Fie \( k \in {\mathbb{N}^*} \). Aratati ca daca \( \exist A, B \in {\cal M_n}({\mathbb R}) \) a.i. \( A^2 + B^2= ctg \frac{\pi}{k}(AB-BA) \) si \( AB - BA \) e inversabila, atunci n este multiplu de k. Pentru k=2 dati exemplu de o matrice care verifica relatiile din enunt.
Etapa judeteana Dolj, 1997
Ordinul unei matrice...
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Radu Titiu
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Trecand la determinanti relatia \( (A-iB)(A+iB)=A^2+B^2+i(AB-BA)=(ctg \frac{\pi}{k}+i)(AB-BA) \) rezulta
\( (i+ctg \frac{\pi}{k})^n \in \mathbb{R} \Leftrightarrow \frac{\cos \frac{n\pi}{k}+i \sin \frac{n \pi}{k}}{\sin ^n \frac{\pi}{k}} \in \mathbb{R} \Rightarrow k|n \)
Ca exemplu fie \( A,B \in \mathcal{M}_2(\mathbb{R}) \)
\( A=\left( \begin{array} 0 \ 1\ \\
6 \ 0\ \end{array} \right) \) si \( B=\left( \begin{array} 0 \ -2\ \\
-3 \ 0\ \end{array} \right) \) respecta cerinta \( A^2+B^2=0 \) , iar\( \det(AB-BA) \neq 0 \)
\( (i+ctg \frac{\pi}{k})^n \in \mathbb{R} \Leftrightarrow \frac{\cos \frac{n\pi}{k}+i \sin \frac{n \pi}{k}}{\sin ^n \frac{\pi}{k}} \in \mathbb{R} \Rightarrow k|n \)
Ca exemplu fie \( A,B \in \mathcal{M}_2(\mathbb{R}) \)
\( A=\left( \begin{array} 0 \ 1\ \\
6 \ 0\ \end{array} \right) \) si \( B=\left( \begin{array} 0 \ -2\ \\
-3 \ 0\ \end{array} \right) \) respecta cerinta \( A^2+B^2=0 \) , iar\( \det(AB-BA) \neq 0 \)
A mathematician is a machine for turning coffee into theorems.