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 $\alpha \neq 1$. 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 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 $f$.

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

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