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
\[ A= \begin{pmatrix}1 & 3 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 3 & -3 & -\dfrac{16}{3} & 0 & 4 & -5 \\ -2 & -\dfrac{11}{2} & 3 & -4 & \dfrac{23}{5} & 5 \\ \dfrac{11}{3} & \dfrac{9}{5} & -\dfrac{7}{2} & 5 & 1 & 1 \\ -\dfrac{5}{3} & -\dfrac{19}{4} & 0 & -\dfrac{10}{3} & \dfrac{17}{4} & -\dfrac{21}{2} \\ -3 & \dfrac{16}{3} & \dfrac{23}{4} & 4 & -\dfrac{24}{5} & -4\end{pmatrix}\]
- Donner les mineurs d'ordre \( (5, 4)\) et \( (6, 3)\) :
\( \widehat{A}_{5, 4}=\)
\( \widehat{A}_{6, 3}=\)
- Expliquer pourquoi
\( \det(A)=\det
\begin{pmatrix}1 & 3 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 0 & -12 & -\dfrac{34}{3} & -\dfrac{3}{2} & -\dfrac{37}{2} & 4 \\ 0 & \dfrac{1}{2} & 7 & -3 & \dfrac{98}{5} & -1 \\ 0 & -\dfrac{46}{5} & -\dfrac{65}{6} & \dfrac{19}{6} & -\dfrac{53}{2} & 12 \\ 0 & \dfrac{1}{4} & \dfrac{10}{3} & -\dfrac{5}{2} & \dfrac{67}{4} & -\dfrac{31}{2} \\ 0 & \dfrac{43}{3} & \dfrac{47}{4} & \dfrac{11}{2} & \dfrac{177}{10} & -13\end{pmatrix}
\)
- 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 \( A^{-1}_{2, 4}\) .
- Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&x&+&3y &+&2z &+&\dfrac{1}{2}t &+&\dfrac{15}{2}u &-&3v &=&-3\\
&3x&-&3y &-&\dfrac{16}{3}z &&&+&4u &-&5v &=&-4\\
&-2x&-&\dfrac{11}{2}y &+&3z &-&4t &+&\dfrac{23}{5}u &+&5v &=&-\dfrac{25}{2}\\
&\dfrac{11}{3}x&+&\dfrac{9}{5}y &-&\dfrac{7}{2}z &+&5t &+&u &+&v &=&1\\
&-\dfrac{5}{3}x&-&\dfrac{19}{4}y &&&-&\dfrac{10}{3}t &+&\dfrac{17}{4}u &-&\dfrac{21}{2}v &=&5\\
&-3x&+&\dfrac{16}{3}y &+&\dfrac{23}{4}z &+&4t &-&\dfrac{24}{5}u &-&4v &=&-6\\
\end{array}
\right.
\)
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{5.1368621996495E+59}{2.1582604592949E+60}x&+&\dfrac{9.1849824501552E+63}{1.0231753288509E+63}y &+&\dfrac{4.4422236500319E+59}{2.1316152684394E+59}z &-&\dfrac{1.1864510924265E+62}{2.3021444899145E+61}t &-&\dfrac{2.236685629075E+55}{4.1633110711706E+54}u &+&\dfrac{5.35480400613E+57}{1.3322595427746E+57}v &=&\dfrac{5}{7}\\
&\dfrac{3.5601797387702E+53}{4.9959732854048E+54}x&-&\dfrac{5.9742036133849E+54}{3.3306488569365E+54}y &-&\dfrac{1.5264656762629E+60}{2.7625733878974E+60}z &+&\dfrac{3.2997410009455E+53}{3.7007209521517E+53}t &+&\dfrac{4.0893884840066E+52}{4.1633110711706E+52}u &-&\dfrac{1.4279873802224E+54}{1.6653244284683E+54}v &=&1\\
&\dfrac{1.7591281407145E+54}{1.2489933213512E+55}x&+&\dfrac{4.2900971066914E+55}{8.3266221423413E+54}y &+&\dfrac{2.2618725621765E+53}{1.7347129463211E+53}z &-&\dfrac{1.6431037288388E+54}{5.5510814282275E+53}t &-&\dfrac{3.2029435198747E+54}{1.0408277677927E+54}u &+&\dfrac{1.6804693669403E+54}{6.9388517852844E+53}v &=&0\\
