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
\[ A= \begin{pmatrix}1 & -3 & \dfrac{7}{2} & 1 & -2 & -\dfrac{17}{3} \\ \dfrac{25}{3} & \dfrac{5}{2} & -5 & -\dfrac{24}{5} & \dfrac{10}{3} & -\dfrac{7}{4} \\ 3 & -\dfrac{5}{2} & \dfrac{14}{5} & \dfrac{6}{5} & 4 & -\dfrac{11}{4} \\ 5 & 2 & \dfrac{12}{5} & -2 & \dfrac{4}{3} & 5 \\ -5 & 2 & 5 & 0 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{10}{3} & -\dfrac{21}{4} & -4 & \dfrac{20}{3} & -2\end{pmatrix}\]
- Donner les mineurs d'ordre \( (2, 4)\) et \( (1, 2)\) :
\( \widehat{A}_{2, 4}=\)
\( \widehat{A}_{1, 2}=\)
- Expliquer pourquoi
\( \det(A)=\det
\begin{pmatrix}1 & -3 & \dfrac{7}{2} & 1 & -2 & -\dfrac{17}{3} \\ 0 & \dfrac{55}{2} & -\dfrac{205}{6} & -\dfrac{197}{15} & 20 & \dfrac{1637}{36} \\ 0 & \dfrac{13}{2} & -\dfrac{77}{10} & -\dfrac{9}{5} & 10 & \dfrac{57}{4} \\ 0 & 17 & -\dfrac{151}{10} & -7 & \dfrac{34}{3} & \dfrac{100}{3} \\ 0 & -13 & \dfrac{45}{2} & 5 & -15 & -35 \\ 0 & \dfrac{157}{15} & -\dfrac{427}{20} & -\dfrac{43}{5} & \dfrac{238}{15} & \dfrac{361}{15}\end{pmatrix}
\)
- Calculer \( \det(A)\) .
- Pourquoi la matrice \( A \) est inversible.
- Donner \( B=A^{-1}\) l'inverse de la matrice \( A \) , en ne détaillant que le calcul de \( A^{-1}_{3, 3}\) .
- Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&x&-&3y &+&\dfrac{7}{2}z &+&t &-&2u &-&\dfrac{17}{3}v &=&-\dfrac{25}{7}\\
&\dfrac{25}{3}x&+&\dfrac{5}{2}y &-&5z &-&\dfrac{24}{5}t &+&\dfrac{10}{3}u &-&\dfrac{7}{4}v &=&8\\
&3x&-&\dfrac{5}{2}y &+&\dfrac{14}{5}z &+&\dfrac{6}{5}t &+&4u &-&\dfrac{11}{4}v &=&1\\
&5x&+&2y &+&\dfrac{12}{5}z &-&2t &+&\dfrac{4}{3}u &+&5v &=&-3\\
&-5x&+&2y &+&5z &&&-&5u &-&\dfrac{20}{3}v &=&\dfrac{40}{3}\\
&\dfrac{23}{5}x&-&\dfrac{10}{3}y &-&\dfrac{21}{4}z &-&4t &+&\dfrac{20}{3}u &-&2v &=&8\\
\end{array}
\right.
