Calcul matriciel

Matrice inverse et pivot de gauss - Exercice 1

20 min
35
Calculer l'inverse (lorsque c'est possible) de la matrice AA dans chacun des cas suivants :
Question 1

A=(1211111−12)A=\left( \begin{array}{ccc}1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right)

Correction
  • Nous disposons à coté de la matrice AA la matrice identité I3I_3. Il faut ensuite appliquer les opérations sur les lignes de AA et ces mêmes opérations sur les lignes de I3I_3 pour transformer AA en I3I_3 et I3I_3 en A−1A^{-1} .
(1211111−12∣∣∣∣∣∣100010001)L1L2L3\left( \begin{array}{ccc}1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)\begin{array}{c} L_1 \\ L_2 \\ L_3\end{array}
(1210−100−31∣∣∣∣∣∣100−110−101)L1L2←L2−L1L3←L3−L1\left( \begin{array}{ccc}1 & 2 & 1 \\ 0 & -1 & 0 \\ 0 & -3 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right) \begin{array}{c}L_1 \\ L_2\leftarrow L_2-L_1 \\ L_3\leftarrow L_3-L_1 \end{array}
(1210−10001∣∣∣∣∣∣100−1102−31)L1L2L3←L3−3L2\left( \begin{array}{ccc}\red{1} & 2 & 1 \\ 0 & \blue{-1} & 0 \\ 0 & 0 & \green{1} \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ 2 & -3 & 1 \end{array}\right) \begin{array}{c}L_1 \\ L_2 \\ L_3\leftarrow L_3-3L_2 \end{array}
Remarque : AA est inversible car les coefficients diagonaux sont non nuls.
(1200−10001∣∣∣∣∣∣−13−1−1102−31)L1←L1−L3L2L3\left( \begin{array}{ccc}1 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}-1 & 3 & -1 \\ -1 & 1 & 0 \\ 2 & -3 & 1 \end{array}\right) \begin{array}{c}L_1\leftarrow L_1-L_3 \\ L_2 \\ L_3 \end{array}
(1000−10001∣∣∣∣∣∣−35−1−1102−31)L1←L1+2L2L2L3\left( \begin{array}{ccc}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}-3 & 5 & -1 \\ -1 & 1 & 0 \\ 2 & -3 & 1 \end{array}\right) \begin{array}{c}L_1\leftarrow L_1+2L_2 \\ L_2 \\ L_3 \end{array}
(100010001∣∣∣∣∣∣−35−11−102−31)L1L2←−L2L3\left( \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}-3 & 5 & -1 \\ 1 & -1 & 0 \\ 2 & -3 & 1 \end{array}\right) \begin{array}{c}L_1 \\ L_2\leftarrow -L_2 \\ L_3 \end{array}
Finalement la matrice inverse de AA est alors
A−1=(−35−11−102−31)A^{-1}=\left( \begin{array}{ccc}-3 & 5 & -1 \\ 1 & -1 & 0 \\ 2 & -3 & 1 \end{array}\right)
Question 2

A=(1−10121110)A=\left( \begin{array}{ccc}1 & -1 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 0 \end{array}\right)

Correction
  • Nous disposons à coté de la matrice AA la matrice identité I3I_3. Il faut ensuite appliquer les opérations sur les lignes de AA et ces mêmes opérations sur les lignes de I3I_3 pour transformer AA en I3I_3 et I3I_3 en A−1A^{-1} .
(1−10121110∣∣∣∣∣∣100010001)L1L2L3\left( \begin{array}{ccc}1 & -1 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 0 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)\begin{array}{c} L_1 \\ L_2 \\ L_3\end{array}
(1−10031020∣∣∣∣∣∣100−110−101)L1L2←L2−L1L3←L3−L1\left( \begin{array}{ccc}1 & -1 & 0 \\ 0 & 3 & 1 \\ 0 & 2 & 0 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right) \begin{array}{c}L_1 \\ L_2\leftarrow L_2-L_1 \\ L_3\leftarrow L_3-L_1 \end{array}
(1−1003100−2∣∣∣∣∣∣100−110−1−23)L1L2L3←3L3−2L2\left( \begin{array}{ccc}\red{1} & -1 & 0 \\ 0 & \blue{3} & 1 \\ 0 & 0 & \green{-2} \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & -2 & 3 \end{array}\right) \begin{array}{c}L_1 \\ L_2 \\ L_3\leftarrow 3L_3-2L_2 \end{array}
Remarque : AA est inversible car les coefficients diagonaux sont non nuls.
(30103100−2∣∣∣∣∣∣210−110−1−23)L1←3L1+L2L2L3\left( \begin{array}{ccc}3 & 0 & 1 \\ 0 & 3 & 1 \\ 0 & 0 & -2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}2 & 1 & 0 \\ -1 & 1 & 0 \\ -1 & -2 & 3 \end{array}\right) \begin{array}{c}L_1\leftarrow 3L_1+L_2 \\ L_2 \\ L_3 \end{array}
(60006000−2∣∣∣∣∣∣303−303−1−23)L1←2L1+L3L2←2L2+L3L3\left( \begin{array}{ccc}6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & -2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}3 & 0 & 3 \\ -3 & 0 & 3 \\ -1 & -2 & 3 \end{array}\right) \begin{array}{c}L_1\leftarrow 2L_1+L_3 \\ L_2\leftarrow 2L_2+L_3 \\ L_3 \end{array}
(100010001∣∣∣∣∣∣12012−12012121−32)L1←16L1L2←16L2L3←−12L3\left( \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}\frac{1}{2} & 0 & \frac{1}{2}\\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & 1 & -\frac{3}{2} \end{array}\right) \begin{array}{c}L_1\leftarrow \frac{1}{6}L_1 \\ L_2\leftarrow \frac{1}{6}L_2 \\ L_3\leftarrow -\frac{1}{2}L_3 \end{array}
Finalement la matrice inverse de AA est alors
A−1=(12012−12012121−32)A^{-1}=\left( \begin{array}{ccc}\frac{1}{2} & 0 & \frac{1}{2}\\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & 1 & -\frac{3}{2}\end{array}\right)
Question 3