&-\dfrac{6.4272626144498E+65}{6.2157901227692E+66}x&-&\dfrac{9.5373109280992E+65}{5.5251467757948E+65}y &-&\dfrac{4.0251508834409E+64}{1.3812866939487E+65}z &+&\dfrac{9.0646070436906E+62}{7.3668623677265E+62}t &+&\dfrac{1.2984807948814E+60}{1.1990335884971E+60}u &-&\dfrac{9.2162143572107E+53}{1.3877703570569E+54}v &=&-\dfrac{11}{6}\\
&\dfrac{4.9274619131908E+52}{1.4987919856214E+54}x&-&\dfrac{2.1036695964099E+54}{9.9919465708095E+53}y &-&\dfrac{8.1261875397735E+63}{1.7901475553575E+64}z &+&\dfrac{4.1625043842745E+53}{3.3306488569365E+53}t &+&\dfrac{3.1581610025888E+54}{2.4979866427024E+54}u &-&\dfrac{8.0491582419464E+53}{8.3266221423413E+53}v &=&-1\\
&-\dfrac{1.4318406807361E+53}{5.9951679424857E+54}x&-&\dfrac{5.4757965957319E+65}{5.9671585178584E+65}y &-&\dfrac{7.6999875468968E+64}{4.4753688883938E+65}z &+&\dfrac{6.5876503167263E+61}{1.2431580245538E+62}t &+&\dfrac{6.00908047019E+54}{1.2489933213512E+55}u &-&\dfrac{9.5596195379701E+52}{2.220432571291E+53}v &=&-8\\
\end{array}
\right.
\)
Cliquer ici pour afficher la solution
Exercice
-
\( \widehat{A}_{5, 4}=\begin{pmatrix}1 & 3 & 2 & \dfrac{15}{2} & -3 \\ 3 & -3 & -\dfrac{16}{3} & 4 & -5 \\ -2 & -\dfrac{11}{2} & 3 & \dfrac{23}{5} & 5 \\ \dfrac{11}{3} & \dfrac{9}{5} & -\dfrac{7}{2} & 1 & 1 \\ -3 & \dfrac{16}{3} & \dfrac{23}{4} & -\dfrac{24}{5} & -4\end{pmatrix}\)
\( \widehat{A}_{6, 3}=\begin{pmatrix}1 & 3 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 3 & -3 & 0 & 4 & -5 \\ -2 & -\dfrac{11}{2} & -4 & \dfrac{23}{5} & 5 \\ \dfrac{11}{3} & \dfrac{9}{5} & 5 & 1 & 1 \\ -\dfrac{5}{3} & -\dfrac{19}{4} & -\dfrac{10}{3} & \dfrac{17}{4} & -\dfrac{21}{2}\end{pmatrix}\)
- 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 : \( L_{2}\leftarrow L_{2}-\left(3\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-2\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(\dfrac{11}{3}\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-\dfrac{5}{3}\right)L_1\) 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 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 0 & -12 & -\dfrac{34}{3} & -\dfrac{3}{2} & -\dfrac{37}{2} & 4 \\ 0 & \dfrac{1}{2} & 7 & -3 & \dfrac{98}{5} & -1 \\ 0 & -\dfrac{46}{5} & -\dfrac{65}{6} & \dfrac{19}{6} & -\dfrac{53}{2} & 12 \\ 0 & \dfrac{1}{4} & \dfrac{10}{3} & -\dfrac{5}{2} & \dfrac{67}{4} & -\dfrac{31}{2} \\ 0 & \dfrac{43}{3} & \dfrac{47}{4} & \dfrac{11}{2} & \dfrac{177}{10} & -13\end{pmatrix}\\
&=&1\times\det\begin{pmatrix}-12 & -\dfrac{34}{3} & -\dfrac{3}{2} & -\dfrac{37}{2} & 4 \\ \dfrac{1}{2} & 7 & -3 & \dfrac{98}{5} & -1 \\ -\dfrac{46}{5} & -\dfrac{65}{6} & \dfrac{19}{6} & -\dfrac{53}{2} & 12 \\ \dfrac{1}{4} & \dfrac{10}{3} & -\dfrac{5}{2} & \dfrac{67}{4} & -\dfrac{31}{2} \\ \dfrac{43}{3} & \dfrac{47}{4} & \dfrac{11}{2} & \dfrac{177}{10} & -13\end{pmatrix}\\
&=&\dfrac{2.3129505950948E+51}{7.0663864180792E+47}
