&\dfrac{2.0011470711266E+54}{6.3604916472471E+54}z &+&\dfrac{2.4089858204877E+47}{1.2269466912128E+49}t &-&\dfrac{4.3935122068713E+60}{1.0015484567421E+61}u &+&\dfrac{5.2298694806267E+48}{2.9446720589107E+49}v &=&3\\
&\dfrac{9.7585724215064E+47}{4.601050092048E+48}x&+&\dfrac{6.1760464675662E+48}{2.7606300552288E+49}y &-&\dfrac{2.7720137167941E+56}{1.5093744826963E+57}z &+&\dfrac{2.3424892735293E+60}{3.6677799929521E+61}t &-&\dfrac{1.5344696447233E+64}{4.0107174222931E+64}u &+&\dfrac{1.0691445123834E+47}{7.3616801472768E+47}v &=&-4\\
\end{array}
\right.
\)
Cliquer ici pour afficher la solution
Exercice
-
\( \widehat{A}_{2, 4}=\begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -4 & 2 \\ \dfrac{7}{2} & 1 & 4 & -5 & 5 \\ -4 & \dfrac{14}{3} & 4 & \dfrac{21}{5} & \dfrac{21}{4} \\ 3 & -2 & -\dfrac{9}{2} & 0 & -3 \\ 3 & 5 & -\dfrac{20}{3} & -\dfrac{19}{3} & 1\end{pmatrix}\)
\( \widehat{A}_{5, 1}=\begin{pmatrix}-3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ -\dfrac{13}{3} & \dfrac{9}{4} & \dfrac{2}{3} & \dfrac{13}{5} & -\dfrac{3}{5} \\ 1 & 4 & \dfrac{23}{3} & -5 & 5 \\ \dfrac{14}{3} & 4 & 3 & \dfrac{21}{5} & \dfrac{21}{4} \\ 5 & -\dfrac{20}{3} & \dfrac{23}{5} & -\dfrac{19}{3} & 1\end{pmatrix}\)
- 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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{25}{4}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(\dfrac{7}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(-4\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(3\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(3\right)L_1\)
- 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 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ 0 & \dfrac{173}{12} & \dfrac{151}{6} & 34 & \dfrac{138}{5} & -\dfrac{131}{10} \\ 0 & \dfrac{23}{2} & \dfrac{101}{6} & \dfrac{79}{3} & 9 & -2 \\ 0 & -\dfrac{22}{3} & -\dfrac{32}{3} & -\dfrac{55}{3} & -\dfrac{59}{5} & \dfrac{53}{4} \\ 0 & 7 & \dfrac{13}{2} & 12 & 12 & -9 \\ 0 & 14 & \dfrac{13}{3} & \dfrac{103}{5} & \dfrac{17}{3} & -5\end{pmatrix}\\
&=&1\times\det\begin{pmatrix}\dfrac{173}{12} & \dfrac{151}{6} & 34 & \dfrac{138}{5} & -\dfrac{131}{10} \\ \dfrac{23}{2} & \dfrac{101}{6} & \dfrac{79}{3} & 9 & -2 \\ -\dfrac{22}{3} & -\dfrac{32}{3} & -\dfrac{55}{3} & -\dfrac{59}{5} & \dfrac{53}{4} \\ 7 & \dfrac{13}{2} & 12 & 12 & -9 \\ 14 & \dfrac{13}{3} & \dfrac{103}{5} & \dfrac{17}{3} & -5\end{pmatrix}\\
&=&\dfrac{2.0449111520213E+46}{1.1914996070292E+42}
\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{2.0449111520213E+46}{1.1914996070292E+42}\right)^{-1}{}^tCo\begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ \dfrac{25}{4} & -\dfrac{13}{3} & \dfrac{9}{4} & \dfrac{2}{3} & \dfrac{13}{5} & -\dfrac{3}{5} \\ \dfrac{7}{2} & 1 & 4 & \dfrac{23}{3} & -5 & 5 \\ -4 & \dfrac{14}{3} & 4 & 3 & \dfrac{21}{5} & \dfrac{21}{4} \\ 3 & -2 & -\dfrac{9}{2} & -4 & 0 & -3 \\ 3 & 5 & -\dfrac{20}{3} & \dfrac{23}{5} & -\dfrac{19}{3} & 1\end{pmatrix}
