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
\[ A= \begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 3 & 0 \\ 2 & -\dfrac{13}{2} & 0 & \dfrac{3}{2} & -2 & 0 \\ -\dfrac{1}{2} & \dfrac{11}{2} & -\dfrac{8}{5} & \dfrac{22}{3} & 3 & -1 \\ 1 & \dfrac{19}{4} & 3 & -2 & \dfrac{8}{5} & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & 5 & -2 & 5 & -\dfrac{17}{5} \\ 4 & 3 & -\dfrac{7}{5} & \dfrac{11}{2} & \dfrac{6}{5} & 2\end{pmatrix}\]
- Donner les mineurs d'ordre \( (3, 3)\) et \( (2, 5)\) :
\( \widehat{A}_{3, 3}=\)
\( \widehat{A}_{2, 5}=\)
- Expliquer pourquoi
\( \det(A)=\det
\begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 3 & 0 \\ 0 & -\dfrac{7}{2} & -25 & \dfrac{3}{2} & -8 & 0 \\ 0 & \dfrac{19}{4} & \dfrac{93}{20} & \dfrac{22}{3} & \dfrac{9}{2} & -1 \\ 0 & \dfrac{25}{4} & -\dfrac{19}{2} & -2 & -\dfrac{7}{5} & 3 \\ 0 & -\dfrac{44}{3} & \dfrac{290}{3} & -2 & 27 & -\dfrac{17}{5} \\ 0 & 9 & -\dfrac{257}{5} & \dfrac{11}{2} & -\dfrac{54}{5} & 2\end{pmatrix}
\)
- 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 \( A^{-1}_{5, 2}\) .
- Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&x&-&\dfrac{3}{2}y &+&\dfrac{25}{2}z &&&+&3u &&&=&-7\\
&2x&-&\dfrac{13}{2}y &&&+&\dfrac{3}{2}t &-&2u &&&=&5\\
&-\dfrac{1}{2}x&+&\dfrac{11}{2}y &-&\dfrac{8}{5}z &+&\dfrac{22}{3}t &+&3u &-&v &=&-7\\
&x&+&\dfrac{19}{4}y &+&3z &-&2t &+&\dfrac{8}{5}u &+&3v &=&9\\
&-\dfrac{22}{3}x&-&\dfrac{11}{3}y &+&5z &-&2t &+&5u &-&\dfrac{17}{5}v &=&7\\
&4x&+&3y &-&\dfrac{7}{5}z &+&\dfrac{11}{2}t &+&\dfrac{6}{5}u &+&2v &=&-2\\
\end{array}
\right.
\)
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{7.53895805625E+24}{5.7406476818531E+25}x&-&\dfrac{1.4599040593809E+31}{2.5832914568339E+31}y &-&\dfrac{2.7549294957563E+27}{5.7406476818531E+27}z &-&\dfrac{1.1736482055188E+27}{2.1527428806949E+27}t &+&\dfrac{1.02585335625E+24}{7.6541969091375E+25}u &+&\dfrac{3.4479243771188E+27}{5.7406476818531E+27}v &=&-9\\
&\dfrac{3.107613549375E+25}{9.5677461364219E+26}x&-&\dfrac{1.9536397325438E+27}{1.7221943045559E+28}y &+&\dfrac{4.36241710125E+25}{7.1758096023164E+26}z &-&\dfrac{2.663835710625E+25}{2.1527428806949E+27}t &-&\dfrac{8.89880731875E+24}{1.1481295363706E+26}u &-&\dfrac{1.5844950320625E+26}{1.9135492272844E+27}v &=&-\dfrac{29}{8}\\
&\dfrac{1.2426037824375E+26}{1.7221943045559E+27}x&+&\dfrac{9.3634971148125E+26}{6.8887772182237E+27}y &+&\dfrac{4.329312755625E+25}{2.8703238409266E+26}z &+&\dfrac{4.1701228025625E+26}{3.4443886091119E+27}t &-&\dfrac{2.51905494375E+24}{3.8270984545688E+25}u &-&\dfrac{5.0078337429375E+26}{2.2962590727412E+27}v &=&-\dfrac{9}{2}\\
&-\dfrac{1.97117863875E+25}{7.1758096023164E+26}x&+&\dfrac{4.0072041511875E+26}{1.1481295363706E+27}y &+&\dfrac{2.2903397139375E+26}{7.1758096023164E+26}z &+&\dfrac{6.7377947563125E+26}{2.8703238409266E+27}t &-&\dfrac{1.74846549375E+24}{3.8270984545688E+25}u &-&\dfrac{3.8776531055625E+26}{1.4351619204633E+27}v &=&2\\