\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{2.3129505950948E+51}{7.0663864180792E+47}\right)^{-1}{}^tCo\begin{pmatrix}1 & 3 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 3 & -3 & -\dfrac{16}{3} & 0 & 4 & -5 \\ -2 & -\dfrac{11}{2} & 3 & -4 & \dfrac{23}{5} & 5 \\ \dfrac{11}{3} & \dfrac{9}{5} & -\dfrac{7}{2} & 5 & 1 & 1 \\ -\dfrac{5}{3} & -\dfrac{19}{4} & 0 & -\dfrac{10}{3} & \dfrac{17}{4} & -\dfrac{21}{2} \\ -3 & \dfrac{16}{3} & \dfrac{23}{4} & 4 & -\dfrac{24}{5} & -4\end{pmatrix}
=\dfrac{7.0663864180792E+47}{2.3129505950948E+51}\begin{pmatrix}\dfrac{1.1881308481599E+111}{1.5251102396239E+108} & \dfrac{2.1244410624022E+115}{7.2301522471057E+110} & \dfrac{1.0274643834885E+111}{1.506281718147E+107} & -\dfrac{2.7442027602787E+113}{1.6267842555988E+109} & -\dfrac{5.173343356809E+106}{2.9419564807559E+102} & \dfrac{1.2385397112594E+109}{9.4142607384189E+104} \\ \dfrac{8.2345198454331E+104}{3.5303477769071E+102} & -\dfrac{1.3818037802796E+106}{2.3535651846047E+102} & -\dfrac{3.530639694304E+111}{1.9521411067185E+108} & \dfrac{7.6321379117956E+104}{2.6150724273386E+101} & \dfrac{9.4585535276569E+103}{2.9419564807559E+100} & -\dfrac{3.3028642608732E+105}{1.1767825923024E+102} \\ \dfrac{4.0687764799137E+105}{8.8258694422677E+102} & \dfrac{9.9227826559363E+106}{5.8839129615118E+102} & \dfrac{5.2315994887149E+104}{1.225815200315E+101} & -\dfrac{3.8004177474202E+105}{3.9226086410079E+101} & -\dfrac{7.4082501203493E+105}{7.3548912018897E+101} & \dfrac{3.8868426223032E+105}{4.9032608012598E+101} \\ -\dfrac{1.4865940888922E+117}{4.3923174901167E+114} & -\dfrac{2.2059328986751E+117}{3.9042822134371E+113} & -\dfrac{9.309975131201E+115}{9.7607055335927E+112} & \dfrac{2.0965988256005E+114}{5.2057096179161E+110} & \dfrac{3.0033219272402E+111}{8.472834664577E+107} & -\dfrac{2.1316648482032E+105}{9.8065216025196E+101} \\ \dfrac{1.1396975964422E+104}{1.0591043330721E+102} & -\dfrac{4.8656838448992E+105}{7.0606955538141E+101} & -\dfrac{1.8795470305971E+115}{1.2649874371536E+112} & \dfrac{9.6276669926924E+104}{2.3535651846047E+101} & \dfrac{7.304670370343E+105}{1.7651738884535E+102} & -\dfrac{1.8617305345722E+105}{5.8839129615118E+101} \\ -\dfrac{3.3117767545896E+104}{4.2364173322885E+102} & -\dfrac{1.2665246994716E+117}{4.216624790512E+113} & -\dfrac{1.7809690778818E+116}{3.162468592884E+113} & \dfrac{1.5236909720349E+113}{8.7846349802334E+109} & \dfrac{1.3898706249499E+106}{8.8258694422677E+102} & -\dfrac{2.2110927699228E+104}{1.5690434564031E+101}\end{pmatrix}
