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