=\dfrac{1.1914996070292E+42}{2.0449111520213E+46}\begin{pmatrix}-\dfrac{1.225067326297E+101}{3.9471475511498E+97} & -\dfrac{1.0445032000753E+95}{2.1928597506388E+91} & \dfrac{6.4523787581872E+102}{8.3260143657066E+98} & \dfrac{1.0086187558137E+95}{3.2892896259581E+91} & \dfrac{7.1063721288623E+106}{4.4960477574815E+102} & -\dfrac{1.7893148366101E+95}{4.3857195012775E+91} \\ -\dfrac{2.6562605745791E+101}{4.3857195012775E+97} & -\dfrac{4.1679514724633E+94}{3.6547662510646E+90} & \dfrac{6.3437940984503E+99}{5.1395150405596E+95} & \dfrac{5.6443966330358E+93}{1.0964298753194E+90} & \dfrac{6.0974292571284E+106}{2.3978921373235E+102} & -\dfrac{1.0506965669644E+93}{1.4619065004258E+89} \\ -\dfrac{4.9849574123129E+94}{1.0964298753194E+91} & -\dfrac{1.6001346330628E+94}{1.8273831255323E+90} & \dfrac{4.4026111890216E+104}{4.1630071828533E+100} & \dfrac{1.335994930055E+94}{5.4821493765969E+90} & \dfrac{2.4447935786976E+109}{1.3488143272445E+105} & -\dfrac{9.083629135721E+93}{1.2182554170215E+90} \\ \dfrac{3.0069930442151E+94}{9.7460433361723E+90} & \dfrac{1.5965731851767E+95}{1.754287800511E+91} & -\dfrac{6.9419847453534E+97}{7.017151202044E+93} & -\dfrac{1.1221326281077E+94}{2.9238130008517E+90} & -\dfrac{5.2081917824215E+98}{2.5261744327359E+94} & \dfrac{1.7816048333397E+102}{2.5261744327359E+98} \\ \dfrac{1.1839793135615E+95}{8.7714390025551E+91} & \dfrac{9.0770188818282E+94}{1.754287800511E+91} & -\dfrac{4.0921679625816E+100}{7.5785232982076E+96} & \dfrac{4.9261619693766E+93}{1.4619065004258E+91} & -\dfrac{8.9843421083729E+106}{1.193344592629E+103} & \dfrac{1.0694618424549E+95}{3.508575601022E+91} \\ \dfrac{1.9955413572546E+94}{5.4821493765969E+90} & \dfrac{1.2629466296928E+95}{3.2892896259581E+91} & -\dfrac{5.6685217630284E+102}{1.7984191029926E+99} & \dfrac{4.7901824389304E+106}{4.3701584202721E+103} & -\dfrac{3.1378540889328E+110}{4.7787682325675E+106} & \dfrac{2.1863055364952E+93}{8.7714390025551E+89}\end{pmatrix}
=\begin{pmatrix}-\dfrac{5.9908095522196E+54}{3.3127560662745E+55} & -\dfrac{5.107817026881E+48}{1.8404200368192E+49} & \dfrac{3.1553345248322E+56}{6.9878448272979E+56} & \dfrac{4.932335347757E+48}{2.7606300552288E+49} & \dfrac{3.4751495789134E+60}{3.7734362067408E+60} & -\dfrac{8.7500859626173E+48}{3.6808400736384E+49} \\ -\dfrac{1.2989613616971E+55}{3.6808400736384E+55} & -\dfrac{2.0382066322751E+48}{3.067366728032E+48} & \dfrac{3.1022345847054E+53}{4.313484461295E+53} & \dfrac{2.7602160746478E+47}{9.2021001840959E+47} & \dfrac{2.981757545359E+60}{2.0124993102618E+60} & -\dfrac{5.1381037553921E+46}{1.2269466912128E+47} \\ -\dfrac{2.437737897505E+48}{9.2021001840959E+48} & -\dfrac{7.8249591992351E+47}{1.533683364016E+48} & \dfrac{2.1529596455425E+58}{3.4939224136489E+58} & \dfrac{6.5332663902508E+47}{4.601050092048E+48} & \dfrac{1.1955500248903E+63}{1.1320308620223E+63} & -\dfrac{4.4420654299538E+47}{1.0224555760107E+48} \\ \dfrac{1.4704761335194E+48}{8.1796446080853E+48} & \dfrac{7.8075430494792E+48}{1.4723360294554E+49} & -\dfrac{3.3947610577073E+51}{5.8893441178214E+51} & -\dfrac{5.487439525177E+47}{2.4538933824256E+48} & -\dfrac{2.5469037015488E+52}{2.1201638824157E+52} & \dfrac{8.7123825970563E+55}{2.1201638824157E+56} \\ \dfrac{5.789881444928E+48}{7.3616801472768E+49} & \dfrac{4.43883289152E+48}{1.4723360294554E+49} & -\dfrac{2.0011470711266E+54}{6.3604916472471E+54} & \dfrac{2.4089858204877E+47}{1.2269466912128E+49} & -\dfrac{4.3935122068713E+60}{1.0015484567421E+61} & \dfrac{5.2298694806267E+48}{2.9446720589107E+49} \\ \dfrac{9.7585724215064E+47}{4.601050092048E+48} & \dfrac{6.1760464675662E+48}{2.7606300552288E+49} & -\dfrac{2.7720137167941E+56}{1.5093744826963E+57} & \dfrac{2.3424892735293E+60}{3.6677799929521E+61} & -\dfrac{1.5344696447233E+64}{4.0107174222931E+64} & \dfrac{1.0691445123834E+47}{7.3616801472768E+47}\end{pmatrix}\) .
Précisément on a calculé \( A^{-1}_{5, 6}=B_{5, 6}=
\left(\dfrac{2.0449111520213E+46}{1.1914996070292E+42}\right)^{-1}Co(A)_{6, 5}=
\left(\dfrac{2.0449111520213E+46}{1.1914996070292E+42}\right)^{-1}\times(-1)^{6+5}\det\begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & 2 \\ \dfrac{25}{4} & -\dfrac{13}{3} & \dfrac{9}{4} & \dfrac{2}{3} & -\dfrac{3}{5} \\ \dfrac{7}{2} & 1 & 4 & \dfrac{23}{3} & 5 \\ -4 & \dfrac{14}{3} & 4 & 3 & \dfrac{21}{4} \\ 3 & -2 & -\dfrac{9}{2} & -4 & -3\end{pmatrix}=\dfrac{5.2298694806267E+48}{2.9446720589107E+49}\) .
- On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul,
\[ B^{-1}=\begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ \dfrac{25}{4} & -\dfrac{13}{3} & \dfrac{9}{4} & \dfrac{2}{3} & \dfrac{13}{5} & -\dfrac{3}{5} \\ \dfrac{7}{2} & 1 & 4 & \dfrac{23}{3} & -5 & 5 \\ -4 & \dfrac{14}{3} & 4 & 3 & \dfrac{21}{5} & \dfrac{21}{4} \\ 3 & -2 & -\dfrac{9}{2} & -4 & 0 & -3 \\ 3 & 5 & -\dfrac{20}{3} & \dfrac{23}{5} & -\dfrac{19}{3} & 1\end{pmatrix}\]
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&x&-&3y &-&\dfrac{11}{3}z &-&\dfrac{16}{3}t &-&4u &+&2v &=&-2\\
&\dfrac{25}{4}x&-&\dfrac{13}{3}y &+&\dfrac{9}{4}z &+&\dfrac{2}{3}t &+&\dfrac{13}{5}u &-&\dfrac{3}{5}v &=&3\\
&\dfrac{7}{2}x&+&y &+&4z &+&\dfrac{23}{3}t &-&5u &+&5v &=&-2\\
&-4x&+&\dfrac{14}{3}y &+&4z &+&3t &+&\dfrac{21}{5}u &+&\dfrac{21}{4}v &=&-7\\
