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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 \[ A= \begin{pmatrix}1 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ \dfrac{7}{4} & -5 & -\dfrac{13}{4} & 1 & \dfrac{1}{4} & 3 \\ -\dfrac{9}{5} & 0 & -1 & -3 & 0 & 0 \\ -\dfrac{25}{4} & -3 & -4 & \dfrac{3}{2} & -2 & -\dfrac{25}{2} \\ -\dfrac{9}{4} & 3 & -3 & 5 & -3 & 4 \\ 4 & -\dfrac{5}{2} & -5 & 1 & -\dfrac{4}{5} & 4\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (3, 1)\) et \( (2, 1)\) : \( \widehat{A}_{3, 1}=\) \( \widehat{A}_{2, 1}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ 0 & \dfrac{67}{16} & \dfrac{101}{12} & \dfrac{1}{8} & \dfrac{1}{4} & -\dfrac{19}{10} \\ 0 & -\dfrac{189}{20} & -13 & -\dfrac{21}{10} & 0 & \dfrac{126}{25} \\ 0 & -\dfrac{573}{16} & -\dfrac{137}{3} & \dfrac{37}{8} & -2 & 5 \\ 0 & -\dfrac{141}{16} & -18 & \dfrac{49}{8} & -3 & \dfrac{103}{10} \\ 0 & \dfrac{37}{2} & \dfrac{65}{3} & -1 & -\dfrac{4}{5} & -\dfrac{36}{5}\end{pmatrix} \)
  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 \( A^{-1}_{5, 3}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{21}{4}y &-&\dfrac{20}{3}z &+&\dfrac{1}{2}t &&&+&\dfrac{14}{5}v &=&0\\ &\dfrac{7}{4}x&-&5y &-&\dfrac{13}{4}z &+&t &+&\dfrac{1}{4}u &+&3v &=&4\\ &-\dfrac{9}{5}x&&&-&z &-&3t &&&&&=&0\\ &-\dfrac{25}{4}x&-&3y &-&4z &+&\dfrac{3}{2}t &-&2u &-&\dfrac{25}{2}v &=&-5\\ &-\dfrac{9}{4}x&+&3y &-&3z &+&5t &-&3u &+&4v &=&7\\ &4x&-&\dfrac{5}{2}y &-&5z &+&t &-&\dfrac{4}{5}u &+&4v &=&5\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{1264016000}{10793595920}x&-&\dfrac{292912000}{5396797960}y &-&\dfrac{26083952000}{323807877600}z &+&\dfrac{91792000}{6745997450}t &-&\dfrac{8485008000}{107935959200}u &+&\dfrac{1052656000}{4317438368}v &=&7\\ &\dfrac{1564016000}{5396797960}x&-&\dfrac{15785904000}{33729987250}y &-&\dfrac{3620560000}{25904630208}z &-&\dfrac{6975056000}{134919949000}t &-&\dfrac{352816000}{269839898000}u &-&\dfrac{2613136000}{215871918400}v &=&-9\\ &-\dfrac{85013616000}{215871918400}x&+&\dfrac{54952464000}{134919949000}y &+&\dfrac{3070832000}{26983989800}z &+&\dfrac{2566704000}{134919949000}t &+&\dfrac{4913328000}{269839898000}u &+&\dfrac{4938864000}{431743836800}v &=&-\dfrac{32}{3}\\ &\dfrac{43506064000}{215871918400}x&-&\dfrac{6961904000}{67459974500}y &-&\dfrac{418275664000}{1295231510400}z &-&\dfrac{1957072000}{134919949000}t &+&\dfrac{22179472000}{539679796000}u &-&\dfrac{64805648000}{431743836800}v &=&-6\\ &\dfrac{46007504000}{43174383680}x&-&\dfrac{89931888000}{107935959200}y &-&\dfrac{39133136000}{64761575520}z &-&\dfrac{9551216000}{53967979600}t &-&\dfrac{15457552000}{107935959200}u &-&\dfrac{36673128960000}{69079013888000}v &=&-2\\ &-\dfrac{6755312000}{215871918400}x&+&\dfrac{1403728000}{10793595920}y &+&\dfrac{4939184000}{51809260416}z &-&\dfrac{11637136000}{215871918400}t &+&\dfrac{13305968000}{215871918400}u &-&\dfrac{4812496000}{86348767360}v &=&-2\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{3, 1}=\begin{pmatrix}-\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ -5 & -\dfrac{13}{4} & 1 & \dfrac{1}{4} & 3 \\ -3 & -4 & \dfrac{3}{2} & -2 & -\dfrac{25}{2} \\ 3 & -3 & 5 & -3 & 4 \\ -\dfrac{5}{2} & -5 & 1 & -\dfrac{4}{5} & 4\end{pmatrix}\) \( \widehat{A}_{2, 1}=\begin{pmatrix}-\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ 0 & -1 & -3 & 0 & 0 \\ -3 & -4 & \dfrac{3}{2} & -2 & -\dfrac{25}{2} \\ 3 & -3 & 5 & -3 & 4 \\ -\dfrac{5}{2} & -5 & 1 & -\dfrac{4}{5} & 4\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{7}{4}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{9}{5}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(-\dfrac{25}{4}\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-\dfrac{9}{4}\right)L_1\) 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 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ 0 & \dfrac{67}{16} & \dfrac{101}{12} & \dfrac{1}{8} & \dfrac{1}{4} & -\dfrac{19}{10} \\ 0 & -\dfrac{189}{20} & -13 & -\dfrac{21}{10} & 0 & \dfrac{126}{25} \\ 0 & -\dfrac{573}{16} & -\dfrac{137}{3} & \dfrac{37}{8} & -2 & 5 \\ 0 & -\dfrac{141}{16} & -18 & \dfrac{49}{8} & -3 & \dfrac{103}{10} \\ 0 & \dfrac{37}{2} & \dfrac{65}{3} & -1 & -\dfrac{4}{5} & -\dfrac{36}{5}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}\dfrac{67}{16} & \dfrac{101}{12} & \dfrac{1}{8} & \dfrac{1}{4} & -\dfrac{19}{10} \\ -\dfrac{189}{20} & -13 & -\dfrac{21}{10} & 0 & \dfrac{126}{25} \\ -\dfrac{573}{16} & -\dfrac{137}{3} & \dfrac{37}{8} & -2 & 5 \\ -\dfrac{141}{16} & -18 & \dfrac{49}{8} & -3 & \dfrac{103}{10} \\ \dfrac{37}{2} & \dfrac{65}{3} & -1 & -\dfrac{4}{5} & -\dfrac{36}{5}\end{pmatrix}\\ &=&\dfrac{134919949}{16000} \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{134919949}{16000}\right)^{-1}{}^tCo\begin{pmatrix}1 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ \dfrac{7}{4} & -5 & -\dfrac{13}{4} & 1 & \dfrac{1}{4} & 3 \\ -\dfrac{9}{5} & 0 & -1 & -3 & 0 & 0 \\ -\dfrac{25}{4} & -3 & -4 & \dfrac{3}{2} & -2 & -\dfrac{25}{2} \\ -\dfrac{9}{4} & 3 & -3 & 5 & -3 & 4 \\ 4 & -\dfrac{5}{2} & -5 & 1 & -\dfrac{4}{5} & 4\end{pmatrix} =\dfrac{16000}{134919949}\begin{pmatrix}-\dfrac{170540974255184000}{172697534720000} & -\dfrac{39519672101488000}{86348767360000} & -\dfrac{3519245473558448000}{5180926041600000} & \dfrac{12384571958608000}{107935959200000} & -\dfrac{1144796846624592000}{1726975347200000} & \dfrac{142024293834544000}{69079013888000} \\ \dfrac{211016958955184000}{86348767360000} & -\dfrac{2129833362598896000}{539679796000000} & -\dfrac{488485770551440000}{414474083328000} & -\dfrac{941074199792144000}{2158719184000000} & -\dfrac{47601916726384000}{4317438368000000} & -\dfrac{352564175850064000}{3453950694400000} \\ -\dfrac{1.1470032735026E+19}{3453950694400000} & \dfrac{7414183640304336000}{2158719184000000} & \dfrac{414316496827568000}{431743836800000} & \dfrac{346299572778096000}{2158719184000000} & \dfrac{662905963180272000}{4317438368000000} & \dfrac{666351278997936000}{6907901388800000} \\ \dfrac{5869835936070736000}{3453950694400000} & -\dfrac{939299732622896000}{1079359592000000} & -\dfrac{5.6433731254821E+19}{20723704166400000} & -\dfrac{264048054429328000}{2158719184000000} & \dfrac{2992453231086928000}{8634876736000000} & -\dfrac{8743574723071952000}{6907901388800000} \\ \dfrac{6207330093297296000}{690790138880000} & -\dfrac{1.2133605742434E+19}{1726975347200000} & -\dfrac{5279840713330064000}{1036185208320000} & -\dfrac{1288649575607984000}{863487673600000} & -\dfrac{2085532127504848000}{1726975347200000} & -\dfrac{4.9479366889536E+21}{1105264222208000000} \\ -\dfrac{911426350519088000}{3453950694400000} & \dfrac{189390910169872000}{172697534720000} & \dfrac{666394453381616000}{828948166656000} & -\dfrac{1570081795626064000}{3453950694400000} & \dfrac{1795240523955632000}{3453950694400000} & -\dfrac{649301714882704000}{1381580277760000}\end{pmatrix} =\begin{pmatrix}-\dfrac{1264016000}{10793595920} & -\dfrac{292912000}{5396797960} & -\dfrac{26083952000}{323807877600} & \dfrac{91792000}{6745997450} & -\dfrac{8485008000}{107935959200} & \dfrac{1052656000}{4317438368} \\ \dfrac{1564016000}{5396797960} & -\dfrac{15785904000}{33729987250} & -\dfrac{3620560000}{25904630208} & -\dfrac{6975056000}{134919949000} & -\dfrac{352816000}{269839898000} & -\dfrac{2613136000}{215871918400} \\ -\dfrac{85013616000}{215871918400} & \dfrac{54952464000}{134919949000} & \dfrac{3070832000}{26983989800} & \dfrac{2566704000}{134919949000} & \dfrac{4913328000}{269839898000} & \dfrac{4938864000}{431743836800} \\ \dfrac{43506064000}{215871918400} & -\dfrac{6961904000}{67459974500} & -\dfrac{418275664000}{1295231510400} & -\dfrac{1957072000}{134919949000} & \dfrac{22179472000}{539679796000} & -\dfrac{64805648000}{431743836800} \\ \dfrac{46007504000}{43174383680} & -\dfrac{89931888000}{107935959200} & -\dfrac{39133136000}{64761575520} & -\dfrac{9551216000}{53967979600} & -\dfrac{15457552000}{107935959200} & -\dfrac{36673128960000}{69079013888000} \\ -\dfrac{6755312000}{215871918400} & \dfrac{1403728000}{10793595920} & \dfrac{4939184000}{51809260416} & -\dfrac{11637136000}{215871918400} & \dfrac{13305968000}{215871918400} & -\dfrac{4812496000}{86348767360}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{5, 3}=B_{5, 3}= \left(\dfrac{134919949}{16000}\right)^{-1}Co(A)_{3, 5}= \left(\dfrac{134919949}{16000}\right)^{-1}\times(-1)^{3+5}\det\begin{pmatrix}1 