Math Matters: Why Do I Need To Know This?Bruce Kessler, Department of Mathematics

Western Kentucky UniversityEpisode Thirteen

1 Matrix Multiplication – Complicated probabilities

Objective: To illustrate how matrix multiplication can turn extremely compli-cated real-world problems into very simple ones. We introduce the concepts ofstochastic matrices and using tree diagrams to calculate probabilities.

Hello and welcome to “Math Matters: Why Do I Need To Know This?” It’s a TV showwhere we take a look at the entry-level mathematics being taught in our general ed. coursesand try to apply that to everyday life. I’ve got some cool stuff to show you today, and I wantto apologize to the 109 students, the general math students, because this is a more algebraicshow. Last time we started talking about matrix multiplication and all the things you coulddo with that, and I gave one example, and I thought “Man, I could do so much more withmatrices and matrix multiplication,” so I’m going to revisit that topic today for a couple ofsegments.

I’d like to talk about how we can use matrix multiplication to calculate probabilities. Thisis a slightly different idea, and I have to go back and recap a little bit of this introductionof what a matrix is. It’s just a rectangular set of values, its something along these lines.I call that a matrix. And we consider the number of rows, the first dimension of this, thenumber of columns the second dimension, and we would call that a 3 by 4 matrix. (Figure 1)We multiply matrices in this fashion. The middle dimension here has to match up. Andit will generate in this case a 3 by 3 matrix in this fashion – it’s messy but I’ll explain it.But here’s what’s happening, you get this first entry in the first row, first column. I look atthe first row of the first matrix and the first column of the second matrix and I take thoseentries, multiply them and add them up. So 3 times 1 plus 1 times 2 gives me that entry.(Figure 2) And to get the next one right here I take the first row, second column and do thesame stunt. (Figure 3) To get this one, I take the second row, first column and add thosethings up. (Figure 4) And you just go through all nine of those entries and that’s how youget the answer with matrix multiplication. (Figure 5)

Now, we showed that in the last episode, but I wanted to recap that because I’m gettingready to use that trick again. And what I’d like to do is take a look at some very, very com-plicated probability situations, and show you how all that stuff built into matrix multiplicationtakes care of all that stuff for you. Now, I’m going to throw a situation at you that’s fairlycontrived, but the way I’m going to solve it is to arrange probabilities in a matrix. So eachone of these would be probabilities and the first entry is going to be the probability that I gofrom one state back into that state. Ok, I’m going to think about different places that I couldbe in my problem. The second entry here will be from the first state to the second state andthe first state to the other states. And then from the second state to the first one and soforth. I don’t mean the United States of America – I mean different places I could be in myproblem. (Figure 6)

1

Figure 1, Segment 1

Figure 2, Segment 1

2

Figure 3, Segment 1

Figure 4, Segment 1

3

Figure 5, Segment 1

Figure 6, Segment 1

4

I’ll give you an example, it’s a really contrived example, but it’s an easy example. Andthat’s what I want to show you first, how you could use this in an easy example and then youcan adapt that if you’d like into tougher situations. Let’s take a look at this idea, I’ve gotthree people, Amy, Brad, and Carol throwing a Frisbee to each other in the backyard. Now,Amy shares the Frisbee equally with Brad and Carol, but Brad is kind of sweet on Carol, sohe throws her the Frisbee 90% of the time. And then Carol and Amy are buds, and Caroldoesn’t think too much of Brad, so she will typically throw the Frisbee to Amy about 75% ofthe time. So let me ask a very complicated sort of question: if Brad is the first guy to holdthe Frisbee and they toss this thing around 12 times, what is the probability that each of thesepeople will have the Frisbee after those 12 tosses? (Figure 7)

Figure 7, Segment 1

Now there’s a hard way, and there’s an easy way. The hard way would be, and I’m justgoing to look at 2 tosses, the hard way would be to list out all the different ways someonecould end up with the Frisbee. And I’ve done this for two tosses. Brad starts and he’s goingto throw it to either Amy or Carol, and then Amy is going to throw it to either Brad orCarol, and Carol’s going to either throw it to Amy or Brad. You can put the probabilities onthis tree diagram, that’s what it’s called. For example, Brad kind of likes Carol, so there’s a910

chance that he’ll throw the Frisbee to her, that means there’s a 110

chance that he’ll throwthe Frisbee to Amy, those have to add up to 1. And then with Amy, she shares the Frisbeeequally so that’s 1

