Applied Mthemtics E-Notes, 10(2010), 151-158 c ISSN 1607-2510 Avilble free t mirror sites of http://wwwmthnthuedutw/ men/ Computing The Determinnts By Reducing The Orders By Four Qefsere Gjonblj, Armend Slihu Received 2 August 2009 Abstrct We present new method to compute the determinnts of n n(n 5) mtrices by reducing their sizes by four To prove our results we use the so-clled cornice determinnts, ie, squre determinnts of order n (n 5) where, with the exception of the first nd lst entries, the entries of the 2nd row nd (n 1)- th row, s well the 2nd column nd (n 1)-th column re ll zero The method introduced here hs the dvntge of reducing the size of determinnt by four, nd thus enbling their quicker nd esier computtion 1 Introduction Let A be n n n mtrix A = 11 12 1n 21 22 2n n1 n2 nn DEFINITION 1 A determinnt of order n, or size n n, (see [5], [9], [10], [11]) is the sum 11 12 1n 21 22 2n D = det(a) = A = = ε j1,j 2,,j n j1 j2 jn, S n n1 n2 nn rnging over the symmetric permuttion group S n, where { +1, ifj 1, j 2,, j n is n even permuttion ε j1,j 2,,j n = 1, ifj 1, j 2,, j n is n odd permution Mthemtics Subject Clssifictions: 15A15, 11C20, 65F40 Deprtment of Mthemtics, Fculty of Electricl nd Computer Engineering, University of Prishtin, Pristin 10000, Kosovo Deprtment of Telecommuniction, Fculty of Electricl nd Computer Engineering, University of Prishtin, Pristin 10000, Kosovo 151
152 A New Method for Computing the Determinnts 11 Properties Chrcterizing Determinnts Let A nd B be ny n n mtrices 1 If A is tringulr mtrix, ie ij = 0 whenever i > j or, lterntively, whenever i < j, then det(a) = 11 22 nn 2 If B results from A by interchnging two rows or columns, then det(b) = det(a) 3 If B results from A by multiplying one row or column with number c, then det(b) = c det(a) 4 If B results from A by dding multiple of one row to nother row, or multiple of one column to nother column, then det(b) = det(a) These four properties (see eg [3], [7], [8]) cn be used to compute the determinnt of ny mtrix, using Gussin elimintion This is n lgorithm tht trnsforms ny given mtrix to tringulr mtrix, only by using the opertions from the lst three items bove Since the effect of these opertions on the determinnt cn be trced, the determinnt of the originl mtrix is known, once Gussin elimintion is performed It is lso possible to expnd determinnt long row or column using Lplce s formul, which is efficient for reltively smll mtrices To do this long the row i, sy, we write n n det(a) = ij C ij = ij ( 1) i+j M ij, j=1 where the C ij represent the mtrix cofctors, ie, C ij is ( 1) i+j times the minor M ij, which is the determinnt of the mtrix tht results from A by removing the i-th row nd the j-th column, nd n is the size of the mtrix 5 If I is identity mtrix, ie ij = j=1 { 1, ifi = j 0, ifi j then the determinnt of the identity mtrix is one, det(i) = 1 6 Multiplying mtrix by number r ffects the determinnt s follows: det(ra) = r n det(a) 7 If A contins zero row (or column), then det(a) = 0 8 If A contins two identicl or proportionl rows (or columns), then det(a) = 0 9 A mtrix nd its trnspose hve the sme determinnt: det(a T ) = det(a), PROPOSITION 1 Let B be obtined from A by one of the following elementry row (column) opertions:
