Inverse d’une matrice par inversion d’un système linéaire - Exercice 1

Soit $$X = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \in \mathcal{M}_{3,1}(\mathbb{R})$$.
Soit $$Y = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \in \mathcal{M}_{3,1}(\mathbb{R})$$.
On pose $$Y = A X$$ ce qui implique que $$A^{-1}Y = X$$.
Ainsi :
$$Y = A X \,\,\, \Longleftrightarrow \,\,\, \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \,\,\, \Longleftrightarrow \,\,\, \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} 0 & -x_2 & 0 \\ x_1 & 0 & -x_3 \\ 0 & x_2 & x_3 \end{pmatrix}$$
Ainsi, on obtient :
$$\left\lbrace \begin{array}{rcl} y_1 & = & -x_2 \\ y_2 & = & x_1 - x_3 \\ y_3 & = & x_2 + x_3\end{array} \right. \,\,\, \Longleftrightarrow \,\,\, \left\lbrace \begin{array}{rcl} -y_1 & = & x_2 \\ y_2 & = & x_1 - x_3 \\ y_3 & = & x_2 + x_3\end{array} \right. \,\,\, \Longleftrightarrow \,\,\, \left\lbrace \begin{array}{rcl} y_2 & = & x_1 - x_3 \\ -1y_1 + 0y_2 + 0y_3 & = & x_2 \\ y_3 & = & x_2 + x_3\end{array} \right.$$
Soit :
$$\left\lbrace \begin{array}{rcl} y_2 & = & x_1 - x_3 \\ -1y_1 + 0y_2 + 0y_3 & = & x_2 \\ y_3 & = & -y_1 + x_3\end{array} \right. \,\,\, \Longleftrightarrow \,\,\, \left\lbrace \begin{array}{rcl} y_2 & = & x_1 - x_3 \\ -1y_1 + 0y_2 + 0y_3 & = & x_2 \\ 1 y_1 + 0 y_2 + 1y_3 & = & x_3\end{array} \right.$$
Ce qui nous permet d'écrire que :
$$\left\lbrace \begin{array}{rcl} y_2 & = & x_1 - (y_1+y_3) \\ -1y_1 + 0y_2 + 0y_3 & = & x_2 \\ 1 y_1 + 0 y_2 + 1y_3 & = & x_3\end{array} \right. \,\,\, \Longleftrightarrow \,\,\, \left\lbrace \begin{array}{rcl} y_2 & = & x_1 - y_1 - y_3 \\ -1y_1 + 0y_2 + 0y_3 & = & x_2 \\ 1 y_1 + 0 y_2 + 1y_3 & = & x_3\end{array} \right.$$
Soit encore :
$$\left\lbrace \begin{array}{rcl} 1y_1 + 1y_2 + 1y_3 & = & x_1 \\ -1y_1 + 0y_2 + 0y_3 & = & x_2 \\ 1 y_1 + 0 y_2 + 1y_3 & = & x_3\end{array} \right.$$
On va écrire ceci sous la forme matricielle suivante :
$$\begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 0 \\ 1 & 0 & 1 \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \,\,\, \Longleftrightarrow \,\,\, A^{-1}Y = X$$
Par identification, on trouve finalement que :