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & \dfrac{14}{5} \\ \dfrac{7}{4} & -5 & -\dfrac{13}{4} & 1 & 3 \\ -\dfrac{25}{4} & -3 & -4 & \dfrac{3}{2} & -\dfrac{25}{2} \\ -\dfrac{9}{4} & 3 & -3 & 5 & 4 \\ 4 & -\dfrac{5}{2} & -5 & 1 & 4\end{pmatrix}=-\dfrac{39133136000}{64761575520}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ \dfrac{7}{4} & -5 & -\dfrac{13}{4} & 1 & \dfrac{1}{4} & 3 \\ -\dfrac{9}{5} & 0 & -1 & -3 & 0 & 0 \\ -\dfrac{25}{4} & -3 & -4 & \dfrac{3}{2} & -2 & -\dfrac{25}{2} \\ -\dfrac{9}{4} & 3 & -3 & 5 & -3 & 4 \\ 4 & -\dfrac{5}{2} & -5 & 1 & -\dfrac{4}{5} & 4\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&-&\dfrac{21}{4}y &-&\dfrac{20}{3}z &+&\dfrac{1}{2}t &&&+&\dfrac{14}{5}v &=&0\\ &\dfrac{7}{4}x&-&5y &-&\dfrac{13}{4}z &+&t &+&\dfrac{1}{4}u &+&3v &=&4\\ &-\dfrac{9}{5}x&&&-&z &-&3t &&&&&=&0\\ &-\dfrac{25}{4}x&-&3y &-&4z &+&\dfrac{3}{2}t &-&2u &-&\dfrac{25}{2}v &=&-5\\ &-\dfrac{9}{4}x&+&3y &-&3z &+&5t &-&3u &+&4v &=&7\\ &4x&-&\dfrac{5}{2}y &-&5z &+&t &-&\dfrac{4}{5}u &+&4v &=&5\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}0 \\ 4 \\ 0 \\ -5 \\ 7 \\ 5\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{1264016000}{10793595920} & -\dfrac{292912000}{5396797960} & -\dfrac{26083952000}{323807877600} & \dfrac{91792000}{6745997450} & -\dfrac{8485008000}{107935959200} & \dfrac{1052656000}{4317438368} \\ \dfrac{1564016000}{5396797960} & -\dfrac{15785904000}{33729987250} & -\dfrac{3620560000}{25904630208} & -\dfrac{6975056000}{134919949000} & -\dfrac{352816000}{269839898000} & -\dfrac{2613136000}{215871918400} \\ -\dfrac{85013616000}{215871918400} & \dfrac{54952464000}{134919949000} & \dfrac{3070832000}{26983989800} & \dfrac{2566704000}{134919949000} & \dfrac{4913328000}{269839898000} & \dfrac{4938864000}{431743836800} \\ \dfrac{43506064000}{215871918400} & -\dfrac{6961904000}{67459974500} & -\dfrac{418275664000}{1295231510400} & -\dfrac{1957072000}{134919949000} & \dfrac{22179472000}{539679796000} & -\dfrac{64805648000}{431743836800} \\ \dfrac{46007504000}{43174383680} & -\dfrac{89931888000}{107935959200} & -\dfrac{39133136000}{64761575520} & -\dfrac{9551216000}{53967979600} & -\dfrac{15457552000}{107935959200} & -\dfrac{36673128960000}{69079013888000} \\ -\dfrac{6755312000}{215871918400} & \dfrac{1403728000}{10793595920} & \dfrac{4939184000}{51809260416} & -\dfrac{11637136000}{215871918400} & \dfrac{13305968000}{215871918400} & -\dfrac{4812496000}{86348767360}\end{pmatrix}\times\begin{pmatrix}0 \\ 4 \\ 0 \\ -5 \\ 7 \\ 5\end{pmatrix}=\begin{pmatrix}\dfrac{6.5090896599166E+39}{1.6965811381203E+40} \\ -\dfrac{4.4620640333297E+44}{2.6509080283129E+44} \\ \dfrac{3.6449403282727E+45}{2.1207264226503E+45} \\ -\dfrac{1.703161833918E+45}{2.1207264226503E+45} \\ -\dfrac{2.6514643845016E+47}{4.3432477135879E+46} \\ \dfrac{4.0937164492576E+43}{4.3432477135879E+43}\end{pmatrix}\) . Ainsi \( x=\dfrac{6.5090896599166E+39}{1.6965811381203E+40}\) , \( y=\dfrac{4.4620640333297E+44}{2.6509080283129E+44}\) , \( z=\dfrac{3.6449403282727E+45}{2.1207264226503E+45}\) , \( t=\dfrac{1.703161833918E+45}{2.1207264226503E+45}\) , \( u=\dfrac{2.6514643845016E+47}{4.3432477135879E+46}\) et \( v=\dfrac{4.0937164492576E+43}{4.3432477135879E+43}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{1264016000}{10793595920}x&-&\dfrac{292912000}{5396797960}y &-&\dfrac{26083952000}{323807877600}z &+&\dfrac{91792000}{6745997450}t &-&\dfrac{8485008000}{107935959200}u &+&\dfrac{1052656000}{4317438368}v &=&7\\ &\dfrac{1564016000}{5396797960}x&-&\dfrac{15785904000}{33729987250}y &-&\dfrac{3620560000}{25904630208}z &-&\dfrac{6975056000}{134919949000}t &-&\dfrac{352816000}{269839898000}u &-&\dfrac{2613136000}{215871918400}v &=&-9\\ &-\dfrac{85013616000}{215871918400}x&+&\dfrac{54952464000}{134919949000}y &+&\dfrac{3070832000}{26983989800}z &+&\dfrac{2566704000}{134919949000}t &+&\dfrac{4913328000}{269839898000}u &+&\dfrac{4938864000}{431743836800}v &=&-\dfrac{32}{3}\\ &\dfrac{43506064000}{215871918400}x&-&\dfrac{6961904000}{67459974500}y &-&\dfrac{418275664000}{1295231510400}z &-&\dfrac{1957072000}{134919949000}t &+&\dfrac{22179472000}{539679796000}u &-&\dfrac{64805648000}{431743836800}v &=&-6\\ &\dfrac{46007504000}{43174383680}x&-&\dfrac{89931888000}{107935959200}y &-&\dfrac{39133136000}{64761575520}z &-&\dfrac{9551216000}{53967979600}t &-&\dfrac{15457552000}{107935959200}u &-&\dfrac{36673128960000}{69079013888000}v &=&-2\\ &-\dfrac{6755312000}{215871918400}x&+&\dfrac{1403728000}{10793595920}y &+&\dfrac{4939184000}{51809260416}z &-&\dfrac{11637136000}{215871918400}t &+&\dfrac{13305968000}{215871918400}u &-&\dfrac{4812496000}{86348767360}v &=&-2\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}7 \\ -9 \\ -\dfrac{32}{3} \\ -6 \\ -2 \\ -2\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -\dfrac{21}{4} & -\dfrac{20}{3} & \dfrac{1}{2} & 0 & \dfrac{14}{5} \\ \dfrac{7}{4} & -5 & -\dfrac{13}{4} & 1 & \dfrac{1}{4} & 3 \\ -\dfrac{9}{5} & 0 & -1 & -3 & 0 & 0 \\ -\dfrac{25}{4} & -3 & -4 & \dfrac{3}{2} & -2 & -\dfrac{25}{2} \\ -\dfrac{9}{4} & 3 & -3 & 5 & -3 & 4 \\ 4 & -\dfrac{5}{2} & -5 & 1 & -\dfrac{4}{5} & 4\end{pmatrix}\times\begin{pmatrix}7 \\ -9 \\ -\dfrac{32}{3} \\ -6 \\ -2 \\ -2\end{pmatrix}=\begin{pmatrix}\dfrac{21017}{180} \\ \dfrac{953}{12} \\ \dfrac{241}{15} \\ \dfrac{551}{12} \\ -\dfrac{171}{4} \\ \dfrac{2743}{30}\end{pmatrix}\) . Ainsi \( x=\dfrac{21017}{180}\) , \( y=\dfrac{953}{12}\) , \( z=\dfrac{241}{15}\) , \( t=\dfrac{551}{12}\) , \( u=\dfrac{171}{4}\) et \( v=\dfrac{2743}{30}\)