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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 \[ A= \begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & 2 & -\dfrac{1}{2} & 4 \\ \dfrac{17}{3} & 2 & -2 & -\dfrac{16}{5} & -5 & 1 \\ -\dfrac{18}{5} & 3 & -2 & -3 & 4 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{6}{5} & -\dfrac{7}{4} & -2 & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -5 & -2 & 0 & -5 \\ -4 & -\dfrac{9}{2} & \dfrac{19}{3} & 0 & -2 & \dfrac{20}{3}\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (6, 5)\) et \( (2, 4)\) : \( \widehat{A}_{6, 5}=\) \( \widehat{A}_{2, 4}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & 2 & -\dfrac{1}{2} & 4 \\ 0 & \dfrac{64}{15} & -\dfrac{353}{15} & -\dfrac{218}{15} & -\dfrac{13}{6} & -\dfrac{65}{3} \\ 0 & \dfrac{39}{25} & \dfrac{292}{25} & \dfrac{21}{5} & \dfrac{11}{5} & \dfrac{77}{5} \\ 0 & -\dfrac{16}{5} & -\dfrac{82}{5} & -\dfrac{39}{4} & 0 & -\dfrac{62}{3} \\ 0 & -\dfrac{22}{15} & -\dfrac{379}{15} & -\dfrac{38}{3} & \dfrac{8}{3} & -\dfrac{79}{3} \\ 0 & -\dfrac{61}{10} & \dfrac{323}{15} & 8 & -4 & \dfrac{68}{3}\end{pmatrix} \)
  3. Calculer \( \det(A)\) .
  4. Pourquoi la matrice \( A \) est inversible.
  5. Donner \( B=A^{-1}\) l'inverse de la matrice \( A \) , en ne détaillant que le calcul de \( A^{-1}_{3, 2}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{2}{5}y &+&\dfrac{19}{5}z &+&2t &-&\dfrac{1}{2}u &+&4v &=&-8\\ &\dfrac{17}{3}x&+&2y &-&2z &-&\dfrac{16}{5}t &-&5u &+&v &=&7\\ &-\dfrac{18}{5}x&+&3y &-&2z &-&3t &+&4u &+&v &=&-7\\ &4x&-&\dfrac{24}{5}y &-&\dfrac{6}{5}z &-&\dfrac{7}{4}t &-&2u &-&\dfrac{14}{3}v &=&-9\\ &\dfrac{16}{3}x&-&\dfrac{18}{5}y &-&5z &-&2t &&&-&5v &=&-3\\ &-4x&-&\dfrac{9}{2}y &+&\dfrac{19}{3}z &&&-&2u &+&\dfrac{20}{3}v &=&-5\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{7.0891404581953E+27}{2.7883560751573E+28}x&+&\dfrac{3.0251740717969E+26}{1.1153424300629E+28}y &+&\dfrac{1.2883293904922E+27}{1.8589040501048E+28}z &+&\dfrac{4.1892605219531E+26}{4.6472601252621E+27}t &+&\dfrac{1.9524198171094E+26}{6.9708901878932E+27}u &-&\dfrac{2.5684300567969E+26}{3.0981734168414E+27}v &=&-\dfrac{1}{3}\\ &\dfrac{8.5234771125E+25}{2.0912670563679E+27}x&+&\dfrac{4.1166579311719E+26}{8.3650682254718E+27}y &+&\dfrac{2.5448442869531E+26}{5.5767121503145E+27}z &+&\dfrac{1.0225364632031E+26}{1.3941780375786E+27}t &-&\dfrac{8.027706331314E+31}{4.7053508768279E+32}u &-&\dfrac{7.1442062648438E+25}{6.1963468336828E+26}v &=&-6\\ &\dfrac{3.3500295311484E+27}{1.3941780375786E+28}x&-&\dfrac{1.0983928999219E+26}{5.5767121503145E+27}y &+&\dfrac{9.5195017856689E+30}{6.2738011691038E+31}z &+&\dfrac{3.1817198809907E+31}{7.8422514613798E+31}t &-&\dfrac{3.8998103470822E+34}{1.2547602338208E+35}u &-&\dfrac{7.0943086680996E+29}{6.2738011691038E+30}v &=&-8\\ &-\dfrac{7.2836956877344E+26}{1.045633528184E+28}x&-&\dfrac{7.5533943065625E+26}{1.045633528184E+28}y &-&\dfrac{1.6326525670313E+26}{6.9708901878932E+26}z &-&\dfrac{3.7547839771172E+27}{1.3941780375786E+28}t &+&\dfrac{2.7791022803203E+27}{2.0912670563679E+28}u &-&\dfrac{1.269861046875E+24}{1.1618150313155E+27}v &=&0\\ &\dfrac{4.5664243406922E+34}{1.7645065788105E+35}x&-&\dfrac{6.4724777578125E+24}{7.4356162004194E+25}y &+&\dfrac{1.391935565383E+36}{7.6226684204612E+36}z &+&\dfrac{1.053200424375E+26}{1.1618150313155E+27}t &+&\dfrac{8.3355813492188E+25}{1.8589040501048E+27}u &-&\dfrac{4.4432906546585E+34}{6.126758954203E+35}v &=&5\\ &\dfrac{1.4420469673883E+33}{4.9014071633624E+34}x&+&\dfrac{1.1736070249922E+27}{2.7883560751573E+28}y &-&\dfrac{9.9870994782806E+36}{5.8816885960349E+38}z &-&\dfrac{4.9923846494626E+32}{1.960562865345E+33}t &+&\dfrac{1.374734142011E+32}{6.5352095511498E+32}u &+&\dfrac{1.6745815089141E+27}{1.5490867084207E+28}v &=&-7\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{6, 5}=\begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & 2 & 4 \\ \dfrac{17}{3} & 2 & -2 & -\dfrac{16}{5} & 1 \\ -\dfrac{18}{5} & 3 & -2 & -3 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{6}{5} & -\dfrac{7}{4} & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -5 & -2 & -5\end{pmatrix}\) \( \widehat{A}_{2, 4}=\begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & -\dfrac{1}{2} & 4 \\ -\dfrac{18}{5} & 3 & -2 & 4 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{6}{5} & -2 & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -5 & 0 & -5 \\ -4 & -\dfrac{9}{2} & \dfrac{19}{3} & -2 & \dfrac{20}{3}\end{pmatrix}\)
  2. 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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{17}{3}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{18}{5}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(4\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(\dfrac{16}{3}\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-4\right)L_1\)
  3. 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{2}{5} & \dfrac{19}{5} & 2 & -\dfrac{1}{2} & 4 \\ 0 & \dfrac{64}{15} & -\dfrac{353}{15} & -\dfrac{218}{15} & -\dfrac{13}{6} & -\dfrac{65}{3} \\ 0 & \dfrac{39}{25} & \dfrac{292}{25} & \dfrac{21}{5} & \dfrac{11}{5} & \dfrac{77}{5} \\ 0 & -\dfrac{16}{5} & -\dfrac{82}{5} & -\dfrac{39}{4} & 0 & -\dfrac{62}{3} \\ 0 & -\dfrac{22}{15} & -\dfrac{379}{15} & -\dfrac{38}{3} & \dfrac{8}{3} & -\dfrac{79}{3} \\ 0 & -\dfrac{61}{10} & \dfrac{323}{15} & 8 & -4 & \dfrac{68}{3}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{64}{15} & -\dfrac{353}{15} & -\dfrac{218}{15} & -\dfrac{13}{6} & -\dfrac{65}{3} \\ \dfrac{39}{25} & \dfrac{292}{25} & \dfrac{21}{5} & \dfrac{11}{5} & \dfrac{77}{5} \\ -\dfrac{16}{5} & -\dfrac{82}{5} & -\dfrac{39}{4} & 0 & -\dfrac{62}{3} \\ -\dfrac{22}{15} & -\dfrac{379}{15} & -\dfrac{38}{3} & \dfrac{8}{3} & -\dfrac{79}{3} \\ -\dfrac{61}{10} & \dfrac{323}{15} & 8 & -4 & \dfrac{68}{3}\end{pmatrix}\\ &=&-\dfrac{3.0981734168414E+24}{1.537734375E+20} \end{eqnarray*}