\)
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{1.2905298739787E+27}{1.1493537842019E+28}x&+&\dfrac{1.0342377692658E+27}{1.1493537842019E+28}y &+&\dfrac{8.8706902790322E+25}{1.1493537842019E+28}z &+&\dfrac{2.9631006668829E+26}{2.2987075684039E+28}t &-&\dfrac{5.0206881978465E+26}{5.7467689210097E+27}u &-&\dfrac{1.6098638340988E+26}{1.9155896403366E+27}v &=&8\\
&-\dfrac{1.4055147543594E+26}{7.6623585613463E+26}x&+&\dfrac{1.6719910680421E+29}{1.3792245410423E+30}y &+&\dfrac{1.7327410207007E+29}{1.3792245410423E+30}z &-&\dfrac{3.0290795952427E+26}{5.7467689210097E+27}t &+&\dfrac{1.1004948446264E+32}{1.5516276086726E+33}u &-&\dfrac{7.3148294311373E+24}{5.7467689210097E+25}v &=&-9\\
&-\dfrac{1.9142002119329E+24}{9.5779482016828E+26}x&-&\dfrac{3.1245290939714E+25}{4.7889741008414E+26}y &+&\dfrac{2.4332677947644E+26}{4.3100766907573E+27}z &+&\dfrac{7.1461377480332E+26}{5.7467689210097E+27}t &+&\dfrac{1.1354756896263E+26}{1.4366922302524E+27}u &+&\dfrac{3.7416942720578E+25}{1.1493537842019E+27}v &=&-2\\
&-\dfrac{2.0310947633217E+27}{6.8961227052117E+28}x&+&\dfrac{1.3823206025366E+26}{1.5324717122693E+27}y &+&\dfrac{3.98195864857E+27}{2.2987075684039E+28}z &-&\dfrac{2.9495833751835E+29}{1.6550694492508E+30}t &-&\dfrac{1.1757990587892E+27}{8.6201533815146E+27}u &-&\dfrac{5.1612945196262E+26}{2.2987075684039E+27}v &=&8\\
&-\dfrac{9.7386700452455E+31}{5.1720920289087E+32}x&-&\dfrac{8.893931375887E+25}{4.7889741008414E+27}y &+&\dfrac{9.633063985526E+31}{5.1720920289087E+32}z &-&\dfrac{6.7320839207742E+29}{3.4480613526058E+31}t &+&\dfrac{4.2202276165768E+29}{6.8961227052117E+30}u &+&\dfrac{3.9126138605916E+25}{9.5779482016828E+26}v &=&5\\
&\dfrac{1.5291187095652E+24}{3.1926494005609E+27}x&-&\dfrac{1.2666750534795E+27}{1.9155896403366E+28}y &-&\dfrac{1.1283030408286E+29}{1.7240306763029E+30}z &+&\dfrac{8.7718676805709E+25}{1.0642164668536E+27}t &-&\dfrac{1.4315401875132E+27}{2.8733844605049E+28}u &+&\dfrac{8.9183659344722E+24}{4.7889741008414E+26}v &=&-\dfrac{23}{2}\\
\end{array}
\right.
\)
Cliquer ici pour afficher la solution
Exercice
-
\( \widehat{A}_{2, 4}=\begin{pmatrix}1 & -3 & \dfrac{7}{2} & -2 & -\dfrac{17}{3} \\ 3 & -\dfrac{5}{2} & \dfrac{14}{5} & 4 & -\dfrac{11}{4} \\ 5 & 2 & \dfrac{12}{5} & \dfrac{4}{3} & 5 \\ -5 & 2 & 5 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{10}{3} & -\dfrac{21}{4} & \dfrac{20}{3} & -2\end{pmatrix}\)
\( \widehat{A}_{1, 2}=\begin{pmatrix}\dfrac{25}{3} & -5 & -\dfrac{24}{5} & \dfrac{10}{3} & -\dfrac{7}{4} \\ 3 & \dfrac{14}{5} & \dfrac{6}{5} & 4 & -\dfrac{11}{4} \\ 5 & \dfrac{12}{5} & -2 & \dfrac{4}{3} & 5 \\ -5 & 5 & 0 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{21}{4} & -4 & \dfrac{20}{3} & -2\end{pmatrix}\)
