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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 \[ A= \begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} & 2 \\ -\dfrac{25}{4} & -3 & -\dfrac{24}{5} & -4 & 5 & 5 \\ -\dfrac{15}{2} & \dfrac{8}{5} & 5 & -\dfrac{18}{5} & 4 & -5 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & \dfrac{3}{2} & -\dfrac{9}{4} & -\dfrac{5}{4} \\ 2 & -2 & 2 & -4 & -2 & -1 \\ -4 & 5 & 5 & 2 & -\dfrac{15}{4} & -5\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (2, 6)\) et \( (5, 5)\) : \( \widehat{A}_{2, 6}=\) \( \widehat{A}_{5, 5}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} & 2 \\ 0 & \dfrac{63}{4} & -\dfrac{269}{30} & -4 & -\dfrac{585}{8} & \dfrac{35}{2} \\ 0 & \dfrac{241}{10} & 0 & -\dfrac{18}{5} & -\dfrac{359}{4} & 10 \\ 0 & -\dfrac{38}{5} & \dfrac{13}{6} & \dfrac{3}{2} & \dfrac{191}{4} & -\dfrac{37}{4} \\ 0 & -8 & \dfrac{10}{3} & -4 & 23 & -5 \\ 0 & 17 & \dfrac{7}{3} & 2 & -\dfrac{215}{4} & 3\end{pmatrix} \)
  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 \( A^{-1}_{4, 3}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&+&3y &-&\dfrac{2}{3}z &&&-&\dfrac{25}{2}u &+&2v &=&-5\\ &-\dfrac{25}{4}x&-&3y &-&\dfrac{24}{5}z &-&4t &+&5u &+&5v &=&1\\ &-\dfrac{15}{2}x&+&\dfrac{8}{5}y &+&5z &-&\dfrac{18}{5}t &+&4u &-&5v &=&-2\\ &4x&+&\dfrac{22}{5}y &-&\dfrac{1}{2}z &+&\dfrac{3}{2}t &-&\dfrac{9}{4}u &-&\dfrac{5}{4}v &=&-3\\ &2x&-&2y &+&2z &-&4t &-&2u &-&v &=&-2\\ &-4x&+&5y &+&5z &+&2t &-&\dfrac{15}{4}u &-&5v &=&4\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &\dfrac{2808612000}{5532963800}x&-&\dfrac{2303844000}{3319778280}y &+&\dfrac{7366524000}{8299445700}z &-&\dfrac{3341120}{27664819}t &-&\dfrac{2487756000}{3319778280}u &-&\dfrac{16577508000}{13832409500}v &=&4\\ &\dfrac{34898988000}{22131855200}x&-&\dfrac{51185100}{27664819}y &+&\dfrac{8955276000}{3319778280}z &-&\dfrac{1041204000}{2766481900}t &-&\dfrac{79378836000}{33197782800}u &-&\dfrac{111036108000}{33197782800}v &=&9\\ &\dfrac{7952172000}{2213185520}x&-&\dfrac{49423212000}{11065927600}y &+&\dfrac{26989332000}{4426371040}z &-&\dfrac{39594900}{27664819}t &-&\dfrac{11956068000}{2213185520}u &-&\dfrac{212691600}{27664819}v &=&8\\ &-\dfrac{3620244000}{22131855200}x&+&\dfrac{2856400}{27664819}y &-&\dfrac{2584812000}{8299445700}z &-&\dfrac{2289468000}{33197782800}t &+&\dfrac{10193532000}{132791131200}u &+&\dfrac{9714560}{27664819}v &=&-9\\ &\dfrac{22210284000}{27664819000}x&-&\dfrac{29061800}{27664819}y &+&\dfrac{24703788000}{16598891400}z &-&\dfrac{7455360}{27664819}t &-&\dfrac{18424116000}{13832409500}u &-&\dfrac{31270308000}{16598891400}v &=&5\\ &\dfrac{226647204000}{55329638000}x&-&\dfrac{81867156000}{16598891400}y &+&\dfrac{75736164000}{11065927600}z &-&\dfrac{6072240}{3952117}t &-&\dfrac{409270092000}{66395565600}u &-&\dfrac{8142421455000000}{933687641250000}v &=&5\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{2, 6}=\begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} \\ -\dfrac{15}{2} & \dfrac{8}{5} & 5 & -\dfrac{18}{5} & 4 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & \dfrac{3}{2} & -\dfrac{9}{4} \\ 2 & -2 & 2 & -4 & -2 \\ -4 & 5 & 5 & 2 & -\dfrac{15}{4}\end{pmatrix}\) \( \widehat{A}_{5, 5}=\begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & 2 \\ -\dfrac{25}{4} & -3 & -\dfrac{24}{5} & -4 & 5 \\ -\dfrac{15}{2} & \dfrac{8}{5} & 5 & -\dfrac{18}{5} & -5 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & \dfrac{3}{2} & -\dfrac{5}{4} \\ -4 & 5 & 5 & 2 & -5\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(-\dfrac{25}{4}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{15}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(4\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(2\right)L_1\) 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 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} & 2 \\ 0 & \dfrac{63}{4} & -\dfrac{269}{30} & -4 & -\dfrac{585}{8} & \dfrac{35}{2} \\ 0 & \dfrac{241}{10} & 0 & -\dfrac{18}{5} & -\dfrac{359}{4} & 10 \\ 0 & -\dfrac{38}{5} & \dfrac{13}{6} & \dfrac{3}{2} & \dfrac{191}{4} & -\dfrac{37}{4} \\ 0 & -8 & \dfrac{10}{3} & -4 & 23 & -5 \\ 0 & 17 & \dfrac{7}{3} & 2 & -\dfrac{215}{4} & 3\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{63}{4} & -\dfrac{269}{30} & -4 & -\dfrac{585}{8} & \dfrac{35}{2} \\ \dfrac{241}{10} & 0 & -\dfrac{18}{5} & -\dfrac{359}{4} & 10 \\ -\dfrac{38}{5} & \dfrac{13}{6} & \dfrac{3}{2} & \dfrac{191}{4} & -\dfrac{37}{4} \\ -8 & \dfrac{10}{3} & -4 & 23 & -5 \\ 17 & \dfrac{7}{3} & 2 & -\dfrac{215}{4} & 3\end{pmatrix}\\ &=&-\dfrac{27664819}{12000} \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{27664819}{12000}\right)^{-1}{}^tCo\begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} & 2 \\ -\dfrac{25}{4} & -3 & -\dfrac{24}{5} & -4 & 5 & 5 \\ -\dfrac{15}{2} & \dfrac{8}{5} & 5 & -\dfrac{18}{5} & 4 & -5 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & \dfrac{3}{2} & -\dfrac{9}{4} & -\dfrac{5}{4} \\ 2 & -2 & 2 & -4 & -2 & -1 \\ -4 & 5 & 5 & 2 & -\dfrac{15}{4} & -5\end{pmatrix} =-\dfrac{12000}{27664819}\begin{pmatrix}-\dfrac{77699742621228000}{66395565600000} & \dfrac{63735427264236000}{39837339360000} & -\dfrac{203793553119156000}{99593348400000} & \dfrac{92431480057280}{331977828000} & \dfrac{68823319456164000}{39837339360000} & \dfrac{458613758291052000}{165988914000000} \\ -\dfrac{965474186303172000}{265582262400000} & \dfrac{1416026526996900}{331977828000} & -\dfrac{247746089635044000}{39837339360000} & \dfrac{28804720202076000}{33197782800000} & \dfrac{2196001130370684000}{398373393600000} & \dfrac{3071793830284452000}{398373393600000} \\ -\dfrac{219995399036868000}{26558226240000} & \dfrac{1367284214378628000}{132791131200000} & -\dfrac{746654984710908000}{53116452480000} & \dfrac{1095385741823100}{331977828000} & \dfrac{330762457171692000}{26558226240000} & \dfrac{5884074616820400}{331977828000} \\ \dfrac{100153394995836000}{265582262400000} & -\dfrac{79021788991600}{331977828000} & \dfrac{71508356129028000}{99593348400000} & \dfrac{63337717826292000}{398373393600000} & -\dfrac{282002217750708000}{1593493574400000} & -\dfrac{268751544064640}{331977828000} \\ -\dfrac{614443486798596000}{331977828000000} & \dfrac{803989436814200}{331977828000} & -\dfrac{683425823634372000}{199186696800000} & \dfrac{206251184979840}{331977828000} & \dfrac{509699834375004000}{165988914000000} & \dfrac{865087410894252000}{199186696800000} \\ -\dfrac{6270153875516076000}{663955656000000} & \dfrac{2264840052784764000}{199186696800000} & -\dfrac{2095227268814316000}{132791131200000} & \dfrac{167987420524560}{47425404000} & \dfrac{1.1322383017293E+19}{796746787200000} & \dfrac{2.2525861577429E+23}{1.1204251695E+19}\end{pmatrix} =\begin{pmatrix}\dfrac{2808612000}{5532963800} & -\dfrac{2303844000}{3319778280} & \dfrac{7366524000}{8299445700} & -\dfrac{3341120}{27664819} & -\dfrac{2487756000}{3319778280} & -\dfrac{16577508000}{13832409500} \\ \dfrac{34898988000}{22131855200} & -\dfrac{51185100}{27664819} & \dfrac{8955276000}{3319778280} & -\dfrac{1041204000}{2766481900} & -\dfrac{79378836000}{33197782800} & -\dfrac{111036108000}{33197782800} \\ \dfrac{7952172000}{2213185520} & -\dfrac{49423212000}{11065927600} & \dfrac{26989332000}{4426371040} & -\dfrac{39594900}{27664819} & -\dfrac{11956068000}{2213185520} & -\dfrac{212691600}{27664819} \\ -\dfrac{3620244000}{22131855200} & \dfrac{2856400}{27664819} & -\dfrac{2584812000}{8299445700} & -\dfrac{2289468000}{33197782800} & \dfrac{10193532000}{132791131200} & \dfrac{9714560}{27664819} \\ \dfrac{22210284000}{27664819000} & -\dfrac{29061800}{27664819} & \dfrac{24703788000}{16598891400} & -\dfrac{7455360}{27664819} & -\dfrac{18424116000}{13832409500} & -\dfrac{31270308000}{16598891400} \\ \dfrac{226647204000}{55329638000} & -\dfrac{81867156000}{16598891400} & \dfrac{75736164000}{11065927600} & -\dfrac{6072240}{3952117} & -\dfrac{409270092000}{66395565600} & -\dfrac{8142421455000000}{933687641250000}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{4, 3}=B_{4, 3}= \left(-\dfrac{27664819}{12000}\right)^{-1}Co(A)_{3, 4}= \left(-\dfrac{27664819}{12000}\right)^{-1}\times(-1)^{3+4}\det\begin{pmatrix}1 & 3 & -\dfrac{2}{3} & -\dfrac{25}{2} & 2 \\ -\dfrac{25}{4} & -3 & -\dfrac{24}{5} & 5 & 5 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & -\dfrac{9}{4} & -\dfrac{5}{4} \\ 2 & -2 & 2 & -2 & -1 \\ -4 & 5 & 5 & -\dfrac{15}{4} & -5\end{pmatrix}=-\dfrac{2584812000}{8299445700}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} & 2 \\ -\dfrac{25}{4} & -3 & -\dfrac{24}{5} & -4 & 5 & 5 \\ -\dfrac{15}{2} & \dfrac{8}{5} & 5 & -\dfrac{18}{5} & 4 & -5 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & \dfrac{3}{2} & -\dfrac{9}{4} & -\dfrac{5}{4} \\ 2 & -2 & 2 & -4 & -2 & -1 \\ -4 & 5 & 5 & 2 & -\dfrac{15}{4} & -5\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&+&3y &-&\dfrac{2}{3}z &&&-&\dfrac{25}{2}u &+&2v &=&-5\\ &-\dfrac{25}{4}x&-&3y &-&\dfrac{24}{5}z &-&4t &+&5u &+&5v &=&1\\ &-\dfrac{15}{2}x&+&\dfrac{8}{5}y &+&5z &-&\dfrac{18}{5}t &+&4u &-&5v &=&-2\\ &4x&+&\dfrac{22}{5}y &-&\dfrac{1}{2}z &+&\dfrac{3}{2}t &-&\dfrac{9}{4}u &-&\dfrac{5}{4}v &=&-3\\ &2x&-&2y &+&2z &-&4t &-&2u &-&v &=&-2\\ &-4x&+&5y &+&5z &+&2t &-&\dfrac{15}{4}u &-&5v &=&4\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-5 \\ 1 \\ -2 \\ -3 \\ -2 \\ 4\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{2808612000}{5532963800} & -\dfrac{2303844000}{3319778280} & \dfrac{7366524000}{8299445700} & -\dfrac{3341120}{27664819} & -\dfrac{2487756000}{3319778280} & -\dfrac{16577508000}{13832409500} \\ \dfrac{34898988000}{22131855200} & -\dfrac{51185100}{27664819} & \dfrac{8955276000}{3319778280} & -\dfrac{1041204000}{2766481900} & -\dfrac{79378836000}{33197782800} & -\dfrac{111036108000}{33197782800} \\ \dfrac{7952172000}{2213185520} & -\dfrac{49423212000}{11065927600} & \dfrac{26989332000}{4426371040} & -\dfrac{39594900}{27664819} & -\dfrac{11956068000}{2213185520} & -\dfrac{212691600}{27664819} \\ -\dfrac{3620244000}{22131855200} & \dfrac{2856400}{27664819} & -\dfrac{2584812000}{8299445700} & -\dfrac{2289468000}{33197782800} & \dfrac{10193532000}{132791131200} & \dfrac{9714560}{27664819} \\ \dfrac{22210284000}{27664819000} & -\dfrac{29061800}{27664819} & \dfrac{24703788000}{16598891400} & -\dfrac{7455360}{27664819} & -\dfrac{18424116000}{13832409500} & -\dfrac{31270308000}{16598891400} \\ \dfrac{226647204000}{55329638000} & -\dfrac{81867156000}{16598891400} & \dfrac{75736164000}{11065927600} & -\dfrac{6072240}{3952117} & -\dfrac{409270092000}{66395565600} & -\dfrac{8142421455000000}{933687641250000}\end{pmatrix}\times\begin{pmatrix}-5 \\ 1 \\ -2 \\ -3 \\ -2 \\ 4\end{pmatrix}=\begin{pmatrix}-\dfrac{1.537695910971E+57}{1.9366482407443E+56} \\ -\dfrac{1.4004040035415E+59}{6.1972743703816E+57} \\ -\dfrac{9.2327496823279E+54}{1.836229443076E+53} \\ \dfrac{1.8604178090708E+57}{6.1972743703816E+56} \\ -\dfrac{9.7674655938709E+56}{8.0693676697677E+55} \\ -\dfrac{1.4204580800271E+65}{2.4899763095283E+63}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.537695910971E+57}{1.9366482407443E+56}\) , \( y=\dfrac{1.4004040035415E+59}{6.1972743703816E+57}\) , \( z=\dfrac{9.2327496823279E+54}{1.836229443076E+53}\) , \( t=\dfrac{1.8604178090708E+57}{6.1972743703816E+56}\) , \( u=\dfrac{9.7674655938709E+56}{8.0693676697677E+55}\) et \( v=\dfrac{1.4204580800271E+65}{2.4899763095283E+63}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{2808612000}{5532963800}x&-&\dfrac{2303844000}{3319778280}y &+&\dfrac{7366524000}{8299445700}z &-&\dfrac{3341120}{27664819}t &-&\dfrac{2487756000}{3319778280}u &-&\dfrac{16577508000}{13832409500}v &=&4\\ &\dfrac{34898988000}{22131855200}x&-&\dfrac{51185100}{27664819}y &+&\dfrac{8955276000}{3319778280}z &-&\dfrac{1041204000}{2766481900}t &-&\dfrac{79378836000}{33197782800}u &-&\dfrac{111036108000}{33197782800}v &=&9\\ &\dfrac{7952172000}{2213185520}x&-&\dfrac{49423212000}{11065927600}y &+&\dfrac{26989332000}{4426371040}z &-&\dfrac{39594900}{27664819}t &-&\dfrac{11956068000}{2213185520}u &-&\dfrac{212691600}{27664819}v &=&8\\ &-\dfrac{3620244000}{22131855200}x&+&\dfrac{2856400}{27664819}y &-&\dfrac{2584812000}{8299445700}z &-&\dfrac{2289468000}{33197782800}t &+&\dfrac{10193532000}{132791131200}u &+&\dfrac{9714560}{27664819}v &=&-9\\ &\dfrac{22210284000}{27664819000}x&-&\dfrac{29061800}{27664819}y &+&\dfrac{24703788000}{16598891400}z &-&\dfrac{7455360}{27664819}t &-&\dfrac{18424116000}{13832409500}u &-&\dfrac{31270308000}{16598891400}v &=&5\\ &\dfrac{226647204000}{55329638000}x&-&\dfrac{81867156000}{16598891400}y &+&\dfrac{75736164000}{11065927600}z &-&\dfrac{6072240}{3952117}t &-&\dfrac{409270092000}{66395565600}u &-&\dfrac{8142421455000000}{933687641250000}v &=&5\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}4 \\ 9 \\ 8 \\ -9 \\ 5 \\ 5\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 3 & -\dfrac{2}{3} & 0 & -\dfrac{25}{2} & 2 \\ -\dfrac{25}{4} & -3 & -\dfrac{24}{5} & -4 & 5 & 5 \\ -\dfrac{15}{2} & \dfrac{8}{5} & 5 & -\dfrac{18}{5} & 4 & -5 \\ 4 & \dfrac{22}{5} & -\dfrac{1}{2} & \dfrac{3}{2} & -\dfrac{9}{4} & -\dfrac{5}{4} \\ 2 & -2 & 2 & -4 & -2 & -1 \\ -4 & 5 & 5 & 2 & -\dfrac{15}{4} & -5\end{pmatrix}\times\begin{pmatrix}4 \\ 9 \\ 8 \\ -9 \\ 5 \\ 5\end{pmatrix}=\begin{pmatrix}-\dfrac{161}{6} \\ -\dfrac{22}{5} \\ \dfrac{259}{5} \\ \dfrac{103}{5} \\ 27 \\ \dfrac{29}{4}\end{pmatrix}\) . Ainsi \( x=\dfrac{161}{6}\) , \( y=\dfrac{22}{5}\) , \( z=\dfrac{259}{5}\) , \( t=\dfrac{103}{5}\) , \( u=27\) et \( v=\dfrac{29}{4}\)