Using matrices to solve word problems using a calculator.

[Pattonias]

Homework Statement

A school has three clubs, and each student is required to belong to exactly one club. One year the students switched club membership as follows:

Club A. 1/5 remain in A, 2/5 switch to B, and 2/5 switch to C.
Club B. 1/4 remain in B, 1/2 switch to A, and 1/4 switch to C.
Club C. 1/6 remain in C, 1/2 switch to A, and 1/3 switch to B.

If the fraction of the student population in each club is unchanged from year to year, find the fraction of the student population in each club every year. Write your answers as common fraction, not decimals.

Homework Equations

I don't think there are any special equations required, just an understanding of matrices.
We are supposed to use a graphing calculator (TI-83) to solve the problem, but I can't figure out how to set up the equation.

The Attempt at a Solution

My first attempt was set up as a 3x3 using the remaining students for each club as in/out flow chart, but that didn't go anywhere.

I am currently trying to set up the equations assuming that the student body is A+B+C=1 or (100%).

A B C
|(1/5) (-2/5) (-2/5) (1)|
|(-1/2) (1/4) (-1/4) (1)|
|(-1/2) (-1/3) (1/6) (1)|
|(1) (1) (1) (1)|

I plugged it into the calculator and it gave me

|1000|
|0100|
|0010|
|0001|

Back to the drawing board... I just really don't know where to start.

 

[Mark44]

Homework Statement

A school has three clubs, and each student is required to belong to exactly one club. One year the students switched club membership as follows:

Club A. 1/5 remain in A, 2/5 switch to B, and 2/5 switch to C.
Club B. 1/4 remain in B, 1/2 switch to A, and 1/4 switch to C.
Club C. 1/6 remain in C, 1/2 switch to A, and 1/3 switch to B.

If the fraction of the student population in each club is unchanged from year to year, find the fraction of the student population in each club every year. Write your answers as common fraction, not decimals.

Homework Equations

I don't think there are any special equations required, just an understanding of matrices.
We are supposed to use a graphing calculator (TI-83) to solve the problem, but I can't figure out how to set up the equation.

The Attempt at a Solution

My first attempt was set up as a 3x3 using the remaining students for each club as in/out flow chart, but that didn't go anywhere.

I am currently trying to set up the equations assuming that the student body is A+B+C=1 or (100%).

A B C
|1/5 -2/5 -2/5 1|
|-1/2 1/4 -1/4 1|
|-1/2 -1/3 1/6 1|
|1 1 1 1|

I'm trying to get that to work now, any suggestions?

A matrix is a representation of a number of equations, but I don't see a single equation anywhere, so I think you skipped an extremely important step.

Write three equations that represent this information:
Club A. 1/5 remain in A, 2/5 switch to B, and 2/5 switch to C.
Club B. 1/4 remain in B, 1/2 switch to A, and 1/4 switch to C.
Club C. 1/6 remain in C, 1/2 switch to A, and 1/3 switch to B.

then, set up your matrix.

 

[Pattonias]

I had tried an equation for the matrix underneath ABC. I just didn't write it out.

The equation for that matrix would have been
(1/5)A+(-2/5)B+(-2/5)C=1
(-1/2)A+(1/4)B+(-1/4)C=1
(-1/2)A+(-1/3)B+(1/6)C=1
1A+1B+1C=1

These are the equations that I plugged into the matrix above that gave me.

|1000|
|0100|
|0010|
|0001|

 

[Pattonias]

It seems to me that somehow I need to set up the equations in such a way that the A, B and C are isolated from the rest of the numbers, but I don't know how to do that without the fractions losing their meaning.

 

[Pattonias]

Am I getting anywhere with this?

1/5A+1/2B+1/2C=A
1/4B+2/5A+1/3C=B
1/6C+2/5A+1/4B=C

I'll try to arrange them better.

Ok here is a better arrangement...

1/2A+1/2B+1/2C=A
2/5A+1/4B+1/3C=B
2/5A+1/4B+1/6C=C

As a matrix...

|(1/2)(1/2)(1/2)1|
|(2/5)(1/4)(1/3)1|
|(2/5)(1/4)(1/6)1|

I'll see what that give me.

 

[Mark44]

I had tried an equation for the matrix underneath ABC. I just didn't write it out.

The equation for that matrix would have been
(1/5)A+(-2/5)B+(-2/5)C=1
(-1/2)A+(1/4)B+(-1/4)C=1
(-1/2)A+(-1/3)B+(1/6)C=1
1A+1B+1C=1

What do A, B, and C represent?
Do you really believe that the last equation is meaningful? It implies that A, B, and C are numbers that are less than 1, which isn't very likely.

I haven't worked this out yet, but the direction I would go is to let a, b, and c be the number of students in clubs A, B, and C before they switched.

