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Matrice de ordin 4 cu elemente intregi si A^3=A+I_n
Posted: Mon Nov 05, 2007 2:05 am
by Cezar Lupu
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
Posted: Tue Mar 25, 2008 10:12 am
by Marius Dragoi
\( {\det A}{\det (A^2-I)}=1 \Rightarrow \det (A^2-I) =1 \) 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
\( \det (A^2-I)=1 \).