=\begin{pmatrix}\dfrac{5.1368621996495E+59}{2.1582604592949E+60} & \dfrac{9.1849824501552E+63}{1.0231753288509E+63} & \dfrac{4.4422236500319E+59}{2.1316152684394E+59} & -\dfrac{1.1864510924265E+62}{2.3021444899145E+61} & -\dfrac{2.236685629075E+55}{4.1633110711706E+54} & \dfrac{5.35480400613E+57}{1.3322595427746E+57} \\ \dfrac{3.5601797387702E+53}{4.9959732854048E+54} & -\dfrac{5.9742036133849E+54}{3.3306488569365E+54} & -\dfrac{1.5264656762629E+60}{2.7625733878974E+60} & \dfrac{3.2997410009455E+53}{3.7007209521517E+53} & \dfrac{4.0893884840066E+52}{4.1633110711706E+52} & -\dfrac{1.4279873802224E+54}{1.6653244284683E+54} \\ \dfrac{1.7591281407145E+54}{1.2489933213512E+55} & \dfrac{4.2900971066914E+55}{8.3266221423413E+54} & \dfrac{2.2618725621765E+53}{1.7347129463211E+53} & -\dfrac{1.6431037288388E+54}{5.5510814282275E+53} & -\dfrac{3.2029435198747E+54}{1.0408277677927E+54} & \dfrac{1.6804693669403E+54}{6.9388517852844E+53} \\ -\dfrac{6.4272626144498E+65}{6.2157901227692E+66} & -\dfrac{9.5373109280992E+65}{5.5251467757948E+65} & -\dfrac{4.0251508834409E+64}{1.3812866939487E+65} & \dfrac{9.0646070436906E+62}{7.3668623677265E+62} & \dfrac{1.2984807948814E+60}{1.1990335884971E+60} & -\dfrac{9.2162143572107E+53}{1.3877703570569E+54} \\ \dfrac{4.9274619131908E+52}{1.4987919856214E+54} & -\dfrac{2.1036695964099E+54}{9.9919465708095E+53} & -\dfrac{8.1261875397735E+63}{1.7901475553575E+64} & \dfrac{4.1625043842745E+53}{3.3306488569365E+53} & \dfrac{3.1581610025888E+54}{2.4979866427024E+54} & -\dfrac{8.0491582419464E+53}{8.3266221423413E+53} \\ -\dfrac{1.4318406807361E+53}{5.9951679424857E+54} & -\dfrac{5.4757965957319E+65}{5.9671585178584E+65} & -\dfrac{7.6999875468968E+64}{4.4753688883938E+65} & \dfrac{6.5876503167263E+61}{1.2431580245538E+62} & \dfrac{6.00908047019E+54}{1.2489933213512E+55} & -\dfrac{9.5596195379701E+52}{2.220432571291E+53}\end{pmatrix}\) .
Précisément on a calculé \( A^{-1}_{2, 4}=B_{2, 4}=
\left(\dfrac{2.3129505950948E+51}{7.0663864180792E+47}\right)^{-1}Co(A)_{4, 2}=
\left(\dfrac{2.3129505950948E+51}{7.0663864180792E+47}\right)^{-1}\times(-1)^{4+2}\det\begin{pmatrix}1 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 3 & -\dfrac{16}{3} & 0 & 4 & -5 \\ -2 & 3 & -4 & \dfrac{23}{5} & 5 \\ -\dfrac{5}{3} & 0 & -\dfrac{10}{3} & \dfrac{17}{4} & -\dfrac{21}{2} \\ -3 & \dfrac{23}{4} & 4 & -\dfrac{24}{5} & -4\end{pmatrix}=\dfrac{3.2997410009455E+53}{3.7007209521517E+53}\) .
- On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul,
\[ B^{-1}=\begin{pmatrix}1 & 3 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 3 & -3 & -\dfrac{16}{3} & 0 & 4 & -5 \\ -2 & -\dfrac{11}{2} & 3 & -4 & \dfrac{23}{5} & 5 \\ \dfrac{11}{3} & \dfrac{9}{5} & -\dfrac{7}{2} & 5 & 1 & 1 \\ -\dfrac{5}{3} & -\dfrac{19}{4} & 0 & -\dfrac{10}{3} & \dfrac{17}{4} & -\dfrac{21}{2} \\ -3 & \dfrac{16}{3} & \dfrac{23}{4} & 4 & -\dfrac{24}{5} & -4\end{pmatrix}\]
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&x&+&3y &+&2z &+&\dfrac{1}{2}t &+&\dfrac{15}{2}u &-&3v &=&-3\\
&3x&-&3y &-&\dfrac{16}{3}z &&&+&4u &-&5v &=&-4\\
&-2x&-&\dfrac{11}{2}y &+&3z &-&4t &+&\dfrac{23}{5}u &+&5v &=&-\dfrac{25}{2}\\