&\dfrac{1.185617806875E+25}{2.2962590727412E+27}x&-&\dfrac{5.3900866448156E+31}{1.2399798992803E+32}y &-&\dfrac{1.6766360758125E+26}{3.8270984545687E+26}z &-&\dfrac{6.1177471765706E+31}{1.8599698489204E+32}t &+&\dfrac{5.305171291875E+25}{2.2962590727412E+26}u &+&\dfrac{3.6763306576425E+29}{5.511021774579E+29}v &=&5\\
&-\dfrac{1.2979930976438E+27}{6.8887772182237E+27}x&+&\dfrac{4.6642317347879E+38}{6.6958914561135E+38}y &+&\dfrac{1.3751318818125E+26}{3.8270984545687E+26}z &+&\dfrac{4.6220261896744E+29}{6.1998994964014E+29}t &+&\dfrac{2.326641316875E+25}{7.6541969091375E+26}u &-&\dfrac{2.3029000346007E+32}{5.9519035165453E+32}v &=&1\\
\end{array}
\right.
\)
Cliquer ici pour afficher la solution
Exercice
-
\( \widehat{A}_{3, 3}=\begin{pmatrix}1 & -\dfrac{3}{2} & 0 & 3 & 0 \\ 2 & -\dfrac{13}{2} & \dfrac{3}{2} & -2 & 0 \\ 1 & \dfrac{19}{4} & -2 & \dfrac{8}{5} & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & -2 & 5 & -\dfrac{17}{5} \\ 4 & 3 & \dfrac{11}{2} & \dfrac{6}{5} & 2\end{pmatrix}\)
\( \widehat{A}_{2, 5}=\begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 0 \\ -\dfrac{1}{2} & \dfrac{11}{2} & -\dfrac{8}{5} & \dfrac{22}{3} & -1 \\ 1 & \dfrac{19}{4} & 3 & -2 & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & 5 & -2 & -\dfrac{17}{5} \\ 4 & 3 & -\dfrac{7}{5} & \dfrac{11}{2} & 2\end{pmatrix}\)
- 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 : \( L_{2}\leftarrow L_{2}-\left(2\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{1}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(1\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-\dfrac{22}{3}\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(4\right)L_1\)
- 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{3}{2} & \dfrac{25}{2} & 0 & 3 & 0 \\ 0 & -\dfrac{7}{2} & -25 & \dfrac{3}{2} & -8 & 0 \\ 0 & \dfrac{19}{4} & \dfrac{93}{20} & \dfrac{22}{3} & \dfrac{9}{2} & -1 \\ 0 & \dfrac{25}{4} & -\dfrac{19}{2} & -2 & -\dfrac{7}{5} & 3 \\ 0 & -\dfrac{44}{3} & \dfrac{290}{3} & -2 & 27 & -\dfrac{17}{5} \\ 0 & 9 & -\dfrac{257}{5} & \dfrac{11}{2} & -\dfrac{54}{5} & 2\end{pmatrix}\\
&=&1\times\det\begin{pmatrix}-\dfrac{7}{2} & -25 & \dfrac{3}{2} & -8 & 0 \\ \dfrac{19}{4} & \dfrac{93}{20} & \dfrac{22}{3} & \dfrac{9}{2} & -1 \\ \dfrac{25}{4} & -\dfrac{19}{2} & -2 & -\dfrac{7}{5} & 3 \\ -\dfrac{44}{3} & \dfrac{290}{3} & -2 & 27 & -\dfrac{17}{5} \\ 9 & -\dfrac{257}{5} & \dfrac{11}{2} & -\dfrac{54}{5} & 2\end{pmatrix}\\
&=&\dfrac{1.9135492272844E+24}{1.2301875E+20}
