S:L22.108

S:L22.108. (Nicolae Mușuroia)

Fie $A,B\in {\mathcal {M}}_{3}\left(\mathbb {R} \right)$ cu $AB=BA$, $A^{2}+B^{2}$ neinversabilă și $\det(A)=\alpha \cdot \det(B)\neq 0$, 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 $x_{1},x_{2},x_{3}\in \mathbb {C}$ sunt rădăcinile polinomului $f$, atunci din relațiile lui Viete avem

$$x_{1}x_{2}x_{3}=-{\frac {\det(A)}{\det(B)}}=-\alpha .$$

Se obține $x_{3}=-\alpha$, ceea ce implică

$$f=\det(B)\cdot \left(X^{2}+1\right)\cdot \left(X+\alpha \right).$$

Atunci

$$f\left(1\right)=\det \left(A+B\right)=2\left(\alpha +1\right)\cdot \det(B)$$

și

$$f\left(-1\right)=\det \left(A-B\right)=2\left(\alpha -1\right)\cdot \det(B).$$

Avem

$${\frac {\det \left(A+B\right)}{\det \left(A+B\right)}}={\frac {\alpha +1}{\alpha -1}}.$$

Cum $\alpha ={\frac {\det(A)}{\det(B)}}$ se obține

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

În concluzie, are loc egalitatea

$${\frac {\det \left(A+B\right)}{\det \left(A+B\right)}}={\frac {\det(A)+\det(B)}{\det(A)-\det(B)}}.$$