Q Gjonblj nd A Slihu 153 1) Two rows (or columns) of A re switched, or 2) A row (or column) of A is multiplied by number α, or 3) A multiplier of row (or column) of A is dded to nother row (or column) Then we hve, respectively, det(b) = det(a), or det(b) = α det(a), or det(b) = det(a) 12 Chio s Condenstion Method Chio s condenstion is method for evluting n n n determinnt in terms of (n 1) (n 1) determinnts; see [1], [4]: A = 11 12 1n 21 22 2n n1 n2 nn = 1 n 2 11 11 12 21 22 11 12 31 32 11 12 n1 n2 11 13 21 23 11 13 31 33 11 13 n1 n3 11 21 11 31 11 n1 1n 2n 1n 3n 1n nn 13 Dodgson Condenstion Method Dodgson s condenstion method computes determinnts of size n n by expressing them in terms of those of size (n 1) (n 1), nd then expresses the ltter in terms of determinnts of size (n 2) (n 2), nd so on (see [2]) 2 A New Method Let A n n be n n n determinnt: 11 12 13 1n 21 22 23 2n A n n = n1 n2 n3 nn (1) Let us now nlyze determinnts of size n n (n 5), where, with the exception of the first nd lst entries, the entries of the 2nd row nd (n 1)-th row, s well the 2nd column nd (n 1)-th column re ll zero We cll such determinnts cornice
154 A New Method for Computing the Determinnts determinnts nd write them down s C n n (n 5): C n n = 11 12 13 1,n 2 1,n 1 1n 21 0 0 0 0 2n 31 0 33 3,n 2 0 3n n 2,1 0 n 2,3 n 2,n 2 0 n 2,n n 1,1 0 0 0 0 n 1,n n1 n2 n3 n,n 2 n,n 1 nn (2) By using elementry row nd column opertions nd bsed on determinnts properties, considerble number of determinnts of size n n cn be trnsformed into cornice determinnts We will prove tht ny cornice determinnt of size n (n 5) cn be computed by reducing its size by four, thus trnsforming it to n (n 4) (n 4) determinnt For exmple, determinnt of size 5 5 is trnsformed to number, one of size 6 6 to determinnt of size 2 2, etc THEOREM 1 Every cornice determinnt C n n of size n n (n 5) cn be computed by reducing the order of the determinnt by four: C n n = ( 12 21 n,n 1 n 1,n 12 2n n,n 1 n 1,1 21 n2 1,n 1 n 1,n + 1,n 1 2n n2 n 1,1 ) C (n 4) (n 4), (3) where C (n 4) (n 4) = 33 3,n 2 n 2,3 n 2,n 2 We illustrte this using the following picture: n n 1,1 11 21 31 n1 n2 n n3 n 2, n 2 1, n 1 Cn! n "!!! "!!! 0 0 2,1 12 0 0 13 33 2,3 1, n 2 0 0 3, n 2 0 0 0 0 n, n 2 0 0 n, n 1 1n 2n 3n n 2, n n 1, n n, n PROOF Let n = 5 We will prove tht Theorem 1 holds for C 5 5 cornice determinnts Bsed on Lplce s formul, the determinnt C 5 5 cn be expnded long
