Se considera multimea \( M=\left\{ \begin{array}{rcl}\left( \begin{array}{cc} a & b \\ b & a \end{array} \right) | a,b \in Z_{7} \end{array} \right \} \) si matricea \( A=\left( \begin{array}{cc} \hat{4} & \hat{3} \\ \hat{3} & \hat{4} \end{array} \right) \in M \). Sa se rezolve in \( M \) ecuatia \( X^{6}=A \).
Concurs Petre Sergescu 2008
Ecuatie matriceala
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bogdanl_yex
- Pitagora
- Posts: 91
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Ecuatie matriceala
"Don't worry about your difficulties in mathematics; I can assure you that mine are still greater"(Albert Einstein)
Din lipsa de idei, putem considera o matrice \( \left( \begin{array}{cc} x & y \\ y & x \end{array} \right)=X \).
Ridicand matricea la puterea a 6-a obtinem un sistem de doua ecuatii:
\( \left{\begin
(x^2+y^2)^3+\hat{5}x^2y^2(x^2+y^2)=\hat{4}\\
2xy(x^2+y^2)^2+x^3y^3+\hat{4}xy(x^2+y^2)^3=\hat{3} \)
dupa rezolvarea lui obtinem solutiile:\( (x,y)\in{(\hat{2}, \hat{3}),(\hat{2}, \hat{5}),(\hat{3}, \hat{4}), (\hat{4}, \hat{5}), (\hat{5}, \hat{2}), (\hat{5}, \hat{4}), (\hat{4}, \hat{6}). \)
PS: metoda mea nu e chiar asa de riguroasa, practic am incercat toate posibilitatile(am experienta din primele variante de bac
).Am stat jumate ora pana am epuizat toate cazurile.As dori sa vad o rezolvare mai riguroasa daca se ofera cineva
Ridicand matricea la puterea a 6-a obtinem un sistem de doua ecuatii:
\( \left{\begin
(x^2+y^2)^3+\hat{5}x^2y^2(x^2+y^2)=\hat{4}\\
2xy(x^2+y^2)^2+x^3y^3+\hat{4}xy(x^2+y^2)^3=\hat{3} \)
dupa rezolvarea lui obtinem solutiile:\( (x,y)\in{(\hat{2}, \hat{3}),(\hat{2}, \hat{5}),(\hat{3}, \hat{4}), (\hat{4}, \hat{5}), (\hat{5}, \hat{2}), (\hat{5}, \hat{4}), (\hat{4}, \hat{6}). \)
PS: metoda mea nu e chiar asa de riguroasa, practic am incercat toate posibilitatile(am experienta din primele variante de bac
).Am stat jumate ora pana am epuizat toate cazurile.As dori sa vad o rezolvare mai riguroasa daca se ofera cineva