&3x&-&2y &-&\dfrac{9}{2}z &-&4t &&&-&3v &=&7\\
&3x&+&5y &-&\dfrac{20}{3}z &+&\dfrac{23}{5}t &-&\dfrac{19}{3}u &+&v &=&\dfrac{43}{6}\\
\end{array}
\right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-2 \\ 3 \\ -2 \\ -7 \\ 7 \\ \dfrac{43}{6}\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{5.9908095522196E+54}{3.3127560662745E+55} & -\dfrac{5.107817026881E+48}{1.8404200368192E+49} & \dfrac{3.1553345248322E+56}{6.9878448272979E+56} & \dfrac{4.932335347757E+48}{2.7606300552288E+49} & \dfrac{3.4751495789134E+60}{3.7734362067408E+60} & -\dfrac{8.7500859626173E+48}{3.6808400736384E+49} \\ -\dfrac{1.2989613616971E+55}{3.6808400736384E+55} & -\dfrac{2.0382066322751E+48}{3.067366728032E+48} & \dfrac{3.1022345847054E+53}{4.313484461295E+53} & \dfrac{2.7602160746478E+47}{9.2021001840959E+47} & \dfrac{2.981757545359E+60}{2.0124993102618E+60} & -\dfrac{5.1381037553921E+46}{1.2269466912128E+47} \\ -\dfrac{2.437737897505E+48}{9.2021001840959E+48} & -\dfrac{7.8249591992351E+47}{1.533683364016E+48} & \dfrac{2.1529596455425E+58}{3.4939224136489E+58} & \dfrac{6.5332663902508E+47}{4.601050092048E+48} & \dfrac{1.1955500248903E+63}{1.1320308620223E+63} & -\dfrac{4.4420654299538E+47}{1.0224555760107E+48} \\ \dfrac{1.4704761335194E+48}{8.1796446080853E+48} & \dfrac{7.8075430494792E+48}{1.4723360294554E+49} & -\dfrac{3.3947610577073E+51}{5.8893441178214E+51} & -\dfrac{5.487439525177E+47}{2.4538933824256E+48} & -\dfrac{2.5469037015488E+52}{2.1201638824157E+52} & \dfrac{8.7123825970563E+55}{2.1201638824157E+56} \\ \dfrac{5.789881444928E+48}{7.3616801472768E+49} & \dfrac{4.43883289152E+48}{1.4723360294554E+49} & -\dfrac{2.0011470711266E+54}{6.3604916472471E+54} & \dfrac{2.4089858204877E+47}{1.2269466912128E+49} & -\dfrac{4.3935122068713E+60}{1.0015484567421E+61} & \dfrac{5.2298694806267E+48}{2.9446720589107E+49} \\ \dfrac{9.7585724215064E+47}{4.601050092048E+48} & \dfrac{6.1760464675662E+48}{2.7606300552288E+49} & -\dfrac{2.7720137167941E+56}{1.5093744826963E+57} & \dfrac{2.3424892735293E+60}{3.6677799929521E+61} & -\dfrac{1.5344696447233E+64}{4.0107174222931E+64} & \dfrac{1.0691445123834E+47}{7.3616801472768E+47}\end{pmatrix}\times\begin{pmatrix}-2 \\ 3 \\ -2 \\ -7 \\ 7 \\ \dfrac{43}{6}\end{pmatrix}=\begin{pmatrix}\dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ -\dfrac{INF}{4.6941191529666E+307} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF}\end{pmatrix}\) . Ainsi \( x=\dfrac{NAN}{INF}\) , \( y=\dfrac{NAN}{INF}\) , \( z=\dfrac{NAN}{INF}\) , \( t=\dfrac{INF}{4.6941191529666E+307}\) , \( u=\dfrac{NAN}{INF}\) et \( v=\dfrac{NAN}{INF}\)