The number of students who left club A must be equal to the number who went into either B or C.
The number of students who left club B must be equal to the number who went into either A or C.
The number of students who left club C must be equal to the number who went into either A or B.

It might be helpful to put this information in a table (not a matrix) and see what equations you get.

These are the equations that I plugged into the matrix above that gave me.

|1000|
|0100|
|0010|
|0001|

 

[Pattonias]

That matrix gave me
|100(3.3)|
|010(1.3)|
|001(0) |

 

[Mark44]

It should be obvious that that can't be right.

 

[Pattonias]

What do A, B, and C represent?
Do you really believe that the last equation is meaningful? It implies that A, B, and C are numbers that are less than 1, which isn't very likely.

I haven't worked this out yet, but the direction I would go is to let a, b, and c be the number of students in clubs A, B, and C before they switched.

The number of students who left club A must be equal to the number who went into either B or C.
The number of students who left club B must be equal to the number who went into either A or C.
The number of students who left club C must be equal to the number who went into either A or B.

It might be helpful to put this information in a table (not a matrix) and see what equations you get.

I know that the fractions of the clubs are a part of the student body as a whole, but that doesn't seem to be of any help.

In my next post I put the numbers into equations set equal to the clubs they represent, but I think it is missing something.

 

[Pattonias]

It should be obvious that that can't be right.

It is obvious that this isn't right. Hence why I am still here

 

[Pattonias]

Let a = the fraction of the student population in A, b = the fraction in B and c = the fraction in C.

So we have

[1/5 1/2 1/2][a]...[a]
[2/5 1/4 1/3] =
[2/5 1/4 1/6][c]...[c]

(dots added to prevent closing up)

[1/5 1/2 1/2][a]...[1 0 0][a]
[2/5 1/4 1/3] = [0 1 0]
[2/5 1/4 1/6][c]...[0 0 1][c]

[- 4/5 1/2 1/2][a]...[0]
[ 2/5 -3/4 1/3] = [0]
[ 2/5 1/4 -5/6][c]...[0]

[- 8...5...5][a]...[0]
[24 -45 20] =[0]
[24 15 -50][c]...[0]

Subtracting line 3 from line 2, we get
- 60b + 70c = 0
30b = 35c
c = 30b/35 = 6b/7

Substituting into line 1, we get
- 8a + 5b + 5(6b/7) = 0
8a = 5b + 30b/7 = 65b/7
a = 65b/56

And a + b + c = 1, so

65b/56 + b + 6b/7 = 1
b(65/56 + 1 + 6/7) = 1
b(169/56) = 1
b = 56/169

c = 6b/7 = (6/7)(56/169) = 48/169
and a = 65b/56 = 65/169

Club A has 65/169 of the student population;
Club B has 56/169 of the student population;
Club C has 48/169 of the student population.

This is what someone else worked out on another website http://answers.yahoo.com/question/index?qid=20110131205744AA5oJbD"


I set up my equations the same way when I got

|100|
|010|
|001|, but I guess I didn't know to subtract that answer from the original matrix. In fact I didn't know you could do that.

 

[Pattonias]

I appreciate your help, and If you can give me any incite into what I was missing I would appreciate it. It seems like I was dancing around the right idea, but I wasn't making it work.

 

[Mark44]

There was some information I missed.

If the fraction of the student population in each club is unchanged from year to year ...

I interpret the above to mean also that the numbers of members in each club don't change. There's no mention of students leaving the school or new students coming in.

Let a, b, and c be the numbers of members in club A, B, and C, respectively.

For each club, the number of members = the number who didn't switch + the number who came in from the other two clubs.

a = 1/5 a + 1/2 b + 1/2 c
b = 1/4 b + 2/5 a + 1/3 c
c = 1/6 c + 2/5 a + 1/4 b

Or, lining the variables up.
a = 1/5 a + 1/2 b + 1/2 c
b = 2/5 a + 1/4 b + 1/3 c
c = 2/5 a + 1/4 b + 1/6 c

This system is equivalent to
4/5 a - 1/2 b - 1/2 c = 0
-2/5 a + 3/4 b - 1/3 c = 0
-2/5 a - 1/4 b + 5/6 c = 0

Solving that system should give you a, b, and c.

EDIT:
I get multiple solutions.
a = 65/48 * c
b = 7/6 * c
c = arbitrary

Letting c = 48, I get a = 65, b = 56, c = 48.

 

[Mark44]

Can the calculators these days solve problems like this, where there are an infinite number of solutions?

 

[Pattonias]

The only function we are using at the moment to solve the matrices are putting them into reduced echelon form. We have to know how to interpret the data once we figure out how to put it in. We haven't really tapped the full potential yet, but we are just getting started.