Q Gjonblj nd A Slihu 155 the first row: C 5 5 is equl to = 11 + 13 + 15 11 12 13 14 15 21 0 0 0 25 31 0 33 0 35 41 0 0 0 45 51 52 53 54 55 0 0 0 25 0 33 0 35 0 0 0 45 52 53 54 55 21 0 0 25 31 0 0 35 41 0 0 45 51 52 54 55 21 0 0 0 31 0 33 0 41 0 0 0 51 52 53 54 21 0 0 25 + ( 1) 1+2 12 31 33 0 35 41 0 0 45 51 53 54 55 21 0 0 25 + +( 1) 1+4 14 31 0 33 35 41 0 0 45 51 52 53 55 Bsed on property 8 from the first section, the first, third nd fifth determinnts re zero, nd hence 33 0 35 C 5 5 = 12 ( 1) 1+1 21 0 0 45 53 54 55 0 33 35 14 ( 1) 1+1 21 0 0 45 52 53 55 + ( 1)1+4 25 + ( 1)1+4 25 31 33 0 41 0 0 51 52 54 31 0 33 41 0 0 51 52 53 = 12 [ 21 ( 33 45 54 ) 25 ( 33 41 54 )] 14 [ 21 ( 33 45 52 ) 25 ( 33 41 52 )] = 12 21 33 45 54 12 25 33 41 54 14 21 33 45 52 + 14 25 33 41 52 = ( 12 21 54 45 12 25 54 41 21 52 14 45 + 41 52 25 14 ) 33, which hs the desired form Next, we prove tht the theorem holds for n 6 Bsed on Lplce s formul, the determinnt C n n cn be expnded long the second column to yield:
156 A New Method for Computing the Determinnts = ( 1) 1+2 12 +( 1) n+2 n2 11 12 13 1,n 2 1,n 1 1n 21 0 0 0 0 2n 31 0 33 3,n 2 0 3n n 2,1 0 n 2,3 n 2,n 2 0 n 2,n n 1,1 0 0 0 0 n 1,n n1 n2 n3 n,n 2 n,n 1 nn 21 0 0 0 2n 31 33 3,n 2 0 3n n 2,1 n 2,3 n 2,n 2 0 n 2,n n 1,1 0 0 0 n 1,n n1 n3 n,n 2 n,n 1 nn 11 13 1,n 2 1,n 1 1n 21 0 0 0 2n 31 33 3,n 2 0 3n n 2,1 n 2,3 n 2,n 2 0 n 2,n n 1,1 0 0 0 n 1,n (4) The determinnts in (4) re expnded long the (n 1)-th column so tht C n n is equl to 21 0 0 2n 31 33 3,n 2 3n ( 1) 1+2 12 ( 1) n 1+n 2 n,n 1 n 2,1 n 2,3 n 2,n 2 n 2,n n 1,1 0 0 n 1,n 21 0 0 2n 31 33 3,n 2 3n +( 1) n+2 n2 ( 1) 1+n 2 1,n 1 n 2,1 n 2,3 n 2,n 2 n 2,n n 1,1 0 0 n 1,n 21 0 0 2n = ( ( 1) 2n 12 n,n 1 + ( 1) 2n+1 ) 31 33 3,n 2 3n n2 1,n 1 n 2,1 n 2,3 n 2,n 2 n 2,n n 1,1 0 0 n 1,n
Q Gjonblj nd A Slihu 157 The determinnt in the right hnd side is expnded long the first row to yield 33 3,n 2 3n ( 12 n,n 1 n2 1,n 1 ) ( 1)1+1 21 + n 2,3 n 2,n 2 n 2,n 0 0 n 1,n 31 33 3,n 2 +( 1) 1+n 2 2n n 2,1 n 2,3 n 2,n 2 (5) n 1,1 0 0 The determinnts in (5) re expnded long the lst row to yield ( 12 n,n 1 n2 1,n 1 ) ( ) 21 ( 1) n 3+n 3 n 1,n + ( 1) n 1 2n ( 1) n 2 n 1,1 33 34 3,n 2 43 44 4,n 2 n 2,3 n 2,4 n 2,n 2 = ( 12 n,n 1 n2 1,n 1 )( 21 n 1,n 2n n 1,1 ) 33 34 3,n 2 43 44 4,n 2 n 2,3 n 2,4 n 2,n 2 = ( 12 21 n,n 1 n 1,n 12 2n n,n 1 n 1,1 21 n2 1,n 1 n 1,n + 1,n 1 2n n2 n 1,1 ) C (n 4) (n 4), which proves tht Theorem 1 holds We remrk tht if mtrix A of size n n (n 5) cn be trnsformed into cornice determinnt C n n, then it cn be computed using the formul (3) To illustrte the usefulness of our method in computing cornice determinnts, we give n exmple: 7 1 3 11 9 6 8 0 0 0 0 2 C 6 6 = 5 0 5 5 0 1 12 0 3 7 0 2 3 0 0 0 0 11 10 9 2 1 5 7 = [8 1 5 11 1 2 3 5 8 9 9 11 + 9 2 3 9] 5 5 3 7 = (440 30 7128 + 486) 50 = 311600
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