S:L22.108: Difference between revisions

Revision as of 08:37, 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 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 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 \det(A) = \alpha \cdot \det(B) \ne 0} , unde 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 \ne 1} . Arătați că 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{\det \left(A+B\right)}{\det \left(A+B\right)} = \frac{\det(A) + \det(B)}{\det(A)-\det(B)}. }

Soluție.

Ipotezele 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 \det(A^2+B^2) = 0} și 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} , 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 A, B \in \mathcal{M}_3 \left( \mathbb{R}\right)} , 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 \det \left( A+iB \right) \cdot \det\left( A- iB\right) =0} Fie polinomul 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 = \det \left( A+X\cdot B\right) \in \mathbb{R}\left[X\right]} . Atunci, există 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 m, n \in \mathbb{R}} pentru careFailed 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( x\right) = \det\left(B\right) \cdot x^3 + mx^2 + nx +\det(A), \forall x\in \mathbb{C}.} 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 f\left( i\right) \cdot f\left( -i \right)=0} , avem 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\left( i\right) = f\left( -i \right) = 0} , deci 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 = i} și 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_2 = -i} sunt rădăcini ale 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} .

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} = \frac{\frac{\det(A)}{\det(B)}+1}{\frac{\det(A)}{\det(B)}-1} = \frac{\det(A) + \det(B)}{\det(A)-\det(B)}. }

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 '''Soluție.'''
-Ipotezele <math>\det(A^2+B^2) = 0</math> și <math>AB=BA</math>, cu ''<math>A, B \in \mathcal{M}_3 \left( \mathbb{R}\right)</math>'', implică <math display="block">\det \left( A+iB \right) \cdot \det\left( A- iB\right) =0</math>Fie polinomul <math>f = \det \left( A+X\cdot B\right) \in \mathbb{R}\left[X\right]</math>. Atunci, există <math>m, n \in \mathbb{R}</math> pentru care<math display="block">f\left( x\right) = \det\left(B\right) \cdot x^3 + mx^2 + nx +\det(A), \forall x\in \mathbb{C}.</math> Cum <math>f\left( i\right) \cdot f\left( -i \right)=0</math>, avem <math>f\left( i\right) = f\left( -i \right) = 0</math>, deci <math>x_1 = i</math> și <math>x_2 = -i</math> sunt rădăcini ale polinomului <math>f</math>.
+Ipotezele <math>\det(A^2+B^2) = 0</math> și <math>AB=BA</math>, cu ''<math>A, B \in \mathcal{M}_3 \left( \mathbb{R}\right)</math>'', implică<math display="block">\det \left( A+iB \right) \cdot \det\left( A- iB\right) =0</math>Fie polinomul <math>f = \det \left( A+X\cdot B\right) \in \mathbb{R}\left[X\right]</math>. Atunci, există <math>m, n \in \mathbb{R}</math> pentru care<math display="block">f\left( x\right) = \det\left(B\right) \cdot x^3 + mx^2 + nx +\det(A), \forall x\in \mathbb{C}.</math>Cum <math>f\left( i\right) \cdot f\left( -i \right)=0</math>, avem <math>f\left( i\right) = f\left( -i \right) = 0</math>, deci <math>x_1 = i</math> și <math>x_2 = -i</math> sunt rădăcini ale polinomului <math>f</math>.
-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{\frac{\det(A)}{\det(B)}+1}{\frac{\det(A)}{\det(B)}-1} = \frac{\det(A) + \det(B)}{\det(A)-\det(B)}. </math>''
+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{\frac{\det(A)}{\det(B)}+1}{\frac{\det(A)}{\det(B)}-1} = \frac{\det(A) + \det(B)}{\det(A)-\det(B)}. </math>''