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&-\dfrac{5.9908095522196E+54}{3.3127560662745E+55}x&-&\dfrac{5.107817026881E+48}{1.8404200368192E+49}y &+&\dfrac{3.1553345248322E+56}{6.9878448272979E+56}z &+&\dfrac{4.932335347757E+48}{2.7606300552288E+49}t &+&\dfrac{3.4751495789134E+60}{3.7734362067408E+60}u &-&\dfrac{8.7500859626173E+48}{3.6808400736384E+49}v &=&-5\\
&-\dfrac{1.2989613616971E+55}{3.6808400736384E+55}x&-&\dfrac{2.0382066322751E+48}{3.067366728032E+48}y &+&\dfrac{3.1022345847054E+53}{4.313484461295E+53}z &+&\dfrac{2.7602160746478E+47}{9.2021001840959E+47}t &+&\dfrac{2.981757545359E+60}{2.0124993102618E+60}u &-&\dfrac{5.1381037553921E+46}{1.2269466912128E+47}v &=&0\\
&-\dfrac{2.437737897505E+48}{9.2021001840959E+48}x&-&\dfrac{7.8249591992351E+47}{1.533683364016E+48}y &+&\dfrac{2.1529596455425E+58}{3.4939224136489E+58}z &+&\dfrac{6.5332663902508E+47}{4.601050092048E+48}t &+&\dfrac{1.1955500248903E+63}{1.1320308620223E+63}u &-&\dfrac{4.4420654299538E+47}{1.0224555760107E+48}v &=&7\\
&\dfrac{1.4704761335194E+48}{8.1796446080853E+48}x&+&\dfrac{7.8075430494792E+48}{1.4723360294554E+49}y &-&\dfrac{3.3947610577073E+51}{5.8893441178214E+51}z &-&\dfrac{5.487439525177E+47}{2.4538933824256E+48}t &-&\dfrac{2.5469037015488E+52}{2.1201638824157E+52}u &+&\dfrac{8.7123825970563E+55}{2.1201638824157E+56}v &=&-7\\
&\dfrac{5.789881444928E+48}{7.3616801472768E+49}x&+&\dfrac{4.43883289152E+48}{1.4723360294554E+49}y &-&\dfrac{2.0011470711266E+54}{6.3604916472471E+54}z &+&\dfrac{2.4089858204877E+47}{1.2269466912128E+49}t &-&\dfrac{4.3935122068713E+60}{1.0015484567421E+61}u &+&\dfrac{5.2298694806267E+48}{2.9446720589107E+49}v &=&3\\
&\dfrac{9.7585724215064E+47}{4.601050092048E+48}x&+&\dfrac{6.1760464675662E+48}{2.7606300552288E+49}y &-&\dfrac{2.7720137167941E+56}{1.5093744826963E+57}z &+&\dfrac{2.3424892735293E+60}{3.6677799929521E+61}t &-&\dfrac{1.5344696447233E+64}{4.0107174222931E+64}u &+&\dfrac{1.0691445123834E+47}{7.3616801472768E+47}v &=&-4\\
\end{array}
\right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-5 \\ 0 \\ 7 \\ -7 \\ 3 \\ -4\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -3 & -\dfrac{11}{3} & -\dfrac{16}{3} & -4 & 2 \\ \dfrac{25}{4} & -\dfrac{13}{3} & \dfrac{9}{4} & \dfrac{2}{3} & \dfrac{13}{5} & -\dfrac{3}{5} \\ \dfrac{7}{2} & 1 & 4 & \dfrac{23}{3} & -5 & 5 \\ -4 & \dfrac{14}{3} & 4 & 3 & \dfrac{21}{5} & \dfrac{21}{4} \\ 3 & -2 & -\dfrac{9}{2} & -4 & 0 & -3 \\ 3 & 5 & -\dfrac{20}{3} & \dfrac{23}{5} & -\dfrac{19}{3} & 1\end{pmatrix}\times\begin{pmatrix}-5 \\ 0 \\ 7 \\ -7 \\ 3 \\ -4\end{pmatrix}=\begin{pmatrix}-\dfrac{40}{3} \\ -\dfrac{299}{30} \\ -\dfrac{469}{6} \\ \dfrac{93}{5} \\ -\dfrac{13}{2} \\ -\dfrac{1753}{15}\end{pmatrix}\) . Ainsi \( x=\dfrac{40}{3}\) , \( y=\dfrac{299}{30}\) , \( z=\dfrac{469}{6}\) , \( t=\dfrac{93}{5}\) , \( u=\dfrac{13}{2}\) et \( v=\dfrac{1753}{15}\)
