Engineering Analysis ENG 3420 Fall Dan C. Marinescu Office: HEC 439 B Office hours: Tu-Th 11:00-12:00
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1 Engineering Anlysis ENG 3420 Fll 2009 Dn C. Mrinescu Office: HEC 439 B Office hours: Tu-Th 11:00-12:00
2 Lecture 13 Lst time: Problem solving in preprtion for the quiz Liner Algebr Concepts Vector Spces, Liner Independence Orthogonl Vectors, Bses Mtrices Tody Solving systems of liner equtions (Chpter 9) Grphicl methods Next Time Guss elimintion Lecture 13 2
3 Solving systems of liner equtions Mtrices provide concise nottion for representing nd solving simultneous liner equtions: 11 x x x 3 = b 1 21 x x x 3 = b 2 31 x x x 3 = b x 1 x 2 x 3 = b 1 b 2 b 3 [A]{x} = {b}
4 Solving systems of liner equtions in Mtlb Two wys to solve systems of liner lgebric equtions [A]{x}={b}: Left-division x = A\b Mtrix inversion x = inv(a)*b Mtrix inversion only works for squre, non-singulr systems; it is less efficient thn left-division.
5 Solving grphiclly systems of liner equtions For smll sets of simultneous equtions, grphing them nd determining the loction of the intersection of the stright line representing ech eqution provides solution. There is no gurntee tht one cn find the solution of system of liner equtions: ) No solution exists b) Infinite solutions exist c) System is ill-conditioned
6 Determinnt of the squre mtrix A=[ ij ] A = [ ij ] 1,1 2,1 =... n n,1 1,1 n 1,2 1,2 2,2... n, , n 1 2, n 1 n 1, n 1... n, n 1 1, n 2, n... n 1, n n, n = det( A) = Ai 1 i 1 + Ai 2i2 + A... A Here the coefficient A ij of ij is clled the cofctor of A A cofctor is polynomil in the remining rows of A nd cn be described s the prtil derivtive of A. The cofctor polynomil contins only entries from n (n-1)x (n-1) mtrix M ij clled minor obtined from A by eliminting row i nd column j. in in
7 Determinnts of severl mtrices Determinnts for 1x1, 2x2, 3x3 mtrices re: = = = Determinnts for squre mtrices lrger thn 3 x 3 re more complicted.
8 Properties of the determinnts If we permute two rows of the rectngulr mtrix A then the sign of the determinnt det(a) chnges. The determinnt of the trnspose of mtrix A is equl to the determinnt of the originl mtrix. If two rows of A re identicl then A =0
9 Crmer s Rule Consider the system of liner equtions: [A]{x}={b} Ech unknown in system of liner lgebric equtions my be expressed s frction of two determinnts with denomintor D nd with the numertor obtined from D by replcing the column of coefficients of the unknown in question by the vector b consisting of constnts b 1, b 2,, b n.
10 Exmple of the Crmer s Rule Find x 2 in the following system of equtions: 0.3x x 2 + x 3 = 0.01 Find the determinnt D D = = Find determinnt D 2 by replcing D s second column with b Divide 0.5x 1 + x x 3 = x x x 3 = = D2 = = = x 2 = D 2 D = = 29.5
11 Guss Elimintion Guss elimintion sequentil process of removing unknowns from equtions using forwrd elimintion followed by bck substitution. Nïve Guss elimintion the process does not check for potentil problems resulting from division by zero.
12 Nïve Guss Elimintion (cont) Forwrd elimintion Strting with the first row, dd or subtrct multiples of tht row to eliminte the first coefficient from the second row nd beyond. Continue this process with the second row to remove the second coefficient from the third row nd beyond. Stop when n upper tringulr mtrix remins. Bck substitution Strting with the lst row, solve for the unknown, then substitute tht vlue into the next highest row. Becuse of the upper-tringulr nture of the mtrix, ech row will contin only one more unknown.
13
14 function x=gussnive(a,b) ExA=[A b]; [m,n]=size(a); q=size(b); if (m~=n) fprintf ('Error: input mtrix is not squre; n = %3.0f, m=%3.0f \n', n,m); End if (n~=q) fprintf ('Error: vector b hs different dimension thn n; q = %2.0f \n', q); end n1=n+1; for k=1:n-1 for i=k+1:n fctor=exa(i,k)/exa(k,k); ExA(i,k:n1)= ExA(i,k:n1)-fctor*ExA(k,k:n1); End End x=zeros(n,1); x(n)=exa(n,n1)/exa(n,n); for i=n-1:-1:1 x(i) = (ExA(i,n1)-ExA(i,i+1:n)*x(i+1:n))/ExA(i,i); end
15 >> C=[ ; ; ] C = >> d= [588.6; 686.7;784.8] d = >> x = GussNive(C,d) x =
16 >> A=[ ; ; ; ; ; ] A = b = >> b=b' b = >> x = GussNive(A,b) x = NN NN NN NN NN NN
17 x=a\b x = >> x=inv(a)*b x =
18 Complexity of Guss elimintion To solve n n x n system of liner equtions by Guss elimintion we crry out the following number of opertions: Flops floting-point opertions. Mflops/sec number of floting point opertion executed by processor per second. Conclusions: Forwrd Elimintion Bck Substitution Totl 2n 3 ( ) 3 + On2 n 2 + On 2n 3 () ( ) 3 + On2 As the system gets lrger, the computtion time increses gretly. Most of the effort is incurred in the elimintion step.
19 Pivoting If coefficient long the digonl is 0 (problem: division by 0) or close to 0 (problem: round-off error) then the Guss elimintion cuses problems. Prtil pivoting determine the coefficient with the lrgest bsolute vlue in the column below the pivot element. The rows cn then be switched so tht the lrgest element is the pivot element. Complete pivoting check lso the rows to the right of the pivot element re lso checked nd switch columns.
20 Prtil Pivoting Progrm
21 Tridigonl systems of liner equtions A tridigonl system of liner equtions bnded system with bndwidth of 3: f 1 g 1 e 2 f 2 g 2 e 3 f 3 g 3 e n 1 f n 1 g n 1 e n Cn be solved using the sme method s Guss elimintion, but with much less effort becuse most of the mtrix elements re lredy 0. f n x 1 x 2 x 3 x n 1 x n = r 1 r 2 r 3 r n 1 r n
22 Tridigonl system solver
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