- 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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{25}{3}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(3\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(5\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-5\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(\dfrac{23}{5}\right)L_1\)
- 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{7}{2} & 1 & -2 & -\dfrac{17}{3} \\ 0 & \dfrac{55}{2} & -\dfrac{205}{6} & -\dfrac{197}{15} & 20 & \dfrac{1637}{36} \\ 0 & \dfrac{13}{2} & -\dfrac{77}{10} & -\dfrac{9}{5} & 10 & \dfrac{57}{4} \\ 0 & 17 & -\dfrac{151}{10} & -7 & \dfrac{34}{3} & \dfrac{100}{3} \\ 0 & -13 & \dfrac{45}{2} & 5 & -15 & -35 \\ 0 & \dfrac{157}{15} & -\dfrac{427}{20} & -\dfrac{43}{5} & \dfrac{238}{15} & \dfrac{361}{15}\end{pmatrix}\\
&=&1\times\det\begin{pmatrix}\dfrac{55}{2} & -\dfrac{205}{6} & -\dfrac{197}{15} & 20 & \dfrac{1637}{36} \\ \dfrac{13}{2} & -\dfrac{77}{10} & -\dfrac{9}{5} & 10 & \dfrac{57}{4} \\ 17 & -\dfrac{151}{10} & -7 & \dfrac{34}{3} & \dfrac{100}{3} \\ -13 & \dfrac{45}{2} & 5 & -15 & -35 \\ \dfrac{157}{15} & -\dfrac{427}{20} & -\dfrac{43}{5} & \dfrac{238}{15} & \dfrac{361}{15}\end{pmatrix}\\
&=&-\dfrac{2.1284329337073E+24}{3.3059881728E+19}
\end{eqnarray*}
- On observe que le déterminant de \( A\) est non nul. D'après le cours, cela signifie que la matrice est inversible.
- D'après le cours
\(
B=A^{-1}=\left(-\dfrac{2.1284329337073E+24}{3.3059881728E+19}\right)^{-1}{}^tCo\begin{pmatrix}1 & -3 & \dfrac{7}{2} & 1 & -2 & -\dfrac{17}{3} \\ \dfrac{25}{3} & \dfrac{5}{2} & -5 & -\dfrac{24}{5} & \dfrac{10}{3} & -\dfrac{7}{4} \\ 3 & -\dfrac{5}{2} & \dfrac{14}{5} & \dfrac{6}{5} & 4 & -\dfrac{11}{4} \\ 5 & 2 & \dfrac{12}{5} & -2 & \dfrac{4}{3} & 5 \\ -5 & 2 & 5 & 0 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{10}{3} & -\dfrac{21}{4} & -4 & \dfrac{20}{3} & -2\end{pmatrix}
=-\dfrac{3.3059881728E+19}{2.1284329337073E+24}\begin{pmatrix}-\dfrac{2.7468062857094E+51}{3.7997500169345E+47} & -\dfrac{2.2013057293894E+51}{3.7997500169345E+47} & -\dfrac{1.8880669334609E+50}{3.7997500169345E+47} & -\dfrac{6.3067610452836E+50}{7.5995000338691E+47} & \dfrac{1.0686198110172E+51}{1.8998750084673E+47} & \dfrac{3.4264872032803E+50}{6.3329166948909E+46} \\ \dfrac{2.99154389199E+50}{2.5331666779564E+46} & -\dfrac{3.5587208540853E+53}{4.5597000203215E+49} & -\dfrac{3.6880230540449E+53}{4.5597000203215E+49} & \dfrac{6.4471927693354E+50}{1.8998750084673E+47} & -\dfrac{2.3423294706779E+56}{5.1296625228616E+52} & \dfrac{1.5569123865684E+49}{1.8998750084673E+45} \\ \dfrac{4.0742467727875E+48}{3.1664583474455E+46} & \dfrac{6.6503506259353E+49}{1.5832291737227E+46} & -\dfrac{5.1790473109058E+50}{1.4249062563505E+47} & -\dfrac{1.5210074931723E+51}{1.8998750084673E+47} & -\dfrac{2.4167838532247E+50}{4.7496875211682E+46} & -\dfrac{7.9639453165118E+49}{3.7997500169345E+46} \\ \dfrac{4.3230489857343E+51}{2.2798500101607E+48} & -\dfrac{2.9421766953811E+50}{5.0663333559127E+46} & -\dfrac{8.4753319282771E+51}{7.5995000338691E+47} & \dfrac{6.2779903964561E+53}{5.4716400243857E+49} & \dfrac{2.5026094401489E+51}{2.8498125127009E+47} & \dfrac{1.0985469236135E+51}{7.5995000338691E+46} \\ \dfrac{2.0728106054809E+56}{1.7098875076205E+52} & \dfrac{1.8930136450571E+50}{1.5832291737227E+47} & -\dfrac{2.0503330639303E+56}{1.7098875076205E+52} & \dfrac{1.4328789129457E+54}{1.1399250050804E+51} & -\dfrac{8.9824714468632E+53}{2.2798500101607E+50} & -\dfrac{8.3277361977628E+49}{3.1664583474455E+46} \\ -\dfrac{3.2546266209865E+48}{1.0554861158152E+47} & \dfrac{2.6960329001313E+51}{6.3329166948909E+47} & \dfrac{2.4015173513016E+53}{5.6996250254018E+49} & -\dfrac{1.867033206145E+50}{3.5182870527172E+46} & \dfrac{3.0469372810285E+51}{9.4993750423364E+47} & -\dfrac{1.8982143769784E+49}{1.5832291737227E+46}\end{pmatrix}
=\begin{pmatrix}\dfrac{1.2905298739787E+27}{1.1493537842019E+28} & \dfrac{1.0342377692658E+27}{1.1493537842019E+28} & \dfrac{8.8706902790322E+25}{1.1493537842019E+28} & \dfrac{2.9631006668829E+26}{2.2987075684039E+28} & -\dfrac{5.0206881978465E+26}{5.7467689210097E+27} & -\dfrac{1.6098638340988E+26}{1.9155896403366E+27} \\ -\dfrac{1.4055147543594E+26}{7.6623585613463E+26} & \dfrac{1.6719910680421E+29}{1.3792245410423E+30} & \dfrac{1.7327410207007E+29}{1.3792245410423E+30} & -\dfrac{3.0290795952427E+26}{5.7467689210097E+27} & \dfrac{1.1004948446264E+32}{1.5516276086726E+33} & -\dfrac{7.3148294311373E+24}{5.7467689210097E+25} \\ -\dfrac{1.9142002119329E+24}{9.5779482016828E+26} & -\dfrac{3.1245290939714E+25}{4.7889741008414E+26} & \dfrac{2.4332677947644E+26}{4.3100766907573E+27} & \dfrac{7.1461377480332E+26}{5.7467689210097E+27} & \dfrac{1.1354756896263E+26}{1.4366922302524E+27} & \dfrac{3.7416942720578E+25}{1.1493537842019E+27} \\ -\dfrac{2.0310947633217E+27}{6.8961227052117E+28} & \dfrac{1.3823206025366E+26}{1.5324717122693E+27} & \dfrac{3.98195864857E+27}{2.2987075684039E+28} & -\dfrac{2.9495833751835E+29}{1.6550694492508E+30} & -\dfrac{1.1757990587892E+27}{8.6201533815146E+27} & -\dfrac{5.1612945196262E+26}{2.2987075684039E+27} \\ -\dfrac{9.7386700452455E+31}{5.1720920289087E+32} & -\dfrac{8.893931375887E+25}{4.7889741008414E+27} & \dfrac{9.633063985526E+31}{5.1720920289087E+32} & -\dfrac{6.7320839207742E+29}{3.4480613526058E+31} & \dfrac{4.2202276165768E+29}{6.8961227052117E+30} & \dfrac{3.9126138605916E+25}{9.5779482016828E+26} \\ \dfrac{1.5291187095652E+24}{3.1926494005609E+27} & -\dfrac{1.2666750534795E+27}{1.9155896403366E+28} & -\dfrac{1.1283030408286E+29}{1.7240306763029E+30} & \dfrac{8.7718676805709E+25}{1.0642164668536E+27} & -\dfrac{1.4315401875132E+27}{2.8733844605049E+28} & \dfrac{8.9183659344722E+24}{4.7889741008414E+26}\end{pmatrix}\) .