&\dfrac{11}{3}x&+&\dfrac{9}{5}y &-&\dfrac{7}{2}z &+&5t &+&u &+&v &=&1\\
&-\dfrac{5}{3}x&-&\dfrac{19}{4}y &&&-&\dfrac{10}{3}t &+&\dfrac{17}{4}u &-&\dfrac{21}{2}v &=&5\\
&-3x&+&\dfrac{16}{3}y &+&\dfrac{23}{4}z &+&4t &-&\dfrac{24}{5}u &-&4v &=&-6\\
\end{array}
\right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-3 \\ -4 \\ -\dfrac{25}{2} \\ 1 \\ 5 \\ -6\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{5.1368621996495E+59}{2.1582604592949E+60} & \dfrac{9.1849824501552E+63}{1.0231753288509E+63} & \dfrac{4.4422236500319E+59}{2.1316152684394E+59} & -\dfrac{1.1864510924265E+62}{2.3021444899145E+61} & -\dfrac{2.236685629075E+55}{4.1633110711706E+54} & \dfrac{5.35480400613E+57}{1.3322595427746E+57} \\ \dfrac{3.5601797387702E+53}{4.9959732854048E+54} & -\dfrac{5.9742036133849E+54}{3.3306488569365E+54} & -\dfrac{1.5264656762629E+60}{2.7625733878974E+60} & \dfrac{3.2997410009455E+53}{3.7007209521517E+53} & \dfrac{4.0893884840066E+52}{4.1633110711706E+52} & -\dfrac{1.4279873802224E+54}{1.6653244284683E+54} \\ \dfrac{1.7591281407145E+54}{1.2489933213512E+55} & \dfrac{4.2900971066914E+55}{8.3266221423413E+54} & \dfrac{2.2618725621765E+53}{1.7347129463211E+53} & -\dfrac{1.6431037288388E+54}{5.5510814282275E+53} & -\dfrac{3.2029435198747E+54}{1.0408277677927E+54} & \dfrac{1.6804693669403E+54}{6.9388517852844E+53} \\ -\dfrac{6.4272626144498E+65}{6.2157901227692E+66} & -\dfrac{9.5373109280992E+65}{5.5251467757948E+65} & -\dfrac{4.0251508834409E+64}{1.3812866939487E+65} & \dfrac{9.0646070436906E+62}{7.3668623677265E+62} & \dfrac{1.2984807948814E+60}{1.1990335884971E+60} & -\dfrac{9.2162143572107E+53}{1.3877703570569E+54} \\ \dfrac{4.9274619131908E+52}{1.4987919856214E+54} & -\dfrac{2.1036695964099E+54}{9.9919465708095E+53} & -\dfrac{8.1261875397735E+63}{1.7901475553575E+64} & \dfrac{4.1625043842745E+53}{3.3306488569365E+53} & \dfrac{3.1581610025888E+54}{2.4979866427024E+54} & -\dfrac{8.0491582419464E+53}{8.3266221423413E+53} \\ -\dfrac{1.4318406807361E+53}{5.9951679424857E+54} & -\dfrac{5.4757965957319E+65}{5.9671585178584E+65} & -\dfrac{7.6999875468968E+64}{4.4753688883938E+65} & \dfrac{6.5876503167263E+61}{1.2431580245538E+62} & \dfrac{6.00908047019E+54}{1.2489933213512E+55} & -\dfrac{9.5596195379701E+52}{2.220432571291E+53}\end{pmatrix}\times\begin{pmatrix}-3 \\ -4 \\ -\dfrac{25}{2} \\ 1 \\ 5 \\ -6\end{pmatrix}=\begin{pmatrix}-\dfrac{INF}{INF} \\ \dfrac{INF}{INF} \\ -\dfrac{INF}{INF} \\ \dfrac{INF}{INF} \\ \dfrac{INF}{INF} \\ \dfrac{INF}{INF}\end{pmatrix}\) . Ainsi \( x=\dfrac{INF}{INF}\) , \( y=\dfrac{INF}{INF}\) , \( z=\dfrac{INF}{INF}\) , \( t=\dfrac{INF}{INF}\) , \( u=\dfrac{INF}{INF}\) et \( v=\dfrac{INF}{INF}\)
