S:L22.108

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S:L22.108. (Nicolae Mușuroia)

Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A, B \in \mathcal{M}_3 \left( \mathbb{R}\right)} cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle AB = BA} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^2+B^2} neinversabilă și , unde . Arătați că

Soluție.

Ipotezele și , cu , implică

Fie polinomul . Atunci, există pentru care
Cum , avem , deci și sunt rădăcini ale polinomului .


Dacă sunt rădăcinile polinomului , atunci din relațiile lui Viete avem

Se obține , ceea ce implică
Atunci
șiFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f\left( -1 \right) = \det \left( A - B \right) = 2\left( \alpha - 1 \right) \cdot \det(B).} AvemFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\det \left(A+B\right)}{\det \left(A+B\right)} = \frac{\alpha +1}{\alpha -1} = \frac{\dfrac{\det(A)}{\det(B)}+1}{\dfrac{\det(A)}{\det(B)}-1} = \frac{\det(A) + \det(B)}{\det(A)-\det(B)}. }