\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{1.9135492272844E+24}{1.2301875E+20}\right)^{-1}{}^tCo\begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 3 & 0 \\ 2 & -\dfrac{13}{2} & 0 & \dfrac{3}{2} & -2 & 0 \\ -\dfrac{1}{2} & \dfrac{11}{2} & -\dfrac{8}{5} & \dfrac{22}{3} & 3 & -1 \\ 1 & \dfrac{19}{4} & 3 & -2 & \dfrac{8}{5} & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & 5 & -2 & 5 & -\dfrac{17}{5} \\ 4 & 3 & -\dfrac{7}{5} & \dfrac{11}{2} & \dfrac{6}{5} & 2\end{pmatrix}
=\dfrac{1.2301875E+20}{1.9135492272844E+24}\begin{pmatrix}\dfrac{1.4426167363067E+49}{7.0620730201197E+45} & -\dfrac{2.7935982847377E+55}{3.1779328590539E+51} & -\dfrac{5.2716932078273E+51}{7.0620730201197E+47} & -\dfrac{2.2458336167741E+51}{2.6482773825449E+47} & \dfrac{1.9630208971593E+48}{9.4160973601596E+45} & \dfrac{6.5977730275705E+51}{7.0620730201197E+47} \\ \dfrac{5.946571506105E+49}{1.1770121700199E+47} & -\dfrac{3.7383858006011E+51}{2.1186219060359E+48} & \dfrac{8.3476998731891E+49}{8.8275912751496E+46} & -\dfrac{5.097380765679E+49}{2.6482773825449E+47} & -\dfrac{1.7028305868547E+49}{1.4124146040239E+46} & -\dfrac{3.0320092442391E+50}{2.3540243400399E+47} \\ \dfrac{2.3777835077039E+50}{2.1186219060359E+47} & \dfrac{1.7917512668729E+51}{8.4744876241436E+47} & \dfrac{8.2843530781986E+49}{3.5310365100598E+46} & \dfrac{7.9797352665244E+50}{4.2372438120718E+47} & -\dfrac{4.8203356410997E+48}{4.7080486800798E+45} & -\dfrac{9.5827363891667E+50}{2.8248292080479E+47} \\ -\dfrac{3.7719473610195E+49}{8.8275912751496E+46} & \dfrac{7.6679824070756E+50}{1.4124146040239E+47} & \dfrac{4.3826777898238E+50}{8.8275912751496E+46} & \dfrac{1.2893101949542E+51}{3.5310365100598E+47} & -\dfrac{3.3457747944987E+48}{4.7080486800798E+45} & -\dfrac{7.420080103826E+50}{1.7655182550299E+47} \\ \dfrac{2.2687380382003E+49}{2.8248292080479E+47} & -\dfrac{1.0314196134183E+56}{1.5254077723459E+52} & -\dfrac{3.2083256673081E+50}{4.7080486800798E+46} & -\dfrac{1.1706610382448E+56}{2.2881116585188E+52} & \dfrac{1.0151706426179E+50}{2.8248292080479E+46} & \dfrac{7.0348396891737E+53}{6.7795900993149E+49} \\ -\dfrac{2.4837736890166E+51}{8.4744876241436E+47} & \dfrac{8.9252370319786E+62}{8.2372019706676E+58} & \dfrac{2.6313825498564E+50}{4.7080486800798E+46} & \dfrac{8.8444746437395E+53}{7.6270388617293E+49} & \dfrac{4.4521426940741E+49}{9.4160973601596E+46} & -\dfrac{4.4067125817233E+56}{7.3219573072601E+52}\end{pmatrix}
=\begin{pmatrix}\dfrac{7.53895805625E+24}{5.7406476818531E+25} & -\dfrac{1.4599040593809E+31}{2.5832914568339E+31} & -\dfrac{2.7549294957563E+27}{5.7406476818531E+27} & -\dfrac{1.1736482055188E+27}{2.1527428806949E+27} & \dfrac{1.02585335625E+24}{7.6541969091375E+25} & \dfrac{3.4479243771188E+27}{5.7406476818531E+27} \\ \dfrac{3.107613549375E+25}{9.5677461364219E+26} & -\dfrac{1.9536397325438E+27}{1.7221943045559E+28} & \dfrac{4.36241710125E+25}{7.1758096023164E+26} & -\dfrac{2.663835710625E+25}{2.1527428806949E+27} & -\dfrac{8.89880731875E+24}{1.1481295363706E+26} & -\dfrac{1.5844950320625E+26}{1.9135492272844E+27} \\ \dfrac{1.2426037824375E+26}{1.7221943045559E+27} & \dfrac{9.3634971148125E+26}{6.8887772182237E+27} & \dfrac{4.329312755625E+25}{2.8703238409266E+26} & \dfrac{4.1701228025625E+26}{3.4443886091119E+27} & -\dfrac{2.51905494375E+24}{3.8270984545688E+25} & -\dfrac{5.0078337429375E+26}{2.2962590727412E+27} \\ -\dfrac{1.97117863875E+25}{7.1758096023164E+26} & \dfrac{4.0072041511875E+26}{1.1481295363706E+27} & \dfrac{2.2903397139375E+26}{7.1758096023164E+26} & \dfrac{6.7377947563125E+26}{2.8703238409266E+27} & -\dfrac{1.74846549375E+24}{3.8270984545688E+25} & -\dfrac{3.8776531055625E+26}{1.4351619204633E+27} \\ \dfrac{1.185617806875E+25}{2.2962590727412E+27} & -\dfrac{5.3900866448156E+31}{1.2399798992803E+32} & -\dfrac{1.6766360758125E+26}{3.8270984545687E+26} & -\dfrac{6.1177471765706E+31}{1.8599698489204E+32} & \dfrac{5.305171291875E+25}{2.2962590727412E+26} & \dfrac{3.6763306576425E+29}{5.511021774579E+29} \\ -\dfrac{1.2979930976438E+27}{6.8887772182237E+27} & \dfrac{4.6642317347879E+38}{6.6958914561135E+38} & \dfrac{1.3751318818125E+26}{3.8270984545687E+26} & \dfrac{4.6220261896744E+29}{6.1998994964014E+29} & \dfrac{2.326641316875E+25}{7.6541969091375E+26} & -\dfrac{2.3029000346007E+32}{5.9519035165453E+32}\end{pmatrix}\) .
Précisément on a calculé \( A^{-1}_{5, 2}=B_{5, 2}=
\left(\dfrac{1.9135492272844E+24}{1.2301875E+20}\right)^{-1}Co(A)_{2, 5}=
\left(\dfrac{1.9135492272844E+24}{1.2301875E+20}\right)^{-1}\times(-1)^{2+5}\det\begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 0 \\ -\dfrac{1}{2} & \dfrac{11}{2} & -\dfrac{8}{5} & \dfrac{22}{3} & -1 \\ 1 & \dfrac{19}{4} & 3 & -2 & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & 5 & -2 & -\dfrac{17}{5} \\ 4 & 3 & -\dfrac{7}{5} & \dfrac{11}{2} & 2\end{pmatrix}=-\dfrac{5.3900866448156E+31}{1.2399798992803E+32}\) .
- On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul,
\[ B^{-1}=\begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 3 & 0 \\ 2 & -\dfrac{13}{2} & 0 & \dfrac{3}{2} & -2 & 0 \\ -\dfrac{1}{2} & \dfrac{11}{2} & -\dfrac{8}{5} & \dfrac{22}{3} & 3 & -1 \\ 1 & \dfrac{19}{4} & 3 & -2 & \dfrac{8}{5} & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & 5 & -2 & 5 & -\dfrac{17}{5} \\ 4 & 3 & -\dfrac{7}{5} & \dfrac{11}{2} & \dfrac{6}{5} & 2\end{pmatrix}\]
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&x&-&\dfrac{3}{2}y &+&\dfrac{25}{2}z &&&+&3u &&&=&-7\\
&2x&-&\dfrac{13}{2}y &&&+&\dfrac{3}{2}t &-&2u &&&=&5\\
&-\dfrac{1}{2}x&+&\dfrac{11}{2}y &-&\dfrac{8}{5}z &+&\dfrac{22}{3}t &+&3u &-&v &=&-7\\
&x&+&\dfrac{19}{4}y &+&3z &-&2t &+&\dfrac{8}{5}u &+&3v &=&9\\
&-\dfrac{22}{3}x&-&\dfrac{11}{3}y &+&5z &-&2t &+&5u &-&\dfrac{17}{5}v &=&7\\
&4x&+&3y &-&\dfrac{7}{5}z &+&\dfrac{11}{2}t &+&\dfrac{6}{5}u &+&2v &=&-2\\
\end{array}
