\( %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 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 ; #2\right]\!\right]} \newcommand{\intEFO}[2]{\left[\!\left[#1 ; #2\right[\!\right[} \newcommand{\intEOF}[2]{\left]\!\left]#1 ; #2\right]\!\right]} \newcommand{\intEOO}[2]{\left]\!\left]#1 ; #2\right[\!\right[} \newcommand{\ou}{\vee} \newcommand{\et}{\wedge} \newcommand{\non}{\neg} \newcommand{\implique}{\Rightarrow} \newcommand{\equivalent}{\Leftrightarrow} \newcommand{\Ab}{\overline{A}} \newcommand{\Bb}{\overline{B}} \newcommand{\Cb}{\overline{C}} \newcommand{\Cl}{\texttt{Cl}} \newcommand{\ab}{\overline{a}} \newcommand{\bb}{\overline{b}} \newcommand{\cb}{\overline{c}} \newcommand{\Rel}{\mathcal{R}} \newcommand{\superepsilon}{\varepsilon\!\!\varepsilon} \newcommand{\supere}{e\!\!e} \makeatletter \newenvironment{console}{\noindent\color{white}\begin{lrbox}{\@tempboxa}\begin{minipage}{\columnwidth} \ttfamily \bfseries\vspace*{0.5cm}} {\vspace*{0.5cm}\end{minipage}\end{lrbox}\colorbox{black}{\usebox{\@tempboxa}} } \makeatother \def\ie{\textit{i.e. }} \def\cf{\textit{c.f. }} \def\vide{ { $ {\text{ }} $ } } %Commande pour les vecteurs \newcommand{\vv}{\overrightarrow{v}} \newcommand{\vu}{\overrightarrow{u}} \newcommand{\vup}{\overrightarrow{u'}} \newcommand{\vx}{\overrightarrow{x}} \newcommand{\vy}{\overrightarrow{y}} \newcommand{\vz}{\overrightarrow{z}} \newcommand{\vt}{\overrightarrow{t}} \newcommand{\va}{\overrightarrow{a}} \newcommand{\vb}{\overrightarrow{b}} \newcommand{\vc}{\overrightarrow{c}} \newcommand{\vd}{\overrightarrow{d}} \newcommand{\ve}[1]{\overrightarrow{e_{#1}}} \newcommand{\vf}[1]{\overrightarrow{f_{#1}}} \newcommand{\vn}{\overrightarrow{0}} \newcommand{\Mat}{Mat} \newcommand{\Pass}{Pass} \newcommand{\mkF}{\mathfrak{F}} \renewcommand{\sp}{Sp} \newcommand{\Co}{Co} \newcommand{\vect}[1]{\dpl{\left\langle #1\right\rangle}} \newcommand{\trans}[1]{{\vphantom{#1}}^{t}{#1}} \SelectTips{cm}{12}%Change le bout des flèches dans un xymatrix \newcommand{\pourDES}[8]{ \begin{itemize} \item Pour la ligne : le premier et dernier caractère forment $#1#2$ soit $#4$ en base 10. \item Pour la colonne : les autres caractères du bloc forment $#3$ soit $#5$ en base 10. \item A l'intersection de la ligne $#4+1$ et de la colonne $#5+1$ de $S_{#8}$ se trouve l'entier $#6$ qui, codé sur $4$ bits, est \textbf{\texttt{$#7$}}. \end{itemize} } \)
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 \[ A= \begin{pmatrix}1 & \dfrac{22}{3} & 0 & 5 & -\dfrac{1}{4} \\ -2 & \dfrac{18}{5} & -\dfrac{21}{4} & -5 & 4 \\ -2 & -2 & -\dfrac{14}{5} & -\dfrac{10}{3} & \dfrac{22}{5} \\ 5 & 5 & 0 & 5 & 2 \\ -5 & 5 & \dfrac{4}{3} & -4 & -4\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (5, 5)\) et \( (3, 3)\) : \( \widehat{A}_{5, 5}=\) \( \widehat{A}_{3, 3}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & \dfrac{22}{3} & 0 & 5 & -\dfrac{1}{4} \\ 0 & \dfrac{274}{15} & -\dfrac{21}{4} & 5 & \dfrac{7}{2} \\ 0 & \dfrac{38}{3} & -\dfrac{14}{5} & \dfrac{20}{3} & \dfrac{39}{10} \\ 0 & -\dfrac{95}{3} & 0 & -20 & \dfrac{13}{4} \\ 0 & \dfrac{125}{3} & \dfrac{4}{3} & 21 & -\dfrac{21}{4}\end{pmatrix} \)
