Calcul matriciel

Soit le système :

\((S) \left\{ \begin{array}{l l l c l}x_{1} & + x_{2} & + x_{3} & = & 2 ~~~~ (\textrm{L}_{1}) \\ 2 x_{1} & & - 3 x_{3} & = & -7 ~~ (\textrm{L}_{2}) \\3x_{1} & -2 x_{2} & -4 x_{3} & = & -5~~ (\textrm{L}_{3}) \\\end{array}\right.\)

Elimination de \(x_{1}\) des lignes \((\textrm{L}_{1})\) et \((\textrm{L}_{2})\) :

\((S) \left\{ \begin{array}{l l l c l}x_{1} & + x_{2} & + x_{3} & = & 2 ~~~~~~ (\textrm{L}_{1}) \\ & -2x_{2} & - 5 x_{3} & = & -11 ~~(\textrm{L}_{2} \leftarrow \textrm{L}_{2} - 2\textrm{L}_{1}) \\ & -5 x_{2} & -7 x_{3} & = & -11~~ (\textrm{L}_{3} \leftarrow \textrm{L}_{3} - 3\textrm{L}_{1}) \\\end{array}\right. \quad\color{red}(2~~\textrm{points})\)

Elimination de \(x_{2}\) des lignes \((\textrm{L}_{1})\) et \((\textrm{L}_{2})\) :

\((S) \left\{ \begin{array}{l l l c l}x_{1} & + x_{2} & + x_{3} & = & 2 ~~~~~~ (\textrm{L}_{1}) \\ & -2x_{2} & - 5 x_{3} & = & -11 ~~(\textrm{L}_{2}) \\ & & 11 x_{3} & = & 33~~~~ (\textrm{L}_{3} \leftarrow 2\textrm{L}_{3} - 3\textrm{L}_{2}) \\\end{array}\right. \quad\color{red}(2~~\textrm{points})\)

De \((\textrm{L}_{3})\) nous tirons : \(x_{3} = 3\).

puis de \((\textrm{L}_{2})\) : \(x_{2} = \frac{11 - 5x^{3}}{2} = \frac{11 -15}{2} = -2\);

enfin de \((\textrm{L}_{1}) : x_{1} = 2 - x_{2} - x_{3} = 2 + 2 - 3 = 1\).

Solution du système \((S)\) : \(\color{blue}x_{1} = 1\) ;\(\color{blue}x_{2} = - 2\) ; \(\color{blue}x_{3} = 3\quad\color{red}(3~~\textrm{points})\)