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
\[ A= \begin{pmatrix}1 & 1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ \dfrac{14}{5} & -\dfrac{10}{3} & -3 & 4 & 0 & 4 \\ \dfrac{13}{4} & -5 & -2 & 5 & 2 & -\dfrac{11}{2} \\ -4 & 4 & -\dfrac{1}{5} & -3 & \dfrac{17}{4} & \dfrac{17}{5} \\ \dfrac{23}{5} & -\dfrac{21}{2} & -\dfrac{22}{5} & \dfrac{5}{2} & \dfrac{19}{2} & -\dfrac{1}{3} \\ 5 & 4 & 0 & 2 & 5 & -3\end{pmatrix}\]
- Donner les mineurs d'ordre \( (6, 1)\) et \( (2, 4)\) :
\( \widehat{A}_{6, 1}=\)
\( \widehat{A}_{2, 4}=\)
- Expliquer pourquoi
\( \det(A)=\det
\begin{pmatrix}1 & 1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ 0 & -\dfrac{92}{15} & -\dfrac{29}{5} & \dfrac{76}{5} & \dfrac{56}{5} & \dfrac{61}{10} \\ 0 & -\dfrac{33}{4} & -\dfrac{21}{4} & 18 & 15 & -\dfrac{49}{16} \\ 0 & 8 & \dfrac{19}{5} & -19 & -\dfrac{47}{4} & \dfrac{2}{5} \\ 0 & -\dfrac{151}{10} & -9 & \dfrac{209}{10} & \dfrac{279}{10} & \dfrac{187}{60} \\ 0 & -1 & -5 & 22 & 25 & \dfrac{3}{4}\end{pmatrix}
\)
- 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 \( A^{-1}_{3, 1}\) .
- Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&x&+&y &+&z &-&4t &-&4u &-&\dfrac{3}{4}v &=&0\\
&\dfrac{14}{5}x&-&\dfrac{10}{3}y &-&3z &+&4t &&&+&4v &=&3\\
&\dfrac{13}{4}x&-&5y &-&2z &+&5t &+&2u &-&\dfrac{11}{2}v &=&5\\
&-4x&+&4y &-&\dfrac{1}{5}z &-&3t &+&\dfrac{17}{4}u &+&\dfrac{17}{5}v &=&-3\\
&\dfrac{23}{5}x&-&\dfrac{21}{2}y &-&\dfrac{22}{5}z &+&\dfrac{5}{2}t &+&\dfrac{19}{2}u &-&\dfrac{1}{3}v &=&-3\\
&5x&+&4y &&&+&2t &+&5u &-&3v &=&-6\\
\end{array}
\right.
\)
- Résoudre le système suivant.
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{7.35015218112E+22}{1.1515782830789E+24}x&+&\dfrac{2.4100949952E+21}{6.397657128216E+22}y &-&\dfrac{2.592781088832E+23}{2.3031565661578E+24}z &-&\dfrac{2.1807154368E+21}{1.9192971384648E+22}t &+&\dfrac{3.4667134272E+21}{7.6771885538592E+22}u &+&\dfrac{8.20479958848E+22}{7.6771885538592E+23}v &=&8\\
&\dfrac{1.857555498816E+23}{2.5590628512864E+24}x&+&\dfrac{1.8566335296E+21}{2.5590628512864E+22}y &+&\dfrac{2.409634360512E+23}{2.5590628512864E+24}z &+&\dfrac{6.72694466112E+22}{5.1181257025728E+23}t &-&\dfrac{8.3675529792E+21}{1.0236251405146E+23}u &+&\dfrac{1.0482091769549E+30}{1.6378002248233E+31}v &=&0\\
&-\dfrac{5.3057156441357E+27}{1.2437045457252E+28}x&-&\dfrac{2.906054667072E+23}{1.0236251405146E+24}y &-&\dfrac{8.2344959831808E+27}{1.6582727276336E+28}z &-&\dfrac{7.67867667264E+22}{1.5354377107718E+23}t &+&\dfrac{9.04250526912E+22}{7.6771885538592E+23}u &+&\dfrac{3.5004578252314E+28}{5.9697818194809E+29}v &=&-1\\
&-\dfrac{1.0310962040898E+30}{4.9748181829008E+30}x&+&\dfrac{1.2792655296E+21}{6.397657128216E+22}y &-&\dfrac{4.230785722176E+23}{2.3031565661578E+25}z &-&\dfrac{4.76015351616E+22}{5.1181257025728E+23}t &-&\dfrac{3.69744647616E+22}{7.6771885538592E+23}u &+&\dfrac{3.5668455714202E+29}{2.9480404046819E+31}v &=&5\\