  4. On observe que le déterminant de \( A\) est non nul. D'après le cours, cela signifie que la matrice est inversible.
  5. D'après le cours \( B=A^{-1}=\left(-\dfrac{3.0981734168414E+24}{1.537734375E+20}\right)^{-1}{}^tCo\begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & 2 & -\dfrac{1}{2} & 4 \\ \dfrac{17}{3} & 2 & -2 & -\dfrac{16}{5} & -5 & 1 \\ -\dfrac{18}{5} & 3 & -2 & -3 & 4 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{6}{5} & -\dfrac{7}{4} & -2 & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -5 & -2 & 0 & -5 \\ -4 & -\dfrac{9}{2} & \dfrac{19}{3} & 0 & -2 & \dfrac{20}{3}\end{pmatrix} =-\dfrac{1.537734375E+20}{3.0981734168414E+24}\begin{pmatrix}-\dfrac{2.1963386515836E+52}{4.2877509865094E+48} & -\dfrac{9.372513890559E+50}{1.7151003946038E+48} & -\dfrac{3.9914678697584E+51}{2.8585006576729E+48} & -\dfrac{1.2979055585338E+51}{7.1462516441824E+47} & -\dfrac{6.0489351758826E+50}{1.0719377466274E+48} & \dfrac{7.9574417249845E+50}{4.7641677627882E+47} \\ -\dfrac{2.6407210209004E+50}{3.2158132398821E+47} & -\dfrac{1.2754120168586E+51}{1.2863252959528E+48} & -\dfrac{7.8843689198389E+50}{8.5755019730188E+47} & -\dfrac{3.167995288047E+50}{2.1438754932547E+47} & \dfrac{2.4871226353886E+56}{7.2355797897346E+52} & \dfrac{2.2133989934171E+50}{9.5283355255765E+46} \\ -\dfrac{1.0378972439038E+52}{2.1438754932547E+48} & \dfrac{3.4030116837853E+50}{8.5755019730188E+47} & -\dfrac{2.9493067373934E+55}{9.6474397196462E+51} & -\dfrac{9.8575199551213E+55}{1.2059299649558E+52} & \dfrac{1.2082288748053E+59}{1.9294879439292E+55} & \dfrac{2.1979398526374E+54}{9.6474397196462E+50} \\ \dfrac{2.2566152356101E+51}{1.607906619941E+48} & \dfrac{2.3401725447513E+51}{1.607906619941E+48} & \dfrac{5.0582407821141E+50}{1.0719377466274E+47} & \dfrac{1.1632971903887E+52}{2.1438754932547E+48} & -\dfrac{8.6101408075717E+51}{3.2158132398821E+48} & \dfrac{3.9342497385105E+48}{1.7865629110456E+47} \\ -\dfrac{1.414757450235E+59}{2.7133424211505E+55} & \dfrac{2.0052858530352E+49}{1.1434002630692E+46} & -\dfrac{4.3124577666256E+60}{1.172163925937E+57} & -\dfrac{3.2629975574047E+50}{1.7865629110456E+47} & -\dfrac{2.5825076550069E+50}{2.8585006576729E+47} & \dfrac{1.3766084989563E+59}{9.421327851217E+55} \\ -\dfrac{4.4677115801992E+57}{7.5370622809736E+54} & -\dfrac{3.6360380866491E+51}{4.2877509865094E+48} & \dfrac{3.094176611496E+61}{9.0444747371683E+58} & \dfrac{1.5467273407612E+57}{3.0148249123894E+53} & -\dfrac{4.2591647740027E+56}{1.0049416374631E+53} & -\dfrac{5.1881439152517E+51}{2.3820838813941E+48}\end{pmatrix} =\begin{pmatrix}\dfrac{7.0891404581953E+27}{2.7883560751573E+28} & \dfrac{3.0251740717969E+26}{1.1153424300629E+28} & \dfrac{1.2883293904922E+27}{1.8589040501048E+28} & \dfrac{4.1892605219531E+26}{4.6472601252621E+27} & \dfrac{1.9524198171094E+26}{6.9708901878932E+27} & -\dfrac{2.5684300567969E+26}{3.0981734168414E+27} \\ \dfrac{8.5234771125E+25}{2.0912670563679E+27} & \dfrac{4.1166579311719E+26}{8.3650682254718E+27} & \dfrac{2.5448442869531E+26}{5.5767121503145E+27} & \dfrac{1.0225364632031E+26}{1.3941780375786E+27} & -\dfrac{8.027706331314E+31}{4.7053508768279E+32} & -\dfrac{7.1442062648438E+25}{6.1963468336828E+26} \\ \dfrac{3.3500295311484E+27}{1.3941780375786E+28} & -\dfrac{1.0983928999219E+26}{5.5767121503145E+27} & \dfrac{9.5195017856689E+30}{6.2738011691038E+31} & \dfrac{3.1817198809907E+31}{7.8422514613798E+31} & -\dfrac{3.8998103470822E+34}{1.2547602338208E+35} & -\dfrac{7.0943086680996E+29}{6.2738011691038E+30} \\ -\dfrac{7.2836956877344E+26}{1.045633528184E+28} & -\dfrac{7.5533943065625E+26}{1.045633528184E+28} & -\dfrac{1.6326525670313E+26}{6.9708901878932E+26} & -\dfrac{3.7547839771172E+27}{1.3941780375786E+28} & \dfrac{2.7791022803203E+27}{2.0912670563679E+28} & -\dfrac{1.269861046875E+24}{1.1618150313155E+27} \\ \dfrac{4.5664243406922E+34}{1.7645065788105E+35} & -\dfrac{6.4724777578125E+24}{7.4356162004194E+25} & \dfrac{1.391935565383E+36}{7.6226684204612E+36} & \dfrac{1.053200424375E+26}{1.1618150313155E+27} & \dfrac{8.3355813492188E+25}{1.8589040501048E+27} & -\dfrac{4.4432906546585E+34}{6.126758954203E+35} \\ \dfrac{1.4420469673883E+33}{4.9014071633624E+34} & \dfrac{1.1736070249922E+27}{2.7883560751573E+28} & -\dfrac{9.9870994782806E+36}{5.8816885960349E+38} & -\dfrac{4.9923846494626E+32}{1.960562865345E+33} & \dfrac{1.374734142011E+32}{6.5352095511498E+32} & \dfrac{1.6745815089141E+27}{1.5490867084207E+28}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{3, 2}=B_{3, 2}= \left(-\dfrac{3.0981734168414E+24}{1.537734375E+20}\right)^{-1}Co(A)_{2, 3}= \left(-\dfrac{3.0981734168414E+24}{1.537734375E+20}\right)^{-1}\times(-1)^{2+3}\det\begin{pmatrix}1 & -\dfrac{2}{5} & 2 & -\dfrac{1}{2} & 4 \\ -\dfrac{18}{5} & 3 & -3 & 4 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{7}{4} & -2 & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -2 & 0 & -5 \\ -4 & -\dfrac{9}{2} & 0 & -2 & \dfrac{20}{3}\end{pmatrix}=-\dfrac{1.0983928999219E+26}{5.5767121503145E+27}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & 2 & -\dfrac{1}{2} & 4 \\ \dfrac{17}{3} & 2 & -2 & -\dfrac{16}{5} & -5 & 1 \\ -\dfrac{18}{5} & 3 & -2 & -3 & 4 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{6}{5} & -\dfrac{7}{4} & -2 & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -5 & -2 & 0 & -5 \\ -4 & -\dfrac{9}{2} & \dfrac{19}{3} & 0 & -2 & \dfrac{20}{3}\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{2}{5}y &+&\dfrac{19}{5}z &+&2t &-&\dfrac{1}{2}u &+&4v &=&-8\\ &\dfrac{17}{3}x&+&2y &-&2z &-&\dfrac{16}{5}t &-&5u &+&v &=&7\\ &-\dfrac{18}{5}x&+&3y &-&2z &-&3t &+&4u &+&v &=&-7\\ &4x&-&\dfrac{24}{5}y &-&\dfrac{6}{5}z &-&\dfrac{7}{4}t &-&2u &-&\dfrac{14}{3}v &=&-9\\ &\dfrac{16}{3}x&-&\dfrac{18}{5}y &-&5z &-&2t &&&-&5v &=&-3\\ &-4x&-&\dfrac{9}{2}y &+&\dfrac{19}{3}z &&&-&2u &+&\dfrac{20}{3}v &=&-5\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-8 \\ 7 \\ -7 \\ -9 \\ -3 \\ -5\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{7.0891404581953E+27}{2.7883560751573E+28} & \dfrac{3.0251740717969E+26}{1.1153424300629E+28} & \dfrac{1.2883293904922E+27}{1.8589040501048E+28} & \dfrac{4.1892605219531E+26}{4.6472601252621E+27} & \dfrac{1.9524198171094E+26}{6.9708901878932E+27} & -\dfrac{2.5684300567969E+26}{3.0981734168414E+27} \\ \dfrac{8.5234771125E+25}{2.0912670563679E+27} & \dfrac{4.1166579311719E+26}{8.3650682254718E+27} & \dfrac{2.5448442869531E+26}{5.5767121503145E+27} & \dfrac{1.0225364632031E+26}{1.3941780375786E+27} & -\dfrac{8.027706331314E+31}{4.7053508768279E+32} & -\dfrac{7.1442062648438E+25}{6.1963468336828E+26} \\ \dfrac{3.3500295311484E+27}{1.3941780375786E+28} & -\dfrac{1.0983928999219E+26}{5.5767121503145E+27} & \dfrac{9.5195017856689E+30}{6.2738011691038E+31} & \dfrac{3.1817198809907E+31}{7.8422514613798E+31} & -\dfrac{3.8998103470822E+34}{1.2547602338208E+35} & -\dfrac{7.0943086680996E+29}{6.2738011691038E+30} \\ -\dfrac{7.2836956877344E+26}{1.045633528184E+28} & -\dfrac{7.5533943065625E+26}{1.045633528184E+28} & -\dfrac{1.6326525670313E+26}{6.9708901878932E+26} & -\dfrac{3.7547839771172E+27}{1.3941780375786E+28} & \dfrac{2.7791022803203E+27}{2.0912670563679E+28} & -\dfrac{1.269861046875E+24}{1.1618150313155E+27} \\ \dfrac{4.5664243406922E+34}{1.7645065788105E+35} & -\dfrac{6.4724777578125E+24}{7.4356162004194E+25} & \dfrac{1.391935565383E+36}{7.6226684204612E+36} & \dfrac{1.053200424375E+26}{1.1618150313155E+27} & \dfrac{8.3355813492188E+25}{1.8589040501048E+27} & -\dfrac{4.4432906546585E+34}{6.126758954203E+35} \\ \dfrac{1.4420469673883E+33}{4.9014071633624E+34} & \dfrac{1.1736070249922E+27}{2.7883560751573E+28} & -\dfrac{9.9870994782806E+36}{5.8816885960349E+38} & -\dfrac{4.9923846494626E+32}{1.960562865345E+33} & \dfrac{1.374734142011E+32}{6.5352095511498E+32} & \dfrac{1.6745815089141E+27}{1.5490867084207E+28}\end{pmatrix}\times\begin{pmatrix}-8 \\ 7 \\ -7 \\ -9 \\ -3 \\ -5\end{pmatrix}=\begin{pmatrix}-\dfrac{1.6304765687851E+168}{5.8023560821861E+167} \\ \dfrac{5.0447000834756E+168}{3.9655477349191E+169} \\ -\dfrac{1.5887664836042E+186}{3.0113378112042E+185} \\ \dfrac{9.6085416341803E+166}{2.5817368065879E+166} \\ -\dfrac{6.0154891951396E+188}{1.3233418115643E+188} \\ \dfrac{2.0713932450757E+196}{1.595464182538E+196}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.6304765687851E+168}{5.8023560821861E+167}\) , \( y=\dfrac{5.0447000834756E+168}{3.9655477349191E+169}\) , \( z=\dfrac{1.5887664836042E+186}{3.0113378112042E+185}\) , \( t=\dfrac{9.6085416341803E+166}{2.5817368065879E+166}\) , \( u=\dfrac{6.0154891951396E+188}{1.3233418115643E+188}\) et \( v=\dfrac{2.0713932450757E+196}{1.595464182538E+196}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{7.0891404581953E+27}{2.7883560751573E+28}x&+&\dfrac{3.0251740717969E+26}{1.1153424300629E+28}y &+&\dfrac{1.2883293904922E+27}{1.8589040501048E+28}z &+&\dfrac{4.1892605219531E+26}{4.6472601252621E+27}t &+&\dfrac{1.9524198171094E+26}{6.9708901878932E+27}u &-&\dfrac{2.5684300567969E+26}{3.0981734168414E+27}v &=&-\dfrac{1}{3}\\ &\dfrac{8.5234771125E+25}{2.0912670563679E+27}x&+&\dfrac{4.1166579311719E+26}{8.3650682254718E+27}y &+&\dfrac{2.5448442869531E+26}{5.5767121503145E+27}z &+&\dfrac{1.0225364632031E+26}{1.3941780375786E+27}t &-&\dfrac{8.027706331314E+31}{4.7053508768279E+32}u &-&\dfrac{7.1442062648438E+25}{6.1963468336828E+26}v &=&-6\\ &\dfrac{3.3500295311484E+27}{1.3941780375786E+28}x&-&\dfrac{1.0983928999219E+26}{5.5767121503145E+27}y &+&\dfrac{9.5195017856689E+30}{6.2738011691038E+31}z &+&\dfrac{3.1817198809907E+31}{7.8422514613798E+31}t &-&\dfrac{3.8998103470822E+34}{1.2547602338208E+35}u &-&\dfrac{7.0943086680996E+29}{6.2738011691038E+30}v &=&-8\\ &-\dfrac{7.2836956877344E+26}{1.045633528184E+28}x&-&\dfrac{7.5533943065625E+26}{1.045633528184E+28}y &-&\dfrac{1.6326525670313E+26}{6.9708901878932E+26}z &-&\dfrac{3.7547839771172E+27}{1.3941780375786E+28}t &+&\dfrac{2.7791022803203E+27}{2.0912670563679E+28}u &-&\dfrac{1.269861046875E+24}{1.1618150313155E+27}v &=&0\\ &\dfrac{4.5664243406922E+34}{1.7645065788105E+35}x&-&\dfrac{6.4724777578125E+24}{7.4356162004194E+25}y &+&\dfrac{1.391935565383E+36}{7.6226684204612E+36}z &+&\dfrac{1.053200424375E+26}{1.1618150313155E+27}t &+&\dfrac{8.3355813492188E+25}{1.8589040501048E+27}u &-&\dfrac{4.4432906546585E+34}{6.126758954203E+35}v &=&5\\ &\dfrac{1.4420469673883E+33}{4.9014071633624E+34}x&+&\dfrac{1.1736070249922E+27}{2.7883560751573E+28}y &-&\dfrac{9.9870994782806E+36}{5.8816885960349E+38}z &-&\dfrac{4.9923846494626E+32}{1.960562865345E+33}t &+&\dfrac{1.374734142011E+32}{6.5352095511498E+32}u &+&\dfrac{1.6745815089141E+27}{1.5490867084207E+28}v &=&-7\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-\dfrac{1}{3} \\ -6 \\ -8 \\ 0 \\ 5 \\ -7\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -\dfrac{2}{5} & \dfrac{19}{5} & 2 & -\dfrac{1}{2} & 4 \\ \dfrac{17}{3} & 2 & -2 & -\dfrac{16}{5} & -5 & 1 \\ -\dfrac{18}{5} & 3 & -2 & -3 & 4 & 1 \\ 4 & -\dfrac{24}{5} & -\dfrac{6}{5} & -\dfrac{7}{4} & -2 & -\dfrac{14}{3} \\ \dfrac{16}{3} & -\dfrac{18}{5} & -5 & -2 & 0 & -5 \\ -4 & -\dfrac{9}{2} & \dfrac{19}{3} & 0 & -2 & \dfrac{20}{3}\end{pmatrix}\times\begin{pmatrix}-\dfrac{1}{3} \\ -6 \\ -8 \\ 0 \\ 5 \\ -7\end{pmatrix}=\begin{pmatrix}-\dfrac{353}{6} \\ -\dfrac{269}{9} \\ \dfrac{61}{5} \\ \dfrac{896}{15} \\ \dfrac{4267}{45} \\ -79\end{pmatrix}\) . Ainsi \( x=\dfrac{353}{6}\) , \( y=\dfrac{269}{9}\) , \( z=\dfrac{61}{5}\) , \( t=\dfrac{896}{15}\) , \( u=\dfrac{4267}{45}\) et \( v=79\)