matrice
Posted: Thu Mar 04, 2010 9:03 pm
Multumesc
Multumesc
Page 1 of 1
Multumesc
Posted: Thu Mar 04, 2010 10:30 pm
\( A_n=\left(\begin{array}{cc}\alpha_1^n&\alpha_2^n&\alpha_1^n\\\alpha_1^{n+1}&\alpha_2^{n+1}&\alpha_3^{n+1}\\\alpha_1^{n+2}&\alpha_2^{n+2}&\alpha_3^{n+2}\end{array}\right)\cdot\left(\begin{array}{cc}\alpha_1^2&\alpha_1&1\\\alpha_2^2&\alpha_2&1\\\alpha_3^2&\alpha_3&1\end{array}\right) \)
Posted: Thu Mar 04, 2010 11:19 pm
Daca scriem ecuatia care are ca radacini cele trei numere \( \alpha_i \) de exemplu \( x^3+ax^2+bx+c=0 \), atunci exista relatia de recurenta \( T_{n+3}+aT_{n+2}+bT_{n+1}+cT_n=0 \). Adunand la prima coloana pe a doua inmultita cu \( a \) si pe a treia inmultita cu \( b \) obtinem \( \det(A_n)=(-c)\det(A_{n-1}) \). Astfel problema se poate reduce la calcularea lui \( \det(A_1) \) ceea ce este mult mai simplu. 