\right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-7 \\ 5 \\ -7 \\ 9 \\ 7 \\ -2\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{7.53895805625E+24}{5.7406476818531E+25} & -\dfrac{1.4599040593809E+31}{2.5832914568339E+31} & -\dfrac{2.7549294957563E+27}{5.7406476818531E+27} & -\dfrac{1.1736482055188E+27}{2.1527428806949E+27} & \dfrac{1.02585335625E+24}{7.6541969091375E+25} & \dfrac{3.4479243771188E+27}{5.7406476818531E+27} \\ \dfrac{3.107613549375E+25}{9.5677461364219E+26} & -\dfrac{1.9536397325438E+27}{1.7221943045559E+28} & \dfrac{4.36241710125E+25}{7.1758096023164E+26} & -\dfrac{2.663835710625E+25}{2.1527428806949E+27} & -\dfrac{8.89880731875E+24}{1.1481295363706E+26} & -\dfrac{1.5844950320625E+26}{1.9135492272844E+27} \\ \dfrac{1.2426037824375E+26}{1.7221943045559E+27} & \dfrac{9.3634971148125E+26}{6.8887772182237E+27} & \dfrac{4.329312755625E+25}{2.8703238409266E+26} & \dfrac{4.1701228025625E+26}{3.4443886091119E+27} & -\dfrac{2.51905494375E+24}{3.8270984545688E+25} & -\dfrac{5.0078337429375E+26}{2.2962590727412E+27} \\ -\dfrac{1.97117863875E+25}{7.1758096023164E+26} & \dfrac{4.0072041511875E+26}{1.1481295363706E+27} & \dfrac{2.2903397139375E+26}{7.1758096023164E+26} & \dfrac{6.7377947563125E+26}{2.8703238409266E+27} & -\dfrac{1.74846549375E+24}{3.8270984545688E+25} & -\dfrac{3.8776531055625E+26}{1.4351619204633E+27} \\ \dfrac{1.185617806875E+25}{2.2962590727412E+27} & -\dfrac{5.3900866448156E+31}{1.2399798992803E+32} & -\dfrac{1.6766360758125E+26}{3.8270984545687E+26} & -\dfrac{6.1177471765706E+31}{1.8599698489204E+32} & \dfrac{5.305171291875E+25}{2.2962590727412E+26} & \dfrac{3.6763306576425E+29}{5.511021774579E+29} \\ -\dfrac{1.2979930976438E+27}{6.8887772182237E+27} & \dfrac{4.6642317347879E+38}{6.6958914561135E+38} & \dfrac{1.3751318818125E+26}{3.8270984545687E+26} & \dfrac{4.6220261896744E+29}{6.1998994964014E+29} & \dfrac{2.326641316875E+25}{7.6541969091375E+26} & -\dfrac{2.3029000346007E+32}{5.9519035165453E+32}\end{pmatrix}\times\begin{pmatrix}-7 \\ 5 \\ -7 \\ 9 \\ 7 \\ -2\end{pmatrix}=\begin{pmatrix}-\dfrac{5.1536082142799E+166}{8.0528179606172E+165} \\ -\dfrac{9.5538722404835E+162}{5.5922346948731E+162} \\ \dfrac{1.8944893043488E+161}{1.030760698959E+162} \\ \dfrac{1.8980007789553E+161}{9.3203911581218E+160} \\ -\dfrac{4.668314430254E+174}{2.5648624624337E+174} \\ \dfrac{4.9775020282947E+183}{4.9860926269711E+182}\end{pmatrix}\) . Ainsi \( x=\dfrac{5.1536082142799E+166}{8.0528179606172E+165}\) , \( y=\dfrac{9.5538722404835E+162}{5.5922346948731E+162}\) , \( z=\dfrac{1.8944893043488E+161}{1.030760698959E+162}\) , \( t=\dfrac{1.8980007789553E+161}{9.3203911581218E+160}\) , \( u=\dfrac{4.668314430254E+174}{2.5648624624337E+174}\) et \( v=\dfrac{4.9775020282947E+183}{4.9860926269711E+182}\)
