Fie \( A\in M_{3}(\mathbb{R}) \) astfel incat \( trA=trA^2=0 \). Sa se arate ca
\( \det(A^2+I_{3})=( \det A )^2+1 \).
Radu Gologan, lista scurta ONM 2003
Matrice ordin 3 cu urma lui A si A^2 nula
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Matrice ordin 3 cu urma lui A si A^2 nula
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\( f(X)=\det(A-XI)=-X^3+Tr(A)X^2-Tr(A^*)X+\det(A)I \)
Dar
\( ab+bc+ca=\frac{1}{2}[(a+b+c)^2-a^2-b^2-c^2] \) \( \Leftrightarrow \) \( Tr(A^*)=\frac{1}{2}[(Tr(A))^2-Tr(A^2)]=0 \) (unde a, b, c sunt radacinile complexe ale ecuatiei P(x)=0).
Asadar \( f(X)=-X^3+d \), unde d=det(A). Astfel identitatea de demonstrat devine: \( f(-i)f(i)=d^2+1 \Leftrightarrow d^2-i^2=d+1 \).
Dar
\( ab+bc+ca=\frac{1}{2}[(a+b+c)^2-a^2-b^2-c^2] \) \( \Leftrightarrow \) \( Tr(A^*)=\frac{1}{2}[(Tr(A))^2-Tr(A^2)]=0 \) (unde a, b, c sunt radacinile complexe ale ecuatiei P(x)=0).
Asadar \( f(X)=-X^3+d \), unde d=det(A). Astfel identitatea de demonstrat devine: \( f(-i)f(i)=d^2+1 \Leftrightarrow d^2-i^2=d+1 \).
A mathematician is a machine for turning coffee into theorems.