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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 \[ A= \begin{pmatrix}1 & 3 & \dfrac{11}{4} & 5 & 3 & -5 \\ 4 & \dfrac{9}{4} & -\dfrac{9}{2} & -5 & \dfrac{21}{5} & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & \dfrac{1}{3} & -\dfrac{14}{3} & -1 & 0 \\ 2 & -\dfrac{11}{2} & 2 & -4 & -2 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & \dfrac{15}{2} & 1 & \dfrac{19}{5} & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & -3 & 4 & -4 & \dfrac{20}{3}\end{pmatrix}\]
  1. Donner les mineurs d'ordre \( (2, 3)\) et \( (1, 5)\) : \( \widehat{A}_{2, 3}=\) \( \widehat{A}_{1, 5}=\)
  2. Expliquer pourquoi \( \det(A)=\det \begin{pmatrix}1 & 3 & \dfrac{11}{4} & 5 & 3 & -5 \\ 0 & -\dfrac{39}{4} & -\dfrac{31}{2} & -25 & -\dfrac{39}{5} & 15 \\ 0 & \dfrac{131}{6} & \dfrac{503}{24} & \dfrac{197}{6} & \dfrac{43}{2} & -\dfrac{75}{2} \\ 0 & -\dfrac{23}{2} & -\dfrac{7}{2} & -14 & -8 & 9 \\ 0 & \dfrac{351}{10} & \dfrac{291}{8} & \dfrac{107}{2} & \dfrac{353}{10} & -\dfrac{113}{2} \\ 0 & \dfrac{29}{4} & \dfrac{95}{16} & \dfrac{81}{4} & \dfrac{23}{4} & -\dfrac{115}{12}\end{pmatrix} \)
  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 \( A^{-1}_{3, 2}\) .
  6. Donner \( B^{-1}\) l'inverse de la matrice \( B\) . Justifier.
  7. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &x&+&3y &+&\dfrac{11}{4}z &+&5t &+&3u &-&5v &=&-\dfrac{8}{9}\\ &4x&+&\dfrac{9}{4}y &-&\dfrac{9}{2}z &-&5t &+&\dfrac{21}{5}u &-&5v &=&-8\\ &-\dfrac{15}{2}x&-&\dfrac{2}{3}y &+&\dfrac{1}{3}z &-&\dfrac{14}{3}t &-&u &&&=&\dfrac{29}{4}\\ &2x&-&\dfrac{11}{2}y &+&2z &-&4t &-&2u &-&v &=&8\\ &-\dfrac{21}{2}x&+&\dfrac{18}{5}y &+&\dfrac{15}{2}z &+&t &+&\dfrac{19}{5}u &-&4v &=&8\\ &-\dfrac{13}{4}x&-&\dfrac{5}{2}y &-&3z &+&4t &-&4u &+&\dfrac{20}{3}v &=&-2\\ \end{array} \right. \)
  8. Résoudre le système suivant. \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{20045035008000000}{488822507424000000}x&-&\dfrac{82141558272000000}{1759761026726400000}y &-&\dfrac{99742938624000000}{1466467522272000000}z &-&\dfrac{32341068288000000}{1759761026726400000}t &-&\dfrac{75148881408000000}{1759761026726400000}u &-&\dfrac{184075835904000000}{1955290029696000000}v &=&-5\\ &\dfrac{4991196672000000}{175976102672640000}x&-&\dfrac{701871340032000000}{3519522053452800000}y &+&\dfrac{38341403136000000}{195529002969600000}z &-&\dfrac{927560719872000000}{3519522053452800000}t &-&\dfrac{628047862272000000}{3519522053452800000}u &-&\dfrac{161249730048000000}{586587008908800000}v &=&2\\ &-\dfrac{1709599882752000000}{1.7597610267264E+19}x&-&\dfrac{71635373568000000}{586587008908800000}y &-&\dfrac{237138513408000000}{2932935044544000000}z &-&\dfrac{4716983808000000}{3519522053452800000}t &+&\dfrac{156388796928000000}{3519522053452800000}u &-&\dfrac{404727860736000000}{2932935044544000000}v &=&8\\ &\dfrac{1404464168448000000}{1.1731740178176E+19}x&+&\dfrac{96044299776000000}{3519522053452800000}y &-&\dfrac{3.26081147904E+20}{9.3853921425408E+21}z &+&\dfrac{238220932608000000}{7039044106905600000}t &+&\dfrac{206153206272000000}{1.4078088213811E+19}u &+&\dfrac{364022678016000000}{2932935044544000000}v &=&8\\ &-\dfrac{88695046656000000}{293293504454400000}x&+&\dfrac{1108560121344000000}{3519522053452800000}y &-&\dfrac{2277945856512000000}{4692696071270400000}z &+&\dfrac{109213733376000000}{703904410690560000}t &+&\dfrac{543019728384000000}{1407808821381120000}u &+&\dfrac{619736375808000000}{2346348035635200000}v &=&8\\ &-\dfrac{898496644608000000}{2932935044544000000}x&+&\dfrac{3146853888000000}{156423202375680000}y &-&\dfrac{1041841046016000000}{3910580059392000000}z &-&\dfrac{1.0025242712064E+21}{2.8156176427622E+22}t &+&\dfrac{363462308352000000}{2346348035635200000}u &+&\dfrac{6.456258772992E+21}{2.8156176427622E+23}v &=&7\\ \end{array} \right. \)
Cliquer ici pour afficher la solution

Exercice


  1. \( \widehat{A}_{2, 3}=\begin{pmatrix}1 & 3 & 5 & 3 & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & -\dfrac{14}{3} & -1 & 0 \\ 2 & -\dfrac{11}{2} & -4 & -2 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & 1 & \dfrac{19}{5} & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & 4 & -4 & \dfrac{20}{3}\end{pmatrix}\) \( \widehat{A}_{1, 5}=\begin{pmatrix}4 & \dfrac{9}{4} & -\dfrac{9}{2} & -5 & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & \dfrac{1}{3} & -\dfrac{14}{3} & 0 \\ 2 & -\dfrac{11}{2} & 2 & -4 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & \dfrac{15}{2} & 1 & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & -3 & 4 & \dfrac{20}{3}\end{pmatrix}\)
  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 : \( L_{2}\leftarrow L_{2}-\left(4\right)L_1\) , \( L_{3}\leftarrow L_{3}-\left(-\dfrac{15}{2}\right)L_1\) , \( L_{4}\leftarrow L_{4}-\left(2\right)L_1\) , \( L_{5}\leftarrow L_{5}-\left(-\dfrac{21}{2}\right)L_1\) et \( L_{6}\leftarrow L_{6}-\left(-\dfrac{13}{4}\right)L_1\)
  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 & 3 & \dfrac{11}{4} & 5 & 3 & -5 \\ 0 & -\dfrac{39}{4} & -\dfrac{31}{2} & -25 & -\dfrac{39}{5} & 15 \\ 0 & \dfrac{131}{6} & \dfrac{503}{24} & \dfrac{197}{6} & \dfrac{43}{2} & -\dfrac{75}{2} \\ 0 & -\dfrac{23}{2} & -\dfrac{7}{2} & -14 & -8 & 9 \\ 0 & \dfrac{351}{10} & \dfrac{291}{8} & \dfrac{107}{2} & \dfrac{353}{10} & -\dfrac{113}{2} \\ 0 & \dfrac{29}{4} & \dfrac{95}{16} & \dfrac{81}{4} & \dfrac{23}{4} & -\dfrac{115}{12}\end{pmatrix}\\ &=&1\times\det\begin{pmatrix}-\dfrac{39}{4} & -\dfrac{31}{2} & -25 & -\dfrac{39}{5} & 15 \\ \dfrac{131}{6} & \dfrac{503}{24} & \dfrac{197}{6} & \dfrac{43}{2} & -\dfrac{75}{2} \\ -\dfrac{23}{2} & -\dfrac{7}{2} & -14 & -8 & 9 \\ \dfrac{351}{10} & \dfrac{291}{8} & \dfrac{107}{2} & \dfrac{353}{10} & -\dfrac{113}{2} \\ \dfrac{29}{4} & \dfrac{95}{16} & \dfrac{81}{4} & \dfrac{23}{4} & -\dfrac{115}{12}\end{pmatrix}\\ &=&\dfrac{4888225074240000}{124416000000} \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{4888225074240000}{124416000000}\right)^{-1}{}^tCo\begin{pmatrix}1 & 3 & \dfrac{11}{4} & 5 & 3 & -5 \\ 4 & \dfrac{9}{4} & -\dfrac{9}{2} & -5 & \dfrac{21}{5} & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & \dfrac{1}{3} & -\dfrac{14}{3} & -1 & 0 \\ 2 & -\dfrac{11}{2} & 2 & -4 & -2 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & \dfrac{15}{2} & 1 & \dfrac{19}{5} & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & -3 & 4 & -4 & \dfrac{20}{3}\end{pmatrix} =\dfrac{124416000000}{4888225074240000}\begin{pmatrix}-\dfrac{9.7984642740124E+31}{6.0817341083664E+28} & -\dfrac{4.0152642478234E+32}{2.1894242790119E+29} & -\dfrac{4.8756593356022E+32}{1.8245202325099E+29} & -\dfrac{1.5809042093311E+32}{2.1894242790119E+29} & -\dfrac{3.6734464639967E+32}{2.1894242790119E+29} & -\dfrac{8.9980411662762E+32}{2.4326936433466E+29} \\ \dfrac{2.4398092722534E+31}{2.1894242790119E+28} & -\dfrac{3.4309050832349E+33}{4.3788485580238E+29} & \dfrac{1.8742140819094E+32}{2.4326936433466E+28} & -\dfrac{4.5341255687584E+33}{4.3788485580238E+29} & -\dfrac{3.0700393081808E+33}{4.3788485580238E+29} & -\dfrac{7.8822497363506E+32}{7.2980809300397E+28} \\ -\dfrac{8.3569090137861E+33}{2.1894242790119E+30} & -\dfrac{3.5016982927765E+32}{7.2980809300397E+28} & -\dfrac{1.159186427309E+33}{3.6490404650199E+29} & -\dfrac{2.305767852505E+31}{4.3788485580238E+29} & \dfrac{7.6446363847368E+32}{4.3788485580238E+29} & -\dfrac{1.9784008770932E+33}{3.6490404650199E+29} \\ \dfrac{6.8653369640791E+33}{1.4596161860079E+30} & \dfrac{4.6948615440287E+32}{4.3788485580238E+29} & -\dfrac{1.5939580434213E+36}{1.1676929488064E+33} & \dfrac{1.1644775359833E+33}{8.7576971160477E+29} & \dfrac{1.0077232720338E+33}{1.7515394232095E+30} & \dfrac{1.7794247822698E+33}{3.6490404650199E+29} \\ -\dfrac{4.3356135102475E+32}{3.6490404650199E+28} & \dfrac{5.4188913814563E+33}{4.3788485580238E+29} & -\dfrac{1.1135112053563E+34}{5.8384647440318E+29} & \dfrac{5.3386130993993E+32}{8.7576971160477E+28} & \dfrac{2.6544026520937E+33}{1.7515394232095E+29} & \dfrac{3.0294108916433E+33}{2.9192323720159E+29} \\ -\dfrac{4.3920538272933E+33}{3.6490404650199E+29} & \dfrac{1.5382530080291E+31}{1.9461549146773E+28} & -\dfrac{5.0927535245078E+33}{4.8653872866932E+29} & -\dfrac{4.9005642800453E+36}{3.5030788464191E+33} & \dfrac{1.7766855692274E+33}{2.9192323720159E+29} & \dfrac{3.1559646019921E+37}{3.5030788464191E+34}\end{pmatrix} =\begin{pmatrix}-\dfrac{20045035008000000}{488822507424000000} & -\dfrac{82141558272000000}{1759761026726400000} & -\dfrac{99742938624000000}{1466467522272000000} & -\dfrac{32341068288000000}{1759761026726400000} & -\dfrac{75148881408000000}{1759761026726400000} & -\dfrac{184075835904000000}{1955290029696000000} \\ \dfrac{4991196672000000}{175976102672640000} & -\dfrac{701871340032000000}{3519522053452800000} & \dfrac{38341403136000000}{195529002969600000} & -\dfrac{927560719872000000}{3519522053452800000} & -\dfrac{628047862272000000}{3519522053452800000} & -\dfrac{161249730048000000}{586587008908800000} \\ -\dfrac{1709599882752000000}{1.7597610267264E+19} & -\dfrac{71635373568000000}{586587008908800000} & -\dfrac{237138513408000000}{2932935044544000000} & -\dfrac{4716983808000000}{3519522053452800000} & \dfrac{156388796928000000}{3519522053452800000} & -\dfrac{404727860736000000}{2932935044544000000} \\ \dfrac{1404464168448000000}{1.1731740178176E+19} & \dfrac{96044299776000000}{3519522053452800000} & -\dfrac{3.26081147904E+20}{9.3853921425408E+21} & \dfrac{238220932608000000}{7039044106905600000} & \dfrac{206153206272000000}{1.4078088213811E+19} & \dfrac{364022678016000000}{2932935044544000000} \\ -\dfrac{88695046656000000}{293293504454400000} & \dfrac{1108560121344000000}{3519522053452800000} & -\dfrac{2277945856512000000}{4692696071270400000} & \dfrac{109213733376000000}{703904410690560000} & \dfrac{543019728384000000}{1407808821381120000} & \dfrac{619736375808000000}{2346348035635200000} \\ -\dfrac{898496644608000000}{2932935044544000000} & \dfrac{3146853888000000}{156423202375680000} & -\dfrac{1041841046016000000}{3910580059392000000} & -\dfrac{1.0025242712064E+21}{2.8156176427622E+22} & \dfrac{363462308352000000}{2346348035635200000} & \dfrac{6.456258772992E+21}{2.8156176427622E+23}\end{pmatrix}\) . Précisément on a calculé \( A^{-1}_{3, 2}=B_{3, 2}= \left(\dfrac{4888225074240000}{124416000000}\right)^{-1}Co(A)_{2, 3}= \left(\dfrac{4888225074240000}{124416000000}\right)^{-1}\times(-1)^{2+3}\det\begin{pmatrix}1 & 3 & 5 & 3 & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & -\dfrac{14}{3} & -1 & 0 \\ 2 & -\dfrac{11}{2} & -4 & -2 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & 1 & \dfrac{19}{5} & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & 4 & -4 & \dfrac{20}{3}\end{pmatrix}=-\dfrac{71635373568000000}{586587008908800000}\) .
  6. On observe que \( B^{-1}=\left(A^{-1}\right)^{-1}=A\) . Trivialement, et sans calcul, \[ B^{-1}=\begin{pmatrix}1 & 3 & \dfrac{11}{4} & 5 & 3 & -5 \\ 4 & \dfrac{9}{4} & -\dfrac{9}{2} & -5 & \dfrac{21}{5} & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & \dfrac{1}{3} & -\dfrac{14}{3} & -1 & 0 \\ 2 & -\dfrac{11}{2} & 2 & -4 & -2 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & \dfrac{15}{2} & 1 & \dfrac{19}{5} & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & -3 & 4 & -4 & \dfrac{20}{3}\end{pmatrix}\]
  7. Le système \( \left\{\begin{array}{*{7}{cr}} &x&+&3y &+&\dfrac{11}{4}z &+&5t &+&3u &-&5v &=&-\dfrac{8}{9}\\ &4x&+&\dfrac{9}{4}y &-&\dfrac{9}{2}z &-&5t &+&\dfrac{21}{5}u &-&5v &=&-8\\ &-\dfrac{15}{2}x&-&\dfrac{2}{3}y &+&\dfrac{1}{3}z &-&\dfrac{14}{3}t &-&u &&&=&\dfrac{29}{4}\\ &2x&-&\dfrac{11}{2}y &+&2z &-&4t &-&2u &-&v &=&8\\ &-\dfrac{21}{2}x&+&\dfrac{18}{5}y &+&\dfrac{15}{2}z &+&t &+&\dfrac{19}{5}u &-&4v &=&8\\ &-\dfrac{13}{4}x&-&\dfrac{5}{2}y &-&3z &+&4t &-&4u &+&\dfrac{20}{3}v &=&-2\\ \end{array} \right.