S:L22.108: Difference between revisions
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Dacă <math>x_1, x_2, x_3 \in \mathbb{C}</math> sunt rădăcinile polinomului <math>f</math>, atunci din [https://ro.wikipedia.org/wiki/Formulele_lui_Vi%C3%A8te relațiile lui Viete] avem<math display="block">x_1x_2x_3 = - \frac{\det(A)}{\det(B)} = - \alpha.</math>Se obține <math>x_3 = -\alpha</math>, ceea ce implică<math display="block">f = \det(B) \cdot \left(X^2 + 1 \right) \cdot \left( X + \alpha \right).</math>Atunci<math display="block">f\left( 1 \right) = \det \left( A + B \right) = 2\left( \alpha +1 \right) \cdot \det(B)</math>și<math display="block">f\left( -1 \right) = \det \left( A - B \right) = 2\left( \alpha - 1 \right) \cdot \det(B).</math>Avem''<math display="block">\frac{\det \left(A+B\right)}{\det \left(A+B\right)} = \frac{\alpha +1}{\alpha -1} = \frac | Dacă <math>x_1, x_2, x_3 \in \mathbb{C}</math> sunt rădăcinile polinomului <math>f</math>, atunci din [https://ro.wikipedia.org/wiki/Formulele_lui_Vi%C3%A8te relațiile lui Viete] avem<math display="block">x_1x_2x_3 = - \frac{\det(A)}{\det(B)} = - \alpha.</math>Se obține <math>x_3 = -\alpha</math>, ceea ce implică<math display="block">f = \det(B) \cdot \left(X^2 + 1 \right) \cdot \left( X + \alpha \right).</math>Atunci<math display="block">f\left( 1 \right) = \det \left( A + B \right) = 2\left( \alpha +1 \right) \cdot \det(B)</math>și<math display="block">f\left( -1 \right) = \det \left( A - B \right) = 2\left( \alpha - 1 \right) \cdot \det(B).</math>Avem''<math display="block">\frac{\det \left(A+B\right)}{\det \left(A+B\right)} = \frac{\alpha +1}{\alpha -1}. </math>''Cum <math>\alpha = \frac{\det(A)}{\det(B)} </math> se obține <math display="block">\frac{\alpha+1}{\alpha -1} = \frac{\det(A) + \det(B)}{\det(A) - \det(B)} </math>În concluzie, are loc egalitatea''<math display="block">\frac{\det \left(A+B\right)}{\det \left(A+B\right)} = \frac{\det(A) + \det(B)}{\det(A)-\det(B)}. </math>'' | ||
Revision as of 08:51, 20 July 2024
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 , neinversabilă și , unde . Arătați că
Soluție.
Ipotezele și , cu , implică
Dacă 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 x_1, x_2, x_3 \in \mathbb{C}}
sunt rădăcinile polinomului 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 f}
, atunci din relațiile lui Viete 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 x_1x_2x_3 = - \frac{\det(A)}{\det(B)} = - \alpha.}
Se obține 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 x_3 = -\alpha}
, ceea ce implicăFailed 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 = \det(B) \cdot \left(X^2 + 1 \right) \cdot \left( X + \alpha \right).}
AtunciFailed 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)}
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}. }
Cum 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 \alpha = \frac{\det(A)}{\det(B)} }
se obține 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 \frac{\alpha+1}{\alpha -1} = \frac{\det(A) + \det(B)}{\det(A) - \det(B)} }
În concluzie, are loc egalitateaFailed 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{\det(A) + \det(B)}{\det(A)-\det(B)}. }