2and 1

2. And then Carol 3

4of the time will throw it to Amy, and that means

14

of the time she will throw it to Brad. (Figure 8)So you can look then and calculate the probabilities for each of these paths. So Brad,

Amy, Brad would be the product of those probabilities – so 120

, and the probability that it ends

5

Figure 8, Segment 1

up in Carol’s hands like this is 120

. And the probability that it ends up in Amy’s hands inthis fashion is 27

40, and that it ends up in Brads hands going through Carol would be 9

40. So

the probability that it ends up in Amy’s hands after two tosses is just this one, so its 2740

, theprobability that Carol has it is just this one, so its 1

20. And then Brad, you’ve got to take

both of these into account and add them together to get 1140

. Okay, I’m sorry I didn’t meanto flash that away. So, that would be the answer for two tosses. To do this for twelve tosses,you’d have to make your table that wide. You’d have to make a huge table. You’d have tohave all these probabilities listed, you’d have to multiply them all out. You’d have all thesethings, its horrible. (Figure 8)

Now watch what I can do with a matrix. We create what’s called a transition matrix,where I put the probabilities in. The probability that Amy throws to Amy is 0, she’s notgoing to throw to herself, but she’s going to divide it equally between Brad and Carol, so I’llput those probabilities in – notice that the sum of those entries is one. Brad will not throwthe Frisbee to himself, there’s a 9 in 10 chance he’ll throw to carol and a 1 in 10 chancehe’ll throw to Amy. Notice again the sum of those entries in the column are one. Carolwill not throw the Frisbee to herself, there’s a 3

4chance she’ll throw to Amy and a 1

4chance

she’ll throw to Brad, the sum of those entries is 1. We call this a stochastic matrix whereall the entries are probabilities between 0 and 1, they could be 0 or 1, and then the sum ofthe columns is 1. (Figure 9)

So to work the problem, after the first toss, what I would do, I would start with Bradholding the Frisbee, that means a 100% chance that Brad has it and just multiply. Thatgives me just really the middle column, there’s a 1

10chance that Amy has the Frisbee, a 9

10

6

Figure 9, Segment 1

chance that Carol has it. After two tosses, well, I do this again, and really what I could doto get this matrix, what I could do is, I multiply this by this. So I could square that one anddo it that way, but I’ll just go through, do the matrix multiplication, and that gives me theprobabilities. The easy way lies right here: to do this is so many tosses, I want to raise thismatrix to a power. (Figure 10) So to do this for twelve tosses what I would do I would putthis in my graphing calculator, I wouldn’t do it by hand – that’s horrible. But these graphingcalculators can do this kind of stuff for you, and I would raise that matrix to a power, whichis fairly messy thing, you can see it here, but then multiply by that original thing. Brad hadthe Frisbee to start out with, so here are my probabilities, this middle column: 33% chancethat Amy has the Frisbee, 27% chance Brad has it, 40% chance Carol has it. And the coolthing that you’ll notice is that each of these columns is pretty much the same, which indicatesthat after a period of time, it really doesn’t matter who started with the Frisbee, there’s justgoing to be a probability that you have the Frisbee in your hand. (Figure 11)

That would be a horrible problem without the matrices and you can adapt the situation tobusiness situations and things like that. I’ve got a couple pages that summarize what I justsaid, I’m going to flash those up really quick and get ready for the next segment.

7

Figure 10, Segment 1

Figure 11, Segment 1

8

Summary page 1, Segment 1

Summary page 2, Segment 1

9

2 Matrix multiplication – Tracking populations by age

group

Objective: To illustrate how matrix multiplication can turn extremely compli-cated real-world problems into very simple ones. We introduce the concepts ofLeslie matrices and using matrices to track populations by age group.

I want to stay with this theme of matrix multiplication and show you another cool applica-tion for matrices and multiplying them and raising them to powers, which would be a horriblething to do by hand, but we are armed with technology today, boys and girls, we’ve got thesecrazy things. And, using these things, we can solve some very complicated problems. Andthe application I’d like to show you now is how to track populations through age groups.