\) est équivalent à l'équation \( AX=a\) où la matrice \( A\) est celle de l'énoncé, \( X\) la matrice colonne des indéterminés et \( a=\begin{pmatrix}-\dfrac{8}{9} \\ -8 \\ \dfrac{29}{4} \\ 8 \\ 8 \\ -2\end{pmatrix}\) de sorte que la solution est \( X=A^{-1}a=\begin{pmatrix}-\dfrac{20045035008000000}{488822507424000000} & -\dfrac{82141558272000000}{1759761026726400000} & -\dfrac{99742938624000000}{1466467522272000000} & -\dfrac{32341068288000000}{1759761026726400000} & -\dfrac{75148881408000000}{1759761026726400000} & -\dfrac{184075835904000000}{1955290029696000000} \\ \dfrac{4991196672000000}{175976102672640000} & -\dfrac{701871340032000000}{3519522053452800000} & \dfrac{38341403136000000}{195529002969600000} & -\dfrac{927560719872000000}{3519522053452800000} & -\dfrac{628047862272000000}{3519522053452800000} & -\dfrac{161249730048000000}{586587008908800000} \\ -\dfrac{1709599882752000000}{1.7597610267264E+19} & -\dfrac{71635373568000000}{586587008908800000} & -\dfrac{237138513408000000}{2932935044544000000} & -\dfrac{4716983808000000}{3519522053452800000} & \dfrac{156388796928000000}{3519522053452800000} & -\dfrac{404727860736000000}{2932935044544000000} \\ \dfrac{1404464168448000000}{1.1731740178176E+19} & \dfrac{96044299776000000}{3519522053452800000} & -\dfrac{3.26081147904E+20}{9.3853921425408E+21} & \dfrac{238220932608000000}{7039044106905600000} & \dfrac{206153206272000000}{1.4078088213811E+19} & \dfrac{364022678016000000}{2932935044544000000} \\ -\dfrac{88695046656000000}{293293504454400000} & \dfrac{1108560121344000000}{3519522053452800000} & -\dfrac{2277945856512000000}{4692696071270400000} & \dfrac{109213733376000000}{703904410690560000} & \dfrac{543019728384000000}{1407808821381120000} & \dfrac{619736375808000000}{2346348035635200000} \\ -\dfrac{898496644608000000}{2932935044544000000} & \dfrac{3146853888000000}{156423202375680000} & -\dfrac{1041841046016000000}{3910580059392000000} & -\dfrac{1.0025242712064E+21}{2.8156176427622E+22} & \dfrac{363462308352000000}{2346348035635200000} & \dfrac{6.456258772992E+21}{2.8156176427622E+23}\end{pmatrix}\times\begin{pmatrix}-\dfrac{8}{9} \\ -8 \\ \dfrac{29}{4} \\ 8 \\ 8 \\ -2\end{pmatrix}=\begin{pmatrix}-\dfrac{1.0548569797717E+110}{2.7497829553647E+110} \\ \dfrac{1.7928612293962E+107}{3.1677499645802E+109} \\ \dfrac{4.3472956760839E+112}{3.9596874557252E+112} \\ -\dfrac{1.7717334097488E+117}{4.0547199546626E+117} \\ -\dfrac{7.994178273506E+110}{4.0547199546626E+110} \\ -\dfrac{1.0952074528615E+120}{1.2013985050852E+120}\end{pmatrix}\) . Ainsi \( x=\dfrac{1.0548569797717E+110}{2.7497829553647E+110}\) , \( y=\dfrac{1.7928612293962E+107}{3.1677499645802E+109}\) , \( z=\dfrac{4.3472956760839E+112}{3.9596874557252E+112}\) , \( t=\dfrac{1.7717334097488E+117}{4.0547199546626E+117}\) , \( u=\dfrac{7.994178273506E+110}{4.0547199546626E+110}\) et \( v=\dfrac{1.0952074528615E+120}{1.2013985050852E+120}\)
