\( %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Mes commandes %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcommand{\multirows}[3]{\multirow{#1}{#2}{$#3$}}%pour rester en mode math \renewcommand{\arraystretch}{1.3}%pour augmenter la taille des case \newcommand{\point}[1]{\marginnote{\small\vspace*{-1em} #1}}%pour indiquer les points ou le temps \newcommand{\dpl}[1]{\displaystyle{#1}}%megamode \newcommand{\A}{\mathscr{A}} \newcommand{\LN}{\mathscr{N}} \newcommand{\LL}{\mathscr{L}} \newcommand{\K}{\mathbb{K}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\M}{\mathcal{M}} \newcommand{\D}{\mathbb{D}} \newcommand{\E}{\mathcal{E}} \renewcommand{\P}{\mathcal{P}} \newcommand{\G}{\mathcal{G}} \newcommand{\Kk}{\mathcal{K}} \newcommand{\Cc}{\mathcal{C}} \newcommand{\Zz}{\mathcal{Z}} \newcommand{\Ss}{\mathcal{S}} \newcommand{\B}{\mathbb{B}} \newcommand{\inde}{\bot\!\!\!\bot} \newcommand{\Proba}{\mathbb{P}} \newcommand{\Esp}[1]{\dpl{\mathbb{E}\left(#1\right)}} \newcommand{\Var}[1]{\dpl{\mathbb{V}\left(#1\right)}} \newcommand{\Cov}[1]{\dpl{Cov\left(#1\right)}} \newcommand{\base}{\mathcal{B}} \newcommand{\Som}{\textbf{Som}} \newcommand{\Chain}{\textbf{Chain}} \newcommand{\Ar}{\textbf{Ar}} \newcommand{\Arc}{\textbf{Arc}} \newcommand{\Min}{\text{Min}} \newcommand{\Max}{\text{Max}} \newcommand{\Ker}{\text{Ker}} \renewcommand{\Im}{\text{Im}} \newcommand{\Sup}{\text{Sup}} \newcommand{\Inf}{\text{Inf}} \renewcommand{\det}{\texttt{det}} \newcommand{\GL}{\text{GL}} \newcommand{\crossmark}{\text{\ding{55}}} \renewcommand{\checkmark}{\text{\ding{51}}} \newcommand{\Card}{\sharp} \newcommand{\Surligne}[2]{\text{\colorbox{#1}{ #2 }}} \newcommand{\SurligneMM}[2]{\text{\colorbox{#1}{ #2 }}} \newcommand{\norm}[1]{\left\lVert#1\right\rVert} \renewcommand{\lim}[1]{\underset{#1}{lim}\,} \newcommand{\nonor}[1]{\left|#1\right|} \newcommand{\Un}{1\!\!1} \newcommand{\sepon}{\setlength{\columnseprule}{0.5pt}} \newcommand{\sepoff}{\setlength{\columnseprule}{0pt}} \newcommand{\flux}{Flux} \newcommand{\Cpp}{\texttt{C++\ }} \newcommand{\Python}{\texttt{Python\ }} %\newcommand{\comb}[2]{\begin{pmatrix} #1\\ #2\end{pmatrix}} \newcommand{\comb}[2]{C_{#1}^{#2}} \newcommand{\arrang}[2]{A_{#1}^{#2}} \newcommand{\supp}[1]{Supp\left(#1\right)} \newcommand{\BB}{\mathcal{B}} \newcommand{\arc}[1]{\overset{\rotatebox{90}{)}}{#1}} \newcommand{\modpi}{\equiv_{2\pi}} \renewcommand{\Re}{Re} \renewcommand{\Im}{Im} \renewcommand{\bar}[1]{\overline{#1}} \newcommand{\mat}{\mathcal{M}} \newcommand{\und}[1]{{\mathbf{\color{red}\underline{#1}}}} \newcommand{\rdots}{\text{\reflectbox{$\ddots$}}} \newcommand{\Compa}{Compa} \newcommand{\dint}{\dpl{\int}} \newcommand{\intEFF}[2]{\left[\!\left[#1 ; 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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.