Let’s say that you’re raising cattle and your herd of cattle, you’ve paid attention to theage distribution throughout the years and you know the following facts about your group ofcattle. You know their birthrates going from one age group to another. Cattle of this age donot actually have any cows of their own, but when they grow up, they’ll grow at a rate of fiftypercent, so however many are in this group, two years later, I’m doing things on a two yearcycle, half of that number will reappear in this group as being born into that group. So yousee as they get older they’re reproducing more and then as they get, and when they get veryold, their reproductive cycle kind of drops off. Also we have studied the survival rates, asyounglings there’s a slightly higher chance that they won’t survive, that means a lower ratethat they do survive, that’s the 0.7. And then about 0.9, there’s about a 100% chance thatthey’ll survive to the next stage, there’s always sickness illnesses and these kind of things.But then as they get older, they start to die of old age, so the survival rate goes down. Andthen I’m assuming here that nothing will live past 12 years. There’s a 0% probability thatthey will live past 12 years. (Figure 1)

So here’s my question: if I start with 40 head, and they’re all the same age, I go tothe stock yards and I buy 40 head of cattle, they’re all 2 years old, what would be the agedistribution of my herd in 30 years, okay? This could be important. Maybe I want to takea certain set of those just like I had at the stock yards and sell them off. Maybe there’sdifferent things you can do with your animals at different age groups. So were just going totry and track them, but were going to track them by groups, and were going to do this on 2year cycles.

Well, let’s think about how you do this year to year, and I should say every 2 years to2 years in my example. If I have a general kind of age distribution, so many in one group,so many in another and so forth, and I think about how you get into that first age group –those are the young ones, so they have to be born into that age group. So to get the total inthe next group, in the next cycle that are in that youngest age group, I look at the birth ratesfor each of the groups and I multiply it by how many are in there. So 0 for the first one, 0.5for the second, 1.1 and so on. And that’s a fairly messy calculation but I can clean it up alittle bit using matrices, watch this. I could say, just take your birth rates and put them ina row in a matrix, take my age distribution, put that in a column and then all of a sudden,that’s matrix multiplication. (Figure 2)

If I worry about how many survive to get into that second age group, well that’s the way

10

Figure 1, Segment 2

Figure 2, Segment 2

11

you get there, you’re not born into that second age group, you live to make it into that secondage group. And I know the survival rates, I know that 70% of those in the first age groupwill then end up in the second age group after a 2 year cycle. And then I can phrase this,now this looks messier then this, but I have a method to my madness. I’m going to take mysurvival rate, fill in the rest with zeros and then I’ll do my matrix multiplication and thatgives me that result. (Figure 3) And then the third age group would be a similar thing, 0.9of those survive from the second age group into the third age group, which I could phrase asmatrix multiplication. (Figure 4)

Figure 3, Segment 2

So the way I would do this in general would be that I would construct a 6 by 6 matrix,if you have more age groups you’d have a bigger matrix, I have six age groups. And I’llput those birth rates across the top, and then when I multiply by my age distribution thatwill give me that number in the first age group in the next cycle. And then 0.7 is the, nowthese are survival rates so I take 70% of the first group, fill in with zeroes. To get the secondone I put 0’s here. You’ll notice that these are falling right below the main diagonal, okay?So I can just fill in the rest of my matrix like that. We call this a Leslie matrix after themathematician who kind of developed this idea, you put the birth rates along the top and thenalong that main diagonal you still have zeroes, you have zeroes everywhere else except alongthat diagonal right below the main diagonal, you put in then the survival rates. And then I’lluse that to show you how you can plot the number of animals in each age group. (Figure 5)

If I do this after one 2 year cycle what I’ll do is I’ll multiply this by my original distribu-tion, which is right here. I have 40 in my second age group. And all that’s going to do is sortof peel this part off right here and multiply by 40, right? Zero times this and all those is going