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{7.53895805625E+24}{5.7406476818531E+25}x&-&\dfrac{1.4599040593809E+31}{2.5832914568339E+31}y &-&\dfrac{2.7549294957563E+27}{5.7406476818531E+27}z &-&\dfrac{1.1736482055188E+27}{2.1527428806949E+27}t &+&\dfrac{1.02585335625E+24}{7.6541969091375E+25}u &+&\dfrac{3.4479243771188E+27}{5.7406476818531E+27}v &=&-9\\
&\dfrac{3.107613549375E+25}{9.5677461364219E+26}x&-&\dfrac{1.9536397325438E+27}{1.7221943045559E+28}y &+&\dfrac{4.36241710125E+25}{7.1758096023164E+26}z &-&\dfrac{2.663835710625E+25}{2.1527428806949E+27}t &-&\dfrac{8.89880731875E+24}{1.1481295363706E+26}u &-&\dfrac{1.5844950320625E+26}{1.9135492272844E+27}v &=&-\dfrac{29}{8}\\
&\dfrac{1.2426037824375E+26}{1.7221943045559E+27}x&+&\dfrac{9.3634971148125E+26}{6.8887772182237E+27}y &+&\dfrac{4.329312755625E+25}{2.8703238409266E+26}z &+&\dfrac{4.1701228025625E+26}{3.4443886091119E+27}t &-&\dfrac{2.51905494375E+24}{3.8270984545688E+25}u &-&\dfrac{5.0078337429375E+26}{2.2962590727412E+27}v &=&-\dfrac{9}{2}\\
&-\dfrac{1.97117863875E+25}{7.1758096023164E+26}x&+&\dfrac{4.0072041511875E+26}{1.1481295363706E+27}y &+&\dfrac{2.2903397139375E+26}{7.1758096023164E+26}z &+&\dfrac{6.7377947563125E+26}{2.8703238409266E+27}t &-&\dfrac{1.74846549375E+24}{3.8270984545688E+25}u &-&\dfrac{3.8776531055625E+26}{1.4351619204633E+27}v &=&2\\
&\dfrac{1.185617806875E+25}{2.2962590727412E+27}x&-&\dfrac{5.3900866448156E+31}{1.2399798992803E+32}y &-&\dfrac{1.6766360758125E+26}{3.8270984545687E+26}z &-&\dfrac{6.1177471765706E+31}{1.8599698489204E+32}t &+&\dfrac{5.305171291875E+25}{2.2962590727412E+26}u &+&\dfrac{3.6763306576425E+29}{5.511021774579E+29}v &=&5\\
&-\dfrac{1.2979930976438E+27}{6.8887772182237E+27}x&+&\dfrac{4.6642317347879E+38}{6.6958914561135E+38}y &+&\dfrac{1.3751318818125E+26}{3.8270984545687E+26}z &+&\dfrac{4.6220261896744E+29}{6.1998994964014E+29}t &+&\dfrac{2.326641316875E+25}{7.6541969091375E+26}u &-&\dfrac{2.3029000346007E+32}{5.9519035165453E+32}v &=&1\\
\end{array}
\right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-9 \\ -\dfrac{29}{8} \\ -\dfrac{9}{2} \\ 2 \\ 5 \\ 1\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -\dfrac{3}{2} & \dfrac{25}{2} & 0 & 3 & 0 \\ 2 & -\dfrac{13}{2} & 0 & \dfrac{3}{2} & -2 & 0 \\ -\dfrac{1}{2} & \dfrac{11}{2} & -\dfrac{8}{5} & \dfrac{22}{3} & 3 & -1 \\ 1 & \dfrac{19}{4} & 3 & -2 & \dfrac{8}{5} & 3 \\ -\dfrac{22}{3} & -\dfrac{11}{3} & 5 & -2 & 5 & -\dfrac{17}{5} \\ 4 & 3 & -\dfrac{7}{5} & \dfrac{11}{2} & \dfrac{6}{5} & 2\end{pmatrix}\times\begin{pmatrix}-9 \\ -\dfrac{29}{8} \\ -\dfrac{9}{2} \\ 2 \\ 5 \\ 1\end{pmatrix}=\begin{pmatrix}-\dfrac{717}{16} \\ -\dfrac{23}{16} \\ \dfrac{4903}{240} \\ -\dfrac{1047}{32} \\ \dfrac{8927}{120} \\ -\dfrac{863}{40}\end{pmatrix}\) . Ainsi \( x=\dfrac{717}{16}\) , \( y=\dfrac{23}{16}\) , \( z=\dfrac{4903}{240}\) , \( t=\dfrac{1047}{32}\) , \( u=\dfrac{8927}{120}\) et \( v=\dfrac{863}{40}\)