&-\dfrac{3.72803228352E+22}{3.8385942769296E+23}x&-&\dfrac{1.86834745152E+22}{2.5590628512864E+23}y &-&\dfrac{1.410277957056E+23}{1.9192971384648E+24}z &-&\dfrac{7.7745365568E+21}{7.6771885538592E+23}t &+&\dfrac{1.9693917504E+21}{3.198828564108E+22}u &+&\dfrac{5.3887346456256E+28}{1.2437045457252E+30}v &=&2\\
&-\dfrac{7.43770216512E+22}{7.6771885538592E+23}x&+&\dfrac{1.30946642496E+22}{2.5590628512864E+23}y &-&\dfrac{7.55391479616E+22}{3.8385942769296E+23}z &-&\dfrac{1.19002053312E+22}{1.2795314256432E+23}t &+&\dfrac{4.7048423616E+21}{1.2795314256432E+23}u &+&\dfrac{5.3236968768E+21}{5.1181257025728E+23}v &=&6\\
\end{array}
\right.
\)
Cliquer ici pour afficher la solution
Exercice
-
\( \widehat{A}_{6, 1}=\begin{pmatrix}1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ -\dfrac{10}{3} & -3 & 4 & 0 & 4 \\ -5 & -2 & 5 & 2 & -\dfrac{11}{2} \\ 4 & -\dfrac{1}{5} & -3 & \dfrac{17}{4} & \dfrac{17}{5} \\ -\dfrac{21}{2} & -\dfrac{22}{5} & \dfrac{5}{2} & \dfrac{19}{2} & -\dfrac{1}{3}\end{pmatrix}\)
\( \widehat{A}_{2, 4}=\begin{pmatrix}1 & 1 & 1 & -4 & -\dfrac{3}{4} \\ \dfrac{13}{4} & -5 & -2 & 2 & -\dfrac{11}{2} \\ -4 & 4 & -\dfrac{1}{5} & \dfrac{17}{4} & \dfrac{17}{5} \\ \dfrac{23}{5} & -\dfrac{21}{2} & -\dfrac{22}{5} & \dfrac{19}{2} & -\dfrac{1}{3} \\ 5 & 4 & 0 & 5 & -3\end{pmatrix}\)
- 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 : \( L_{2}\leftarrow L_{2}-\left(\dfrac{14}{5}\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(\dfrac{13}{4}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(-4\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(\dfrac{23}{5}\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(5\right)L_1\)
- 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 & 1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ 0 & -\dfrac{92}{15} & -\dfrac{29}{5} & \dfrac{76}{5} & \dfrac{56}{5} & \dfrac{61}{10} \\ 0 & -\dfrac{33}{4} & -\dfrac{21}{4} & 18 & 15 & -\dfrac{49}{16} \\ 0 & 8 & \dfrac{19}{5} & -19 & -\dfrac{47}{4} & \dfrac{2}{5} \\ 0 & -\dfrac{151}{10} & -9 & \dfrac{209}{10} & \dfrac{279}{10} & \dfrac{187}{60} \\ 0 & -1 & -5 & 22 & 25 & \dfrac{3}{4}\end{pmatrix}\\
&=&1\times\det\begin{pmatrix}-\dfrac{92}{15} & -\dfrac{29}{5} & \dfrac{76}{5} & \dfrac{56}{5} & \dfrac{61}{10} \\ -\dfrac{33}{4} & -\dfrac{21}{4} & 18 & 15 & -\dfrac{49}{16} \\ 8 & \dfrac{19}{5} & -19 & -\dfrac{47}{4} & \dfrac{2}{5} \\ -\dfrac{151}{10} & -9 & \dfrac{209}{10} & \dfrac{279}{10} & \dfrac{187}{60} \\ -1 & -5 & 22 & 25 & \dfrac{3}{4}\end{pmatrix}\\
&=&-\dfrac{6.397657128216E+20}{13996800000000000}
