Exercice

1. \( \widehat{A}_{3, 6}=\begin{pmatrix}1 & 3 & 1 & -\dfrac{8}{3} & -4 \\ -5 & -5 & \dfrac{5}{3} & 5 & 0 \\ -\dfrac{16}{5} & \dfrac{1}{2} & 3 & \dfrac{1}{5} & -3 \\ 5 & \dfrac{1}{3} & -\dfrac{7}{5} & -\dfrac{5}{2} & \dfrac{13}{5} \\ -5 & -1 & -\dfrac{17}{2} & -4 & -4\end{pmatrix}\) \( \widehat{A}_{2, 6}=\begin{pmatrix}1 & 3 & 1 & -\dfrac{8}{3} & -4 \\ \dfrac{2}{3} & -3 & -2 & \dfrac{7}{4} & 3 \\ -\dfrac{16}{5} & \dfrac{1}{2} & 3 & \dfrac{1}{5} & -3 \\ 5 & \dfrac{1}{3} & -\dfrac{7}{5} & -\dfrac{5}{2} & \dfrac{13}{5} \\ -5 & -1 & -\dfrac{17}{2} & -4 & -4\end{pmatrix}\)
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 : \( L_{2}\leftarrow L_{2}-\left(-5\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(\dfrac{2}{3}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(-\dfrac{16}{5}\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(5\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-5\right)L_1\)
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 & 3 & 1 & -\dfrac{8}{3} & -4 & 1 \\ 0 & 10 & \dfrac{20}{3} & -\dfrac{25}{3} & -20 & 4 \\ 0 & -5 & -\dfrac{8}{3} & \dfrac{127}{36} & \dfrac{17}{3} & \dfrac{1}{3} \\ 0 & \dfrac{101}{10} & \dfrac{31}{5} & -\dfrac{25}{3} & -\dfrac{79}{5} & \dfrac{1}{5} \\ 0 & -\dfrac{44}{3} & -\dfrac{32}{5} & \dfrac{65}{6} & \dfrac{113}{5} & -2 \\ 0 & 14 & -\dfrac{7}{2} & -\dfrac{52}{3} & -24 & 0\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}10 & \dfrac{20}{3} & -\dfrac{25}{3} & -20 & 4 \\ -5 & -\dfrac{8}{3} & \dfrac{127}{36} & \dfrac{17}{3} & \dfrac{1}{3} \\ \dfrac{101}{10} & \dfrac{31}{5} & -\dfrac{25}{3} & -\dfrac{79}{5} & \dfrac{1}{5} \\ -\dfrac{44}{3} & -\dfrac{32}{5} & \dfrac{65}{6} & \dfrac{113}{5} & -2 \\ 14 & -\dfrac{7}{2} & -\dfrac{52}{3} & -24 & 0\end{pmatrix}\\ &=&-\dfrac{3063383}{2700} \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{3063383}{2700}\right)^{-1}{}^tCo\begin{pmatrix}1 & 3 & 1 & -\dfrac{8}{3} & -4 & 1 \\ -5 & -5 & \dfrac{5}{3} & 5 & 0 & -1 \\ \dfrac{2}{3} & -3 & -2 & \dfrac{7}{4} & 3 & 1 \\ -\dfrac{16}{5} & \dfrac{1}{2} & 3 & \dfrac{1}{5} & -3 & -3 \\ 5 & \dfrac{1}{3} & -\dfrac{7}{5} & -\dfrac{5}{2} & \dfrac{13}{5} & 3 \\ -5 & -1 & -\dfrac{17}{2} & -4 & -4 & -5\end{pmatrix} =-\dfrac{2700}{3063383}\begin{pmatrix}-\dfrac{20169962802857700}{29776082760000} & \dfrac{23949813188619}{33084536400} & -\dfrac{16835093917587}{8271134100} & -\dfrac{7687207349455}{8271134100} & \dfrac{218128186515}{486537300} & \dfrac{1152168980130}{8271134100} \\ \dfrac{3367517808608100}{14888041380000} & \dfrac{1406837199069}{16542268200} & \dfrac{4710227066970}{8271134100} & \dfrac{4109589562160}{8271134100} & \dfrac{153107882340}{486537300} & \dfrac{271477001460}{8271134100} \\ \dfrac{1248683924928}{8271134100} & -\dfrac{1517679586158}{8271134100} & \dfrac{3231771036744}{8271134100} & -\dfrac{1029786829280}{8271134100} & -\dfrac{152924079360}{486537300} & \dfrac{257673397662}{8271134100} \\ -\dfrac{3434116980381300}{7444020690000} & \dfrac{4317339007071}{8271134100} & -\dfrac{10997741026512}{8271134100} & -\dfrac{1837110785100}{8271134100} & \dfrac{389294711640}{486537300} & \dfrac{1246919416320}{8271134100} \\ \dfrac{7587086496527700}{11166031035000} & -\dfrac{2237780144157300}{7444020690000} & \dfrac{13795846393229100}{11166031035000} & \dfrac{11622720172640}{24813402300} & -\dfrac{169542932135}{486537300} & -\dfrac{388755556232}{8271134100} \\ \dfrac{3590639922089700}{17865649656000} & -\dfrac{18046134073196100}{29776082760000} & \dfrac{11021914181765}{8271134100} & \dfrac{20958670886525}{24813402300} & -\dfrac{164580251675}{486537300} & -\dfrac{454177163580}{8271134100}\end{pmatrix} =\begin{pmatrix}\dfrac{6584211900}{11028178800} & -\dfrac{7818093}{12253532} & \dfrac{5495589}{3063383} & \dfrac{2509385}{3063383} & -\dfrac{71205}{180199} & -\dfrac{376110}{3063383} \\ -\dfrac{1099280700}{5514089400} & -\dfrac{459243}{6126766} & -\dfrac{1537590}{3063383} & -\dfrac{1341520}{3063383} & -\dfrac{49980}{180199} & -\dfrac{88620}{3063383} \\ -\dfrac{407616}{3063383} & \dfrac{495426}{3063383} & -\dfrac{1054968}{3063383} & \dfrac{336160}{3063383} & \dfrac{49920}{180199} & -\dfrac{84114}{3063383} \\ \dfrac{1121021100}{2757044700} & -\dfrac{1409337}{3063383} & \dfrac{3590064}{3063383} & \dfrac{599700}{3063383} & -\dfrac{127080}{180199} & -\dfrac{407040}{3063383} \\ -\dfrac{2476701900}{4135567050} & \dfrac{730493100}{2757044700} & -\dfrac{4503467700}{4135567050} & -\dfrac{3794080}{9190149} & \dfrac{55345}{180199} & \dfrac{126904}{3063383} \\ -\dfrac{1172115900}{6616907280} & \dfrac{5890916700}{11028178800} & -\dfrac{3597955}{3063383} & -\dfrac{6841675}{9190149} & \dfrac{53725}{180199} & \dfrac{148260}{3063383}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{5, 6}=B_{5, 6}= \left(-\dfrac{3063383}{2700}\right)^{-1}Co(A)_{6, 5}= \left(-\dfrac{3063383}{2700}\right)^{-1}\times(-1)^{6+5}\det\begin{pmatrix}1 & 3 & 1 & -\dfrac{8}{3} & 1 \\ -5 & -5 & \dfrac{5}{3} & 5 & -1 \\ \dfrac{2}{3} & -3 & -2 & \dfrac{7}{4} & 1 \\ -\dfrac{16}{5} & \dfrac{1}{2} & 3 & \dfrac{1}{5} & -3 \\ 5 & \dfrac{1}{3} & -\dfrac{7}{5} & -\dfrac{5}{2} & 3\end{pmatrix}=\dfrac{126904}{3063383}\) .
