Bonjour
Design971,
Meilleurs voeux ! :)
Exercice 6
[tex]1)\ A\times B=\begin{pmatrix}\cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&\cos(\alpha+\beta)\end{pmatrix}[/tex]
[tex]2)\ A\times B=I_2\\\\\begin{pmatrix}\cos(\alpha+\beta)&-\sin(\alpha+\beta)\\\sin(\alpha+\beta)&\cos(\alpha+\beta)\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\\\\\\\left\{\begin{matrix}\cos(\alpha+\beta)=1\\\sin(\alpha+\beta)=0\end{matrix}\right.\ \ \ \ \Longrightarrow\boxed{\alpha+\beta=2k\pi\ \ (k\in\matbbb{Z})}[/tex]
[tex]3)\ A^2=\begin{pmatrix}\cos(2\alpha)&-\sin(2\alpha)\\\sin(2\alpha)&\cos(2\alpha)\end{pmatrix}\\\\\\A^2=I_2\\\\\begin{pmatrix}\cos(2\alpha)&-\sin(2\alpha)\\\sin(2\alpha)&\cos(2\alpha)\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\\\\\\\left\{\begin{matrix}\cos(2\alpha)=1\\\sin(2\alpha)=0\end{matrix}\right.\ \ \ \ \Longrightarrow2\alpha=2k\pi\ \ (k\in\matbbb{Z})\\\\\\\Longrightarrow\boxed{\alpha=k\pi\ \ (k\in\matbbb{Z})}[/tex]
[tex]4)\ A=\begin{pmatrix}\cos\dfrac{\pi}{6}&-\sin\dfrac{\pi}{6}\\\\\sin\dfrac{\pi}{6}&\cos\dfrac{\pi}{6}\end{pmatrix}=\begin{pmatrix}\dfrac{\sqrt{3}}{2}&-\dfrac{1}{2}\\\\\dfrac{1}{2}&\dfrac{\sqrt{3}}{2}\end{pmatrix}\\\\\\A^2=\begin{pmatrix}\dfrac{1}{2}&-\dfrac{\sqrt{3}}{2}\\\\\dfrac{\sqrt{3}}{2}&\dfrac{1}{2}\end{pmatrix}\ \ \ \ A^3=\begin{pmatrix}0&-1\\\\1&0\end{pmatrix}\\\\\\A^4=A^3\times A=\begin{pmatrix}-\dfrac{1}{2}&-\dfrac{\sqrt{3}}{2}\\\\\dfrac{\sqrt{3}}{2}&-\dfrac{1}{2}\end{pmatrix}[/tex]
[tex]A^5=A^3\times A^2=\begin{pmatrix}-\dfrac{\sqrt{3}}{2}&-\dfrac{1}{2}\\\\\dfrac{1}{2}&-\dfrac{\sqrt{3}}{2}\end{pmatrix}\ \ \ A^6=(A^3)^2=\begin{pmatrix}-1&0\\\\0&-1\end{pmatrix}\\\\\\==\ \textgreater \ A^6=-I_2[/tex]
Par conséquent,
[tex]Si\ k\in\mathbb{N^*},\ alors\ A^{6k}=(-1)^kI_2\\\\Si\ k\in\mathbb{N},\ alors\\\\A^{6k+i}=(-1)^kA^i\ \ (i\in\mathbb{N},1\le i\le5)[/tex]
