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det(A+B)=detA+tr(A*B)+detB

 
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R4DIC4L



Joined: 02 Feb 2011
Posts: 2
Location: Craiova

PostPosted: Fri Mar 11, 2011 5:51 pm    Post subject: det(A+B)=detA+tr(A*B)+detB Reply with quote

Stie cineva cum se poate demonstra ca pentru orice matrici A{\normal\ ,\ }B\in\mathcal{M}_2(\mathbb{R})
det(A+B)=detA+tr(A*B)+detB ?
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bogdy92y
Arhimede


Joined: 09 Mar 2011
Posts: 7

PostPosted: Fri Mar 11, 2011 6:18 pm    Post subject: Reply with quote

Poti incerca luand doua matrici de ordin doi la forma lor generala.
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Radu Titiu
Thales


Joined: 28 Sep 2007
Posts: 181
Location: Mures \Bucuresti

PostPosted: Fri Mar 11, 2011 6:32 pm    Post subject: Reply with quote

Daca A e inversabila atunci cred ca formula e clara. (*)

Daca A nu e inversabila atunci consideram functia polinomiala
P(x)=\det(B+(A-xI)) .Daca notez cu S=\{x \in \mathbb{R} | \det(A-xI)=0\} , atunci din observatia (*) avem :

P(x)=\det(B)+tr(B(A-xI)^*)+\det(A-xI) . \forall x \in \mathbb{R} - S

Deoarece egalitatea de mai sus reprezinta o egalitate de functii polinomiale care e adevarata pentru o infinitate de valori (mai precis oricare ar fi x real , fara multimea S), rezulta ca egalitatea e adevarata pentru orice x real.In particular si pentru x=0.
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R4DIC4L



Joined: 02 Feb 2011
Posts: 2
Location: Craiova

PostPosted: Fri Mar 11, 2011 7:35 pm    Post subject: Reply with quote

Dar daca consideram x=0 nu inseamna ca det(A-xI)=0, deci A-xI neinversabila(x din S) si astfel nu putem aplica formula (*) ?
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turcas
Pitagora


Joined: 28 Sep 2007
Posts: 86
Location: Simleu Silvaniei, jud Salaj

PostPosted: Fri Mar 11, 2011 7:58 pm    Post subject: Reply with quote

Pai, dupa cum a scris si Radu, tu consideri acei  x din  \mathbb{R} \setmin S , pentru care  A-xI este inversabila. Acestia sunt o infinitate.
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Radu Titiu
Thales


Joined: 28 Sep 2007
Posts: 181
Location: Mures \Bucuresti

PostPosted: Fri Mar 11, 2011 8:53 pm    Post subject: Reply with quote

R4DIC4L wrote:
Dar daca consideram x=0 nu inseamna ca det(A-xI)=0, deci A-xI neinversabila(x din S) si astfel nu putem aplica formula (*) ?


Folosesc proprietatea ca daca P,Q si sunt doua polinoame si P(x)=Q(x) pentru o infinitate de x , atunci $P(x)=Q(x)$ pentru orice x.Poate acuma e mai clar ce se intampla acolo.

In problema egalitatea are loc pt orice x real mai putin multimea S, deci pentru o infinitate de x.Deci egalitatea are loc pentru orice x real (din proprietatea de ce-am scris-o mai sus), in particular si pentru x=0.
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