  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 \( A^{-1}_{4, 5}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{6}{cr}} &x&+&\dfrac{22}{3}y &&&+&5t &-&\dfrac{1}{4}u &=&7\\ &-2x&+&\dfrac{18}{5}y &-&\dfrac{21}{4}z &-&5t &+&4u &=&5\\ &-2x&-&2y &-&\dfrac{14}{5}z &-&\dfrac{10}{3}t &+&\dfrac{22}{5}u &=&-6\\ &5x&+&5y &&&+&5t &+&2u &=&-6\\ &-5x&+&5y &+&\dfrac{4}{3}z &-&4t &-&4u &=&-8\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{6}{cr}} &-\dfrac{21066093792000}{89638446528000}x&+&\dfrac{5938632288000}{53783067916800}y &-&\dfrac{2038703904000}{9561434296320}z &+&\dfrac{207481774752000}{1075661358336000}t &-&\dfrac{155294496000}{11951792870400}u &=&4\\ &-\dfrac{46119456000}{5975896435200}x&+&\dfrac{188886816000}{17927689305600}y &+&\dfrac{87270048000}{3187144765440}z &+&\dfrac{8296813152000}{71710757222400}t &+&\dfrac{78871968000}{796786191360}u &=&-9\\ &-\dfrac{557352576000}{2987948217600}x&-&\dfrac{577228032000}{1792768930560}y &+&\dfrac{774232992000}{1991965478400}z &+&\dfrac{16243916256000}{44819223264000}t &+&\dfrac{891736128000}{2987948217600}u &=&-6\\ &\dfrac{153334944000}{597589643520}x&-&\dfrac{83910816000}{1792768930560}y &+&\dfrac{546365088000}{23903585740800}z &-&\dfrac{7070413536000}{35855378611200}t &-&\dfrac{1357479648000}{9959827392000}u &=&6\\ &-\dfrac{310588992000}{8963844652800}x&-&\dfrac{996992064000}{5378306791680}y &+&\dfrac{324655776000}{796786191360}z &+&\dfrac{59569191072000}{268915339584000}t &+&\dfrac{375744096000}{2987948217600}u &=&-3\\ \end{array} \right. \)
Cliquer ici pour afficher la solution
  1. \( \widehat{A}_{5, 5}=\begin{pmatrix}1 & \dfrac{22}{3} & 0 & 5 \\ -2 & \dfrac{18}{5} & -\dfrac{21}{4} & -5 \\ -2 & -2 & -\dfrac{14}{5} & -\dfrac{10}{3} \\ 5 & 5 & 0 & 5\end{pmatrix}\) \( \widehat{A}_{3, 3}=\begin{pmatrix}1 & \dfrac{22}{3} & 5 & -\dfrac{1}{4} \\ -2 & \dfrac{18}{5} & -5 & 4 \\ 5 & 5 & 5 & 2 \\ -5 & 5 & -4 & -4\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(-2\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-2\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(5\right)L_1\) et \( L_{5}\leftarrow L_{5}-\left(-5\right)L_1\)
  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 & \dfrac{22}{3} & 0 & 5 & -\dfrac{1}{4} \\ 0 & \dfrac{274}{15} & -\dfrac{21}{4} & 5 & \dfrac{7}{2} \\ 0 & \dfrac{38}{3} & -\dfrac{14}{5} & \dfrac{20}{3} & \dfrac{39}{10} \\ 0 & -\dfrac{95}{3} & 0 & -20 & \dfrac{13}{4} \\ 0 & \dfrac{125}{3} & \dfrac{4}{3} & 21 & -\dfrac{21}{4}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{274}{15} & -\dfrac{21}{4} & 5 & \dfrac{7}{2} \\ \dfrac{38}{3} & -\dfrac{14}{5} & \dfrac{20}{3} & \dfrac{39}{10} \\ -\dfrac{95}{3} & 0 & -20 & \dfrac{13}{4} \\ \dfrac{125}{3} & \dfrac{4}{3} & 21 & -\dfrac{21}{4}\end{pmatrix}\\ &=&\dfrac{199196547840}{69984000} \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{199196547840}{69984000}\right)^{-1}{}^tCo\begin{pmatrix}1 & \dfrac{22}{3} & 0 & 5 & -\dfrac{1}{4} \\ -2 & \dfrac{18}{5} & -\dfrac{21}{4} & -5 & 4 \\ -2 & -2 & -\dfrac{14}{5} & -\dfrac{10}{3} & \dfrac{22}{5} \\ 5 & 5 & 0 & 5 & 2 \\ -5 & 5 & \dfrac{4}{3} & -4 & -4\end{pmatrix} =\dfrac{69984000}{199196547840}\begin{pmatrix}-\dfrac{4.1962931598401E+24}{6.2732570418156E+21} & \dfrac{1.1829550506608E+24}{3.7639542250893E+21} & -\dfrac{4.0610277974473E+23}{6.6914741779366E+20} & \dfrac{4.1329653270315E+25}{7.5279084501787E+22} & -\dfrac{3.0934127501753E+22}{8.3643427224207E+20} \\ -\dfrac{9.1868364234588E+21}{4.1821713612104E+20} & \dfrac{3.7625601679689E+22}{1.2546514083631E+21} & \dfrac{1.7383892291431E+22}{2.2304913926455E+20} & \dfrac{1.6526965379519E+24}{5.0186056334524E+21} & \dfrac{1.5711023746947E+22}{5.5762284816138E+19} \\ -\dfrac{1.1102270906893E+23}{2.0910856806052E+20} & -\dfrac{1.1498183129088E+23}{1.2546514083631E+20} & \dfrac{1.5422453923023E+23}{1.3940571204035E+20} & \dfrac{3.2357320415973E+24}{3.1366285209078E+21} & \dfrac{1.7763075828181E+23}{2.0910856806052E+20} \\ \dfrac{3.054379150804E+22}{4.1821713612104E+19} & -\dfrac{1.6714744873637E+22}{1.2546514083631E+20} & \dfrac{1.088340393899E+23}{1.6728685444841E+21} & -\dfrac{1.4084019681724E+24}{2.5093028167262E+21} & -\dfrac{2.7040525964466E+23}{6.9702856020173E+20} \\ -\dfrac{6.1868255003505E+22}{6.2732570418156E+20} & -\dfrac{1.9859737737268E+23}{3.7639542250893E+20} & \dfrac{6.4670309815516E+22}{5.5762284816138E+19} & \dfrac{1.1865977219164E+25}{1.8819771125447E+22} & \dfrac{7.4846926794462E+22}{2.0910856806052E+20}\end{pmatrix} =\begin{pmatrix}-\dfrac{21066093792000}{89638446528000} & \dfrac{5938632288000}{53783067916800} & -\dfrac{2038703904000}{9561434296320} & \dfrac{207481774752000}{1075661358336000} & -\dfrac{155294496000}{11951792870400} \\ -\dfrac{46119456000}{5975896435200} & \dfrac{188886816000}{17927689305600} & \dfrac{87270048000}{3187144765440} & \dfrac{8296813152000}{71710757222400} & \dfrac{78871968000}{796786191360} \\ -\dfrac{557352576000}{2987948217600} & -\dfrac{577228032000}{1792768930560} & \dfrac{774232992000}{1991965478400} & \dfrac{16243916256000}{44819223264000} & \dfrac{891736128000}{2987948217600} \\ \dfrac{153334944000}{597589643520} & -\dfrac{83910816000}{1792768930560} & \dfrac{546365088000}{23903585740800} & -\dfrac{7070413536000}{35855378611200} & -\dfrac{1357479648000}{9959827392000} \\ -\dfrac{310588992000}{8963844652800} & -\dfrac{996992064000}{5378306791680} & \dfrac{324655776000}{796786191360} & \dfrac{59569191072000}{268915339584000} & \dfrac{375744096000}{2987948217600}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{4, 5}=B_{4, 5}= \left(\dfrac{199196547840}{69984000}\right)^{-1}Co(A)_{5, 4}= \left(\dfrac{199196547840}{69984000}\right)^{-1}\times(-1)^{5+4}\det\begin{pmatrix}1 & \dfrac{22}{3} & 0 & -\dfrac{1}{4} \\ -2 & \dfrac{18}{5} & -\dfrac{21}{4} & 4 \\ -2 & -2 & -\dfrac{14}{5} & \dfrac{22}{5} \\ 5 & 5 & 0 & 2\end{pmatrix}=-\dfrac{1357479648000}{9959827392000}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & \dfrac{22}{3} & 0 & 5 & -\dfrac{1}{4} \\ -2 & \dfrac{18}{5} & -\dfrac{21}{4} & -5 & 4 \\ -2 & -2 & -\dfrac{14}{5} & -\dfrac{10}{3} & \dfrac{22}{5} \\ 5 & 5 & 0 & 5 & 2 \\ -5 & 5 & \dfrac{4}{3} & -4 & -4\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{6}{cr}} &x&+&\dfrac{22}{3}y &&&+&5t &-&\dfrac{1}{4}u &=&7\\ &-2x&+&\dfrac{18}{5}y &-&\dfrac{21}{4}z &-&5t &+&4u &=&5\\ &-2x&-&2y &-&\dfrac{14}{5}z &-&\dfrac{10}{3}t &+&\dfrac{22}{5}u &=&-6\\ &5x&+&5y &&&+&5t &+&2u &=&-6\\ &-5x&+&5y &+&\dfrac{4}{3}z &-&4t &-&4u &=&-8\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}7 \\ 5 \\ -6 \\ -6 \\ -8\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{21066093792000}{89638446528000} & \dfrac{5938632288000}{53783067916800} & -\dfrac{2038703904000}{9561434296320} & \dfrac{207481774752000}{1075661358336000} & -\dfrac{155294496000}{11951792870400} \\ -\dfrac{46119456000}{5975896435200} & \dfrac{188886816000}{17927689305600} & \dfrac{87270048000}{3187144765440} & \dfrac{8296813152000}{71710757222400} & \dfrac{78871968000}{796786191360} \\ -\dfrac{557352576000}{2987948217600} & -\dfrac{577228032000}{1792768930560} & \dfrac{774232992000}{1991965478400} & \dfrac{16243916256000}{44819223264000} & \dfrac{891736128000}{2987948217600} \\ \dfrac{153334944000}{597589643520} & -\dfrac{83910816000}{1792768930560} & \dfrac{546365088000}{23903585740800} & -\dfrac{7070413536000}{35855378611200} & -\dfrac{1357479648000}{9959827392000} \\ -\dfrac{310588992000}{8963844652800} & -\dfrac{996992064000}{5378306791680} & \dfrac{324655776000}{796786191360} & \dfrac{59569191072000}{268915339584000} & \dfrac{375744096000}{2987948217600}\end{pmatrix}\times\begin{pmatrix}7 \\ 5 \\ -6 \\ -6 \\ -8\end{pmatrix}=\begin{pmatrix}-\dfrac{5.1382002297211E+68}{5.9261353148209E+68} \\ -\dfrac{3.2225006885719E+64}{1.9509910501468E+64} \\ -\dfrac{1.4017737318535E+64}{1.4289485230567E+63} \\ \dfrac{3.3823550090487E+64}{9.1452705475632E+63} \\ -\dfrac{1.8362599954639E+65}{3.0865288098026E+64}\end{pmatrix}\) . Ainsi \( x=\dfrac{5.1382002297211E+68}{5.9261353148209E+68}\) , \( y=\dfrac{3.2225006885719E+64}{1.9509910501468E+64}\) , \( z=\dfrac{1.4017737318535E+64}{1.4289485230567E+63}\) , \( t=\dfrac{3.3823550090487E+64}{9.1452705475632E+63}\) et \( u=\dfrac{1.8362599954639E+65}{3.0865288098026E+64}\)
  8. Le système \( \left\{\begin{array}{*{6}{cr}} &-\dfrac{21066093792000}{89638446528000}x&+&\dfrac{5938632288000}{53783067916800}y &-&\dfrac{2038703904000}{9561434296320}z &+&\dfrac{207481774752000}{1075661358336000}t &-&\dfrac{155294496000}{11951792870400}u &=&4\\ &-\dfrac{46119456000}{5975896435200}x&+&\dfrac{188886816000}{17927689305600}y &+&\dfrac{87270048000}{3187144765440}z &+&\dfrac{8296813152000}{71710757222400}t &+&\dfrac{78871968000}{796786191360}u &=&-9\\ &-\dfrac{557352576000}{2987948217600}x&-&\dfrac{577228032000}{1792768930560}y &+&\dfrac{774232992000}{1991965478400}z &+&\dfrac{16243916256000}{44819223264000}t &+&\dfrac{891736128000}{2987948217600}u &=&-6\\ &\dfrac{153334944000}{597589643520}x&-&\dfrac{83910816000}{1792768930560}y &+&\dfrac{546365088000}{23903585740800}z &-&\dfrac{7070413536000}{35855378611200}t &-&\dfrac{1357479648000}{9959827392000}u &=&6\\ &-\dfrac{310588992000}{8963844652800}x&-&\dfrac{996992064000}{5378306791680}y &+&\dfrac{324655776000}{796786191360}z &+&\dfrac{59569191072000}{268915339584000}t &+&\dfrac{375744096000}{2987948217600}u &=&-3\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}4 \\ -9 \\ -6 \\ 6 \\ -3\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & \dfrac{22}{3} & 0 & 5 & -\dfrac{1}{4} \\ -2 & \dfrac{18}{5} & -\dfrac{21}{4} & -5 & 4 \\ -2 & -2 & -\dfrac{14}{5} & -\dfrac{10}{3} & \dfrac{22}{5} \\ 5 & 5 & 0 & 5 & 2 \\ -5 & 5 & \dfrac{4}{3} & -4 & -4\end{pmatrix}\times\begin{pmatrix}4 \\ -9 \\ -6 \\ 6 \\ -3\end{pmatrix}=\begin{pmatrix}-\dfrac{125}{4} \\ -\dfrac{509}{10} \\ -\dfrac{32}{5} \\ -1 \\ -85\end{pmatrix}\) . Ainsi \( x=\dfrac{125}{4}\) , \( y=\dfrac{509}{10}\) , \( z=\dfrac{32}{5}\) , \( t=1\) et \( u=85\)