Question

Anyone good at linear algebra, msg me please!
Need help in finding value of f to stabalize population matrix!!
A =  [0.7 − f 0.2]
[3 0 ]

Answer 1

what does it mean to stabalize a population?

Answer 2

cause I dont recall anything about stabalizing a matrix :)

Answer 3

i believe that stabalizing means that if you start with a population, Xo = 80, and Yo = 30, if you keep multiplying the product of A and (Xo, Yo), then it willl converge towards a point .. if f is the right value, if not then the population will grow or die off.

Answer 4

like a marchov chain

Answer 5

uhm maybe?

Answer 6

a markov chain takes the results of the the matrix and reapplies the matrix to it in a cycle

Answer 7

Px = a ; Pa = b ; Pb = c ; ..... to stabilize i think would be to get to the point where Pn=n

Answer 8

hmm ok. not sure how i would go about doing that.

Answer 9

http://www.sosmath.com/matrix/markov/markov.html

this might help a little, im reading it now to refresh the brain

Answer 10

lol, eugene value of 1

Answer 11

ahahh ugh i suck.

Answer 12

[0.7 − f 0.2] [x] = [x]
[3 0 ] [y] [y]

this creates a system of equations such that

x(.7-f) + .2y = x

3x + 0y = y

Answer 13

x(.7-f) + .2y = x
.7x -fx -x +.2y = 0

3x + 0y = y
3x -y = 0

(.7-1-f)x +.2y = 0
3x - y = 0

Answer 14

-.3-f .2
3 -1


3 -1
-.3-f .2



1 -1/3
-1 .2/(.3-f)



1 -1/3
0 .2/(.3-f) - 1/3


.2/(.3-f) - 1/3 = 1

.2/(.3-f) = 4/3

1/(.3-f) = 4/.6

.3-f = .6/4

-f = .6/4 - .3

f = .3- .6/4

maybe

Answer 15

with any luck, that means that when f - .15 its stabalized

Answer 16

when f = .15 ....

Answer 17

ok, i got f = .5 but i did it a completely different way, yours looks a little more right ..

Answer 18

lets try them out, cause mine could have errors in it :)

Answer 19

ok, how are we supposed to test the value? ahah.

Answer 20

just put (80, 30) through a munch of times i'm guessing.

Answer 21

hmm, yeah that should work :) does the vector need to equal 100% tho? i cant recall

Answer 22

i have no idea to be honest, i dont even know what i'm looking for.

Answer 23

i remember someting :) maybe

Answer 24

the eugene vector associated with a eugen value of 1 will give us Px = x

Answer 25

so we want the determinant of this given matrix to be zero

Answer 26

since 1 times anything is itself; we dont need to modify anything

Answer 27

well, -1 the diags

Answer 28

ok! thats how i got f = .5 for my answer, but as i'm running it through it, the pop'n doesn't seem to be stabalizing

Answer 29

[0.7−f-1 0.2 ]
[ 3 0-1 ]

thats what I did too in another fashion lol

-(-0.3-f) - (0.6) = 0

0.3+f = 0.6

f = 0.3

Answer 30

0.7 - 0.3 = 0.4

Answer 31

you sure you got f=.5?

Answer 32

i probably did it wrong, but i'm doing trial and analysis, and .4 would probably work! i'm going to try it!!!

Answer 33

ok so i think it works, stabalizing at (54, 161)

Answer 34

yay!! :)

Answer 35

does that seem right to you?!

Answer 36

dunno, i aint got a program to simulate it with :/

Answer 37

we could try crossing lines is my thought about it, but then lines are not "feedback" loops

Answer 38

hmmm, i dont even know how to do that haha, i suck.
can i ask you one more question that is another part of this question?

If initially there are no fish in our lake, and we wish to stock it with adult fish (and
no young fish) so that the total number of fish is 10, 000 when it hits the stable
level (counting both adult and young fish together, and with the f from above),
how many adult fish should we introduce into the lake to make this happen?

Answer 39

i would have to know the information that the matrix was made with in order to know how to use this new information

Answer 40

the matrix is just the new one that we made i think where it stabalizes
with
.4 .2
3 0

Answer 41

right, and we have to use a vector that resembles [adult,young] or is it [young, adult] to run a scenario with

Answer 42

the matrix itself does not exist in a vaccuum; it was created by some information that tells me how to use this new information; otherwise I have no direction to go with

Answer 43

(adult, young), i'm not sure then .. would it be the first matrix we were working with without the f?

Answer 44

no it wouldn't be that matrix becuase the pop'n doesnt stabalize in that matrix.

Answer 45

is the left column vector represent adult information or young information?

Answer 46

the numbers are evidently a yearly cycle of how many live and die