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Kevin Guenther Numerical Analysis Homework #6 6.1.15 D with no species 2 the matrix reduces to: 1 0 3 1 2 2 0 1 1 In order to find the amount of increase a species could have the equations can be solved holding two of three populations constant and solving each equation for the third. The smallest of these solutions will be the largest possible increase for the species. Species 1: 200 increase Species 2: 150 increase Species 3: 150 increase 6.2.9 D GAUSSIAN ELIMINATION WITH SCALED PARTIAL PIVOTING THE REDUCED SYSTEM: 0.00000000E+00 0.00000000E+00 -.47648979E+04 0.00000000E+00 0.24081841E+02 0.72129354E+01 -.15611000E+01 0.51792002E+01 -.16855000E+01 0.00000000E+00 0.00000000E+00 0.00000000E+00 HAS SOLUTION VECTOR 0.99999982E+00 0.50000000E+00 -.99999982E+00 6.3.11 FROM MAPLE A: 0 A 0 1 / 6 1 0 3 A 0 1 0 0 2 2 0 0 3 0 0 0 0 1 Since A^3 is the identity matrix the patern will keep repeating starting with A. The population is cyclical with the smallest populations at time periods A. B: Given 6000 one year old … 6000 two year old … and 6000 three year old: After one year: 6000*6 = 36000 one year old 6000*.5 = 3000 two year old 6000*.33 = 2000 three year old After two years: 6000*2 = 12000 one year old 6000*3=18000 two year old 6000*(1/6) = 1000 three year old After three years: The original populations C: 0 2 0 A 0 0 3 1 / 6 0 0 1 This can be used to determine the amount need alive to have a certain population in the future. 6.5.3 A ENTRIES OF L BELOW/ON DIAGONAL AND ENTRIES OF U ABOVE/ ON DIAGONAL 0.20000000E+01 -0.10000000E+01 0.10000000E+01 0.15000000E+01 0.45000000E+01 0.75000000E+01 0.15000000E+01 0.10000000E+01 -0.40000000E+01 1 0 0 2 1 1 LU = 1.5 1 0 0 4.5 7.5 1.5 1 1 0 0 4 Y1 = -1 Y2 = 1.5 Y3 = 4 X1 = 1 X2 = 2 X3 = -1 6.6.6 D THE SOLUTION VECTOR IS -0.93577921E-01 0.15871558E+01 -0.11674311E+01 0.54128444E+00