Précisément on a calculé \( A^{-1}_{3, 3}=B_{3, 3}=
\left(-\dfrac{2.1284329337073E+24}{3.3059881728E+19}\right)^{-1}Co(A)_{3, 3}=
\left(-\dfrac{2.1284329337073E+24}{3.3059881728E+19}\right)^{-1}\times(-1)^{3+3}\det\begin{pmatrix}1 & -3 & 1 & -2 & -\dfrac{17}{3} \\ \dfrac{25}{3} & \dfrac{5}{2} & -\dfrac{24}{5} & \dfrac{10}{3} & -\dfrac{7}{4} \\ 5 & 2 & -2 & \dfrac{4}{3} & 5 \\ -5 & 2 & 0 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{10}{3} & -4 & \dfrac{20}{3} & -2\end{pmatrix}=\dfrac{2.4332677947644E+26}{4.3100766907573E+27}\) .
- On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul,
\[ B^{-1}=\begin{pmatrix}1 & -3 & \dfrac{7}{2} & 1 & -2 & -\dfrac{17}{3} \\ \dfrac{25}{3} & \dfrac{5}{2} & -5 & -\dfrac{24}{5} & \dfrac{10}{3} & -\dfrac{7}{4} \\ 3 & -\dfrac{5}{2} & \dfrac{14}{5} & \dfrac{6}{5} & 4 & -\dfrac{11}{4} \\ 5 & 2 & \dfrac{12}{5} & -2 & \dfrac{4}{3} & 5 \\ -5 & 2 & 5 & 0 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{10}{3} & -\dfrac{21}{4} & -4 & \dfrac{20}{3} & -2\end{pmatrix}\]
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&x&-&3y &+&\dfrac{7}{2}z &+&t &-&2u &-&\dfrac{17}{3}v &=&-\dfrac{25}{7}\\
&\dfrac{25}{3}x&+&\dfrac{5}{2}y &-&5z &-&\dfrac{24}{5}t &+&\dfrac{10}{3}u &-&\dfrac{7}{4}v &=&8\\
&3x&-&\dfrac{5}{2}y &+&\dfrac{14}{5}z &+&\dfrac{6}{5}t &+&4u &-&\dfrac{11}{4}v &=&1\\
&5x&+&2y &+&\dfrac{12}{5}z &-&2t &+&\dfrac{4}{3}u &+&5v &=&-3\\
&-5x&+&2y &+&5z &&&-&5u &-&\dfrac{20}{3}v &=&\dfrac{40}{3}\\
&\dfrac{23}{5}x&-&\dfrac{10}{3}y &-&\dfrac{21}{4}z &-&4t &+&\dfrac{20}{3}u &-&2v &=&8\\
\end{array}
\right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-\dfrac{25}{7} \\ 8 \\ 1 \\ -3 \\ \dfrac{40}{3} \\ 8\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{1.2905298739787E+27}{1.1493537842019E+28} & \dfrac{1.0342377692658E+27}{1.1493537842019E+28} & \dfrac{8.8706902790322E+25}{1.1493537842019E+28} & \dfrac{2.9631006668829E+26}{2.2987075684039E+28} & -\dfrac{5.0206881978465E+26}{5.7467689210097E+27} & -\dfrac{1.6098638340988E+26}{1.9155896403366E+27} \\ -\dfrac{1.4055147543594E+26}{7.6623585613463E+26} & \dfrac{1.6719910680421E+29}{1.3792245410423E+30} & \dfrac{1.7327410207007E+29}{1.3792245410423E+30} & -\dfrac{3.0290795952427E+26}{5.7467689210097E+27} & \dfrac{1.1004948446264E+32}{1.5516276086726E+33} & -\dfrac{7.3148294311373E+24}{5.7467689210097E+25} \\ -\dfrac{1.9142002119329E+24}{9.5779482016828E+26} & -\dfrac{3.1245290939714E+25}{4.7889741008414E+26} & \dfrac{2.4332677947644E+26}{4.3100766907573E+27} & \dfrac{7.1461377480332E+26}{5.7467689210097E+27} & \dfrac{1.1354756896263E+26}{1.4366922302524E+27} & \dfrac{3.7416942720578E+25}{1.1493537842019E+27} \\ -\dfrac{2.0310947633217E+27}{6.8961227052117E+28} & \dfrac{1.3823206025366E+26}{1.5324717122693E+27} & \dfrac{3.98195864857E+27}{2.2987075684039E+28} & -\dfrac{2.9495833751835E+29}{1.6550694492508E+30} & -\dfrac{1.1757990587892E+27}{8.6201533815146E+27} & -\dfrac{5.1612945196262E+26}{2.2987075684039E+27} \\ -\dfrac{9.7386700452455E+31}{5.1720920289087E+32} & -\dfrac{8.893931375887E+25}{4.7889741008414E+27} & \dfrac{9.633063985526E+31}{5.1720920289087E+32} & -\dfrac{6.7320839207742E+29}{3.4480613526058E+31} & \dfrac{4.2202276165768E+29}{6.8961227052117E+30} & \dfrac{3.9126138605916E+25}{9.5779482016828E+26} \\ \dfrac{1.5291187095652E+24}{3.1926494005609E+27} & -\dfrac{1.2666750534795E+27}{1.9155896403366E+28} & -\dfrac{1.1283030408286E+29}{1.7240306763029E+30} & \dfrac{8.7718676805709E+25}{1.0642164668536E+27} & -\dfrac{1.4315401875132E+27}{2.8733844605049E+28} & \dfrac{8.9183659344722E+24}{4.7889741008414E+26}\end{pmatrix}\times\begin{pmatrix}-\dfrac{25}{7} \\ 8 \\ 1 \\ -3 \\ \dfrac{40}{3} \\ 8\end{pmatrix}=\begin{pmatrix}-\dfrac{1.2500319163287E+169}{8.0684558091191E+168} \\ \dfrac{2.8798856060216E+175}{1.5685078092928E+175} \\ \dfrac{1.902125498426E+164}{3.9396756880464E+164} \\ -\dfrac{3.4804123806918E+171}{1.6730749965789E+171} \\ \dfrac{1.1711650834048E+184}{6.1269836300498E+183} \\ -\dfrac{4.4057042992098E+169}{3.2425314304909E+169}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.2500319163287E+169}{8.0684558091191E+168}\) , \( y=\dfrac{2.8798856060216E+175}{1.5685078092928E+175}\) , \( z=\dfrac{1.902125498426E+164}{3.9396756880464E+164}\) , \( t=\dfrac{3.4804123806918E+171}{1.6730749965789E+171}\) , \( u=\dfrac{1.1711650834048E+184}{6.1269836300498E+183}\) et \( v=\dfrac{4.4057042992098E+169}{3.2425314304909E+169}\)
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{1.2905298739787E+27}{1.1493537842019E+28}x&+&\dfrac{1.0342377692658E+27}{1.1493537842019E+28}y &+&\dfrac{8.8706902790322E+25}{1.1493537842019E+28}z &+&\dfrac{2.9631006668829E+26}{2.2987075684039E+28}t &-&\dfrac{5.0206881978465E+26}{5.7467689210097E+27}u &-&\dfrac{1.6098638340988E+26}{1.9155896403366E+27}v &=&8\\
