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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 \[ A= \begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & -\dfrac{17}{3} & \dfrac{24}{5} & -5 & -1 \\ \dfrac{9}{2} & 3 & 4 & -\dfrac{7}{5} & -\dfrac{9}{4} & -2 \\ 2 & 5 & \dfrac{1}{5} & 3 & 1 & 0 \\ -\dfrac{2}{5} & 0 & -\dfrac{23}{2} & 4 & 0 & -5 \\ -4 & -5 & -\dfrac{19}{4} & 1 & -\dfrac{18}{5} & -1\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (3, 4)\) et \( (5, 3)\) : \( \widehat{A}_{3, 4}=\) \( \widehat{A}_{5, 3}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ 0 & -24 & -\dfrac{778}{15} & -\dfrac{183}{10} & -\dfrac{281}{10} & 20 \\ 0 & -6 & -\dfrac{79}{5} & -\dfrac{113}{10} & -\dfrac{243}{20} & 7 \\ 0 & 1 & -\dfrac{43}{5} & -\dfrac{7}{5} & -\dfrac{17}{5} & 4 \\ 0 & \dfrac{4}{5} & -\dfrac{487}{50} & \dfrac{122}{25} & \dfrac{22}{25} & -\dfrac{29}{5} \\ 0 & 3 & \dfrac{257}{20} & \dfrac{49}{5} & \dfrac{26}{5} & -9\end{pmatrix} \)
  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 \( A^{-1}_{4, 6}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&+&2y &+&\dfrac{22}{5}z &+&\dfrac{11}{5}t &+&\dfrac{11}{5}u &-&2v &=&-\dfrac{11}{2}\\ &\dfrac{21}{2}x&-&3y &-&\dfrac{17}{3}z &+&\dfrac{24}{5}t &-&5u &-&v &=&-1\\ &\dfrac{9}{2}x&+&3y &+&4z &-&\dfrac{7}{5}t &-&\dfrac{9}{4}u &-&2v &=&9\\ &2x&+&5y &+&\dfrac{1}{5}z &+&3t &+&u &&&=&-2\\ &-\dfrac{2}{5}x&&&-&\dfrac{23}{2}z &+&4t &&&-&5v &=&-8\\ &-4x&-&5y &-&\dfrac{19}{4}z &+&t &-&\dfrac{18}{5}u &-&v &=&-1\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{73806400000}{10019875552000}x&+&\dfrac{656001600000}{10019875552000}y &+&\dfrac{4004800000}{2504968888000}z &-&\dfrac{765300800000}{10019875552000}t &+&\dfrac{638385600000}{50099377760000}u &-&\dfrac{589406400000}{5009937776000}v &=&-2\\ &-\dfrac{146546100800000}{1502981332800000}x&-&\dfrac{13241508800000}{250496888800000}y &+&\dfrac{100756800000}{1252484444000}z &+&\dfrac{102031053440000000}{601192533120000000}t &+&\dfrac{286619200000}{18787266660000}u &+&\dfrac{1339771200000}{125248444400000}v &=&5\\ &\dfrac{132078400000}{1252484444000}x&-&\dfrac{974400000}{1252484444000}y &+&\dfrac{25636800000}{1252484444000}z &-&\dfrac{76241600000}{25049688880000}t &-&\dfrac{303553600000}{5009937776000}u &+&\dfrac{51988800000}{1001987555200}v &=&2\\ &\dfrac{16113504128000000}{120238506624000000}x&+&\dfrac{10728212800000}{250496888800000}y &-&\dfrac{11564132800000}{125248444400000}z &+&\dfrac{6998118400000}{46968166650000}t &-&\dfrac{3722916800000}{75149066640000}u &+&\dfrac{7609332800000}{62624222200000}v &=&-8\\ &\dfrac{29729300800000}{375745333200000}x&+&\dfrac{630900800000}{125248444400000}y &-&\dfrac{16599217600000}{125248444400000}z &-&\dfrac{13357566400000}{93936333300000}t &+&\dfrac{4431870400000}{75149066640000}u &-&\dfrac{12092126400000}{62624222200000}v &=&8\\ &-\dfrac{506289164800000000}{3757453332000000000}x&+&\dfrac{19296977600000}{626242222000000}y &-&\dfrac{37909473600000}{313121111000000}z &+&\dfrac{4971419536000000000}{3.757453332E+19}t &-&\dfrac{1.141823076E+26}{1.1272359996E+27}u &-&\dfrac{15930347200000}{1252484444000000}v &=&4\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{3, 4}=\begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & -\dfrac{17}{3} & -5 & -1 \\ 2 & 5 & \dfrac{1}{5} & 1 & 0 \\ -\dfrac{2}{5} & 0 & -\dfrac{23}{2} & 0 & -5 \\ -4 & -5 & -\dfrac{19}{4} & -\dfrac{18}{5} & -1\end{pmatrix}\) \( \widehat{A}_{5, 3}=\begin{pmatrix}1 & 2 & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & \dfrac{24}{5} & -5 & -1 \\ \dfrac{9}{2} & 3 & -\dfrac{7}{5} & -\dfrac{9}{4} & -2 \\ 2 & 5 & 3 & 1 & 0 \\ -4 & -5 & 1 & -\dfrac{18}{5} & -1\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{21}{2}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(\dfrac{9}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(2\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-\dfrac{2}{5}\right)L_1\) 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 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ 0 & -24 & -\dfrac{778}{15} & -\dfrac{183}{10} & -\dfrac{281}{10} & 20 \\ 0 & -6 & -\dfrac{79}{5} & -\dfrac{113}{10} & -\dfrac{243}{20} & 7 \\ 0 & 1 & -\dfrac{43}{5} & -\dfrac{7}{5} & -\dfrac{17}{5} & 4 \\ 0 & \dfrac{4}{5} & -\dfrac{487}{50} & \dfrac{122}{25} & \dfrac{22}{25} & -\dfrac{29}{5} \\ 0 & 3 & \dfrac{257}{20} & \dfrac{49}{5} & \dfrac{26}{5} & -9\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-24 & -\dfrac{778}{15} & -\dfrac{183}{10} & -\dfrac{281}{10} & 20 \\ -6 & -\dfrac{79}{5} & -\dfrac{113}{10} & -\dfrac{243}{20} & 7 \\ 1 & -\dfrac{43}{5} & -\dfrac{7}{5} & -\dfrac{17}{5} & 4 \\ \dfrac{4}{5} & -\dfrac{487}{50} & \dfrac{122}{25} & \dfrac{22}{25} & -\dfrac{29}{5} \\ 3 & \dfrac{257}{20} & \dfrac{49}{5} & \dfrac{26}{5} & -9\end{pmatrix}\\ &=&-\dfrac{125248444400}{1600000} \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{125248444400}{1600000}\right)^{-1}{}^tCo\begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & -\dfrac{17}{3} & \dfrac{24}{5} & -5 & -1 \\ \dfrac{9}{2} & 3 & 4 & -\dfrac{7}{5} & -\dfrac{9}{4} & -2 \\ 2 & 5 & \dfrac{1}{5} & 3 & 1 & 0 \\ -\dfrac{2}{5} & 0 & -\dfrac{23}{2} & 4 & 0 & -5 \\ -4 & -5 & -\dfrac{19}{4} & 1 & -\dfrac{18}{5} & -1\end{pmatrix} =-\dfrac{1600000}{125248444400}\begin{pmatrix}\dfrac{9.2441367867642E+21}{1.60318008832E+19} & -\dfrac{8.2163179923911E+22}{1.60318008832E+19} & -\dfrac{5.0159497013312E+20}{4007950220800000000} & \dfrac{9.5852734698076E+22}{1.60318008832E+19} & -\dfrac{7.9956803327361E+22}{8.0159004416E+19} & \dfrac{7.3822234719404E+22}{8015900441600000000} \\ \dfrac{1.8354671158086E+25}{2.40477013248E+21} & \dfrac{1.6584783787089E+24}{4.0079502208E+20} & -\dfrac{1.2619632462722E+22}{2003975110400000000} & -\dfrac{1.2779230723853E+28}{9.61908052992E+23} & -\dfrac{3.5898608935172E+22}{3.0059626656E+19} & -\dfrac{1.6780425865192E+23}{2.0039751104E+20} \\ -\dfrac{1.6542614138841E+22}{2003975110400000000} & \dfrac{1.2204208422336E+20}{2003975110400000000} & -\dfrac{3.2109693193939E+21}{2003975110400000000} & \dfrac{9.549141798567E+21}{4.0079502208E+19} & \dfrac{3.801961619202E+22}{8015900441600000000} & -\dfrac{6.5115163262227E+21}{1603180088320000000} \\ -\dfrac{2.018191325865E+27}{1.923816105984E+23} & -\dfrac{1.3436919643922E+24}{4.0079502208E+20} & \dfrac{1.448389644035E+24}{2.0039751104E+20} & -\dfrac{8.7650344332702E+23}{7.514906664E+19} & \dfrac{4.6628953783063E+23}{1.20238506624E+20} & -\dfrac{9.530570961219E+23}{1.0019875552E+20} \\ -\dfrac{3.7235486782997E+24}{6.0119253312E+20} & -\dfrac{7.9019343770716E+22}{2.0039751104E+20} & \dfrac{2.0790261826571E+24}{2.0039751104E+20} & \dfrac{1.6730144125697E+24}{1.5029813328E+20} & -\dfrac{5.5508487338241E+23}{1.20238506624E+20} & \dfrac{1.5145200210882E+24}{1.0019875552E+20} \\ \dfrac{6.3411930307775E+28}{6.0119253312E+24} & -\dfrac{2.4169164260216E+24}{1.0019875552E+21} & \dfrac{4.7481025964229E+24}{5.009937776E+20} & -\dfrac{6.2266256334377E+29}{6.0119253312E+25} & \dfrac{1.4301156404902E+37}{1.80357759936E+33} & \dfrac{1.9952512055519E+24}{2.0039751104E+21}\end{pmatrix} =\begin{pmatrix}-\dfrac{73806400000}{10019875552000} & \dfrac{656001600000}{10019875552000} & \dfrac{4004800000}{2504968888000} & -\dfrac{765300800000}{10019875552000} & \dfrac{638385600000}{50099377760000} & -\dfrac{589406400000}{5009937776000} \\ -\dfrac{146546100800000}{1502981332800000} & -\dfrac{13241508800000}{250496888800000} & \dfrac{100756800000}{1252484444000} & \dfrac{102031053440000000}{601192533120000000} & \dfrac{286619200000}{18787266660000} & \dfrac{1339771200000}{125248444400000} \\ \dfrac{132078400000}{1252484444000} & -\dfrac{974400000}{1252484444000} & \dfrac{25636800000}{1252484444000} & -\dfrac{76241600000}{25049688880000} & -\dfrac{303553600000}{5009937776000} & \dfrac{51988800000}{1001987555200} \\ \dfrac{16113504128000000}{120238506624000000} & \dfrac{10728212800000}{250496888800000} & -\dfrac{11564132800000}{125248444400000} & \dfrac{6998118400000}{46968166650000} & -\dfrac{3722916800000}{75149066640000} & \dfrac{7609332800000}{62624222200000} \\ \dfrac{29729300800000}{375745333200000} & \dfrac{630900800000}{125248444400000} & -\dfrac{16599217600000}{125248444400000} & -\dfrac{13357566400000}{93936333300000} & \dfrac{4431870400000}{75149066640000} & -\dfrac{12092126400000}{62624222200000} \\ -\dfrac{506289164800000000}{3757453332000000000} & \dfrac{19296977600000}{626242222000000} & -\dfrac{37909473600000}{313121111000000} & \dfrac{4971419536000000000}{3.757453332E+19} & -\dfrac{1.141823076E+26}{1.1272359996E+27} & -\dfrac{15930347200000}{1252484444000000}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{4, 6}=B_{4, 6}= \left(-\dfrac{125248444400}{1600000}\right)^{-1}Co(A)_{6, 4}= \left(-\dfrac{125248444400}{1600000}\right)^{-1}\times(-1)^{6+4}\det\begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & -\dfrac{17}{3} & -5 & -1 \\ \dfrac{9}{2} & 3 & 4 & -\dfrac{9}{4} & -2 \\ 2 & 5 & \dfrac{1}{5} & 1 & 0 \\ -\dfrac{2}{5} & 0 & -\dfrac{23}{2} & 0 & -5\end{pmatrix}=\dfrac{7609332800000}{62624222200000}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & -\dfrac{17}{3} & \dfrac{24}{5} & -5 & -1 \\ \dfrac{9}{2} & 3 & 4 & -\dfrac{7}{5} & -\dfrac{9}{4} & -2 \\ 2 & 5 & \dfrac{1}{5} & 3 & 1 & 0 \\ -\dfrac{2}{5} & 0 & -\dfrac{23}{2} & 4 & 0 & -5 \\ -4 & -5 & -\dfrac{19}{4} & 1 & -\dfrac{18}{5} & -1\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&+&2y &+&\dfrac{22}{5}z &+&\dfrac{11}{5}t &+&\dfrac{11}{5}u &-&2v &=&-\dfrac{11}{2}\\ &\dfrac{21}{2}x&-&3y &-&\dfrac{17}{3}z &+&\dfrac{24}{5}t &-&5u &-&v &=&-1\\ &\dfrac{9}{2}x&+&3y &+&4z &-&\dfrac{7}{5}t &-&\dfrac{9}{4}u &-&2v &=&9\\ &2x&+&5y &+&\dfrac{1}{5}z &+&3t &+&u &&&=&-2\\ &-\dfrac{2}{5}x&&&-&\dfrac{23}{2}z &+&4t &&&-&5v &=&-8\\ &-4x&-&5y &-&\dfrac{19}{4}z &+&t &-&\dfrac{18}{5}u &-&v &=&-1\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-\dfrac{11}{2} \\ -1 \\ 9 \\ -2 \\ -8 \\ -1\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{73806400000}{10019875552000} & \dfrac{656001600000}{10019875552000} & \dfrac{4004800000}{2504968888000} & -\dfrac{765300800000}{10019875552000} & \dfrac{638385600000}{50099377760000} & -\dfrac{589406400000}{5009937776000} \\ -\dfrac{146546100800000}{1502981332800000} & -\dfrac{13241508800000}{250496888800000} & \dfrac{100756800000}{1252484444000} & \dfrac{102031053440000000}{601192533120000000} & \dfrac{286619200000}{18787266660000} & \dfrac{1339771200000}{125248444400000} \\ \dfrac{132078400000}{1252484444000} & -\dfrac{974400000}{1252484444000} & \dfrac{25636800000}{1252484444000} & -\dfrac{76241600000}{25049688880000} & -\dfrac{303553600000}{5009937776000} & \dfrac{51988800000}{1001987555200} \\ \dfrac{16113504128000000}{120238506624000000} & \dfrac{10728212800000}{250496888800000} & -\dfrac{11564132800000}{125248444400000} & \dfrac{6998118400000}{46968166650000} & -\dfrac{3722916800000}{75149066640000} & \dfrac{7609332800000}{62624222200000} \\ \dfrac{29729300800000}{375745333200000} & \dfrac{630900800000}{125248444400000} & -\dfrac{16599217600000}{125248444400000} & -\dfrac{13357566400000}{93936333300000} & \dfrac{4431870400000}{75149066640000} & -\dfrac{12092126400000}{62624222200000} \\ -\dfrac{506289164800000000}{3757453332000000000} & \dfrac{19296977600000}{626242222000000} & -\dfrac{37909473600000}{313121111000000} & \dfrac{4971419536000000000}{3.757453332E+19} & -\dfrac{1.141823076E+26}{1.1272359996E+27} & -\dfrac{15930347200000}{1252484444000000}\end{pmatrix}\times\begin{pmatrix}-\dfrac{11}{2} \\ -1 \\ 9 \\ -2 \\ -8 \\ -1\end{pmatrix}=\begin{pmatrix}\dfrac{1.9973613211996E+77}{1.2649809301264E+78} \\ \dfrac{1.1219837624117E+87}{1.3341595747427E+87} \\ \dfrac{2.170652807105E+73}{4.9413317583064E+74} \\ -\dfrac{2.7250923970552E+87}{1.6676994684284E+87} \\ -\dfrac{8.4806880369533E+84}{5.2115608388388E+84} \\ \dfrac{1.4000736914307E+109}{7.8173412582582E+109}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.9973613211996E+77}{1.2649809301264E+78}\) , \( y=\dfrac{1.1219837624117E+87}{1.3341595747427E+87}\) , \( z=\dfrac{2.170652807105E+73}{4.9413317583064E+74}\) , \( t=\dfrac{2.7250923970552E+87}{1.6676994684284E+87}\) , \( u=\dfrac{8.4806880369533E+84}{5.2115608388388E+84}\) et \( v=\dfrac{1.4000736914307E+109}{7.8173412582582E+109}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{73806400000}{10019875552000}x&+&\dfrac{656001600000}{10019875552000}y &+&\dfrac{4004800000}{2504968888000}z &-&\dfrac{765300800000}{10019875552000}t &+&\dfrac{638385600000}{50099377760000}u &-&\dfrac{589406400000}{5009937776000}v &=&-2\\ &-\dfrac{146546100800000}{1502981332800000}x&-&\dfrac{13241508800000}{250496888800000}y &+&\dfrac{100756800000}{1252484444000}z &+&\dfrac{102031053440000000}{601192533120000000}t &+&\dfrac{286619200000}{18787266660000}u &+&\dfrac{1339771200000}{125248444400000}v &=&5\\ &\dfrac{132078400000}{1252484444000}x&-&\dfrac{974400000}{1252484444000}y &+&\dfrac{25636800000}{1252484444000}z &-&\dfrac{76241600000}{25049688880000}t &-&\dfrac{303553600000}{5009937776000}u &+&\dfrac{51988800000}{1001987555200}v &=&2\\ &\dfrac{16113504128000000}{120238506624000000}x&+&\dfrac{10728212800000}{250496888800000}y &-&\dfrac{11564132800000}{125248444400000}z &+&\dfrac{6998118400000}{46968166650000}t &-&\dfrac{3722916800000}{75149066640000}u &+&\dfrac{7609332800000}{62624222200000}v &=&-8\\ &\dfrac{29729300800000}{375745333200000}x&+&\dfrac{630900800000}{125248444400000}y &-&\dfrac{16599217600000}{125248444400000}z &-&\dfrac{13357566400000}{93936333300000}t &+&\dfrac{4431870400000}{75149066640000}u &-&\dfrac{12092126400000}{62624222200000}v &=&8\\ &-\dfrac{506289164800000000}{3757453332000000000}x&+&\dfrac{19296977600000}{626242222000000}y &-&\dfrac{37909473600000}{313121111000000}z &+&\dfrac{4971419536000000000}{3.757453332E+19}t &-&\dfrac{1.141823076E+26}{1.1272359996E+27}u &-&\dfrac{15930347200000}{1252484444000000}v &=&4\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-2 \\ 5 \\ 2 \\ -8 \\ 8 \\ 4\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 2 & \dfrac{22}{5} & \dfrac{11}{5} & \dfrac{11}{5} & -2 \\ \dfrac{21}{2} & -3 & -\dfrac{17}{3} & \dfrac{24}{5} & -5 & -1 \\ \dfrac{9}{2} & 3 & 4 & -\dfrac{7}{5} & -\dfrac{9}{4} & -2 \\ 2 & 5 & \dfrac{1}{5} & 3 & 1 & 0 \\ -\dfrac{2}{5} & 0 & -\dfrac{23}{2} & 4 & 0 & -5 \\ -4 & -5 & -\dfrac{19}{4} & 1 & -\dfrac{18}{5} & -1\end{pmatrix}\times\begin{pmatrix}-2 \\ 5 \\ 2 \\ -8 \\ 8 \\ 4\end{pmatrix}=\begin{pmatrix}\dfrac{44}{5} \\ -\dfrac{1946}{15} \\ -\dfrac{4}{5} \\ \dfrac{27}{5} \\ -\dfrac{371}{5} \\ -\dfrac{673}{10}\end{pmatrix}\) . Ainsi \( x=\dfrac{44}{5}\) , \( y=\dfrac{1946}{15}\) , \( z=\dfrac{4}{5}\) , \( t=\dfrac{27}{5}\) , \( u=\dfrac{371}{5}\) et \( v=\dfrac{673}{10}\)