\( %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 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.


Exercice


On considère la matrice \[ A= \begin{pmatrix}1 & 0 & 5 & -4 & 3 & 3 \\ 5 & \dfrac{1}{2} & 3 & \dfrac{1}{3} & \dfrac{9}{2} & 1 \\ 4 & \dfrac{14}{3} & 0 & \dfrac{21}{4} & -5 & 4 \\ \dfrac{22}{3} & -1 & -\dfrac{21}{2} & -5 & -3 & -5 \\ 4 & -3 & -\dfrac{1}{2} & 3 & \dfrac{21}{4} & 0 \\ -2 & 2 & \dfrac{13}{4} & 1 & -\dfrac{13}{4} & -2\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (3, 5)\) et \( (3, 6)\) : \( \widehat{A}_{3, 5}=\) \( \widehat{A}_{3, 6}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & 0 & 5 & -4 & 3 & 3 \\ 0 & \dfrac{1}{2} & -22 & \dfrac{61}{3} & -\dfrac{21}{2} & -14 \\ 0 & \dfrac{14}{3} & -20 & \dfrac{85}{4} & -17 & -8 \\ 0 & -1 & -\dfrac{283}{6} & \dfrac{73}{3} & -25 & -27 \\ 0 & -3 & -\dfrac{41}{2} & 19 & -\dfrac{27}{4} & -12 \\ 0 & 2 & \dfrac{53}{4} & -7 & \dfrac{11}{4} & 4\end{pmatrix} \)
  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 \( A^{-1}_{3, 3}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&&&+&5z &-&4t &+&3u &+&3v &=&-2\\ &5x&+&\dfrac{1}{2}y &+&3z &+&\dfrac{1}{3}t &+&\dfrac{9}{2}u &+&v &=&-9\\ &4x&+&\dfrac{14}{3}y &&&+&\dfrac{21}{4}t &-&5u &+&4v &=&-\dfrac{25}{3}\\ &\dfrac{22}{3}x&-&y &-&\dfrac{21}{2}z &-&5t &-&3u &-&5v &=&-7\\ &4x&-&3y &-&\dfrac{1}{2}z &+&3t &+&\dfrac{21}{4}u &&&=&-2\\ &-2x&+&2y &+&\dfrac{13}{4}z &+&t &-&\dfrac{13}{4}u &-&2v &=&-1\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{58485335176249344}{480925441176109056}x&-&\dfrac{3801405110353920}{80154240196018176}y &+&\dfrac{847607437983744}{13359040032669696}z &+&\dfrac{11400050815008768}{160308480392036352}t &+&\dfrac{33415994252722176}{240462720588054528}u &+&\dfrac{8641856975929344}{80154240196018176}v &=&7\\ &-\dfrac{85183552253067264}{240462720588054528}x&+&\dfrac{40621854544625664}{80154240196018176}y &-&\dfrac{1203279123972096}{20038560049004544}z &-&\dfrac{2821331204702208}{40077120098009088}t &-&\dfrac{56131356236709888}{120231360294027264}u &-&\dfrac{2225053227810816}{10019280024502272}v &=&9\\ &\dfrac{1679740121382912}{10019280024502272}x&-&\dfrac{697776589504512}{6679520016334848}y &+&\dfrac{34352670965760}{1669880004083712}z &+&\dfrac{124320894418944}{10019280024502272}t &+&\dfrac{18038546245877760}{120231360294027264}u &+&\dfrac{8390830221950976}{40077120098009088}v &=&-9\\ &-\dfrac{1336818825953280}{15028920036753408}x&-&\dfrac{59229691969536}{10019280024502272}y &+&\dfrac{443355846672384}{13359040032669696}z &-&\dfrac{98343120273408}{2504820006125568}t &+&\dfrac{2479565700071424}{30057840073506816}u &+&\dfrac{141015650402304}{5009640012251136}v &=&6\\ &-\dfrac{41171020287049728}{180347040441040896}x&+&\dfrac{19186576735076352}{60115680147013632}y &-&\dfrac{332840231239680}{3339760008167424}z &-&\dfrac{1063923776225280}{15028920036753408}t &-&\dfrac{77560121256837120}{360694080882081792}u &-&\dfrac{4111744112787456}{20038560049004544}v &=&4\\ &\dfrac{44389457360584704}{360694080882081792}x&-&\dfrac{32973656602705920}{240462720588054528}y &+&\dfrac{3545791884951552}{40077120098009088}z &-&\dfrac{12476000319307776}{480925441176109056}t &+&\dfrac{20664888204460032}{721388161764163584}u &-&\dfrac{34184228216242176}{240462720588054528}v &=&-\dfrac{18}{5}\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{3, 5}=\begin{pmatrix}1 & 0 & 5 & -4 & 3 \\ 5 & \dfrac{1}{2} & 3 & \dfrac{1}{3} & 1 \\ \dfrac{22}{3} & -1 & -\dfrac{21}{2} & -5 & -5 \\ 4 & -3 & -\dfrac{1}{2} & 3 & 0 \\ -2 & 2 & \dfrac{13}{4} & 1 & -2\end{pmatrix}\) \( \widehat{A}_{3, 6}=\begin{pmatrix}1 & 0 & 5 & -4 & 3 \\ 5 & \dfrac{1}{2} & 3 & \dfrac{1}{3} & \dfrac{9}{2} \\ \dfrac{22}{3} & -1 & -\dfrac{21}{2} & -5 & -3 \\ 4 & -3 & -\dfrac{1}{2} & 3 & \dfrac{21}{4} \\ -2 & 2 & \dfrac{13}{4} & 1 & -\dfrac{13}{4}\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(5\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(4\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(\dfrac{22}{3}\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(4\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-2\right)L_1\)
  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 & 5 & -4 & 3 & 3 \\ 0 & \dfrac{1}{2} & -22 & \dfrac{61}{3} & -\dfrac{21}{2} & -14 \\ 0 & \dfrac{14}{3} & -20 & \dfrac{85}{4} & -17 & -8 \\ 0 & -1 & -\dfrac{283}{6} & \dfrac{73}{3} & -25 & -27 \\ 0 & -3 & -\dfrac{41}{2} & 19 & -\dfrac{27}{4} & -12 \\ 0 & 2 & \dfrac{53}{4} & -7 & \dfrac{11}{4} & 4\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{1}{2} & -22 & \dfrac{61}{3} & -\dfrac{21}{2} & -14 \\ \dfrac{14}{3} & -20 & \dfrac{85}{4} & -17 & -8 \\ -1 & -\dfrac{283}{6} & \dfrac{73}{3} & -25 & -27 \\ -3 & -\dfrac{41}{2} & 19 & -\dfrac{27}{4} & -12 \\ 2 & \dfrac{53}{4} & -7 & \dfrac{11}{4} & 4\end{pmatrix}\\ &=&-\dfrac{417470001020928}{9172942848} \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{417470001020928}{9172942848}\right)^{-1}{}^tCo\begin{pmatrix}1 & 0 & 5 & -4 & 3 & 3 \\ 5 & \dfrac{1}{2} & 3 & \dfrac{1}{3} & \dfrac{9}{2} & 1 \\ 4 & \dfrac{14}{3} & 0 & \dfrac{21}{4} & -5 & 4 \\ \dfrac{22}{3} & -1 & -\dfrac{21}{2} & -5 & -3 & -5 \\ 4 & -3 & -\dfrac{1}{2} & 3 & \dfrac{21}{4} & 0 \\ -2 & 2 & \dfrac{13}{4} & 1 & -\dfrac{13}{4} & -2\end{pmatrix} =-\dfrac{9172942848}{417470001020928}\begin{pmatrix}-\dfrac{2.4415872935738E+31}{4.4115015860576E+27} & \dfrac{1.5869725953004E+30}{7.3525026434294E+26} & -\dfrac{3.5385067800042E+29}{1.2254171072382E+26} & -\dfrac{4.7591792253803E+30}{1.4705005286859E+27} & -\dfrac{1.3950175154799E+31}{2.2057507930288E+27} & -\dfrac{3.6077160405639E+30}{7.3525026434294E+26} \\ \dfrac{3.5561577646054E+31}{2.2057507930288E+27} & -\dfrac{1.6958405658217E+31}{7.3525026434294E+26} & \dfrac{5.0233293711309E+29}{1.8381256608573E+26} & \dfrac{1.1778211409074E+30}{3.6762513217147E+26} & \dfrac{2.3433157345445E+31}{1.1028753965144E+27} & \dfrac{9.288929732858E+29}{9.1906283042867E+25} \\ -\dfrac{7.0124111018862E+29}{9.1906283042867E+25} & \dfrac{2.9130079353283E+29}{6.1270855361912E+25} & -\dfrac{1.4341209583147E+28}{1.5317713840478E+25} & -\dfrac{5.1900243919999E+28}{9.1906283042867E+25} & -\dfrac{7.5305519196826E+30}{1.1028753965144E+27} & -\dfrac{3.5029199013243E+30}{3.6762513217147E+26} \\ \dfrac{5.5808175663551E+29}{1.378594245643E+26} & \dfrac{2.4726619566991E+28}{9.1906283042867E+25} & -\dfrac{1.8508776576295E+29}{1.2254171072382E+26} & \dfrac{4.1055302520941E+28}{2.2976570760717E+25} & -\dfrac{1.0351442953403E+30}{2.757188491286E+26} & -\dfrac{5.8869803717417E+28}{4.5953141521434E+25} \\ \dfrac{1.7187665881267E+31}{1.6543130947716E+27} & -\dfrac{8.0098202091804E+30}{5.514376982572E+26} & \dfrac{1.3895081167544E+29}{3.0635427680956E+25} & \dfrac{4.4415625994696E+29}{1.378594245643E+26} & \dfrac{3.2379023900275E+31}{3.3086261895432E+27} & \dfrac{1.7165298189632E+30}{1.8381256608573E+26} \\ -\dfrac{1.8531266809642E+31}{3.3086261895432E+27} & \dfrac{1.3765512455595E+31}{2.2057507930288E+27} & -\dfrac{1.4802617418307E+30}{3.6762513217147E+26} & \dfrac{5.2083558660385E+30}{4.4115015860576E+27} & -\dfrac{8.6269708998133E+30}{6.6172523790865E+27} & \dfrac{1.4270889788334E+31}{2.2057507930288E+27}\end{pmatrix} =\begin{pmatrix}\dfrac{58485335176249344}{480925441176109056} & -\dfrac{3801405110353920}{80154240196018176} & \dfrac{847607437983744}{13359040032669696} & \dfrac{11400050815008768}{160308480392036352} & \dfrac{33415994252722176}{240462720588054528} & \dfrac{8641856975929344}{80154240196018176} \\ -\dfrac{85183552253067264}{240462720588054528} & \dfrac{40621854544625664}{80154240196018176} & -\dfrac{1203279123972096}{20038560049004544} & -\dfrac{2821331204702208}{40077120098009088} & -\dfrac{56131356236709888}{120231360294027264} & -\dfrac{2225053227810816}{10019280024502272} \\ \dfrac{1679740121382912}{10019280024502272} & -\dfrac{697776589504512}{6679520016334848} & \dfrac{34352670965760}{1669880004083712} & \dfrac{124320894418944}{10019280024502272} & \dfrac{18038546245877760}{120231360294027264} & \dfrac{8390830221950976}{40077120098009088} \\ -\dfrac{1336818825953280}{15028920036753408} & -\dfrac{59229691969536}{10019280024502272} & \dfrac{443355846672384}{13359040032669696} & -\dfrac{98343120273408}{2504820006125568} & \dfrac{2479565700071424}{30057840073506816} & \dfrac{141015650402304}{5009640012251136} \\ -\dfrac{41171020287049728}{180347040441040896} & \dfrac{19186576735076352}{60115680147013632} & -\dfrac{332840231239680}{3339760008167424} & -\dfrac{1063923776225280}{15028920036753408} & -\dfrac{77560121256837120}{360694080882081792} & -\dfrac{4111744112787456}{20038560049004544} \\ \dfrac{44389457360584704}{360694080882081792} & -\dfrac{32973656602705920}{240462720588054528} & \dfrac{3545791884951552}{40077120098009088} & -\dfrac{12476000319307776}{480925441176109056} & \dfrac{20664888204460032}{721388161764163584} & -\dfrac{34184228216242176}{240462720588054528}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{3, 3}=B_{3, 3}= \left(-\dfrac{417470001020928}{9172942848}\right)^{-1}Co(A)_{3, 3}= \left(-\dfrac{417470001020928}{9172942848}\right)^{-1}\times(-1)^{3+3}\det\begin{pmatrix}1 & 0 & -4 & 3 & 3 \\ 5 & \dfrac{1}{2} & \dfrac{1}{3} & \dfrac{9}{2} & 1 \\ \dfrac{22}{3} & -1 & -5 & -3 & -5 \\ 4 & -3 & 3 & \dfrac{21}{4} & 0 \\ -2 & 2 & 1 & -\dfrac{13}{4} & -2\end{pmatrix}=\dfrac{34352670965760}{1669880004083712}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 0 & 5 & -4 & 3 & 3 \\ 5 & \dfrac{1}{2} & 3 & \dfrac{1}{3} & \dfrac{9}{2} & 1 \\ 4 & \dfrac{14}{3} & 0 & \dfrac{21}{4} & -5 & 4 \\ \dfrac{22}{3} & -1 & -\dfrac{21}{2} & -5 & -3 & -5 \\ 4 & -3 & -\dfrac{1}{2} & 3 & \dfrac{21}{4} & 0 \\ -2 & 2 & \dfrac{13}{4} & 1 & -\dfrac{13}{4} & -2\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&&&+&5z &-&4t &+&3u &+&3v &=&-2\\ &5x&+&\dfrac{1}{2}y &+&3z &+&\dfrac{1}{3}t &+&\dfrac{9}{2}u &+&v &=&-9\\ &4x&+&\dfrac{14}{3}y &&&+&\dfrac{21}{4}t &-&5u &+&4v &=&-\dfrac{25}{3}\\ &\dfrac{22}{3}x&-&y &-&\dfrac{21}{2}z &-&5t &-&3u &-&5v &=&-7\\ &4x&-&3y &-&\dfrac{1}{2}z &+&3t &+&\dfrac{21}{4}u &&&=&-2\\ &-2x&+&2y &+&\dfrac{13}{4}z &+&t &-&\dfrac{13}{4}u &-&2v &=&-1\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-2 \\ -9 \\ -\dfrac{25}{3} \\ -7 \\ -2 \\ -1\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{58485335176249344}{480925441176109056} & -\dfrac{3801405110353920}{80154240196018176} & \dfrac{847607437983744}{13359040032669696} & \dfrac{11400050815008768}{160308480392036352} & \dfrac{33415994252722176}{240462720588054528} & \dfrac{8641856975929344}{80154240196018176} \\ -\dfrac{85183552253067264}{240462720588054528} & \dfrac{40621854544625664}{80154240196018176} & -\dfrac{1203279123972096}{20038560049004544} & -\dfrac{2821331204702208}{40077120098009088} & -\dfrac{56131356236709888}{120231360294027264} & -\dfrac{2225053227810816}{10019280024502272} \\ \dfrac{1679740121382912}{10019280024502272} & -\dfrac{697776589504512}{6679520016334848} & \dfrac{34352670965760}{1669880004083712} & \dfrac{124320894418944}{10019280024502272} & \dfrac{18038546245877760}{120231360294027264} & \dfrac{8390830221950976}{40077120098009088} \\ -\dfrac{1336818825953280}{15028920036753408} & -\dfrac{59229691969536}{10019280024502272} & \dfrac{443355846672384}{13359040032669696} & -\dfrac{98343120273408}{2504820006125568} & \dfrac{2479565700071424}{30057840073506816} & \dfrac{141015650402304}{5009640012251136} \\ -\dfrac{41171020287049728}{180347040441040896} & \dfrac{19186576735076352}{60115680147013632} & -\dfrac{332840231239680}{3339760008167424} & -\dfrac{1063923776225280}{15028920036753408} & -\dfrac{77560121256837120}{360694080882081792} & -\dfrac{4111744112787456}{20038560049004544} \\ \dfrac{44389457360584704}{360694080882081792} & -\dfrac{32973656602705920}{240462720588054528} & \dfrac{3545791884951552}{40077120098009088} & -\dfrac{12476000319307776}{480925441176109056} & \dfrac{20664888204460032}{721388161764163584} & -\dfrac{34184228216242176}{240462720588054528}\end{pmatrix}\times\begin{pmatrix}-2 \\ -9 \\ -\dfrac{25}{3} \\ -7 \\ -2 \\ -1\end{pmatrix}=\begin{pmatrix}-\dfrac{5.8649290663843E+102}{4.7734405177336E+102} \\ -\dfrac{9.5301952787624E+100}{5.5938756067191E+100} \\ -\dfrac{2.6356744706539E+96}{1.6185982658331E+97} \\ \dfrac{8.2476874625041E+94}{2.2761538113278E+96} \\ -\dfrac{5.3638441476484E+99}{1.1799581357923E+100} \\ \dfrac{4.4992024299886E+104}{8.6995953435695E+104}\end{pmatrix}\) . Ainsi \( x=\dfrac{5.8649290663843E+102}{4.7734405177336E+102}\) , \( y=\dfrac{9.5301952787624E+100}{5.5938756067191E+100}\) , \( z=\dfrac{2.6356744706539E+96}{1.6185982658331E+97}\) , \( t=\dfrac{8.2476874625041E+94}{2.2761538113278E+96}\) , \( u=\dfrac{5.3638441476484E+99}{1.1799581357923E+100}\) et \( v=\dfrac{4.4992024299886E+104}{8.6995953435695E+104}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{58485335176249344}{480925441176109056}x&-&\dfrac{3801405110353920}{80154240196018176}y &+&\dfrac{847607437983744}{13359040032669696}z &+&\dfrac{11400050815008768}{160308480392036352}t &+&\dfrac{33415994252722176}{240462720588054528}u &+&\dfrac{8641856975929344}{80154240196018176}v &=&7\\ &-\dfrac{85183552253067264}{240462720588054528}x&+&\dfrac{40621854544625664}{80154240196018176}y &-&\dfrac{1203279123972096}{20038560049004544}z &-&\dfrac{2821331204702208}{40077120098009088}t &-&\dfrac{56131356236709888}{120231360294027264}u &-&\dfrac{2225053227810816}{10019280024502272}v &=&9\\ &\dfrac{1679740121382912}{10019280024502272}x&-&\dfrac{697776589504512}{6679520016334848}y &+&\dfrac{34352670965760}{1669880004083712}z &+&\dfrac{124320894418944}{10019280024502272}t &+&\dfrac{18038546245877760}{120231360294027264}u &+&\dfrac{8390830221950976}{40077120098009088}v &=&-9\\ &-\dfrac{1336818825953280}{15028920036753408}x&-&\dfrac{59229691969536}{10019280024502272}y &+&\dfrac{443355846672384}{13359040032669696}z &-&\dfrac{98343120273408}{2504820006125568}t &+&\dfrac{2479565700071424}{30057840073506816}u &+&\dfrac{141015650402304}{5009640012251136}v &=&6\\ &-\dfrac{41171020287049728}{180347040441040896}x&+&\dfrac{19186576735076352}{60115680147013632}y &-&\dfrac{332840231239680}{3339760008167424}z &-&\dfrac{1063923776225280}{15028920036753408}t &-&\dfrac{77560121256837120}{360694080882081792}u &-&\dfrac{4111744112787456}{20038560049004544}v &=&4\\ &\dfrac{44389457360584704}{360694080882081792}x&-&\dfrac{32973656602705920}{240462720588054528}y &+&\dfrac{3545791884951552}{40077120098009088}z &-&\dfrac{12476000319307776}{480925441176109056}t &+&\dfrac{20664888204460032}{721388161764163584}u &-&\dfrac{34184228216242176}{240462720588054528}v &=&-\dfrac{18}{5}\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}7 \\ 9 \\ -9 \\ 6 \\ 4 \\ -\dfrac{18}{5}\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 0 & 5 & -4 & 3 & 3 \\ 5 & \dfrac{1}{2} & 3 & \dfrac{1}{3} & \dfrac{9}{2} & 1 \\ 4 & \dfrac{14}{3} & 0 & \dfrac{21}{4} & -5 & 4 \\ \dfrac{22}{3} & -1 & -\dfrac{21}{2} & -5 & -3 & -5 \\ 4 & -3 & -\dfrac{1}{2} & 3 & \dfrac{21}{4} & 0 \\ -2 & 2 & \dfrac{13}{4} & 1 & -\dfrac{13}{4} & -2\end{pmatrix}\times\begin{pmatrix}7 \\ 9 \\ -9 \\ 6 \\ 4 \\ -\dfrac{18}{5}\end{pmatrix}=\begin{pmatrix}-\dfrac{304}{5} \\ \dfrac{289}{10} \\ \dfrac{671}{10} \\ \dfrac{677}{6} \\ \dfrac{89}{2} \\ -\dfrac{501}{20}\end{pmatrix}\) . Ainsi \( x=\dfrac{304}{5}\) , \( y=\dfrac{289}{10}\) , \( z=\dfrac{671}{10}\) , \( t=\dfrac{677}{6}\) , \( u=\dfrac{89}{2}\) et \( v=\dfrac{501}{20}\)