# Tom O'Keeffe NAE HW6

```Tom O'Keeffe
NAE
HW6
6.1.15
a.
yes, Ax &lt;= b.
b.
Ax &lt;= b, such that xi of (x1,x2,x3) where i =1,2,3 is changed one at time. Xi is increased until Ax is less
than
to equal to b.
X1= 1200, increase of 200.
X2= 650, increase of 150
X3= 450, increase of 100
X4= 500, increase of 100
c.
Eliminating the column from the A matrix associated with the extinct species. Xi is increased until Ax is
less
than o equal to b.
X2= increase of 650
X3= increase of 150
X4= increase of 150
d.
Eliminating both column one and two, extenct species and then xi is increased until Ax is less than o equal
to b.
X3= increase of 150
X4= increase of 150
6.2.9.d
A=((3.33 15920 10.333 | 7953)(-2.22 16.710 9.6210 | 0.965)(-1.5611 5.1792 1.6855 | 2.714))
S1=15920, S2=16.710, S3=5.1792
E1-&gt;E3, E3-&gt; E1, E2-(2.222/-1.5611)E1-&gt;E2, E3-(3.333/-1.5611)E1-&gt;E3,E3-(1591.06/24.08184)E2-&gt;E3
A'x=((-1.5611 5.1792 -1.6855)(0 24.08184 7.21.2935)(0 0 -4764.0))=(2.714 4.827 4764.987)
Backsubstitution yields
x=(1 0.5 -1)
6.3.11
a.
A^2=AA=((0 2 0)(0 0 3)(1/6 0 0))
A^3=AAA=((100)(010)(001))
Every three years the cycle repeats, period=3.
b.
6000 initial beetle population of ages groups 1,2 and 3 over sequential years.
Age group 1 6000-&gt;36000-&gt;12000-&gt;6000
Age group 2 6000-&gt;3000-&gt;18000-&gt;6000
Age group 3 6000-&gt;2000-&gt;1000-&gt;6000
c.
A^-1=((0 2 0)(0 0 3)(1/6 0 0))
The aij matrix denotes the contribution of a single beetle to the next year population based on age. The
inverse of aij = A^-1 denotes the number of beetles needed to produce a single beetle for the next years
population based on the beetle's age.
6.5.3.a
PA=LU, by making the row change E2-&gt;E3 and E3-&gt;E2 allows the permuted matrix (PA) to be solved by
factoring.
P=((1 0 0)(0 0 1)(0 1 0))
PA=((2 -1 1)(3 3 5)(3 3 9))
Ly=b, where L=((1 0 0)(1.5 1 0)(1.5 1 1)), y=(-1 1.5 4)
Ux=y where U=((2 -1 1)(0 4.5 7.5)(0 0 -4)), X=(1 2 -1)
6.6.6.d
A=((0.5 0.25 0 0)(0.35 0.8 0.4 0)(0 0.25 1 0.5)(0 0 1 -2))
L=((0.5 0 0 0)(0.35 0.625 0 0)(0 0.25 0.84 0)(0 0 1 -2.595))
U=((1 0.5 0 0)(0 1 0.64 0)(0 0 1 0.595)(0 0 0 1))
Ly=b, y=0.7 0.84 -0.845 0.541)
Ux=y where x=(-0.0917 1.518 -1.16 0.541)
```