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{5.1368621996495E+59}{2.1582604592949E+60}x&+&\dfrac{9.1849824501552E+63}{1.0231753288509E+63}y &+&\dfrac{4.4422236500319E+59}{2.1316152684394E+59}z &-&\dfrac{1.1864510924265E+62}{2.3021444899145E+61}t &-&\dfrac{2.236685629075E+55}{4.1633110711706E+54}u &+&\dfrac{5.35480400613E+57}{1.3322595427746E+57}v &=&\dfrac{5}{7}\\
&\dfrac{3.5601797387702E+53}{4.9959732854048E+54}x&-&\dfrac{5.9742036133849E+54}{3.3306488569365E+54}y &-&\dfrac{1.5264656762629E+60}{2.7625733878974E+60}z &+&\dfrac{3.2997410009455E+53}{3.7007209521517E+53}t &+&\dfrac{4.0893884840066E+52}{4.1633110711706E+52}u &-&\dfrac{1.4279873802224E+54}{1.6653244284683E+54}v &=&1\\
&\dfrac{1.7591281407145E+54}{1.2489933213512E+55}x&+&\dfrac{4.2900971066914E+55}{8.3266221423413E+54}y &+&\dfrac{2.2618725621765E+53}{1.7347129463211E+53}z &-&\dfrac{1.6431037288388E+54}{5.5510814282275E+53}t &-&\dfrac{3.2029435198747E+54}{1.0408277677927E+54}u &+&\dfrac{1.6804693669403E+54}{6.9388517852844E+53}v &=&0\\
&-\dfrac{6.4272626144498E+65}{6.2157901227692E+66}x&-&\dfrac{9.5373109280992E+65}{5.5251467757948E+65}y &-&\dfrac{4.0251508834409E+64}{1.3812866939487E+65}z &+&\dfrac{9.0646070436906E+62}{7.3668623677265E+62}t &+&\dfrac{1.2984807948814E+60}{1.1990335884971E+60}u &-&\dfrac{9.2162143572107E+53}{1.3877703570569E+54}v &=&-\dfrac{11}{6}\\
&\dfrac{4.9274619131908E+52}{1.4987919856214E+54}x&-&\dfrac{2.1036695964099E+54}{9.9919465708095E+53}y &-&\dfrac{8.1261875397735E+63}{1.7901475553575E+64}z &+&\dfrac{4.1625043842745E+53}{3.3306488569365E+53}t &+&\dfrac{3.1581610025888E+54}{2.4979866427024E+54}u &-&\dfrac{8.0491582419464E+53}{8.3266221423413E+53}v &=&-1\\
&-\dfrac{1.4318406807361E+53}{5.9951679424857E+54}x&-&\dfrac{5.4757965957319E+65}{5.9671585178584E+65}y &-&\dfrac{7.6999875468968E+64}{4.4753688883938E+65}z &+&\dfrac{6.5876503167263E+61}{1.2431580245538E+62}t &+&\dfrac{6.00908047019E+54}{1.2489933213512E+55}u &-&\dfrac{9.5596195379701E+52}{2.220432571291E+53}v &=&-8\\
\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 \( b=\begin{pmatrix}\dfrac{5}{7} \\ 1 \\ 0 \\ -\dfrac{11}{6} \\ -1 \\ -8\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 3 & 2 & \dfrac{1}{2} & \dfrac{15}{2} & -3 \\ 3 & -3 & -\dfrac{16}{3} & 0 & 4 & -5 \\ -2 & -\dfrac{11}{2} & 3 & -4 & \dfrac{23}{5} & 5 \\ \dfrac{11}{3} & \dfrac{9}{5} & -\dfrac{7}{2} & 5 & 1 & 1 \\ -\dfrac{5}{3} & -\dfrac{19}{4} & 0 & -\dfrac{10}{3} & \dfrac{17}{4} & -\dfrac{21}{2} \\ -3 & \dfrac{16}{3} & \dfrac{23}{4} & 4 & -\dfrac{24}{5} & -4\end{pmatrix}\times\begin{pmatrix}\dfrac{5}{7} \\ 1 \\ 0 \\ -\dfrac{11}{6} \\ -1 \\ -8\end{pmatrix}=\begin{pmatrix}\dfrac{1621}{84} \\ \dfrac{246}{7} \\ -\dfrac{9281}{210} \\ -\dfrac{2887}{210} \\ \dfrac{5035}{63} \\ \dfrac{1143}{35}\end{pmatrix}\) . Ainsi \( x=\dfrac{1621}{84}\) , \( y=\dfrac{246}{7}\) , \( z=\dfrac{9281}{210}\) , \( t=\dfrac{2887}{210}\) , \( u=\dfrac{5035}{63}\) et \( v=\dfrac{1143}{35}\)