\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{6.397657128216E+20}{13996800000000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & 1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ \dfrac{14}{5} & -\dfrac{10}{3} & -3 & 4 & 0 & 4 \\ \dfrac{13}{4} & -5 & -2 & 5 & 2 & -\dfrac{11}{2} \\ -4 & 4 & -\dfrac{1}{5} & -3 & \dfrac{17}{4} & \dfrac{17}{5} \\ \dfrac{23}{5} & -\dfrac{21}{2} & -\dfrac{22}{5} & \dfrac{5}{2} & \dfrac{19}{2} & -\dfrac{1}{3} \\ 5 & 4 & 0 & 2 & 5 & -3\end{pmatrix}
=-\dfrac{13996800000000000}{6.397657128216E+20}\begin{pmatrix}-\dfrac{4.7023753495015E+43}{1.6118410912598E+40} & -\dfrac{1.5418961425719E+42}{8.9546727292214E+38} & \dfrac{1.658772441487E+44}{3.2236821825197E+40} & \dfrac{1.3951469658854E+42}{2.6864018187664E+38} & -\dfrac{2.2178843869008E+42}{1.0745607275066E+39} & -\dfrac{5.2491494572823E+43}{1.0745607275066E+40} \\ -\dfrac{1.1884003178057E+44}{3.5818690916885E+40} & -\dfrac{1.187810473513E+42}{3.5818690916885E+38} & -\dfrac{1.5416014442924E+44}{3.5818690916885E+40} & -\dfrac{4.3036685462329E+43}{7.1637381833771E+39} & \dfrac{5.3532734963104E+42}{1.4327476366754E+39} & -\dfrac{6.7060829128068E+50}{2.2923962186807E+47} \\ \dfrac{3.3944149510992E+48}{1.7407883785606E+44} & \dfrac{1.8591941355779E+44}{1.4327476366754E+40} & \dfrac{5.2681481924063E+48}{2.3210511714142E+44} & \dfrac{4.9125540549981E+43}{2.1491214550131E+39} & -\dfrac{5.7850848291916E+43}{1.0745607275066E+40} & -\dfrac{2.2394728957611E+49}{8.355784217091E+45} \\ \dfrac{6.5965999799716E+50}{6.9631535142425E+46} & -\dfrac{8.1843022343265E+41}{8.9546727292214E+38} & \dfrac{2.7067116433434E+44}{3.2236821825197E+41} & \dfrac{3.0453830074063E+43}{7.1637381833771E+39} & \dfrac{2.3654994804402E+43}{1.0745607275066E+40} & -\dfrac{2.2819454995242E+50}{4.1263131936252E+47} \\ \dfrac{2.3850672312881E+43}{5.3728036375328E+39} & \dfrac{1.1953046391201E+43}{3.5818690916885E+39} & \dfrac{9.0224748247252E+43}{2.6864018187664E+40} & \dfrac{4.9738819221187E+42}{1.0745607275066E+40} & -\dfrac{1.2599493170196E+42}{4.4773363646107E+38} & -\dfrac{3.4475276617651E+49}{1.7407883785606E+46} \\ \dfrac{4.7583868274228E+43}{1.0745607275066E+40} & -\dfrac{8.3775172078049E+42}{3.5818690916885E+39} & \dfrac{4.8327356841589E+43}{5.3728036375328E+39} & \dfrac{7.6133433464386E+42}{1.7909345458443E+39} & -\dfrac{3.0099968271823E+42}{1.7909345458443E+39} & -\dfrac{3.4059187272321E+42}{7.1637381833771E+39}\end{pmatrix}
=\begin{pmatrix}\dfrac{7.35015218112E+22}{1.1515782830789E+24} & \dfrac{2.4100949952E+21}{6.397657128216E+22} & -\dfrac{2.592781088832E+23}{2.3031565661578E+24} & -\dfrac{2.1807154368E+21}{1.9192971384648E+22} & \dfrac{3.4667134272E+21}{7.6771885538592E+22} & \dfrac{8.20479958848E+22}{7.6771885538592E+23} \\ \dfrac{1.857555498816E+23}{2.5590628512864E+24} & \dfrac{1.8566335296E+21}{2.5590628512864E+22} & \dfrac{2.409634360512E+23}{2.5590628512864E+24} & \dfrac{6.72694466112E+22}{5.1181257025728E+23} & -\dfrac{8.3675529792E+21}{1.0236251405146E+23} & \dfrac{1.0482091769549E+30}{1.6378002248233E+31} \\ -\dfrac{5.3057156441357E+27}{1.2437045457252E+28} & -\dfrac{2.906054667072E+23}{1.0236251405146E+24} & -\dfrac{8.2344959831808E+27}{1.6582727276336E+28} & -\dfrac{7.67867667264E+22}{1.5354377107718E+23} & \dfrac{9.04250526912E+22}{7.6771885538592E+23} & \dfrac{3.5004578252314E+28}{5.9697818194809E+29} \\ -\dfrac{1.0310962040898E+30}{4.9748181829008E+30} & \dfrac{1.2792655296E+21}{6.397657128216E+22} & -\dfrac{4.230785722176E+23}{2.3031565661578E+25} & -\dfrac{4.76015351616E+22}{5.1181257025728E+23} & -\dfrac{3.69744647616E+22}{7.6771885538592E+23} & \dfrac{3.5668455714202E+29}{2.9480404046819E+31} \\ -\dfrac{3.72803228352E+22}{3.8385942769296E+23} & -\dfrac{1.86834745152E+22}{2.5590628512864E+23} & -\dfrac{1.410277957056E+23}{1.9192971384648E+24} & -\dfrac{7.7745365568E+21}{7.6771885538592E+23} & \dfrac{1.9693917504E+21}{3.198828564108E+22} & \dfrac{5.3887346456256E+28}{1.2437045457252E+30} \\ -\dfrac{7.43770216512E+22}{7.6771885538592E+23} & \dfrac{1.30946642496E+22}{2.5590628512864E+23} & -\dfrac{7.55391479616E+22}{3.8385942769296E+23} & -\dfrac{1.19002053312E+22}{1.2795314256432E+23} & \dfrac{4.7048423616E+21}{1.2795314256432E+23} & \dfrac{5.3236968768E+21}{5.1181257025728E+23}\end{pmatrix}\) .
Précisément on a calculé \( A^{-1}_{3, 1}=B_{3, 1}=
\left(-\dfrac{6.397657128216E+20}{13996800000000000}\right)^{-1}Co(A)_{1, 3}=
\left(-\dfrac{6.397657128216E+20}{13996800000000000}\right)^{-1}\times(-1)^{1+3}\det\begin{pmatrix}\dfrac{14}{5} & -\dfrac{10}{3} & 4 & 0 & 4 \\ \dfrac{13}{4} & -5 & 5 & 2 & -\dfrac{11}{2} \\ -4 & 4 & -3 & \dfrac{17}{4} & \dfrac{17}{5} \\ \dfrac{23}{5} & -\dfrac{21}{2} & \dfrac{5}{2} & \dfrac{19}{2} & -\dfrac{1}{3} \\ 5 & 4 & 2 & 5 & -3\end{pmatrix}=-\dfrac{5.3057156441357E+27}{1.2437045457252E+28}\) .
- On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul,
\[ B^{-1}=\begin{pmatrix}1 & 1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ \dfrac{14}{5} & -\dfrac{10}{3} & -3 & 4 & 0 & 4 \\ \dfrac{13}{4} & -5 & -2 & 5 & 2 & -\dfrac{11}{2} \\ -4 & 4 & -\dfrac{1}{5} & -3 & \dfrac{17}{4} & \dfrac{17}{5} \\ \dfrac{23}{5} & -\dfrac{21}{2} & -\dfrac{22}{5} & \dfrac{5}{2} & \dfrac{19}{2} & -\dfrac{1}{3} \\ 5 & 4 & 0 & 2 & 5 & -3\end{pmatrix}\]
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&x&+&y &+&z &-&4t &-&4u &-&\dfrac{3}{4}v &=&0\\
&\dfrac{14}{5}x&-&\dfrac{10}{3}y &-&3z &+&4t &&&+&4v &=&3\\
&\dfrac{13}{4}x&-&5y &-&2z &+&5t &+&2u &-&\dfrac{11}{2}v &=&5\\
&-4x&+&4y &-&\dfrac{1}{5}z &-&3t &+&\dfrac{17}{4}u &+&\dfrac{17}{5}v &=&-3\\
&\dfrac{23}{5}x&-&\dfrac{21}{2}y &-&\dfrac{22}{5}z &+&\dfrac{5}{2}t &+&\dfrac{19}{2}u &-&\dfrac{1}{3}v &=&-3\\
