# Tema 3P - CNA-TEMA 3 1.Se considera matricea 3 3 2 A = 1 5...

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CNA-TEMA 3 1.Se considera matricea: - - - - = 0 3 1 2 5 1 2 3 3 A . Sa se calculeze « de mana« fara a folosi functii Matlab : Comparati rezultatele obtinute cu cele date prin aplicarea functiilor Matlab corespunzatoare a) detA= -6+(-6)-[-10-18]=16 - - - - - = 0 2 2 3 5 3 1 1 3 T A iar - - = 12 6 2 4 2 2 4 6 6 * A - - - - = = - 12 6 2 4 2 2 4 6 6 16 1 * det 1 1 A A A In Matlab >> F=A^(-1) sau F=inv(A) F = 0.3750 0.3750 -0.2500 0.1250 0.1250 0.2500 0.1250 -0.3750 0.7500 b) 2 ln A e G = ) 8 cos( ) ) ) 2 ln 1 A H c e G b A F a A π = = = -

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X e G A X = - - - - = - - - - = = 0 2 ln 3 2 ln 2 ln 2 2 ln 5 2 ln 2 ln 2 2 ln 3 2 ln 3 2 ln 0 3 1 2 5 1 2 3 3 2 ln Pentru determinarea valorii lui X se foloseste un algorim(algoritmul 3.1) in care matricea X este descompusa in vectori si valorii proprii. In Matlab >> [G]=alg31('exp',A*log(2)) G = 10.0000 -18.0000 12.0000 -6.0000 22.0000 -12.0000 -6.0000 18.0000 -8.0000 unde algoritmul de calcul este urmatorul : function [G]=alg31(f,A) [n,n]=size(A); [V,L]=eig(A); D=zeros(n,n); for i=1:n D(i,i)=feval(f,L(i,i)); end G=V*D/V; G=real(F); c) = A H 8 cos π Y e H A Y = - - - - = - - - - = = 0 8 3 8 8 2 8 5 8 8 2 8 3 8 3
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