  8. Le système \( \left\{\begin{array}{*{7}{cr}} &-\dfrac{20045035008000000}{488822507424000000}x&-&\dfrac{82141558272000000}{1759761026726400000}y &-&\dfrac{99742938624000000}{1466467522272000000}z &-&\dfrac{32341068288000000}{1759761026726400000}t &-&\dfrac{75148881408000000}{1759761026726400000}u &-&\dfrac{184075835904000000}{1955290029696000000}v &=&-5\\ &\dfrac{4991196672000000}{175976102672640000}x&-&\dfrac{701871340032000000}{3519522053452800000}y &+&\dfrac{38341403136000000}{195529002969600000}z &-&\dfrac{927560719872000000}{3519522053452800000}t &-&\dfrac{628047862272000000}{3519522053452800000}u &-&\dfrac{161249730048000000}{586587008908800000}v &=&2\\ &-\dfrac{1709599882752000000}{1.7597610267264E+19}x&-&\dfrac{71635373568000000}{586587008908800000}y &-&\dfrac{237138513408000000}{2932935044544000000}z &-&\dfrac{4716983808000000}{3519522053452800000}t &+&\dfrac{156388796928000000}{3519522053452800000}u &-&\dfrac{404727860736000000}{2932935044544000000}v &=&8\\ &\dfrac{1404464168448000000}{1.1731740178176E+19}x&+&\dfrac{96044299776000000}{3519522053452800000}y &-&\dfrac{3.26081147904E+20}{9.3853921425408E+21}z &+&\dfrac{238220932608000000}{7039044106905600000}t &+&\dfrac{206153206272000000}{1.4078088213811E+19}u &+&\dfrac{364022678016000000}{2932935044544000000}v &=&8\\ &-\dfrac{88695046656000000}{293293504454400000}x&+&\dfrac{1108560121344000000}{3519522053452800000}y &-&\dfrac{2277945856512000000}{4692696071270400000}z &+&\dfrac{109213733376000000}{703904410690560000}t &+&\dfrac{543019728384000000}{1407808821381120000}u &+&\dfrac{619736375808000000}{2346348035635200000}v &=&8\\ &-\dfrac{898496644608000000}{2932935044544000000}x&+&\dfrac{3146853888000000}{156423202375680000}y &-&\dfrac{1041841046016000000}{3910580059392000000}z &-&\dfrac{1.0025242712064E+21}{2.8156176427622E+22}t &+&\dfrac{363462308352000000}{2346348035635200000}u &+&\dfrac{6.456258772992E+21}{2.8156176427622E+23}v &=&7\\ \end{array} \right.\) est équivalent à l'équation \( BX=b\) où la matrice \( B=A^{-1}\) déterminée précédement, \( X\) la matrice colonne des indéterminés et \( b=\begin{pmatrix}-5 \\ 2 \\ 8 \\ 8 \\ 8 \\ 7\end{pmatrix}\) de sorte que la solution est \( X=B^{-1}b=Ab=\begin{pmatrix}1 & 3 & \dfrac{11}{4} & 5 & 3 & -5 \\ 4 & \dfrac{9}{4} & -\dfrac{9}{2} & -5 & \dfrac{21}{5} & -5 \\ -\dfrac{15}{2} & -\dfrac{2}{3} & \dfrac{1}{3} & -\dfrac{14}{3} & -1 & 0 \\ 2 & -\dfrac{11}{2} & 2 & -4 & -2 & -1 \\ -\dfrac{21}{2} & \dfrac{18}{5} & \dfrac{15}{2} & 1 & \dfrac{19}{5} & -4 \\ -\dfrac{13}{4} & -\dfrac{5}{2} & -3 & 4 & -4 & \dfrac{20}{3}\end{pmatrix}\times\begin{pmatrix}-5 \\ 2 \\ 8 \\ 8 \\ 8 \\ 7\end{pmatrix}=\begin{pmatrix}52 \\ -\dfrac{929}{10} \\ -\dfrac{13}{2} \\ -60 \\ \dfrac{1301}{10} \\ \dfrac{407}{12}\end{pmatrix}\) . Ainsi \( x=52\) , \( y=\dfrac{929}{10}\) , \( z=\dfrac{13}{2}\) , \( t=60\) , \( u=\dfrac{1301}{10}\) et \( v=\dfrac{407}{12}\)