&5x&+&4y &&&+&2t &+&5u &-&3v &=&-6\\
\end{array}
\right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}0 \\ 3 \\ 5 \\ -3 \\ -3 \\ -6\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}\dfrac{7.35015218112E+22}{1.1515782830789E+24} & \dfrac{2.4100949952E+21}{6.397657128216E+22} & -\dfrac{2.592781088832E+23}{2.3031565661578E+24} & -\dfrac{2.1807154368E+21}{1.9192971384648E+22} & \dfrac{3.4667134272E+21}{7.6771885538592E+22} & \dfrac{8.20479958848E+22}{7.6771885538592E+23} \\ \dfrac{1.857555498816E+23}{2.5590628512864E+24} & \dfrac{1.8566335296E+21}{2.5590628512864E+22} & \dfrac{2.409634360512E+23}{2.5590628512864E+24} & \dfrac{6.72694466112E+22}{5.1181257025728E+23} & -\dfrac{8.3675529792E+21}{1.0236251405146E+23} & \dfrac{1.0482091769549E+30}{1.6378002248233E+31} \\ -\dfrac{5.3057156441357E+27}{1.2437045457252E+28} & -\dfrac{2.906054667072E+23}{1.0236251405146E+24} & -\dfrac{8.2344959831808E+27}{1.6582727276336E+28} & -\dfrac{7.67867667264E+22}{1.5354377107718E+23} & \dfrac{9.04250526912E+22}{7.6771885538592E+23} & \dfrac{3.5004578252314E+28}{5.9697818194809E+29} \\ -\dfrac{1.0310962040898E+30}{4.9748181829008E+30} & \dfrac{1.2792655296E+21}{6.397657128216E+22} & -\dfrac{4.230785722176E+23}{2.3031565661578E+25} & -\dfrac{4.76015351616E+22}{5.1181257025728E+23} & -\dfrac{3.69744647616E+22}{7.6771885538592E+23} & \dfrac{3.5668455714202E+29}{2.9480404046819E+31} \\ -\dfrac{3.72803228352E+22}{3.8385942769296E+23} & -\dfrac{1.86834745152E+22}{2.5590628512864E+23} & -\dfrac{1.410277957056E+23}{1.9192971384648E+24} & -\dfrac{7.7745365568E+21}{7.6771885538592E+23} & \dfrac{1.9693917504E+21}{3.198828564108E+22} & \dfrac{5.3887346456256E+28}{1.2437045457252E+30} \\ -\dfrac{7.43770216512E+22}{7.6771885538592E+23} & \dfrac{1.30946642496E+22}{2.5590628512864E+23} & -\dfrac{7.55391479616E+22}{3.8385942769296E+23} & -\dfrac{1.19002053312E+22}{1.2795314256432E+23} & \dfrac{4.7048423616E+21}{1.2795314256432E+23} & \dfrac{5.3236968768E+21}{5.1181257025728E+23}\end{pmatrix}\times\begin{pmatrix}0 \\ 3 \\ 5 \\ -3 \\ -3 \\ -6\end{pmatrix}=\begin{pmatrix}-\dfrac{1.4763141475501E+116}{1.6668290197771E+116} \\ \dfrac{8.7313092129653E+123}{5.6192030164667E+124} \\ -\dfrac{3.0333713606691E+129}{1.1945099481096E+129} \\ \dfrac{5.445608251841E+126}{1.7068329162518E+127} \\ -\dfrac{1.5012078748474E+124}{1.5001461177994E+124} \\ -\dfrac{5.9605838756016E+116}{8.2312544186525E+116}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.4763141475501E+116}{1.6668290197771E+116}\) , \( y=\dfrac{8.7313092129653E+123}{5.6192030164667E+124}\) , \( z=\dfrac{3.0333713606691E+129}{1.1945099481096E+129}\) , \( t=\dfrac{5.445608251841E+126}{1.7068329162518E+127}\) , \( u=\dfrac{1.5012078748474E+124}{1.5001461177994E+124}\) et \( v=\dfrac{5.9605838756016E+116}{8.2312544186525E+116}\)
