Fie matricea \( A\in M_{4}(\mathbb{Z}) \) astfel incat \( A^{3}=A+I_{4} \). Aratati ca \( \det(A^{2}-I_{4})=1 \).
Dorin Arventiev, Tudorel Lupu, R.M.I. C-ta, 2004
Matrice de ordin 4 cu elemente intregi si A^3=A+I_n
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Matrice de ordin 4 cu elemente intregi si A^3=A+I_n
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Marius Dragoi
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\( {\det A}{\det (A^2-I)}=1 \Rightarrow \det (A^2-I) =1 \) sau \( \det (A^2-I)= -1 \).
Fie \( \det (A^2-I)=-1 \Rightarrow \det A=-1 \).
Cum \( \det A^3=\det (A+I) \Rightarrow \det (A+I)=-1 \Rightarrow \det (A-I)=1 \).
Din \( {\det (A-I)}{\det (A^2+A+I)}=\det A=-1 \Rightarrow \det (A^2+A+I)=-1 \) fals, deoarece \( \det (A^2+A+I) \geq 0 \).
Asadar \( \det (A^2-I)=1 \).
Fie \( \det (A^2-I)=-1 \Rightarrow \det A=-1 \).
Cum \( \det A^3=\det (A+I) \Rightarrow \det (A+I)=-1 \Rightarrow \det (A-I)=1 \).
Din \( {\det (A-I)}{\det (A^2+A+I)}=\det A=-1 \Rightarrow \det (A^2+A+I)=-1 \) fals, deoarece \( \det (A^2+A+I) \geq 0 \).
Asadar \( \det (A^2-I)=1 \).
Last edited by Marius Dragoi on Tue Mar 25, 2008 12:06 pm, edited 1 time in total.
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers