Sa consideram \( A,B\in \mathcal{M}_2(\mathbb{Z}) \) cu proprietatea ca \( AB+BA=I_2 \)
a) Sa se calculeze \( \tr(AB) \).
b) Sa se demonstreze ca \( \tr A=\tr B=0. \)
Marius Dragan, Concursul Arhimede, 28.02.2009
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Marius Mainea
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a) \( \tr(AB+BA)=\tr(AB)+\tr(BA)=2\tr(AB)=2 \Rightarrow \tr(AB)=1. \)
b) \( AB+BA=I_2 \Rightarrow
a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}+a_{22}b_{22}=1 \)
\( \tr(AB)=1 \Rightarrow
2a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}=1 \) deci \( a_{11}b_{11}=a_{22}b_{22} \) si \( a_{11}b_{12}+a_{12}b_{22}+a_{12}b_{11}+a_{22}b_{12}=0=b_{12}\tr(A)+a_{12}\tr(B) \)
\( a_{21}b_{11}+a_{22}b_{21}+a_{11}b_{21}+a_{21}b_{22}=0=b_{21}\tr(A)+a_{21}\tr(B) \Rightarrow a_{12}b_{21}\tr(A)\tr(B)=a_{21}b_{12}\tr(A)\tr(B) \).
Daca \( \tr(A)\tr(B) \neq 0 \Rightarrow a_{12}b_{21}=a_{21}b_{12} \Rightarrow 2(a_{11}b_{11}+a_{12}b_{21})=1 \) fals, deci \( \tr(A)\tr(B)=0 \). Fie \( \tr(A)=0,\ a_{11}=-a_{22} \) si din \( a_{11}b_{11}=a_{22}b_{22} \) avem ca si \( \tr(B)=0. \)
b) \( AB+BA=I_2 \Rightarrow
a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}+a_{22}b_{22}=1 \)
\( \tr(AB)=1 \Rightarrow
2a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}=1 \) deci \( a_{11}b_{11}=a_{22}b_{22} \) si \( a_{11}b_{12}+a_{12}b_{22}+a_{12}b_{11}+a_{22}b_{12}=0=b_{12}\tr(A)+a_{12}\tr(B) \)
\( a_{21}b_{11}+a_{22}b_{21}+a_{11}b_{21}+a_{21}b_{22}=0=b_{21}\tr(A)+a_{21}\tr(B) \Rightarrow a_{12}b_{21}\tr(A)\tr(B)=a_{21}b_{12}\tr(A)\tr(B) \).
Daca \( \tr(A)\tr(B) \neq 0 \Rightarrow a_{12}b_{21}=a_{21}b_{12} \Rightarrow 2(a_{11}b_{11}+a_{12}b_{21})=1 \) fals, deci \( \tr(A)\tr(B)=0 \). Fie \( \tr(A)=0,\ a_{11}=-a_{22} \) si din \( a_{11}b_{11}=a_{22}b_{22} \) avem ca si \( \tr(B)=0. \)