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matrice

 
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elena_romina
Euclid


Joined: 15 Nov 2008
Posts: 47

PostPosted: Thu Mar 04, 2010 9:03 pm    Post subject: matrice Reply with quote


Multumesc Very Happy


Last edited by elena_romina on Fri Mar 05, 2010 11:27 am; edited 1 time in total
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Marius Mainea
Gauss


Joined: 26 May 2008
Posts: 1099
Location: Gaesti (Dambovita)

PostPosted: Thu Mar 04, 2010 10:30 pm    Post subject: Reply with quote

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)

Last edited by Marius Mainea on Fri Mar 05, 2010 12:06 am; edited 1 time in total
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Beniamin Bogosel
Co-admin


Joined: 07 Mar 2008
Posts: 760
Location: Chambery, Franta

PostPosted: Thu Mar 04, 2010 11:19 pm    Post subject: Reply with quote

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. Smile
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