S:L22.108

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

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

Soluție.

Ipotezele $\det(A^{2}+B^{2})=0$ și $AB=BA$, cu $A,B\in {\mathcal {M}}_{3}\left(\mathbb {R} \right)$, implică

$$\det \left(A+iB\right)\cdot \det \left(A-iB\right)=0$$

Fie polinomul $f=\det \left(A+X\cdot B\right)\in \mathbb {R} \left[X\right]$. Atunci, există $m,n\in \mathbb {R}$ pentru care

$$f\left(x\right)=\det \left(B\right)\cdot x^{3}+mx^{2}+nx+\det(A),\forall x\in \mathbb {C} .$$

Cum $f\left(i\right)\cdot f\left(-i\right)=0$ , avem $f\left(i\right)=f\left(-i\right)=0$ , deci $x_{1}=i$ și $x_{2}=-i$ sunt rădăcini ale polinomului $f$.

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

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