A=(211−12−11−11)A=\left( \begin{array}{ccc}2 & 1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 1 \end{array}\right)

Correction
  • Nous disposons à coté de la matrice AA la matrice identité I3I_3. Il faut ensuite appliquer les opérations sur les lignes de AA et ces mêmes opérations sur les lignes de I3I_3 pour transformer AA en I3I_3 et I3I_3 en A−1A^{-1} .
(211−12−11−11∣∣∣∣∣∣100010001)L1L2L3\left( \begin{array}{ccc}2 & 1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right)\begin{array}{c} L_1 \\ L_2 \\ L_3\end{array}
(21105−10−31∣∣∣∣∣∣100120−102)L1L2←2L2+L1L3←2L3−L1\left( \begin{array}{ccc}2 & 1 & 1 \\ 0 & 5 & -1 \\ 0 & -3 & 1 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ 1 & 2 & 0\\ -1 & 0 & 2 \end{array}\right)\begin{array}{c} L_1 \\ L_2\leftarrow 2L_2+L_1 \\ L_3\leftarrow 2L_3-L_1 \end{array}
(21105−1002∣∣∣∣∣∣100120−2610)L1L2←2L2+L1L3←5L3+3L2\left( \begin{array}{ccc}2 & 1 & 1 \\ 0 & 5 & -1 \\ 0 & 0 & 2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & 0 & 0 \\ 1 & 2 & 0\\ -2 & 6 & 10 \end{array}\right)\begin{array}{c} L_1 \\ L_2\leftarrow 2L_2+L_1 \\ L_3\leftarrow 5L_3+3L_2 \end{array}
Remarque : AA est inversible car les coefficients diagonaux sont non nuls.
(4200100002∣∣∣∣∣∣4−6−1001010−2610)L1←2L1−L3L2←2L2+L3L3\left( \begin{array}{ccc}4 & 2 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}4 & -6 & -10 \\ 0 & 10 & 10\\ -2 & 6 & 10 \end{array}\right)\begin{array}{c} L_1\leftarrow 2L_1-L_3 \\ L_2\leftarrow 2L_2+L_3 \\ L_3 \end{array}
(20000100002∣∣∣∣∣∣20−40−6001010−2610)L1←5L1−L2L2L3\left( \begin{array}{ccc}20 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}20 & -40 & -60 \\ 0 & 10 & 10\\ -2 & 6 & 10 \end{array}\right)\begin{array}{c} L_1\leftarrow 5L_1-L_2 \\ L_2 \\ L_3 \end{array}
(100010002∣∣∣∣∣∣1−2−3011−135)L1←120L1L2←110L2L3←12L3\left( \begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{c}| \\ | \\ | \end{array} \begin{array}{ccc}1 & -2 & -3 \\ 0 & 1 & 1\\ -1 & 3 & 5 \end{array}\right)\begin{array}{c} L_1\leftarrow \frac{1}{20}L_1 \\ L_2\leftarrow \frac{1}{10}L_2 \\ L_3\leftarrow \frac{1}{2}L_3 \end{array}
Finalement la matrice inverse de AA est alors
A−1=(1−2−3011−135)A^{-1}=\left( \begin{array}{ccc}1 & -2 & -3 \\ 0 & 1 & 1\\ -1 & 3 & 5\end{array}\right)