Déterminant et systèmes linéaires

La méthode du pivot de Gauss - Exercice 2

10 min
20
En appliquant la méthode du pivot de Gauss, résoudre, dans R3\mathbb{R}^3, les systèmes linéaires suivants :
Question 1

{x−y+4z=112x+y+2z=10−x+2y−3z=−6\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 2x+y+2z & = & 10 \\ -x+2y-3z & = & -6 \end{array}\right.

Correction
{x−y+4z=112x+y+2z=10−x+2y−3z=−6L1L2L3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 2x+y+2z & = & 10 \\ -x+2y-3z & = & -6 \end{array} \begin{array}{c} \end{array} \begin{array}{c}L_1 \\ L_2 \\ L_3 \end{array}\right.
Notons L1L_1 la première ligne de notre système linéaire qui correspondra à la ligne pivot.
Maintenant, nous allons supprimer le coefficient de xx de la ligne L2L_2 (respectivement de la ligne L3L_3) à l'aide d'une combinaison linéaire de L1L_1 avec L2L_2 (respectivement à l'aide d'une combinaison de L1L_1 avec L3L_3 ) .
{x−y+4z=113y−6z=−12y+z=5L1L2⟵L2−2L1L3⟵L3+L1\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y-6z & = & -12 \\ y+z & = & 5 \end{array} \begin{array}{c} \end{array} \begin{array}{c}L_1 \\ L_2\longleftarrow L_2-2L_1 \\ L_3\longleftarrow L_3+L_1 \end{array}\right.
La ligne L2L_2 est désormais la ligne pivot de notre système linéaire.
Maintenant, nous allons supprimer le coefficient de yy de la ligne L3L_3 à l'aide d'une combinaison linéaire de L2L_2 avec L3L_3.
{x−y+4z=113y−6z=−129z=27L1L2L3⟵3L3−L2\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y-6z & = & -12 \\ 9z & = & 27 \end{array} \begin{array}{c} \end{array} \begin{array}{c}L_1 \\ L_2 \\ L_3\longleftarrow 3L_3-L_2 \end{array}\right.
{x−y+4z=113y−6z=−12z=279\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y-6z & = & -12 \\ z & = & \frac{27}{9} \end{array}\right.
{x−y+4z=113y−6z=−12z=3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y-6z & = & -12 \\ z & = & 3 \end{array}\right.
{x−y+4z=113y−6×3=−12z=3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y-6\times 3 & = & -12 \\ z & = & 3 \end{array}\right.
{x−y+4z=113y=−12+18z=3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y & = & -12+18 \\ z & = & 3 \end{array}\right.
{x−y+4z=113y=6z=3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ 3y & = & 6 \\ z & = & 3 \end{array}\right.
{x−y+4z=11y=63z=3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ y & = & \frac{6}{3} \\ z & = & 3 \end{array}\right.
{x−y+4z=11y=2z=3\left\{ \begin{array}{ccc}x-y+4z & = & 11 \\ y & = & 2 \\ z & = & 3 \end{array}\right.
{x−2+4×3=11y=2z=3\left\{ \begin{array}{ccc}x-2+4\times 3 & = & 11 \\ y & = & 2 \\ z & = & 3 \end{array}\right.
{x=11−10y=2z=3\left\{ \begin{array}{ccc}x & = & 11-10 \\ y & = & 2 \\ z & = & 3 \end{array}\right.
{x=1y=2z=3\left\{ \begin{array}{ccc}x & = & 1 \\ y & = & 2 \\ z & = & 3 \end{array}\right.
L’unique solution du système est le triplet
(x;y;z)={(1;2;3)}\left(x;y;z\right)=\left\{\left(1;2;3\right)\right\}
Question 2

{x−y+2z=22x+3y−3z=−15x−y+2z=6\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 2x+3y-3z & = & -1 \\ 5x-y+2z & = & 6 \end{array}\right.

Correction
{x−y+2z=22x+3y−3z=−15x−y+2z=6L1L2L3\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 2x+3y-3z & = & -1 \\ 5x-y+2z & = & 6 \end{array} \begin{array}{cc} & L_1 \\ & L_2 \\ & L_3 \end{array}\right.
{x−y+2z=25y−7z=−54y−8z=−4L1L2⟵L2−2L1L3⟵L3−5L1\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 5y-7z & = & -5 \\ 4y-8z & = & -4 \end{array} \begin{array}{cc} & L_1 \\ & L_2\longleftarrow L_2-2L_1 \\ & L_3\longleftarrow L_3-5L_1\end{array}\right.
{x−y+2z=25y−7z=−5y−4z=−1L1L2L3⟵12L3\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 5y-7z & = & -5 \\ y-4z & = & -1 \end{array} \begin{array}{cc} & L_1 \\ & L_2 \\ & L_3\longleftarrow \frac{1}{2}L_3 \end{array}\right.
{x−y+2z=25y−7z=−5−13z=0L1L2L3⟵5L3−L2\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 5y-7z & = & -5 \\ -13z & = & 0 \end{array} \begin{array}{cc} & L_1 \\ & L_2 \\ & L_3\longleftarrow 5L_3-L_2 \end{array}\right.
{x−y+2z=25y−7z=−5z=0\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 5y-7z & = & -5 \\ z & = & 0 \end{array}\right.
{x−y+2z=25y−7×0=−5z=0\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 5y-7\times 0 & = & -5 \\ z & = & 0 \end{array}\right.
{x−y+2z=25y=−5z=0\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ 5y & = & -5 \\ z & = & 0 \end{array}\right.
{x−y+2z=2y=−1z=0\left\{ \begin{array}{ccc}x-y+2z & = & 2 \\ y & = & -1 \\ z & = & 0 \end{array}\right.
{x−(−1)+2×0=2y=−1z=0\left\{ \begin{array}{ccc}x-\left(-1\right)+2\times 0 & = & 2 \\ y & = & -1 \\ z & = & 0 \end{array}\right.
{x+1=2y=−1z=0\left\{ \begin{array}{ccc}x+1 & = & 2 \\ y & = & -1 \\ z & = & 0 \end{array}\right.
{x=1y=−1z=0\left\{ \begin{array}{ccc}x & = & 1 \\ y & = & -1 \\ z & = & 0 \end{array}\right.