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MATEMATIKA 1 Senka Banic PREDAVANJA (grupa G1): utorak i ...

MATEMATIKA 1 Senka Banic PREDAVANJA (grupa G1): utorak i ...

MATEMATIKA 1 Senka Banic PREDAVANJA (grupa G1): utorak i ...

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Primjer. Odredimo inverznu matricu matrice2 32 1 3A = 4 2 3 4 5:5 0 3Kako jedet A =(razvoj po 2. stupcu) = = 1 ( 1) 1+2 2 4 5 3 + 3 ( 2 31)2+2 5 3 + 0 = 1 6= 0to je matrica A regularna, tj ima inverz.23A 1 1= Adet (A) ~ A T 1 11 A 21 A 31= 4 A 12 A 22 A 325 : ()det (A)A 13 A 23 A 33Racunamo sad A ij algebarske komplemente elemenataa ij zadane matrice:A 11 = 3 40 3 = 9; A 12 =2 4 5 3 = 26A 13 =2 3 5 0 = 15; A 21 = 1 30 3 = 3A 22 = 9; A 23 = 5;A 31 = 5; A 32 = 14; A 33 = 8:

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