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 \[ A= \begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ -\dfrac{23}{3} & \dfrac{21}{2} & -2 & -3 & -\dfrac{7}{2} & -\dfrac{19}{5} \\ \dfrac{11}{2} & -1 & 5 & 5 & -3 & -1 \\ -3 & \dfrac{19}{5} & \dfrac{19}{2} & -5 & -\dfrac{22}{5} & \dfrac{17}{2} \\ \dfrac{17}{2} & 5 & -1 & -\dfrac{7}{5} & \dfrac{21}{4} & 0 \\ \dfrac{20}{3} & 0 & -\dfrac{13}{2} & -1 & -4 & \dfrac{9}{4}\end{pmatrix}\]

Donner les mineurs d'ordre $ (6, 5)$ et $ (4, 5)$ : $ \widehat{A}_{6, 5}=$ $ \widehat{A}_{4, 5}=$
Expliquer pourquoi $ \det(A)=\det \begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ 0 & -\dfrac{283}{30} & \dfrac{86}{3} & -\dfrac{413}{15} & -\dfrac{151}{30} & \dfrac{1}{30} \\ 0 & \dfrac{133}{10} & -17 & \dfrac{113}{5} & -\dfrac{19}{10} & -\dfrac{15}{4} \\ 0 & -4 & \dfrac{43}{2} & -\dfrac{73}{5} & -5 & 10 \\ 0 & \dfrac{271}{10} & -35 & \dfrac{129}{5} & \dfrac{139}{20} & -\dfrac{17}{4} \\ 0 & \dfrac{52}{3} & -\dfrac{199}{6} & \dfrac{61}{3} & -\dfrac{8}{3} & -\dfrac{13}{12}\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}_{1, 2}$ .
Donner $ B^{-1}$ l'inverse de la matrice $ B$ . Justifier.
Résoudre le système suivant. $ \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{13}{5}y &+&4z &-&\dfrac{16}{5}t &-&\dfrac{1}{5}u &+&\dfrac{1}{2}v &=&9\\ &-\dfrac{23}{3}x&+&\dfrac{21}{2}y &-&2z &-&3t &-&\dfrac{7}{2}u &-&\dfrac{19}{5}v &=&7\\ &\dfrac{11}{2}x&-&y &+&5z &+&5t &-&3u &-&v &=&6\\ &-3x&+&\dfrac{19}{5}y &+&\dfrac{19}{2}z &-&5t &-&\dfrac{22}{5}u &+&\dfrac{17}{2}v &=&6\\ &\dfrac{17}{2}x&+&5y &-&z &-&\dfrac{7}{5}t &+&\dfrac{21}{4}u &&&=&-5\\ &\dfrac{20}{3}x&&&-&\dfrac{13}{2}z &-&t &-&4u &+&\dfrac{9}{4}v &=&9\\ \end{array} \right. $
Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{4.5488628878915E+96}{9.1982625723763E+97}x&-&\dfrac{1.7920879855845E+92}{7.186142634669E+94}y &+&\dfrac{2.9801737734111E+98}{7.186142634669E+99}z &-&\dfrac{1.444487826481E+93}{1.4372285269338E+95}t &+&\dfrac{8.4231353848092E+93}{1.7965356586672E+95}u &+&\dfrac{2.9606017937115E+96}{7.186142634669E+97}v &=&4\\ &-\dfrac{1.233649551974E+99}{1.6556872630277E+100}x&+&\dfrac{4.9430548667038E+94}{1.149782821547E+96}y &+&\dfrac{2.5292501716699E+100}{1.2935056742404E+102}z &+&\dfrac{4.6311306454775E+92}{1.6168820928005E+94}t &+&\dfrac{8.5882095740126E+92}{1.6168820928005E+94}u &-&\dfrac{1.4873753580345E+93}{1.4372285269338E+95}v &=&-9\\ &\dfrac{1.6532904728712E+94}{2.8744570538676E+95}x&+&\dfrac{5.5049122004121E+84}{8.9826782933362E+86}y &+&\dfrac{7.1524637020086E+104}{1.3474017440004E+106}z &+&\dfrac{8.7977674288835E+91}{4.4913391466681E+93}t &+&\dfrac{1.4337082925916E+85}{1.122834786667E+87}u &-&\dfrac{2.3731767160298E+102}{4.4913391466681E+103}v &=&0\\ &-\dfrac{1.527841065918E+93}{8.2784363151387E+93}x&-&\dfrac{2.6190182255671E+84}{7.186142634669E+85}y &+&\dfrac{3.07553813278E+90}{5.1740226969617E+91}z &+&\dfrac{3.704110604529E+87}{3.233764185601E+89}t &-&\dfrac{2.1896878388822E+85}{1.3474017440004E+87}u &-&\dfrac{5.3742997821447E+92}{1.4372285269338E+94}v &=&8\\ &-\dfrac{1.5316860605658E+93}{3.233764185601E+94}x&-&\dfrac{4.0833343358247E+88}{8.9826782933362E+89}y &-&\dfrac{2.4173905341E+97}{4.0422052320013E+98}z &-&\dfrac{3.412061608661E+88}{8.0844104640026E+90}t &+&\dfrac{1.0037340969183E+93}{1.6168820928005E+94}u &-&\dfrac{6.90613558205E+91}{8.9826782933362E+92}v &=&8\\ &-\dfrac{7.5851265540482E+90}{5.1740226969617E+91}x&-&\dfrac{8.0752836335199E+84}{1.122834786667E+86}y &-&\dfrac{1.6648686672372E+85}{3.3685043600011E+86}z &+&\dfrac{2.8280407868084E+85}{3.3685043600011E+86}t &+&\dfrac{9.3156430526577E+82}{8.4212609000027E+85}u &+&\dfrac{1.8446246214844E+84}{1.122834786667E+86}v &=&7\\ \end{array} \right. \)

Cliquer ici pour afficher la solution

Exercice

$ \widehat{A}_{6, 5}=\begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & \dfrac{1}{2} \\ -\dfrac{23}{3} & \dfrac{21}{2} & -2 & -3 & -\dfrac{19}{5} \\ \dfrac{11}{2} & -1 & 5 & 5 & -1 \\ -3 & \dfrac{19}{5} & \dfrac{19}{2} & -5 & \dfrac{17}{2} \\ \dfrac{17}{2} & 5 & -1 & -\dfrac{7}{5} & 0\end{pmatrix}$ $ \widehat{A}_{4, 5}=\begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & \dfrac{1}{2} \\ -\dfrac{23}{3} & \dfrac{21}{2} & -2 & -3 & -\dfrac{19}{5} \\ \dfrac{11}{2} & -1 & 5 & 5 & -1 \\ \dfrac{17}{2} & 5 & -1 & -\dfrac{7}{5} & 0 \\ \dfrac{20}{3} & 0 & -\dfrac{13}{2} & -1 & \dfrac{9}{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{23}{3}\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(\dfrac{17}{2}\right)L_1$ et $ L_{6}\leftarrow L_{6}-\left(\dfrac{20}{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 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ 0 & -\dfrac{283}{30} & \dfrac{86}{3} & -\dfrac{413}{15} & -\dfrac{151}{30} & \dfrac{1}{30} \\ 0 & \dfrac{133}{10} & -17 & \dfrac{113}{5} & -\dfrac{19}{10} & -\dfrac{15}{4} \\ 0 & -4 & \dfrac{43}{2} & -\dfrac{73}{5} & -5 & 10 \\ 0 & \dfrac{271}{10} & -35 & \dfrac{129}{5} & \dfrac{139}{20} & -\dfrac{17}{4} \\ 0 & \dfrac{52}{3} & -\dfrac{199}{6} & \dfrac{61}{3} & -\dfrac{8}{3} & -\dfrac{13}{12}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-\dfrac{283}{30} & \dfrac{86}{3} & -\dfrac{413}{15} & -\dfrac{151}{30} & \dfrac{1}{30} \\ \dfrac{133}{10} & -17 & \dfrac{113}{5} & -\dfrac{19}{10} & -\dfrac{15}{4} \\ -4 & \dfrac{43}{2} & -\dfrac{73}{5} & -5 & 10 \\ \dfrac{271}{10} & -35 & \dfrac{129}{5} & \dfrac{139}{20} & -\dfrac{17}{4} \\ \dfrac{52}{3} & -\dfrac{199}{6} & \dfrac{61}{3} & -\dfrac{8}{3} & -\dfrac{13}{12}\end{pmatrix}\\ &=&-\dfrac{1.122834786667E+84}{1.5832967439974E+78} \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{1.122834786667E+84}{1.5832967439974E+78}\right)^{-1}{}^tCo\begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ -\dfrac{23}{3} & \dfrac{21}{2} & -2 & -3 & -\dfrac{7}{2} & -\dfrac{19}{5} \\ \dfrac{11}{2} & -1 & 5 & 5 & -3 & -1 \\ -3 & \dfrac{19}{5} & \dfrac{19}{2} & -5 & -\dfrac{22}{5} & \dfrac{17}{2} \\ \dfrac{17}{2} & 5 & -1 & -\dfrac{7}{5} & \dfrac{21}{4} & 0 \\ \dfrac{20}{3} & 0 & -\dfrac{13}{2} & -1 & -4 & \dfrac{9}{4}\end{pmatrix} =-\dfrac{1.5832967439974E+78}{1.122834786667E+84}\begin{pmatrix}-\dfrac{5.1076214903032E+180}{1.4563579181277E+176} & \dfrac{2.0122187309823E+176}{1.1377796235373E+173} & -\dfrac{3.3462427830987E+182}{1.1377796235373E+178} & \dfrac{1.62192118049E+177}{2.2755592470745E+173} & -\dfrac{9.4577894228697E+177}{2.8444490588431E+173} & -\dfrac{3.324266683448E+180}{1.1377796235373E+176} \\ \dfrac{1.3851846315126E+183}{2.6214442526298E+178} & -\dfrac{5.5502339567388E+178}{1.8204473976596E+174} & -\dfrac{2.8399300769345E+184}{2.0480033223671E+180} & -\dfrac{5.1999945903418E+176}{2.5600041529588E+172} & -\dfrac{9.6431404648882E+176}{2.5600041529588E+172} & \dfrac{1.6700767928325E+177}{2.2755592470745E+173} \\ -\dfrac{1.8563720554049E+178}{4.551118494149E+173} & -\dfrac{6.1811069161705E+168}{1.4222245294216E+165} & -\dfrac{8.0310350549885E+188}{2.1333367941324E+184} & -\dfrac{9.8784393141566E+175}{7.1111226471079E+171} & -\dfrac{1.6098175448549E+169}{1.777780661777E+165} & \dfrac{2.6646853716665E+186}{7.1111226471079E+181} \\ \dfrac{1.7155130973112E+177}{1.3107221263149E+172} & \dfrac{2.9407247705817E+168}{1.1377796235373E+164} & -\dfrac{3.4533212032063E+174}{8.1920132894683E+169} & -\dfrac{4.1591042404273E+171}{5.1200083059177E+167} & \dfrac{2.4586576774387E+169}{2.1333367941324E+165} & \dfrac{6.0344507493691E+176}{2.2755592470745E+172} \\ \dfrac{1.7198303910562E+177}{5.1200083059177E+172} & \dfrac{4.5849098378558E+172}{1.4222245294216E+168} & \dfrac{2.7143301846471E+181}{6.4000103823971E+176} & \dfrac{3.8311814684556E+172}{1.2800020764794E+169} & -\dfrac{1.1270275605836E+177}{2.5600041529588E+172} & \dfrac{7.7544492729647E+175}{1.4222245294216E+171} \\ \dfrac{8.5168439561571E+174}{8.1920132894683E+169} & \dfrac{9.0672093759191E+168}{1.777780661777E+164} & \dfrac{1.8693724548059E+169}{5.3333419853309E+164} & -\dfrac{3.1754225735417E+169}{5.3333419853309E+164} & -\dfrac{1.0459928079697E+167}{1.3333354963327E+164} & -\dfrac{2.0712086933452E+168}{1.777780661777E+164}\end{pmatrix} =\begin{pmatrix}\dfrac{4.5488628878915E+96}{9.1982625723763E+97} & -\dfrac{1.7920879855845E+92}{7.186142634669E+94} & \dfrac{2.9801737734111E+98}{7.186142634669E+99} & -\dfrac{1.444487826481E+93}{1.4372285269338E+95} & \dfrac{8.4231353848092E+93}{1.7965356586672E+95} & \dfrac{2.9606017937115E+96}{7.186142634669E+97} \\ -\dfrac{1.233649551974E+99}{1.6556872630277E+100} & \dfrac{4.9430548667038E+94}{1.149782821547E+96} & \dfrac{2.5292501716699E+100}{1.2935056742404E+102} & \dfrac{4.6311306454775E+92}{1.6168820928005E+94} & \dfrac{8.5882095740126E+92}{1.6168820928005E+94} & -\dfrac{1.4873753580345E+93}{1.4372285269338E+95} \\ \dfrac{1.6532904728712E+94}{2.8744570538676E+95} & \dfrac{5.5049122004121E+84}{8.9826782933362E+86} & \dfrac{7.1524637020086E+104}{1.3474017440004E+106} & \dfrac{8.7977674288835E+91}{4.4913391466681E+93} & \dfrac{1.4337082925916E+85}{1.122834786667E+87} & -\dfrac{2.3731767160298E+102}{4.4913391466681E+103} \\ -\dfrac{1.527841065918E+93}{8.2784363151387E+93} & -\dfrac{2.6190182255671E+84}{7.186142634669E+85} & \dfrac{3.07553813278E+90}{5.1740226969617E+91} & \dfrac{3.704110604529E+87}{3.233764185601E+89} & -\dfrac{2.1896878388822E+85}{1.3474017440004E+87} & -\dfrac{5.3742997821447E+92}{1.4372285269338E+94} \\ -\dfrac{1.5316860605658E+93}{3.233764185601E+94} & -\dfrac{4.0833343358247E+88}{8.9826782933362E+89} & -\dfrac{2.4173905341E+97}{4.0422052320013E+98} & -\dfrac{3.412061608661E+88}{8.0844104640026E+90} & \dfrac{1.0037340969183E+93}{1.6168820928005E+94} & -\dfrac{6.90613558205E+91}{8.9826782933362E+92} \\ -\dfrac{7.5851265540482E+90}{5.1740226969617E+91} & -\dfrac{8.0752836335199E+84}{1.122834786667E+86} & -\dfrac{1.6648686672372E+85}{3.3685043600011E+86} & \dfrac{2.8280407868084E+85}{3.3685043600011E+86} & \dfrac{9.3156430526577E+82}{8.4212609000027E+85} & \dfrac{1.8446246214844E+84}{1.122834786667E+86}\end{pmatrix}\) . Précisément on a calculé $ A^{-1}_{1, 2}=B_{1, 2}= \left(-\dfrac{1.122834786667E+84}{1.5832967439974E+78}\right)^{-1}Co(A)_{2, 1}= \left(-\dfrac{1.122834786667E+84}{1.5832967439974E+78}\right)^{-1}\times(-1)^{2+1}\det\begin{pmatrix}-\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ -1 & 5 & 5 & -3 & -1 \\ \dfrac{19}{5} & \dfrac{19}{2} & -5 & -\dfrac{22}{5} & \dfrac{17}{2} \\ 5 & -1 & -\dfrac{7}{5} & \dfrac{21}{4} & 0 \\ 0 & -\dfrac{13}{2} & -1 & -4 & \dfrac{9}{4}\end{pmatrix}=-\dfrac{1.7920879855845E+92}{7.186142634669E+94}$ .
