# Matrix problem

1. Oct 26, 2010

### Nope

1. The problem statement, all variables and given/known data
Suppose that a country is divided into three regions: Upper, Lower and Central. Each year, one-quarter of the residents of the Upper region move to the Lower region and the remaining residents stay in the Upper region. One-half of the residents of the Lower region move to the Central region and the remaining residents remain in the Lower region. Three-quarters of the residents of the Central region move to the Lower region, and the remaining residents stay in the Central region.
In the long run, what proportion of the residents settle in each region?
of the total residents settle in the Upper region,
of the total residents settle in the Lower region, and
of the total residents settle in the Central region.

2. Relevant equations

3. The attempt at a solution
I have
U=3/4U
L=1/4 U+1/2L+3/4C
C=0 +1/2L+1/4C

2. Oct 26, 2010

### Dick

Express that linear system in matrix form. (U,L,C)=M(U,L,C) where M is a 3x3 matrix. What's M? If there is a steady state then M has a eigenvalue of 1. What's the corresponding eigenvector?

3. Oct 26, 2010

### Nope

M is
3/4 0 0
1/4 1/2 3/4
0 1/2 1/4
What do u mean "steady state" and "eigenvalue"? I don't think I learn these yet...
btw, is there a simple way to do it?
ty

4. Oct 27, 2010

### Dick

Yes, actually, there is. Just solve the equations you have for U, L and C. What does the first equation tell you about U?

5. Oct 27, 2010

### Nope

what about the total population, do i assume 1? or 3?
If 3,
then i got 25%=U , 50%= L, 25%=C
but I don't know how to determine the long run, but I think U is going to be 0

6. Oct 27, 2010

### Dick

I hope you concluded U=0 from one of your equations. You don't have to assume anything for the total population. Call it P. So U+L+C=P. Or put P=1 if you just want to work with percentages.

7. Oct 27, 2010

### Nope

no, I just guessing, cause i tried to cube the matrix and multiply it by 1, so in the third year ,U is decreasing, L and C is increasing
but how do i get the long run?

8. Oct 27, 2010

### Dick

If you don't know 'eigenvalue' and 'eigenvector' forget about the matrix. Just solve the original equations you wrote down.

U=3/4U
L=1/4 U+1/2L+3/4C
C=0 +1/2L+1/4C

As I said before, what does the first equation tell you about U?

9. Oct 27, 2010

### Nope

the final population of Uf is 3/4 of the initial population(Ui)?

10. Oct 27, 2010

### Dick

That's true if you mean U_i is the population at the beginning of the year and U_f is the population at the end of the year. The problem says "In the long run". They are implying that the population of each area will settle down to a constant value. So that equation becomes U=(3/4)U.

Last edited: Oct 27, 2010
11. Oct 27, 2010

### Nope

if constant,do you mean the answer for U is 3/4?

12. Oct 27, 2010

### Dick

Does U=3/4 satisfy U=(3/4)U?

13. Oct 27, 2010

### Nope

no...
sorry , i still don't get it..

14. Oct 27, 2010

### Dick

U=(3/4)U. Solve for U. Use algebra. How would you do that?

15. Oct 27, 2010

### Nope

1=3/4?
that's what i am confusing
or do you mean Uf/Ui=3/4? Uf=4 Ui=3?

16. Oct 27, 2010

### Nope

oh, I think I got it....
U=0
0=(3/4)*0?

17. Oct 27, 2010

### Dick

That's it. U=0. Now put that into the other equations and find a relation between L and C.

18. Oct 27, 2010

### Nope

ok ,i got 60% for L and 40% for C
But I don't understand why L=L,C=C in the long run...

19. Oct 27, 2010

### Dick

Suppose U=0, L=60 and C=40 in one year. Following the instruction in the problem what are U, L and C in the next year?

20. Oct 27, 2010