On considère la matrice \[ A= \begin{pmatrix}1 & -4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ \dfrac{7}{2} & \dfrac{13}{2} & \dfrac{11}{2} & 3 & -\dfrac{16}{3} & -3 \\ -1 & -\dfrac{14}{3} & -2 & 5 & 0 & 4 \\ \dfrac{19}{2} & \dfrac{13}{2} & -\dfrac{1}{3} & \dfrac{5}{2} & 2 & -\dfrac{1}{5} \\ -3 & 1 & -5 & 0 & 0 & 1 \\ \dfrac{16}{5} & 2 & -\dfrac{7}{4} & 3 & -4 & -\dfrac{21}{5}\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (5, 3)\) et \( (6, 1)\) : \( \widehat{A}_{5, 3}=\) \( \widehat{A}_{6, 1}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & -4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ 0 & \dfrac{41}{2} & \dfrac{11}{2} & -\dfrac{121}{4} & -\dfrac{58}{3} & \dfrac{9}{4} \\ 0 & -\dfrac{26}{3} & -2 & \dfrac{29}{2} & 4 & \dfrac{5}{2} \\ 0 & \dfrac{89}{2} & -\dfrac{1}{3} & -\dfrac{351}{4} & -36 & \dfrac{281}{20} \\ 0 & -11 & -5 & \dfrac{57}{2} & 12 & -\dfrac{7}{2} \\ 0 & \dfrac{74}{5} & -\dfrac{7}{4} & -\dfrac{137}{5} & -\dfrac{84}{5} & \dfrac{3}{5}\end{pmatrix} \)
  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 \( A^{-1}_{5, 4}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&-&4y &&&+&\dfrac{19}{2}t &+&4u &-&\dfrac{3}{2}v &=&9\\ &\dfrac{7}{2}x&+&\dfrac{13}{2}y &+&\dfrac{11}{2}z &+&3t &-&\dfrac{16}{3}u &-&3v &=&-4\\ &-x&-&\dfrac{14}{3}y &-&2z &+&5t &&&+&4v &=&-8\\ &\dfrac{19}{2}x&+&\dfrac{13}{2}y &-&\dfrac{1}{3}z &+&\dfrac{5}{2}t &+&2u &-&\dfrac{1}{5}v &=&\dfrac{54}{5}\\ &-3x&+&y &-&5z &&&&&+&v &=&-3\\ &\dfrac{16}{5}x&+&2y &-&\dfrac{7}{4}z &+&3t &-&4u &-&\dfrac{21}{5}v &=&-5\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{4.8401741595456E+20}{9.813705408322E+21}x&-&\dfrac{6.9981069210048E+20}{9.813705408322E+21}y &+&\dfrac{9.9963660959424E+20}{1.9627410816644E+22}z &+&\dfrac{3.5641443541824E+20}{4.906852704161E+21}t &-&\dfrac{2.5983165754886E+21}{1.9627410816644E+22}u &+&\dfrac{6.0410845669824E+20}{7.3602790562415E+21}v &=&1\\ &\dfrac{3.377175757632E+19}{1.9627410816644E+21}x&+&\dfrac{7.642818830592E+19}{8.1780878402683E+20}y &-&\dfrac{2.1001629989952E+20}{3.9254821633288E+21}z &+&\dfrac{5.965029832896E+19}{1.6356175680537E+21}t &+&\dfrac{1.5012727317153E+22}{9.813705408322E+22}u &-&\dfrac{4.3752066501312E+20}{4.906852704161E+21}v &=&-4\\ &\dfrac{4.659130492416E+19}{2.4534263520805E+21}x&+&\dfrac{1.902439737216E+19}{2.7260292800894E+20}y &-&\dfrac{1.656871479936E+19}{1.3630146400447E+21}z &-&\dfrac{6.551332550208E+19}{2.4534263520805E+21}t &-&\dfrac{2.1098636254022E+21}{2.4534263520805E+22}u &-&\dfrac{2.859441683904E+19}{3.2712351361073E+20}v &=&-4\\ &\dfrac{1.0299181347804E+22}{1.4720558112483E+23}x&+&\dfrac{3.6873774441792E+20}{5.4520585601789E+21}y &+&\dfrac{9.5632816453056E+20}{2.4534263520805E+22}z &-&\dfrac{2234262994560000000}{4.906852704161E+21}t &+&\dfrac{8.634581270208E+19}{1.3084940544429E+21}u &-&\dfrac{3.00889139616E+19}{1.4720558112483E+21}v &=&6\\ &\dfrac{5.6945261938349E+21}{6.5424702722147E+22}x&-&\dfrac{3.3132997805933E+21}{9.813705408322E+22}y &-&\dfrac{2.8416634944192E+20}{2.7260292800894E+21}z &+&\dfrac{4.8890225156544E+20}{1.3084940544429E+22}t &+&\dfrac{1.6055065593734E+21}{4.3616468481431E+22}u &-&\dfrac{1.9482715491782E+21}{1.9627410816644E+22}v &=&3\\ &-\dfrac{1.7227123431552E+20}{2.4534263520805E+21}x&+&\dfrac{4.531346566848E+19}{1.0904117120358E+21}y &+&\dfrac{2.8560438627264E+20}{1.9627410816644E+21}z &+&\dfrac{1.1757957812352E+20}{2.4534263520805E+21}t &+&\dfrac{1.1713833754157E+21}{5.8882232449932E+22}u &-&\dfrac{4.988436845184E+19}{4.906852704161E+20}v &=&2\\ \end{array} \right. \)
Cliquer ici pour afficher la solution
  1. \( \widehat{A}_{5, 3}=\begin{pmatrix}1 & -4 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ \dfrac{7}{2} & \dfrac{13}{2} & 3 & -\dfrac{16}{3} & -3 \\ -1 & -\dfrac{14}{3} & 5 & 0 & 4 \\ \dfrac{19}{2} & \dfrac{13}{2} & \dfrac{5}{2} & 2 & -\dfrac{1}{5} \\ \dfrac{16}{5} & 2 & 3 & -4 & -\dfrac{21}{5}\end{pmatrix}\) \( \widehat{A}_{6, 1}=\begin{pmatrix}-4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ \dfrac{13}{2} & \dfrac{11}{2} & 3 & -\dfrac{16}{3} & -3 \\ -\dfrac{14}{3} & -2 & 5 & 0 & 4 \\ \dfrac{13}{2} & -\dfrac{1}{3} & \dfrac{5}{2} & 2 & -\dfrac{1}{5} \\ 1 & -5 & 0 & 0 & 1\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{7}{2}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-1\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(\dfrac{19}{2}\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-3\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(\dfrac{16}{5}\right)L_1\)
  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 & -4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ 0 & \dfrac{41}{2} & \dfrac{11}{2} & -\dfrac{121}{4} & -\dfrac{58}{3} & \dfrac{9}{4} \\ 0 & -\dfrac{26}{3} & -2 & \dfrac{29}{2} & 4 & \dfrac{5}{2} \\ 0 & \dfrac{89}{2} & -\dfrac{1}{3} & -\dfrac{351}{4} & -36 & \dfrac{281}{20} \\ 0 & -11 & -5 & \dfrac{57}{2} & 12 & -\dfrac{7}{2} \\ 0 & \dfrac{74}{5} & -\dfrac{7}{4} & -\dfrac{137}{5} & -\dfrac{84}{5} & \dfrac{3}{5}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{41}{2} & \dfrac{11}{2} & -\dfrac{121}{4} & -\dfrac{58}{3} & \dfrac{9}{4} \\ -\dfrac{26}{3} & -2 & \dfrac{29}{2} & 4 & \dfrac{5}{2} \\ \dfrac{89}{2} & -\dfrac{1}{3} & -\dfrac{351}{4} & -36 & \dfrac{281}{20} \\ -11 & -5 & \dfrac{57}{2} & 12 & -\dfrac{7}{2} \\ \dfrac{74}{5} & -\dfrac{7}{4} & -\dfrac{137}{5} & -\dfrac{84}{5} & \dfrac{3}{5}\end{pmatrix}\\ &=&-\dfrac{5.4520585601789E+19}{340122240000000} \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{5.4520585601789E+19}{340122240000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & -4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ \dfrac{7}{2} & \dfrac{13}{2} & \dfrac{11}{2} & 3 & -\dfrac{16}{3} & -3 \\ -1 & -\dfrac{14}{3} & -2 & 5 & 0 & 4 \\ \dfrac{19}{2} & \dfrac{13}{2} & -\dfrac{1}{3} & \dfrac{5}{2} & 2 & -\dfrac{1}{5} \\ -3 & 1 & -5 & 0 & 0 & 1 \\ \dfrac{16}{5} & 2 & -\dfrac{7}{4} & 3 & -4 & -\dfrac{21}{5}\end{pmatrix} =-\dfrac{340122240000000}{5.4520585601789E+19}\begin{pmatrix}\dfrac{2.6388912959307E+40}{3.3378594661786E+36} & \dfrac{3.8154088743711E+40}{3.3378594661786E+36} & -\dfrac{5.4500773344065E+40}{6.6757189323572E+36} & -\dfrac{1.9431923735933E+40}{1.6689297330893E+36} & \dfrac{1.4166174127448E+41}{6.6757189323572E+36} & -\dfrac{3.2936346826181E+40}{2.5033945996339E+36} \\ -\dfrac{1.8412559998626E+39}{6.6757189323572E+35} & -\dfrac{4.1669095829225E+39}{2.7815495551488E+35} & \dfrac{1.1450211656443E+40}{1.3351437864714E+36} & -\dfrac{3.2521691962163E+39}{5.5630991102976E+35} & -\dfrac{8.1850268481113E+41}{3.3378594661786E+37} & \dfrac{2.3853882869399E+40}{1.6689297330893E+36} \\ -\dfrac{2.5401852284167E+39}{8.3446486654465E+35} & -\dfrac{1.0372212854513E+39}{9.2718318504961E+34} & \dfrac{9.0333603353013E+38}{4.635915925248E+35} & \dfrac{3.571824871094E+39}{8.3446486654465E+35} & \dfrac{1.1503100039684E+41}{8.3446486654465E+36} & \dfrac{1.5589843510061E+39}{1.1126198220595E+35} \\ -\dfrac{5.6151739830127E+41}{5.0067891992679E+37} & -\dfrac{2.0103797759148E+40}{1.8543663700992E+36} & -\dfrac{5.213957155769E+40}{8.3446486654465E+36} & \dfrac{1.2181332685182E+38}{1.6689297330893E+36} & -\dfrac{4.7076242727798E+39}{4.4504792882381E+35} & \dfrac{1.6404652093083E+39}{5.0067891992679E+35} \\ -\dfrac{3.104689028126E+41}{2.2252396441191E+37} & \dfrac{1.8064304431222E+41}{3.3378594661786E+37} & \dfrac{1.5492915779896E+40}{9.2718318504961E+35} & -\dfrac{2.6655237057381E+40}{4.4504792882381E+36} & -\dfrac{8.7533157804553E+40}{1.4834930960794E+37} & \dfrac{1.062209057725E+41}{6.6757189323572E+36} \\ \dfrac{9.3923285772251E+39}{8.3446486654465E+35} & -\dfrac{2.4705166838921E+39}{3.7087327401984E+35} & -\dfrac{1.5571318390024E+40}{6.6757189323572E+35} & -\dfrac{6.4105074541056E+39}{8.3446486654465E+35} & -\dfrac{6.3864507591863E+40}{2.0027156797072E+37} & \dfrac{2.7197249803697E+39}{1.6689297330893E+35}\end{pmatrix} =\begin{pmatrix}-\dfrac{4.8401741595456E+20}{9.813705408322E+21} & -\dfrac{6.9981069210048E+20}{9.813705408322E+21} & \dfrac{9.9963660959424E+20}{1.9627410816644E+22} & \dfrac{3.5641443541824E+20}{4.906852704161E+21} & -\dfrac{2.5983165754886E+21}{1.9627410816644E+22} & \dfrac{6.0410845669824E+20}{7.3602790562415E+21} \\ \dfrac{3.377175757632E+19}{1.9627410816644E+21} & \dfrac{7.642818830592E+19}{8.1780878402683E+20} & -\dfrac{2.1001629989952E+20}{3.9254821633288E+21} & \dfrac{5.965029832896E+19}{1.6356175680537E+21} & \dfrac{1.5012727317153E+22}{9.813705408322E+22} & -\dfrac{4.3752066501312E+20}{4.906852704161E+21} \\ \dfrac{4.659130492416E+19}{2.4534263520805E+21} & \dfrac{1.902439737216E+19}{2.7260292800894E+20} & -\dfrac{1.656871479936E+19}{1.3630146400447E+21} & -\dfrac{6.551332550208E+19}{2.4534263520805E+21} & -\dfrac{2.1098636254022E+21}{2.4534263520805E+22} & -\dfrac{2.859441683904E+19}{3.2712351361073E+20} \\ \dfrac{1.0299181347804E+22}{1.4720558112483E+23} & \dfrac{3.6873774441792E+20}{5.4520585601789E+21} & \dfrac{9.5632816453056E+20}{2.4534263520805E+22} & -\dfrac{2234262994560000000}{4.906852704161E+21} & \dfrac{8.634581270208E+19}{1.3084940544429E+21} & -\dfrac{3.00889139616E+19}{1.4720558112483E+21} \\ \dfrac{5.6945261938349E+21}{6.5424702722147E+22} & -\dfrac{3.3132997805933E+21}{9.813705408322E+22} & -\dfrac{2.8416634944192E+20}{2.7260292800894E+21} & \dfrac{4.8890225156544E+20}{1.3084940544429E+22} & \dfrac{1.6055065593734E+21}{4.3616468481431E+22} & -\dfrac{1.9482715491782E+21}{1.9627410816644E+22} \\ -\dfrac{1.7227123431552E+20}{2.4534263520805E+21} & \dfrac{4.531346566848E+19}{1.0904117120358E+21} & \dfrac{2.8560438627264E+20}{1.9627410816644E+21} & \dfrac{1.1757957812352E+20}{2.4534263520805E+21} & \dfrac{1.1713833754157E+21}{5.8882232449932E+22} & -\dfrac{4.988436845184E+19}{4.906852704161E+20}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{5, 4}=B_{5, 4}= \left(-\dfrac{5.4520585601789E+19}{340122240000000}\right)^{-1}Co(A)_{4, 5}= \left(-\dfrac{5.4520585601789E+19}{340122240000000}\right)^{-1}\times(-1)^{4+5}\det\begin{pmatrix}1 & -4 & 0 & \dfrac{19}{2} & -\dfrac{3}{2} \\ \dfrac{7}{2} & \dfrac{13}{2} & \dfrac{11}{2} & 3 & -3 \\ -1 & -\dfrac{14}{3} & -2 & 5 & 4 \\ -3 & 1 & -5 & 0 & 1 \\ \dfrac{16}{5} & 2 & -\dfrac{7}{4} & 3 & -\dfrac{21}{5}\end{pmatrix}=\dfrac{4.8890225156544E+20}{1.3084940544429E+22}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ \dfrac{7}{2} & \dfrac{13}{2} & \dfrac{11}{2} & 3 & -\dfrac{16}{3} & -3 \\ -1 & -\dfrac{14}{3} & -2 & 5 & 0 & 4 \\ \dfrac{19}{2} & \dfrac{13}{2} & -\dfrac{1}{3} & \dfrac{5}{2} & 2 & -\dfrac{1}{5} \\ -3 & 1 & -5 & 0 & 0 & 1 \\ \dfrac{16}{5} & 2 & -\dfrac{7}{4} & 3 & -4 & -\dfrac{21}{5}\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&-&4y &&&+&\dfrac{19}{2}t &+&4u &-&\dfrac{3}{2}v &=&9\\ &\dfrac{7}{2}x&+&\dfrac{13}{2}y &+&\dfrac{11}{2}z &+&3t &-&\dfrac{16}{3}u &-&3v &=&-4\\ &-x&-&\dfrac{14}{3}y &-&2z &+&5t &&&+&4v &=&-8\\ &\dfrac{19}{2}x&+&\dfrac{13}{2}y &-&\dfrac{1}{3}z &+&\dfrac{5}{2}t &+&2u &-&\dfrac{1}{5}v &=&\dfrac{54}{5}\\ &-3x&+&y &-&5z &&&&&+&v &=&-3\\ &\dfrac{16}{5}x&+&2y &-&\dfrac{7}{4}z &+&3t &-&4u &-&\dfrac{21}{5}v &=&-5\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}9 \\ -4 \\ -8 \\ \dfrac{54}{5} \\ -3 \\ -5\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{4.8401741595456E+20}{9.813705408322E+21} & -\dfrac{6.9981069210048E+20}{9.813705408322E+21} & \dfrac{9.9963660959424E+20}{1.9627410816644E+22} & \dfrac{3.5641443541824E+20}{4.906852704161E+21} & -\dfrac{2.5983165754886E+21}{1.9627410816644E+22} & \dfrac{6.0410845669824E+20}{7.3602790562415E+21} \\ \dfrac{3.377175757632E+19}{1.9627410816644E+21} & \dfrac{7.642818830592E+19}{8.1780878402683E+20} & -\dfrac{2.1001629989952E+20}{3.9254821633288E+21} & \dfrac{5.965029832896E+19}{1.6356175680537E+21} & \dfrac{1.5012727317153E+22}{9.813705408322E+22} & -\dfrac{4.3752066501312E+20}{4.906852704161E+21} \\ \dfrac{4.659130492416E+19}{2.4534263520805E+21} & \dfrac{1.902439737216E+19}{2.7260292800894E+20} & -\dfrac{1.656871479936E+19}{1.3630146400447E+21} & -\dfrac{6.551332550208E+19}{2.4534263520805E+21} & -\dfrac{2.1098636254022E+21}{2.4534263520805E+22} & -\dfrac{2.859441683904E+19}{3.2712351361073E+20} \\ \dfrac{1.0299181347804E+22}{1.4720558112483E+23} & \dfrac{3.6873774441792E+20}{5.4520585601789E+21} & \dfrac{9.5632816453056E+20}{2.4534263520805E+22} & -\dfrac{2234262994560000000}{4.906852704161E+21} & \dfrac{8.634581270208E+19}{1.3084940544429E+21} & -\dfrac{3.00889139616E+19}{1.4720558112483E+21} \\ \dfrac{5.6945261938349E+21}{6.5424702722147E+22} & -\dfrac{3.3132997805933E+21}{9.813705408322E+22} & -\dfrac{2.8416634944192E+20}{2.7260292800894E+21} & \dfrac{4.8890225156544E+20}{1.3084940544429E+22} & \dfrac{1.6055065593734E+21}{4.3616468481431E+22} & -\dfrac{1.9482715491782E+21}{1.9627410816644E+22} \\ -\dfrac{1.7227123431552E+20}{2.4534263520805E+21} & \dfrac{4.531346566848E+19}{1.0904117120358E+21} & \dfrac{2.8560438627264E+20}{1.9627410816644E+21} & \dfrac{1.1757957812352E+20}{2.4534263520805E+21} & \dfrac{1.1713833754157E+21}{5.8882232449932E+22} & -\dfrac{4.988436845184E+19}{4.906852704161E+20}\end{pmatrix}\times\begin{pmatrix}9 \\ -4 \\ -8 \\ \dfrac{54}{5} \\ -3 \\ -5\end{pmatrix}=\begin{pmatrix}\dfrac{1.3743843778218E+132}{6.6997618495507E+132} \\ \dfrac{1.4635526550939E+130}{2.4813932776114E+130} \\ \dfrac{3.5510862000067E+127}{8.9749467506198E+127} \\ -\dfrac{4.9659394997846E+130}{9.3052247910426E+131} \\ \dfrac{2.4916870238607E+135}{9.8030351708104E+134} \\ -\dfrac{1.8537128789276E+129}{1.8610449582085E+129}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.3743843778218E+132}{6.6997618495507E+132}\) , \( y=\dfrac{1.4635526550939E+130}{2.4813932776114E+130}\) , \( z=\dfrac{3.5510862000067E+127}{8.9749467506198E+127}\) , \( t=\dfrac{4.9659394997846E+130}{9.3052247910426E+131}\) , \( u=\dfrac{2.4916870238607E+135}{9.8030351708104E+134}\) et \( v=\dfrac{1.8537128789276E+129}{1.8610449582085E+129}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{4.8401741595456E+20}{9.813705408322E+21}x&-&\dfrac{6.9981069210048E+20}{9.813705408322E+21}y &+&\dfrac{9.9963660959424E+20}{1.9627410816644E+22}z &+&\dfrac{3.5641443541824E+20}{4.906852704161E+21}t &-&\dfrac{2.5983165754886E+21}{1.9627410816644E+22}u &+&\dfrac{6.0410845669824E+20}{7.3602790562415E+21}v &=&1\\ &\dfrac{3.377175757632E+19}{1.9627410816644E+21}x&+&\dfrac{7.642818830592E+19}{8.1780878402683E+20}y &-&\dfrac{2.1001629989952E+20}{3.9254821633288E+21}z &+&\dfrac{5.965029832896E+19}{1.6356175680537E+21}t &+&\dfrac{1.5012727317153E+22}{9.813705408322E+22}u &-&\dfrac{4.3752066501312E+20}{4.906852704161E+21}v &=&-4\\ &\dfrac{4.659130492416E+19}{2.4534263520805E+21}x&+&\dfrac{1.902439737216E+19}{2.7260292800894E+20}y &-&\dfrac{1.656871479936E+19}{1.3630146400447E+21}z &-&\dfrac{6.551332550208E+19}{2.4534263520805E+21}t &-&\dfrac{2.1098636254022E+21}{2.4534263520805E+22}u &-&\dfrac{2.859441683904E+19}{3.2712351361073E+20}v &=&-4\\ &\dfrac{1.0299181347804E+22}{1.4720558112483E+23}x&+&\dfrac{3.6873774441792E+20}{5.4520585601789E+21}y &+&\dfrac{9.5632816453056E+20}{2.4534263520805E+22}z &-&\dfrac{2234262994560000000}{4.906852704161E+21}t &+&\dfrac{8.634581270208E+19}{1.3084940544429E+21}u &-&\dfrac{3.00889139616E+19}{1.4720558112483E+21}v &=&6\\ &\dfrac{5.6945261938349E+21}{6.5424702722147E+22}x&-&\dfrac{3.3132997805933E+21}{9.813705408322E+22}y &-&\dfrac{2.8416634944192E+20}{2.7260292800894E+21}z &+&\dfrac{4.8890225156544E+20}{1.3084940544429E+22}t &+&\dfrac{1.6055065593734E+21}{4.3616468481431E+22}u &-&\dfrac{1.9482715491782E+21}{1.9627410816644E+22}v &=&3\\ &-\dfrac{1.7227123431552E+20}{2.4534263520805E+21}x&+&\dfrac{4.531346566848E+19}{1.0904117120358E+21}y &+&\dfrac{2.8560438627264E+20}{1.9627410816644E+21}z &+&\dfrac{1.1757957812352E+20}{2.4534263520805E+21}t &+&\dfrac{1.1713833754157E+21}{5.8882232449932E+22}u &-&\dfrac{4.988436845184E+19}{4.906852704161E+20}v &=&2\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}1 \\ -4 \\ -4 \\ 6 \\ 3 \\ 2\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -4 & 0 & \dfrac{19}{2} & 4 & -\dfrac{3}{2} \\ \dfrac{7}{2} & \dfrac{13}{2} & \dfrac{11}{2} & 3 & -\dfrac{16}{3} & -3 \\ -1 & -\dfrac{14}{3} & -2 & 5 & 0 & 4 \\ \dfrac{19}{2} & \dfrac{13}{2} & -\dfrac{1}{3} & \dfrac{5}{2} & 2 & -\dfrac{1}{5} \\ -3 & 1 & -5 & 0 & 0 & 1 \\ \dfrac{16}{5} & 2 & -\dfrac{7}{4} & 3 & -4 & -\dfrac{21}{5}\end{pmatrix}\times\begin{pmatrix}1 \\ -4 \\ -4 \\ 6 \\ 3 \\ 2\end{pmatrix}=\begin{pmatrix}83 \\ -\dfrac{97}{2} \\ \dfrac{191}{3} \\ \dfrac{163}{30} \\ 15 \\ -\dfrac{1}{5}\end{pmatrix}\) . Ainsi \( x=83\) , \( y=\dfrac{97}{2}\) , \( z=\dfrac{191}{3}\) , \( t=\dfrac{163}{30}\) , \( u=15\) et \( v=\dfrac{1}{5}\)