On observe que $ B^{-1}=\left(A^{-1}\right)^{-1}=A$ . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ -\dfrac{23}{3} & \dfrac{21}{2} & -2 & -3 & -\dfrac{7}{2} & -\dfrac{19}{5} \\ \dfrac{11}{2} & -1 & 5 & 5 & -3 & -1 \\ -3 & \dfrac{19}{5} & \dfrac{19}{2} & -5 & -\dfrac{22}{5} & \dfrac{17}{2} \\ \dfrac{17}{2} & 5 & -1 & -\dfrac{7}{5} & \dfrac{21}{4} & 0 \\ \dfrac{20}{3} & 0 & -\dfrac{13}{2} & -1 & -4 & \dfrac{9}{4}\end{pmatrix}\]
Le système $ \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{13}{5}y &+&4z &-&\dfrac{16}{5}t &-&\dfrac{1}{5}u &+&\dfrac{1}{2}v &=&9\\ &-\dfrac{23}{3}x&+&\dfrac{21}{2}y &-&2z &-&3t &-&\dfrac{7}{2}u &-&\dfrac{19}{5}v &=&7\\ &\dfrac{11}{2}x&-&y &+&5z &+&5t &-&3u &-&v &=&6\\ &-3x&+&\dfrac{19}{5}y &+&\dfrac{19}{2}z &-&5t &-&\dfrac{22}{5}u &+&\dfrac{17}{2}v &=&6\\ &\dfrac{17}{2}x&+&5y &-&z &-&\dfrac{7}{5}t &+&\dfrac{21}{4}u &&&=&-5\\ &\dfrac{20}{3}x&&&-&\dfrac{13}{2}z &-&t &-&4u &+&\dfrac{9}{4}v &=&9\\ \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}9 \\ 7 \\ 6 \\ 6 \\ -5 \\ 9\end{pmatrix}$ de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{4.5488628878915E+96}{9.1982625723763E+97} & -\dfrac{1.7920879855845E+92}{7.186142634669E+94} & \dfrac{2.9801737734111E+98}{7.186142634669E+99} & -\dfrac{1.444487826481E+93}{1.4372285269338E+95} & \dfrac{8.4231353848092E+93}{1.7965356586672E+95} & \dfrac{2.9606017937115E+96}{7.186142634669E+97} \\ -\dfrac{1.233649551974E+99}{1.6556872630277E+100} & \dfrac{4.9430548667038E+94}{1.149782821547E+96} & \dfrac{2.5292501716699E+100}{1.2935056742404E+102} & \dfrac{4.6311306454775E+92}{1.6168820928005E+94} & \dfrac{8.5882095740126E+92}{1.6168820928005E+94} & -\dfrac{1.4873753580345E+93}{1.4372285269338E+95} \\ \dfrac{1.6532904728712E+94}{2.8744570538676E+95} & \dfrac{5.5049122004121E+84}{8.9826782933362E+86} & \dfrac{7.1524637020086E+104}{1.3474017440004E+106} & \dfrac{8.7977674288835E+91}{4.4913391466681E+93} & \dfrac{1.4337082925916E+85}{1.122834786667E+87} & -\dfrac{2.3731767160298E+102}{4.4913391466681E+103} \\ -\dfrac{1.527841065918E+93}{8.2784363151387E+93} & -\dfrac{2.6190182255671E+84}{7.186142634669E+85} & \dfrac{3.07553813278E+90}{5.1740226969617E+91} & \dfrac{3.704110604529E+87}{3.233764185601E+89} & -\dfrac{2.1896878388822E+85}{1.3474017440004E+87} & -\dfrac{5.3742997821447E+92}{1.4372285269338E+94} \\ -\dfrac{1.5316860605658E+93}{3.233764185601E+94} & -\dfrac{4.0833343358247E+88}{8.9826782933362E+89} & -\dfrac{2.4173905341E+97}{4.0422052320013E+98} & -\dfrac{3.412061608661E+88}{8.0844104640026E+90} & \dfrac{1.0037340969183E+93}{1.6168820928005E+94} & -\dfrac{6.90613558205E+91}{8.9826782933362E+92} \\ -\dfrac{7.5851265540482E+90}{5.1740226969617E+91} & -\dfrac{8.0752836335199E+84}{1.122834786667E+86} & -\dfrac{1.6648686672372E+85}{3.3685043600011E+86} & \dfrac{2.8280407868084E+85}{3.3685043600011E+86} & \dfrac{9.3156430526577E+82}{8.4212609000027E+85} & \dfrac{1.8446246214844E+84}{1.122834786667E+86}\end{pmatrix}\times\begin{pmatrix}9 \\ 7 \\ 6 \\ 6 \\ -5 \\ 9\end{pmatrix}=\begin{pmatrix}\dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ \dfrac{NAN}{INF} \\ -\dfrac{INF}{INF} \\ \dfrac{NAN}{INF}\end{pmatrix}\) . Ainsi $ x=\dfrac{NAN}{INF}$ , $ y=\dfrac{NAN}{INF}$ , $ z=\dfrac{NAN}{INF}$ , $ t=\dfrac{NAN}{INF}$ , $ u=\dfrac{INF}{INF}$ et $ v=\dfrac{NAN}{INF}$
Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{4.5488628878915E+96}{9.1982625723763E+97}x&-&\dfrac{1.7920879855845E+92}{7.186142634669E+94}y &+&\dfrac{2.9801737734111E+98}{7.186142634669E+99}z &-&\dfrac{1.444487826481E+93}{1.4372285269338E+95}t &+&\dfrac{8.4231353848092E+93}{1.7965356586672E+95}u &+&\dfrac{2.9606017937115E+96}{7.186142634669E+97}v &=&4\\ &-\dfrac{1.233649551974E+99}{1.6556872630277E+100}x&+&\dfrac{4.9430548667038E+94}{1.149782821547E+96}y &+&\dfrac{2.5292501716699E+100}{1.2935056742404E+102}z &+&\dfrac{4.6311306454775E+92}{1.6168820928005E+94}t &+&\dfrac{8.5882095740126E+92}{1.6168820928005E+94}u &-&\dfrac{1.4873753580345E+93}{1.4372285269338E+95}v &=&-9\\ &\dfrac{1.6532904728712E+94}{2.8744570538676E+95}x&+&\dfrac{5.5049122004121E+84}{8.9826782933362E+86}y &+&\dfrac{7.1524637020086E+104}{1.3474017440004E+106}z &+&\dfrac{8.7977674288835E+91}{4.4913391466681E+93}t &+&\dfrac{1.4337082925916E+85}{1.122834786667E+87}u &-&\dfrac{2.3731767160298E+102}{4.4913391466681E+103}v &=&0\\ &-\dfrac{1.527841065918E+93}{8.2784363151387E+93}x&-&\dfrac{2.6190182255671E+84}{7.186142634669E+85}y &+&\dfrac{3.07553813278E+90}{5.1740226969617E+91}z &+&\dfrac{3.704110604529E+87}{3.233764185601E+89}t &-&\dfrac{2.1896878388822E+85}{1.3474017440004E+87}u &-&\dfrac{5.3742997821447E+92}{1.4372285269338E+94}v &=&8\\ &-\dfrac{1.5316860605658E+93}{3.233764185601E+94}x&-&\dfrac{4.0833343358247E+88}{8.9826782933362E+89}y &-&\dfrac{2.4173905341E+97}{4.0422052320013E+98}z &-&\dfrac{3.412061608661E+88}{8.0844104640026E+90}t &+&\dfrac{1.0037340969183E+93}{1.6168820928005E+94}u &-&\dfrac{6.90613558205E+91}{8.9826782933362E+92}v &=&8\\ &-\dfrac{7.5851265540482E+90}{5.1740226969617E+91}x&-&\dfrac{8.0752836335199E+84}{1.122834786667E+86}y &-&\dfrac{1.6648686672372E+85}{3.3685043600011E+86}z &+&\dfrac{2.8280407868084E+85}{3.3685043600011E+86}t &+&\dfrac{9.3156430526577E+82}{8.4212609000027E+85}u &+&\dfrac{1.8446246214844E+84}{1.122834786667E+86}v &=&7\\ \end{array} \right.\) 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}4 \\ -9 \\ 0 \\ 8 \\ 8 \\ 7\end{pmatrix}$ de sorte que la solution est $ X=B^{-1}b=Ab=\begin{pmatrix}1 & -\dfrac{13}{5} & 4 & -\dfrac{16}{5} & -\dfrac{1}{5} & \dfrac{1}{2} \\ -\dfrac{23}{3} & \dfrac{21}{2} & -2 & -3 & -\dfrac{7}{2} & -\dfrac{19}{5} \\ \dfrac{11}{2} & -1 & 5 & 5 & -3 & -1 \\ -3 & \dfrac{19}{5} & \dfrac{19}{2} & -5 & -\dfrac{22}{5} & \dfrac{17}{2} \\ \dfrac{17}{2} & 5 & -1 & -\dfrac{7}{5} & \dfrac{21}{4} & 0 \\ \dfrac{20}{3} & 0 & -\dfrac{13}{2} & -1 & -4 & \dfrac{9}{4}\end{pmatrix}\times\begin{pmatrix}4 \\ -9 \\ 0 \\ 8 \\ 8 \\ 7\end{pmatrix}=\begin{pmatrix}\dfrac{37}{10} \\ -\dfrac{6113}{30} \\ 40 \\ -\dfrac{619}{10} \\ \dfrac{99}{5} \\ \dfrac{29}{12}\end{pmatrix}$ . Ainsi $ x=\dfrac{37}{10}$ , $ y=\dfrac{6113}{30}$ , $ z=40$ , $ t=\dfrac{619}{10}$ , $ u=\dfrac{99}{5}$ et $ v=\dfrac{29}{12}$