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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 \[ A= \begin{pmatrix}1 & 0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 1 & -1 & 4 & 4 & -\dfrac{11}{3} & -\dfrac{23}{3} \\ -2 & 4 & 4 & 1 & 2 & \dfrac{9}{5} \\ -1 & -2 & -1 & -\dfrac{8}{5} & -3 & -\dfrac{3}{5} \\ -4 & \dfrac{3}{4} & 2 & \dfrac{18}{5} & -\dfrac{17}{5} & -\dfrac{22}{5} \\ \dfrac{25}{2} & -2 & 0 & 2 & -5 & \dfrac{21}{4}\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (2, 4)\) et \( (2, 1)\) : \( \widehat{A}_{2, 4}=\) \( \widehat{A}_{2, 1}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & 0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 0 & -1 & 6 & 6 & \dfrac{1}{3} & -13 \\ 0 & 4 & 0 & -3 & -6 & \dfrac{187}{15} \\ 0 & -2 & -3 & -\dfrac{18}{5} & -7 & \dfrac{71}{15} \\ 0 & \dfrac{3}{4} & -6 & -\dfrac{22}{5} & -\dfrac{97}{5} & \dfrac{254}{15} \\ 0 & -2 & 25 & 27 & 45 & -\dfrac{737}{12}\end{pmatrix} \)
  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 \( A^{-1}_{1, 1}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&&&-&2z &-&2t &-&4u &+&\dfrac{16}{3}v &=&0\\ &x&-&y &+&4z &+&4t &-&\dfrac{11}{3}u &-&\dfrac{23}{3}v &=&-2\\ &-2x&+&4y &+&4z &+&t &+&2u &+&\dfrac{9}{5}v &=&-9\\ &-x&-&2y &-&z &-&\dfrac{8}{5}t &-&3u &-&\dfrac{3}{5}v &=&0\\ &-4x&+&\dfrac{3}{4}y &+&2z &+&\dfrac{18}{5}t &-&\dfrac{17}{5}u &-&\dfrac{22}{5}v &=&-1\\ &\dfrac{25}{2}x&-&2y &&&+&2t &-&5u &+&\dfrac{21}{4}v &=&-5\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{9.8045580708E+22}{3.85566399972E+23}x&+&\dfrac{1.6448715684E+22}{6.4261066662E+22}y &-&\dfrac{1.310091732E+21}{1.28522133324E+22}z &-&\dfrac{3.9309216732E+22}{1.156699199916E+23}t &-&\dfrac{1.293732972E+21}{5.78349599958E+21}u &-&\dfrac{1.0974094812E+22}{1.445873999895E+23}v &=&-8\\ &\dfrac{1.7718960276E+22}{1.92783199986E+22}x&+&\dfrac{1.2256464132E+22}{1.92783199986E+22}y &-&\dfrac{3.586435056E+21}{1.445873999895E+22}z &-&\dfrac{1.5688173312E+22}{1.445873999895E+22}t &-&\dfrac{1.537784676E+21}{3.85566399972E+21}u &-&\dfrac{9.121250916E+21}{2.409789999825E+22}v &=&1\\ &-\dfrac{9.303401772E+21}{2.409789999825E+22}x&-&\dfrac{5.8371476148E+22}{3.85566399972E+23}y &+&\dfrac{1.311561396E+21}{3.85566399972E+21}z &+&\dfrac{1.41574194036E+23}{2.313398399832E+23}t &+&\dfrac{2.72298996E+20}{6.4261066662E+21}u &+&\dfrac{1.1562703992E+22}{7.229369999475E+22}v &=&-\dfrac{33}{7}\\ &-\dfrac{1.5649760844E+22}{3.85566399972E+22}x&-&\dfrac{5.056247772E+21}{1.28522133324E+22}y &+&\dfrac{3.082244076E+21}{2.313398399832E+23}z &+&\dfrac{4.856600916E+21}{2.89174799979E+22}t &+&\dfrac{2.534706756E+21}{5.78349599958E+21}u &+&\dfrac{1.412180892E+21}{6.4261066662E+21}v &=&2\\ &-\dfrac{2.1518566596E+22}{7.71132799944E+22}x&-&\dfrac{1.3305121884E+22}{7.71132799944E+22}y &+&\dfrac{3.25311876E+20}{7.71132799944E+21}z &+&\dfrac{4.643517132E+21}{3.85566399972E+22}t &+&\dfrac{2.69079732E+20}{6.4261066662E+21}u &+&\dfrac{4.23551916E+20}{6.4261066662E+21}v &=&3\\ &-\dfrac{1.4129148492E+22}{3.85566399972E+22}x&-&\dfrac{2.45285172E+20}{6.4261066662E+20}y &+&\dfrac{3.533920812E+21}{1.92783199986E+22}z &+&\dfrac{1.721615148E+21}{3.85566399972E+21}t &+&\dfrac{8.1478872E+19}{3.2130533331E+20}u &+&\dfrac{1.324106028E+21}{6.4261066662E+21}v &=&-9\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{2, 4}=\begin{pmatrix}1 & 0 & -2 & -4 & \dfrac{16}{3} \\ -2 & 4 & 4 & 2 & \dfrac{9}{5} \\ -1 & -2 & -1 & -3 & -\dfrac{3}{5} \\ -4 & \dfrac{3}{4} & 2 & -\dfrac{17}{5} & -\dfrac{22}{5} \\ \dfrac{25}{2} & -2 & 0 & -5 & \dfrac{21}{4}\end{pmatrix}\) \( \widehat{A}_{2, 1}=\begin{pmatrix}0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 4 & 4 & 1 & 2 & \dfrac{9}{5} \\ -2 & -1 & -\dfrac{8}{5} & -3 & -\dfrac{3}{5} \\ \dfrac{3}{4} & 2 & \dfrac{18}{5} & -\dfrac{17}{5} & -\dfrac{22}{5} \\ -2 & 0 & 2 & -5 & \dfrac{21}{4}\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(1\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-2\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(-1\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-4\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(\dfrac{25}{2}\right)L_1\)
  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 & 0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 0 & -1 & 6 & 6 & \dfrac{1}{3} & -13 \\ 0 & 4 & 0 & -3 & -6 & \dfrac{187}{15} \\ 0 & -2 & -3 & -\dfrac{18}{5} & -7 & \dfrac{71}{15} \\ 0 & \dfrac{3}{4} & -6 & -\dfrac{22}{5} & -\dfrac{97}{5} & \dfrac{254}{15} \\ 0 & -2 & 25 & 27 & 45 & -\dfrac{737}{12}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-1 & 6 & 6 & \dfrac{1}{3} & -13 \\ 4 & 0 & -3 & -6 & \dfrac{187}{15} \\ -2 & -3 & -\dfrac{18}{5} & -7 & \dfrac{71}{15} \\ \dfrac{3}{4} & -6 & -\dfrac{22}{5} & -\dfrac{97}{5} & \dfrac{254}{15} \\ -2 & 25 & 27 & 45 & -\dfrac{737}{12}\end{pmatrix}\\ &=&-\dfrac{6.4261066662E+19}{8748000000000000} \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{6.4261066662E+19}{8748000000000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & 0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 1 & -1 & 4 & 4 & -\dfrac{11}{3} & -\dfrac{23}{3} \\ -2 & 4 & 4 & 1 & 2 & \dfrac{9}{5} \\ -1 & -2 & -1 & -\dfrac{8}{5} & -3 & -\dfrac{3}{5} \\ -4 & \dfrac{3}{4} & 2 & \dfrac{18}{5} & -\dfrac{17}{5} & -\dfrac{22}{5} \\ \dfrac{25}{2} & -2 & 0 & 2 & -5 & \dfrac{21}{4}\end{pmatrix} =-\dfrac{8748000000000000}{6.4261066662E+19}\begin{pmatrix}-\dfrac{6.3005135977913E+42}{3.3729348669551E+39} & -\dfrac{1.0570120150738E+42}{5.6215581115918E+38} & \dfrac{8.4187892123387E+40}{1.1243116223184E+38} & \dfrac{2.5260521968461E+42}{1.0118804600865E+39} & \dfrac{8.3136660756519E+40}{5.0594023004326E+37} & \dfrac{7.0520703826904E+41}{1.2648505751081E+39} \\ -\dfrac{1.1386392874774E+42}{1.6864674334775E+38} & -\dfrac{7.8761345862686E+41}{1.6864674334775E+38} & \dfrac{2.3046814221255E+41}{1.2648505751081E+38} & \dfrac{1.0081387510074E+42}{1.2648505751081E+38} & \dfrac{9.8819683576238E+40}{3.3729348669551E+37} & \dfrac{5.861413131539E+41}{2.1080842918469E+38} \\ \dfrac{5.9784652145386E+41}{2.1080842918469E+38} & \dfrac{3.751013319906E+42}{3.3729348669551E+39} & -\dfrac{8.4282334299662E+40}{3.3729348669551E+37} & -\dfrac{9.0977087205663E+42}{2.023760920173E+39} & -\dfrac{1.7498223933952E+40}{5.6215581115918E+37} & -\dfrac{7.4303169202289E+41}{6.3242528755407E+38} \\ \dfrac{1.0056703248406E+42}{3.3729348669551E+38} & \dfrac{3.2491987513608E+41}{1.1243116223184E+38} & -\dfrac{1.9806829203639E+41}{2.023760920173E+39} & -\dfrac{3.1209035521381E+41}{2.5297011502163E+38} & -\dfrac{1.6288295981594E+41}{5.0594023004326E+37} & -\dfrac{9.0748250439615E+40}{5.6215581115918E+37} \\ \dfrac{1.3828060424962E+42}{6.7458697339101E+38} & \dfrac{8.5500132433376E+41}{6.7458697339101E+38} & -\dfrac{2.0904888149576E+40}{6.7458697339101E+37} & -\dfrac{2.9839736396559E+41}{3.3729348669551E+38} & -\dfrac{1.7291350595445E+40}{5.6215581115918E+37} & -\dfrac{2.7217897908894E+40}{5.6215581115918E+37} \\ \dfrac{9.0795415312171E+41}{3.3729348669551E+38} & \dfrac{1.5762286789092E+40}{5.6215581115918E+36} & -\dfrac{2.2709352087816E+41}{1.6864674334775E+38} & -\dfrac{1.1063282579194E+41}{3.3729348669551E+37} & -\dfrac{5.2359192251366E+39}{2.8107790557959E+36} & -\dfrac{8.5088465732864E+40}{5.6215581115918E+37}\end{pmatrix} =\begin{pmatrix}\dfrac{9.8045580708E+22}{3.85566399972E+23} & \dfrac{1.6448715684E+22}{6.4261066662E+22} & -\dfrac{1.310091732E+21}{1.28522133324E+22} & -\dfrac{3.9309216732E+22}{1.156699199916E+23} & -\dfrac{1.293732972E+21}{5.78349599958E+21} & -\dfrac{1.0974094812E+22}{1.445873999895E+23} \\ \dfrac{1.7718960276E+22}{1.92783199986E+22} & \dfrac{1.2256464132E+22}{1.92783199986E+22} & -\dfrac{3.586435056E+21}{1.445873999895E+22} & -\dfrac{1.5688173312E+22}{1.445873999895E+22} & -\dfrac{1.537784676E+21}{3.85566399972E+21} & -\dfrac{9.121250916E+21}{2.409789999825E+22} \\ -\dfrac{9.303401772E+21}{2.409789999825E+22} & -\dfrac{5.8371476148E+22}{3.85566399972E+23} & \dfrac{1.311561396E+21}{3.85566399972E+21} & \dfrac{1.41574194036E+23}{2.313398399832E+23} & \dfrac{2.72298996E+20}{6.4261066662E+21} & \dfrac{1.1562703992E+22}{7.229369999475E+22} \\ -\dfrac{1.5649760844E+22}{3.85566399972E+22} & -\dfrac{5.056247772E+21}{1.28522133324E+22} & \dfrac{3.082244076E+21}{2.313398399832E+23} & \dfrac{4.856600916E+21}{2.89174799979E+22} & \dfrac{2.534706756E+21}{5.78349599958E+21} & \dfrac{1.412180892E+21}{6.4261066662E+21} \\ -\dfrac{2.1518566596E+22}{7.71132799944E+22} & -\dfrac{1.3305121884E+22}{7.71132799944E+22} & \dfrac{3.25311876E+20}{7.71132799944E+21} & \dfrac{4.643517132E+21}{3.85566399972E+22} & \dfrac{2.69079732E+20}{6.4261066662E+21} & \dfrac{4.23551916E+20}{6.4261066662E+21} \\ -\dfrac{1.4129148492E+22}{3.85566399972E+22} & -\dfrac{2.45285172E+20}{6.4261066662E+20} & \dfrac{3.533920812E+21}{1.92783199986E+22} & \dfrac{1.721615148E+21}{3.85566399972E+21} & \dfrac{8.1478872E+19}{3.2130533331E+20} & \dfrac{1.324106028E+21}{6.4261066662E+21}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{1, 1}=B_{1, 1}= \left(-\dfrac{6.4261066662E+19}{8748000000000000}\right)^{-1}Co(A)_{1, 1}= \left(-\dfrac{6.4261066662E+19}{8748000000000000}\right)^{-1}\times(-1)^{1+1}\det\begin{pmatrix}-1 & 4 & 4 & -\dfrac{11}{3} & -\dfrac{23}{3} \\ 4 & 4 & 1 & 2 & \dfrac{9}{5} \\ -2 & -1 & -\dfrac{8}{5} & -3 & -\dfrac{3}{5} \\ \dfrac{3}{4} & 2 & \dfrac{18}{5} & -\dfrac{17}{5} & -\dfrac{22}{5} \\ -2 & 0 & 2 & -5 & \dfrac{21}{4}\end{pmatrix}=\dfrac{9.8045580708E+22}{3.85566399972E+23}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 1 & -1 & 4 & 4 & -\dfrac{11}{3} & -\dfrac{23}{3} \\ -2 & 4 & 4 & 1 & 2 & \dfrac{9}{5} \\ -1 & -2 & -1 & -\dfrac{8}{5} & -3 & -\dfrac{3}{5} \\ -4 & \dfrac{3}{4} & 2 & \dfrac{18}{5} & -\dfrac{17}{5} & -\dfrac{22}{5} \\ \dfrac{25}{2} & -2 & 0 & 2 & -5 & \dfrac{21}{4}\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&&&-&2z &-&2t &-&4u &+&\dfrac{16}{3}v &=&0\\ &x&-&y &+&4z &+&4t &-&\dfrac{11}{3}u &-&\dfrac{23}{3}v &=&-2\\ &-2x&+&4y &+&4z &+&t &+&2u &+&\dfrac{9}{5}v &=&-9\\ &-x&-&2y &-&z &-&\dfrac{8}{5}t &-&3u &-&\dfrac{3}{5}v &=&0\\ &-4x&+&\dfrac{3}{4}y &+&2z &+&\dfrac{18}{5}t &-&\dfrac{17}{5}u &-&\dfrac{22}{5}v &=&-1\\ &\dfrac{25}{2}x&-&2y &&&+&2t &-&5u &+&\dfrac{21}{4}v &=&-5\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}0 \\ -2 \\ -9 \\ 0 \\ -1 \\ -5\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{9.8045580708E+22}{3.85566399972E+23} & \dfrac{1.6448715684E+22}{6.4261066662E+22} & -\dfrac{1.310091732E+21}{1.28522133324E+22} & -\dfrac{3.9309216732E+22}{1.156699199916E+23} & -\dfrac{1.293732972E+21}{5.78349599958E+21} & -\dfrac{1.0974094812E+22}{1.445873999895E+23} \\ \dfrac{1.7718960276E+22}{1.92783199986E+22} & \dfrac{1.2256464132E+22}{1.92783199986E+22} & -\dfrac{3.586435056E+21}{1.445873999895E+22} & -\dfrac{1.5688173312E+22}{1.445873999895E+22} & -\dfrac{1.537784676E+21}{3.85566399972E+21} & -\dfrac{9.121250916E+21}{2.409789999825E+22} \\ -\dfrac{9.303401772E+21}{2.409789999825E+22} & -\dfrac{5.8371476148E+22}{3.85566399972E+23} & \dfrac{1.311561396E+21}{3.85566399972E+21} & \dfrac{1.41574194036E+23}{2.313398399832E+23} & \dfrac{2.72298996E+20}{6.4261066662E+21} & \dfrac{1.1562703992E+22}{7.229369999475E+22} \\ -\dfrac{1.5649760844E+22}{3.85566399972E+22} & -\dfrac{5.056247772E+21}{1.28522133324E+22} & \dfrac{3.082244076E+21}{2.313398399832E+23} & \dfrac{4.856600916E+21}{2.89174799979E+22} & \dfrac{2.534706756E+21}{5.78349599958E+21} & \dfrac{1.412180892E+21}{6.4261066662E+21} \\ -\dfrac{2.1518566596E+22}{7.71132799944E+22} & -\dfrac{1.3305121884E+22}{7.71132799944E+22} & \dfrac{3.25311876E+20}{7.71132799944E+21} & \dfrac{4.643517132E+21}{3.85566399972E+22} & \dfrac{2.69079732E+20}{6.4261066662E+21} & \dfrac{4.23551916E+20}{6.4261066662E+21} \\ -\dfrac{1.4129148492E+22}{3.85566399972E+22} & -\dfrac{2.45285172E+20}{6.4261066662E+20} & \dfrac{3.533920812E+21}{1.92783199986E+22} & \dfrac{1.721615148E+21}{3.85566399972E+21} & \dfrac{8.1478872E+19}{3.2130533331E+20} & \dfrac{1.324106028E+21}{6.4261066662E+21}\end{pmatrix}\times\begin{pmatrix}0 \\ -2 \\ -9 \\ 0 \\ -1 \\ -5\end{pmatrix}=\begin{pmatrix}\dfrac{6.9662173040248E+89}{6.9063207361123E+89} \\ \dfrac{8.4229463423832E+88}{2.5898702760421E+88} \\ -\dfrac{2.4868127882634E+90}{6.9063207361123E+89} \\ -\dfrac{9.6150527829735E+88}{1.105011317778E+89} \\ -\dfrac{9.9702643356927E+87}{2.4555807061733E+88} \\ -\dfrac{5.5512386406711E+85}{2.5578965689305E+85}\end{pmatrix}\) . Ainsi \( x=\dfrac{6.9662173040248E+89}{6.9063207361123E+89}\) , \( y=\dfrac{8.4229463423832E+88}{2.5898702760421E+88}\) , \( z=\dfrac{2.4868127882634E+90}{6.9063207361123E+89}\) , \( t=\dfrac{9.6150527829735E+88}{1.105011317778E+89}\) , \( u=\dfrac{9.9702643356927E+87}{2.4555807061733E+88}\) et \( v=\dfrac{5.5512386406711E+85}{2.5578965689305E+85}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{9.8045580708E+22}{3.85566399972E+23}x&+&\dfrac{1.6448715684E+22}{6.4261066662E+22}y &-&\dfrac{1.310091732E+21}{1.28522133324E+22}z &-&\dfrac{3.9309216732E+22}{1.156699199916E+23}t &-&\dfrac{1.293732972E+21}{5.78349599958E+21}u &-&\dfrac{1.0974094812E+22}{1.445873999895E+23}v &=&-8\\ &\dfrac{1.7718960276E+22}{1.92783199986E+22}x&+&\dfrac{1.2256464132E+22}{1.92783199986E+22}y &-&\dfrac{3.586435056E+21}{1.445873999895E+22}z &-&\dfrac{1.5688173312E+22}{1.445873999895E+22}t &-&\dfrac{1.537784676E+21}{3.85566399972E+21}u &-&\dfrac{9.121250916E+21}{2.409789999825E+22}v &=&1\\ &-\dfrac{9.303401772E+21}{2.409789999825E+22}x&-&\dfrac{5.8371476148E+22}{3.85566399972E+23}y &+&\dfrac{1.311561396E+21}{3.85566399972E+21}z &+&\dfrac{1.41574194036E+23}{2.313398399832E+23}t &+&\dfrac{2.72298996E+20}{6.4261066662E+21}u &+&\dfrac{1.1562703992E+22}{7.229369999475E+22}v &=&-\dfrac{33}{7}\\ &-\dfrac{1.5649760844E+22}{3.85566399972E+22}x&-&\dfrac{5.056247772E+21}{1.28522133324E+22}y &+&\dfrac{3.082244076E+21}{2.313398399832E+23}z &+&\dfrac{4.856600916E+21}{2.89174799979E+22}t &+&\dfrac{2.534706756E+21}{5.78349599958E+21}u &+&\dfrac{1.412180892E+21}{6.4261066662E+21}v &=&2\\ &-\dfrac{2.1518566596E+22}{7.71132799944E+22}x&-&\dfrac{1.3305121884E+22}{7.71132799944E+22}y &+&\dfrac{3.25311876E+20}{7.71132799944E+21}z &+&\dfrac{4.643517132E+21}{3.85566399972E+22}t &+&\dfrac{2.69079732E+20}{6.4261066662E+21}u &+&\dfrac{4.23551916E+20}{6.4261066662E+21}v &=&3\\ &-\dfrac{1.4129148492E+22}{3.85566399972E+22}x&-&\dfrac{2.45285172E+20}{6.4261066662E+20}y &+&\dfrac{3.533920812E+21}{1.92783199986E+22}z &+&\dfrac{1.721615148E+21}{3.85566399972E+21}t &+&\dfrac{8.1478872E+19}{3.2130533331E+20}u &+&\dfrac{1.324106028E+21}{6.4261066662E+21}v &=&-9\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-8 \\ 1 \\ -\dfrac{33}{7} \\ 2 \\ 3 \\ -9\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 0 & -2 & -2 & -4 & \dfrac{16}{3} \\ 1 & -1 & 4 & 4 & -\dfrac{11}{3} & -\dfrac{23}{3} \\ -2 & 4 & 4 & 1 & 2 & \dfrac{9}{5} \\ -1 & -2 & -1 & -\dfrac{8}{5} & -3 & -\dfrac{3}{5} \\ -4 & \dfrac{3}{4} & 2 & \dfrac{18}{5} & -\dfrac{17}{5} & -\dfrac{22}{5} \\ \dfrac{25}{2} & -2 & 0 & 2 & -5 & \dfrac{21}{4}\end{pmatrix}\times\begin{pmatrix}-8 \\ 1 \\ -\dfrac{33}{7} \\ 2 \\ 3 \\ -9\end{pmatrix}=\begin{pmatrix}-\dfrac{438}{7} \\ \dfrac{267}{7} \\ -\dfrac{247}{35} \\ \dfrac{137}{35} \\ \dfrac{8389}{140} \\ -\dfrac{641}{4}\end{pmatrix}\) . Ainsi \( x=\dfrac{438}{7}\) , \( y=\dfrac{267}{7}\) , \( z=\dfrac{247}{35}\) , \( t=\dfrac{137}{35}\) , \( u=\dfrac{8389}{140}\) et \( v=\dfrac{641}{4}\)