- Le système
\( \left\{\begin{array}{*{7}{cr}}
&\dfrac{7.35015218112E+22}{1.1515782830789E+24}x&+&\dfrac{2.4100949952E+21}{6.397657128216E+22}y &-&\dfrac{2.592781088832E+23}{2.3031565661578E+24}z &-&\dfrac{2.1807154368E+21}{1.9192971384648E+22}t &+&\dfrac{3.4667134272E+21}{7.6771885538592E+22}u &+&\dfrac{8.20479958848E+22}{7.6771885538592E+23}v &=&8\\
&\dfrac{1.857555498816E+23}{2.5590628512864E+24}x&+&\dfrac{1.8566335296E+21}{2.5590628512864E+22}y &+&\dfrac{2.409634360512E+23}{2.5590628512864E+24}z &+&\dfrac{6.72694466112E+22}{5.1181257025728E+23}t &-&\dfrac{8.3675529792E+21}{1.0236251405146E+23}u &+&\dfrac{1.0482091769549E+30}{1.6378002248233E+31}v &=&0\\
&-\dfrac{5.3057156441357E+27}{1.2437045457252E+28}x&-&\dfrac{2.906054667072E+23}{1.0236251405146E+24}y &-&\dfrac{8.2344959831808E+27}{1.6582727276336E+28}z &-&\dfrac{7.67867667264E+22}{1.5354377107718E+23}t &+&\dfrac{9.04250526912E+22}{7.6771885538592E+23}u &+&\dfrac{3.5004578252314E+28}{5.9697818194809E+29}v &=&-1\\
&-\dfrac{1.0310962040898E+30}{4.9748181829008E+30}x&+&\dfrac{1.2792655296E+21}{6.397657128216E+22}y &-&\dfrac{4.230785722176E+23}{2.3031565661578E+25}z &-&\dfrac{4.76015351616E+22}{5.1181257025728E+23}t &-&\dfrac{3.69744647616E+22}{7.6771885538592E+23}u &+&\dfrac{3.5668455714202E+29}{2.9480404046819E+31}v &=&5\\
&-\dfrac{3.72803228352E+22}{3.8385942769296E+23}x&-&\dfrac{1.86834745152E+22}{2.5590628512864E+23}y &-&\dfrac{1.410277957056E+23}{1.9192971384648E+24}z &-&\dfrac{7.7745365568E+21}{7.6771885538592E+23}t &+&\dfrac{1.9693917504E+21}{3.198828564108E+22}u &+&\dfrac{5.3887346456256E+28}{1.2437045457252E+30}v &=&2\\
&-\dfrac{7.43770216512E+22}{7.6771885538592E+23}x&+&\dfrac{1.30946642496E+22}{2.5590628512864E+23}y &-&\dfrac{7.55391479616E+22}{3.8385942769296E+23}z &-&\dfrac{1.19002053312E+22}{1.2795314256432E+23}t &+&\dfrac{4.7048423616E+21}{1.2795314256432E+23}u &+&\dfrac{5.3236968768E+21}{5.1181257025728E+23}v &=&6\\
\end{array}
\right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}8 \\ 0 \\ -1 \\ 5 \\ 2 \\ 6\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 1 & 1 & -4 & -4 & -\dfrac{3}{4} \\ \dfrac{14}{5} & -\dfrac{10}{3} & -3 & 4 & 0 & 4 \\ \dfrac{13}{4} & -5 & -2 & 5 & 2 & -\dfrac{11}{2} \\ -4 & 4 & -\dfrac{1}{5} & -3 & \dfrac{17}{4} & \dfrac{17}{5} \\ \dfrac{23}{5} & -\dfrac{21}{2} & -\dfrac{22}{5} & \dfrac{5}{2} & \dfrac{19}{2} & -\dfrac{1}{3} \\ 5 & 4 & 0 & 2 & 5 & -3\end{pmatrix}\times\begin{pmatrix}8 \\ 0 \\ -1 \\ 5 \\ 2 \\ 6\end{pmatrix}=\begin{pmatrix}-\dfrac{51}{2} \\ \dfrac{347}{5} \\ 24 \\ -\dfrac{179}{10} \\ \dfrac{707}{10} \\ 42\end{pmatrix}\) . Ainsi \( x=\dfrac{51}{2}\) , \( y=\dfrac{347}{5}\) , \( z=24\) , \( t=\dfrac{179}{10}\) , \( u=\dfrac{707}{10}\) et \( v=42\)