Jeffrey E. Dahan
MEAE4960h01
Homework #6
October 27, 1999
6.1.15
The following three equation matrix was entered into Maple via and external *.dta file
A=[1,2,0,3;1,0,2,2;0,0,1,1]
After running the Gaussian Elimination with Backward Substitution, the following output was
obtained:
GAUSSIAN ELIMINATION
The reduced system - output by rows:
1.00000000 2.00000000 0.00000000 3.00000000
0.00000000 -2.00000000 2.00000000 -1.00000000
0.00000000 0.00000000 1.00000000 1.00000000
Has solution vector:
0.00000000 1.50000000 1.00000000
with 0 row interchange(s)
a. For the given supply of food and given species, there will be more than enough food to
supply the species.
Specifically, there will be 650 extra food units
b. While meeting the amount of food given by b = [3500,2700,900] there could be a total x
species
Of x = [200,150,100,100] This answer is obtained by setting all the zeroes to infinity to
avoid undefined.
c. species two can be increased by 650, or three by 150, or four by 150 using the same type
of process
in part b.
d. species three could be increased by 150, or four by 150.
6.2.9d
The Matrix A has been entered in via a *.dta file and is as follows:
3.333
15920
10.333
7953
2.222
16.71
9.612
0.965
5.179 -1.6855
2.714
-1.5611
The Matrix is computed using the Guassian Elimination with scaled partial pivoting and obtains
the
Following output from Maple:
GAUSSIAN ELIMINATION WITH SCALED PARTIAL PIVOTING
The reduced system - output by rows:
0.00000000 0.00000000 -4764.95347400
0.00000000 24.08155723 7.21293460
-1.56110000 5.17900000 -1.68550000
Has solution vector:
.99991467 .50000001 -.99998028
with 1 row interchange(s)
The rows have been logically re-ordered to:
321
The actual solutions are compared to the results of computing the matrix using the Guassian
Elimination
With scaled partial pivoting:
Algorithm
Exact
Difference
0.99991467 1.00000000 0.00008533
0.50000001 0.50000000 0.00000001
-0.99998082 -1.00000000 0.00001918
Scaled Partial Pivoting is a highly accurate alternative to manually obtaining the actual answers
(if possible).
6.3.11
The study of a female basis on a probabilistic basis reveals that A = [0 0 6, .5 0 0, 0 1/3 0]
a. A2 is constructed by doing matrix multiplication for A*A to equal [0 2 0, 0 0 3, 1/6 0 0]
A3 is constructed by A2*A to equal [1 0 0, 0 1 0, 0 0 1]
Another iteration is carried out to determine the trend:
A4 = [0 0 6, 1/2 0 0, 0 1/3 0] which is identical to a. Therefore, the cycle repeats itself every four
years.
b. for 6000 beetles, the results will simply be scaled by 6000, which is accurate of scaling a
matrix:
Age 1
Age 2
Age 3
1st year
6000
6000
6000
2nd year
36000
3000
2000
3rd year
12000
18000
1000
4th year
6000
6000
6000
c. the inverse matrix A-1 is found to be [0 2 0, 0 0 3, 1/6 0 0], which is identical to A2.
6.5.3a
The initial matrix [2 -1 -1, 3 3 9, 3 3 5] using LU factorization yeilds
L = [1 0 0, 1.5 1 0, 1.5 1 1] using diagonal ones and everything below, and
U = [2 -.5 .5, 0 4.5 1.66667, 0 0 -4] using the upper triangle
Then, Ly=b is used to determine the y's. y = [-1, 1.5, 4]
Next, knowing Ux=y, backwards substitution can be used to find x:
x = [4.5 9 -1]