6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 3 & 1 & -\dfrac{8}{3} & -4 & 1 \\ -5 & -5 & \dfrac{5}{3} & 5 & 0 & -1 \\ \dfrac{2}{3} & -3 & -2 & \dfrac{7}{4} & 3 & 1 \\ -\dfrac{16}{5} & \dfrac{1}{2} & 3 & \dfrac{1}{5} & -3 & -3 \\ 5 & \dfrac{1}{3} & -\dfrac{7}{5} & -\dfrac{5}{2} & \dfrac{13}{5} & 3 \\ -5 & -1 & -\dfrac{17}{2} & -4 & -4 & -5\end{pmatrix}\]
7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&+&3y &+&z &-&\dfrac{8}{3}t &-&4u &+&v &=&-8\\ &-5x&-&5y &+&\dfrac{5}{3}z &+&5t &&&-&v &=&8\\ &\dfrac{2}{3}x&-&3y &-&2z &+&\dfrac{7}{4}t &+&3u &+&v &=&-2\\ &-\dfrac{16}{5}x&+&\dfrac{1}{2}y &+&3z &+&\dfrac{1}{5}t &-&3u &-&3v &=&-8\\ &5x&+&\dfrac{1}{3}y &-&\dfrac{7}{5}z &-&\dfrac{5}{2}t &+&\dfrac{13}{5}u &+&3v &=&-6\\ &-5x&-&y &-&\dfrac{17}{2}z &-&4t &-&4u &-&5v &=&-3\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-8 \\ 8 \\ -2 \\ -8 \\ -6 \\ -3\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{6584211900}{11028178800} & -\dfrac{7818093}{12253532} & \dfrac{5495589}{3063383} & \dfrac{2509385}{3063383} & -\dfrac{71205}{180199} & -\dfrac{376110}{3063383} \\ -\dfrac{1099280700}{5514089400} & -\dfrac{459243}{6126766} & -\dfrac{1537590}{3063383} & -\dfrac{1341520}{3063383} & -\dfrac{49980}{180199} & -\dfrac{88620}{3063383} \\ -\dfrac{407616}{3063383} & \dfrac{495426}{3063383} & -\dfrac{1054968}{3063383} & \dfrac{336160}{3063383} & \dfrac{49920}{180199} & -\dfrac{84114}{3063383} \\ \dfrac{1121021100}{2757044700} & -\dfrac{1409337}{3063383} & \dfrac{3590064}{3063383} & \dfrac{599700}{3063383} & -\dfrac{127080}{180199} & -\dfrac{407040}{3063383} \\ -\dfrac{2476701900}{4135567050} & \dfrac{730493100}{2757044700} & -\dfrac{4503467700}{4135567050} & -\dfrac{3794080}{9190149} & \dfrac{55345}{180199} & \dfrac{126904}{3063383} \\ -\dfrac{1172115900}{6616907280} & \dfrac{5890916700}{11028178800} & -\dfrac{3597955}{3063383} & -\dfrac{6841675}{9190149} & \dfrac{53725}{180199} & \dfrac{148260}{3063383}\end{pmatrix}\times\begin{pmatrix}-8 \\ 8 \\ -2 \\ -8 \\ -6 \\ -3\end{pmatrix}=\begin{pmatrix}-\dfrac{3.0245924274217E+42}{1.7500939644718E+41} \\ \dfrac{6.3470479987792E+41}{8.7504698223592E+40} \\ \dfrac{2.8651912610359E+37}{4.8613721235329E+37} \\ -\dfrac{2.7184730491347E+41}{4.3752349111796E+40} \\ \dfrac{2.4936472521088E+48}{2.3921596876875E+47} \\ \dfrac{1.3678012150058E+46}{1.1340608889778E+45}\end{pmatrix}\) . Ainsi \( x=\dfrac{3.0245924274217E+42}{1.7500939644718E+41}\) , \( y=\dfrac{6.3470479987792E+41}{8.7504698223592E+40}\) , \( z=\dfrac{2.8651912610359E+37}{4.8613721235329E+37}\) , \( t=\dfrac{2.7184730491347E+41}{4.3752349111796E+40}\) , \( u=\dfrac{2.4936472521088E+48}{2.3921596876875E+47}\) et \( v=\dfrac{1.3678012150058E+46}{1.1340608889778E+45}\)
8. Le système \( \left\{\begin{array}{*{7}{cr}} &\dfrac{6584211900}{11028178800}x&-&\dfrac{7818093}{12253532}y &+&\dfrac{5495589}{3063383}z &+&\dfrac{2509385}{3063383}t &-&\dfrac{71205}{180199}u &-&\dfrac{376110}{3063383}v &=&-5\\ &-\dfrac{1099280700}{5514089400}x&-&\dfrac{459243}{6126766}y &-&\dfrac{1537590}{3063383}z &-&\dfrac{1341520}{3063383}t &-&\dfrac{49980}{180199}u &-&\dfrac{88620}{3063383}v &=&\dfrac{53}{4}\\ &-\dfrac{407616}{3063383}x&+&\dfrac{495426}{3063383}y &-&\dfrac{1054968}{3063383}z &+&\dfrac{336160}{3063383}t &+&\dfrac{49920}{180199}u &-&\dfrac{84114}{3063383}v &=&-1\\ &\dfrac{1121021100}{2757044700}x&-&\dfrac{1409337}{3063383}y &+&\dfrac{3590064}{3063383}z &+&\dfrac{599700}{3063383}t &-&\dfrac{127080}{180199}u &-&\dfrac{407040}{3063383}v &=&-7\\ &-\dfrac{2476701900}{4135567050}x&+&\dfrac{730493100}{2757044700}y &-&\dfrac{4503467700}{4135567050}z &-&\dfrac{3794080}{9190149}t &+&\dfrac{55345}{180199}u &+&\dfrac{126904}{3063383}v &=&-9\\ &-\dfrac{1172115900}{6616907280}x&+&\dfrac{5890916700}{11028178800}y &-&\dfrac{3597955}{3063383}z &-&\dfrac{6841675}{9190149}t &+&\dfrac{53725}{180199}u &+&\dfrac{148260}{3063383}v &=&3\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-5 \\ \dfrac{53}{4} \\ -1 \\ -7 \\ -9 \\ 3\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 3 & 1 & -\dfrac{8}{3} & -4 & 1 \\ -5 & -5 & \dfrac{5}{3} & 5 & 0 & -1 \\ \dfrac{2}{3} & -3 & -2 & \dfrac{7}{4} & 3 & 1 \\ -\dfrac{16}{5} & \dfrac{1}{2} & 3 & \dfrac{1}{5} & -3 & -3 \\ 5 & \dfrac{1}{3} & -\dfrac{7}{5} & -\dfrac{5}{2} & \dfrac{13}{5} & 3 \\ -5 & -1 & -\dfrac{17}{2} & -4 & -4 & -5\end{pmatrix}\times\begin{pmatrix}-5 \\ \dfrac{53}{4} \\ -1 \\ -7 \\ -9 \\ 3\end{pmatrix}=\begin{pmatrix}\dfrac{1097}{12} \\ -\dfrac{971}{12} \\ -\dfrac{232}{3} \\ \dfrac{1449}{40} \\ -\dfrac{193}{12} \\ \dfrac{277}{4}\end{pmatrix}\) . Ainsi \( x=\dfrac{1097}{12}\) , \( y=\dfrac{971}{12}\) , \( z=\dfrac{232}{3}\) , \( t=\dfrac{1449}{40}\) , \( u=\dfrac{193}{12}\) et \( v=\dfrac{277}{4}\)