&-\dfrac{1.4055147543594E+26}{7.6623585613463E+26}x&+&\dfrac{1.6719910680421E+29}{1.3792245410423E+30}y &+&\dfrac{1.7327410207007E+29}{1.3792245410423E+30}z &-&\dfrac{3.0290795952427E+26}{5.7467689210097E+27}t &+&\dfrac{1.1004948446264E+32}{1.5516276086726E+33}u &-&\dfrac{7.3148294311373E+24}{5.7467689210097E+25}v &=&-9\\
&-\dfrac{1.9142002119329E+24}{9.5779482016828E+26}x&-&\dfrac{3.1245290939714E+25}{4.7889741008414E+26}y &+&\dfrac{2.4332677947644E+26}{4.3100766907573E+27}z &+&\dfrac{7.1461377480332E+26}{5.7467689210097E+27}t &+&\dfrac{1.1354756896263E+26}{1.4366922302524E+27}u &+&\dfrac{3.7416942720578E+25}{1.1493537842019E+27}v &=&-2\\
&-\dfrac{2.0310947633217E+27}{6.8961227052117E+28}x&+&\dfrac{1.3823206025366E+26}{1.5324717122693E+27}y &+&\dfrac{3.98195864857E+27}{2.2987075684039E+28}z &-&\dfrac{2.9495833751835E+29}{1.6550694492508E+30}t &-&\dfrac{1.1757990587892E+27}{8.6201533815146E+27}u &-&\dfrac{5.1612945196262E+26}{2.2987075684039E+27}v &=&8\\
&-\dfrac{9.7386700452455E+31}{5.1720920289087E+32}x&-&\dfrac{8.893931375887E+25}{4.7889741008414E+27}y &+&\dfrac{9.633063985526E+31}{5.1720920289087E+32}z &-&\dfrac{6.7320839207742E+29}{3.4480613526058E+31}t &+&\dfrac{4.2202276165768E+29}{6.8961227052117E+30}u &+&\dfrac{3.9126138605916E+25}{9.5779482016828E+26}v &=&5\\
&\dfrac{1.5291187095652E+24}{3.1926494005609E+27}x&-&\dfrac{1.2666750534795E+27}{1.9155896403366E+28}y &-&\dfrac{1.1283030408286E+29}{1.7240306763029E+30}z &+&\dfrac{8.7718676805709E+25}{1.0642164668536E+27}t &-&\dfrac{1.4315401875132E+27}{2.8733844605049E+28}u &+&\dfrac{8.9183659344722E+24}{4.7889741008414E+26}v &=&-\dfrac{23}{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 \( b=\begin{pmatrix}8 \\ -9 \\ -2 \\ 8 \\ 5 \\ -\dfrac{23}{2}\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & -3 & \dfrac{7}{2} & 1 & -2 & -\dfrac{17}{3} \\ \dfrac{25}{3} & \dfrac{5}{2} & -5 & -\dfrac{24}{5} & \dfrac{10}{3} & -\dfrac{7}{4} \\ 3 & -\dfrac{5}{2} & \dfrac{14}{5} & \dfrac{6}{5} & 4 & -\dfrac{11}{4} \\ 5 & 2 & \dfrac{12}{5} & -2 & \dfrac{4}{3} & 5 \\ -5 & 2 & 5 & 0 & -5 & -\dfrac{20}{3} \\ \dfrac{23}{5} & -\dfrac{10}{3} & -\dfrac{21}{4} & -4 & \dfrac{20}{3} & -2\end{pmatrix}\times\begin{pmatrix}8 \\ -9 \\ -2 \\ 8 \\ 5 \\ -\dfrac{23}{2}\end{pmatrix}=\begin{pmatrix}\dfrac{547}{6} \\ \dfrac{6307}{120} \\ \dfrac{817}{8} \\ -\dfrac{1489}{30} \\ -\dfrac{49}{3} \\ \dfrac{3049}{30}\end{pmatrix}\) . Ainsi \( x=\dfrac{547}{6}\) , \( y=\dfrac{6307}{120}\) , \( z=\dfrac{817}{8}\) , \( t=\dfrac{1489}{30}\) , \( u=\dfrac{49}{3}\) et \( v=\dfrac{3049}{30}\)
