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Numerical Mathematics - A Collection of Solved Problems

Numerical Mathematics - A Collection of Solved Problems

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88 NUMERIČKI METODI U LINEARNOJ ALGEBRIgde suA =2645 −1 1 30 5 2 −11 −2 3 11 −1 3 43 2 3 2x 175 , x = 6 x 274 x 35 , b = 64x 420410375 .Ako stavimoA = C 1 + D + C 2 ,gde suC 1 =2640 0 0 00 0 0 01 −2 0 01 −1 3 0375 , D = diag (A), C 2 =može se obrazovati Gauss–Seidelov metod (varijanta Nekrasova)2640 −1 1 30 0 2 −10 0 0 10 0 0 0(2) x (k) = −D −1 C 1 x (k) − D −1 C 2 x (k−1) + D −1 b (k = 1,2, . . . ) .Primetimo da na osnovu (1) možemo direktno formirati metod Nekrasova, takošto i–tu jednačinu rešimo po x i (i = 1, 2,3,4) i tada formiramo iterativni procespo ideji Gauss–Seidela (videti zadatak 4.2.5). Tako, ili na osnovu (2), dobijamo375 ,(3)x (k)1=15 x(k−1) 2− 1 5 x(k−1) 3− 3 5 x(k) 4+ 2 5 ,x (k)2= − 2 5 x(k−1) 3+ 1 5 x(k−1) 4,x (k)3= − 1 3 x(k) 1+ 2 3 x(k) 2− 1 3 x(k−1) 4+ 4 3 ,x (k)4= − 1 4 x(k) 1+ 1 4 x(k) 2− 3 4 x(k) 3+ 5 2 .Pri proizvoljnom vektoru x (0) , iterativni proces (2), tj. (3), konvergira ako isamo ako su svi koreni jednačine5λ −1 1 3(4) P(λ) = det [C 2 + (D + C 1 )λ] =0 5λ 2 −1λ −2λ 3λ 1= 0˛ λ −λ 3λ 4λ ˛po modulu manji od jedinice (videti [1, str. 265]).

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