Bonjour Charline661
Exercice 26
[tex]\begin{pmatrix}x+y&1\\0&x-y\end{pmatrix}=\begin{pmatrix}2&1-z\\z&-3\end{pmatrix}[/tex]
Egalisons les termes correspondants des deux matrices.
[tex]\left\{\begin{matrix}x+y=2\\1=1-z\\z=0\\x-y=-3\end{matrix}\right.\ \ \ \ \ \ \left\{\begin{matrix}\left\{\begin{matrix}x+y=2\\x-y=-3\end{matrix}\right.\\\left\{\begin{matrix}1=1-z\\z=0\end{matrix}\right.\end{matrix}\right.\\\\\\\left\{\begin{matrix}x+y=2\\x-y=-3\end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=2-y\\x-y=-3\end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=2-y\\2-y-y=-3\end{matrix}\right.[/tex]
[tex]\Longrightarrow\left\{\begin{matrix}x=2-y\\-2y=-5\end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=2-y\\\\y=\dfrac{5}{2}\end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=2-\dfrac{5}{2}\\\\y=\dfrac{5}{2}\end{matrix}\right.\Longrightarrow\boxed{\left\{\begin{matrix}x=-\dfrac{1}{2}\\\\y=\dfrac{5}{2}\end{matrix}\right.}[/tex]
[tex]\left\{\begin{matrix}1=1-z\\z=0\end{matrix}\right.\Longrightarrow\left\{\begin{matrix}z=1-1\\z=0\end{matrix}\right.\Longrightarrow\left\{\begin{matrix}z=0\\z=0\end{matrix}\right.\Longrightarrow\boxed{z=0}[/tex]
Par conséquent, [tex]\boxed{x=-\dfrac{1}{2}\ ;\ y=\dfrac{5}{2}\ ;\ z=0}[/tex]
Exercice 27
[tex]A=\begin{pmatrix}0&z+t+1\\x-y+1&2z-t\\x+y-1&0\end{pmatrix}[/tex]
La matrice A est une matrice nulle si tous ses termes sont
égaux à 0.
Donc
[tex]\left\{\begin{matrix}z+t+1=0\\2z-t=0 \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}z+t+1=0\\t=2z \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}z+2z+1=0\\t=2z \end{matrix}\right.\\\\\\\Longrightarrow\left\{\begin{matrix}3z+1=0\\t=2z \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}z=-\dfrac{1}{3}\\\\t=2z \end{matrix}\right.\Longrightarrow\boxed{\left\{\begin{matrix}z=-\dfrac{1}{3}\\\\t=-\dfrac{2}{3} \end{matrix}\right.}[/tex]
De plus,
[tex]\left\{\begin{matrix}x-y+1=0\\x+y-1=0 \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=y-1\\x+y-1=0 \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=y-1\\y-1+y-1=0 \end{matrix}\right.\\\\\\\left\{\begin{matrix}x=y-1\\2y-2=0 \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=y-1\\2y=2 \end{matrix}\right.\Longrightarrow\left\{\begin{matrix}x=y-1\\y=1 \end{matrix}\right.\Longrightarrow\boxed{\left\{\begin{matrix}x=0\\y=1 \end{matrix}\right.}[/tex]
Par conséquent,
[tex]\boxed{x=0\ ;\ y=1\ ;\ z=-\dfrac{1}{3}\ ;\ t=-\dfrac{2}{3}}[/tex]
