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Metode izračunavanja determinanti matrica n-tog reda

Authors
Publisher
Department of Mathematics University of Osijek and Croatian Mathematical Society-Divison Osijek
Publication Date
Keywords
  • Determinante
  • Algebarski Komplement
  • Laplaceov Razvoj Determinante
  • Determinants
  • Cofactor
  • Laplace Expansion Of The Determinant

Abstract

03kece.DVI Osjecˇki matematicˇki list 10(2010), 31–42 31 Metode izracˇunavanja determinanti matrica n-tog reda Damira Kecˇek∗ STUDENTSKA RUBRIKA Sazˇetak. U cˇlanku su opisane metode izracˇunavanja determinanti matrica n-tog reda. Svaka od njih je ilustrirana primjerima. Kljucˇne rijecˇi: determinante, algebarski komplement, Laplaceov razvoj determinante Methods of computing determinants of the n-th order Abstract. In the paper are described methods of computing deter- minants of the n-th order. Each of them is illustrated by examples. Key words: determinants, cofactor, Laplace expansion of the de- terminant 1. Uvod Determinantu kvadratne matrice A = [aij] reda n definiramo kao broj detA = ∑ (−1)i(p)a1p(1)a2p(2)...anp(n) gdje p(1), p(2), ..., p(n) prolaze svih n! moguc´ih permutacija brojeva 1, 2, ..., n. Predznak svakog sumanda u detA ovisi o broju inverzija u permutaciji, i(p), tj. o broju situacija kad u permutaciji vrijedi i < j i p(i) > p(j). Determinantu matrice obicˇno oznacˇavamo sa detA = det  a11 a12 . . . a1n a21 a22 . . . a2n ... ... . . . ... an1 an2 . . . ann  = ∣∣∣∣∣∣∣∣∣ a11 a12 . . . a1n a21 a22 . . . a2n ... ... . . . ... an1 an2 . . . ann ∣∣∣∣∣∣∣∣∣ . ∗Veleucˇiliˇste u Varazˇdinu Jurja Krizˇanic´a 33, 42000 Varazˇdin, [email protected] 32 Damira Kecˇek Ako je n = 1, detA = |a| = a. Za n = 2 imamo 2! = 2 permutacije, p1 = ( 1 2 1 2 ) i p2 = ( 1 2 2 1 ) . Kako je i(p1) = 0, a i(p2) = 1 imamo∣∣∣∣ a11 a12a21 a22 ∣∣∣∣ = (−1)0a11a22 + (−1)1a12a21 = a11a22 − a12a21. Za n = 3 racˇun postaje kompliciraniji, imamo 3! = 6 permutacija, p1 = ( 1 2 3 1 2 3 ) , p2 = ( 1 2 3 1 3 2 ) , p3 = ( 1 2 3 2 1 3 ) , p4 = ( 1 2 3 2 3 1 ) , p5 = ( 1 2 3 3 1 2 ) i p6 = ( 1 2 3 3 2 1 ) . Dobivamo∣∣∣∣∣∣ a11 a12 a13 a21 a22 a23 a31 a32 a33 ∣∣∣∣∣∣ = (−1)0a11a22a33 + (−1)1a11a23a32 + (−1)1a12a21a33 +(−1)2a12a23a31 + (−1)2a13a21a32 + (−1)3a13a22a31 = a11a22a33 + a12a23a31 + a13a21a32 − a1

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