12

Figure 4, Segment 2

Figure 5, Segment 2

13

to be zero. But 0.5 times 40 will give me 20 and .9 times 40 will give me 36, so that’s howmany I would have in the age groups after the first 2 year cycle. (Figure 6) Then if I wantedto do this after 4 years, that would be the second 2 year cycle. So again I would multiply bythis Leslie matrix, let me show you this times my original result – do the arithmetic. This isjust a little matrix multiplication. I’m not showing you the zero entries, the zero products.And then, I’m starting to get fractional parts of cows, which is not always a good thing, butyou realize we’re going to have to do a little bit of rounding when were done. The quick wayto do this is not to do the multiplication 15 times, but to simply raise that Leslie matrix, ifyou notice here, to get the answer, D1, I got that by multiplying L times the original. So Icould raise L to the second power. And just like the problem before where I was dealing withthe probabilities and the transition matrices, that’s the quick way to do it. (Figure 7)

Figure 6, Segment 2

So to get the age distribution after 30 years, okay, that’s 15 two-year cycles, I wouldjust raise that matrix to the 15th power. And again that’s something we do on a graphingcalculator, we could use a computer algebra system to do that. And it’s fairly messy, butwhat do I care? The calculators doing it, not me. Multiply that out and it gives me the agedistribution for each of the age groups. And as you can see here, things have grown very veryfast, and in fact I’ve got over a thousand cows now. That could be an issue. I may not haveenough grass and hay and corn and things to feed this many cows. So what you’d have to dois go back and say, well, probably, I’m going to start selling some off and you’d reflect thatsell off in your survival rates. It’s not necessarily survival “Are they alive?”, it’s survival“Do they make it to the next group?” And at some point there, you’d start selling them offand that would lower the so-called “survival rate.” (Figure 8)

14

Figure 7, Segment 2

Figure 8, Segment 2

15

Okay, again I have a couple slides to show you kind of what we talked about, and we’llget ready for the next segment.

Summary page 1, Segment 2

16

Summary page 2, Segment 2

17

3 Population models – Unrestricted and restricted growth

models

Objective: To illustrate how we can use functions to model population growth,and then use these models to predict future population sizes. We show examplesusing the unrestricted exponential growth model, and the generally more realisticrestricted-growth logistic population model.

The last thing I’d like to talk to you today is exponential growth and how we model thatwith a continuous model. What we’ve been using in the past two examples is really a discretemodel. We’re taking snippets of what happens at the end of a year cycle, a 2-year cyclesomething like that. We can use functions and then a graphing calculator to draw the answerand then realize that that’s an approximation of what’s going on. And what I’d like to talkto you about is, okay, how do you really do that. If you’ve got some kind of populationgrowth, how do you do that, okay? And I want to talk about two situations, unrestricted andrestricted models.

What we just looked at with that Leslie matrix was an unrestricted growth model and itwas actually exponential growth because the rate of growth depended on how many cows orwhatever it was I had. And that’s what we call exponential growth. That’s what we havewhen the rate of growth of an amount is proportional to how many you have. Now that kindof situation, we’ve talked about before, can be modeled with an exponential function, A0e

kt,where k is the rate of growth and A0 is the initial amount that you start with. In fact, in aprevious episode, (Episode 1, Segment 3) we talked about how when your step gets smallerand smaller and smaller you get this function ekt. So we have kind of talked about this in thepast, but the problem with this and I discussed it in the last example is that you’re assuminghere that they can grow without bounds, that I have unlimited resource for sustaining growth.(Figure 1)

The model in Figure 1, which is stated without proof in many algebra texts, is actuallythe solution of the differential equation mentioned in Figure 1:

dA

dt= kA, A(0) = A0.

The differential equation is separable, so we may integrate with respect to both variables:∫dA

A=

∫k dt + C

ln |A| = kt + C

|A| = ekt+C = eCekt.

By letting K = sign(A)eC , we may remove the absolute value signs, and our model becomes

A(t) = Kekt.

By letting t = 0 and using our initial condition A(0) = A0, we see that K = A0 and thegiven model is derived.

18

Figure 1, Segment 3

Now I just want to take a second and prove to you, show you that this model does whatI say it does. If I let the initial amount be 1 and the growth rate be 20% or 0.2, here is agraph of that curve. You’ll notice that the initial amount is 1 and if I take little snippets,if I take, the rate of change can be construed, and we’ve talked about this previously, as theslope of a tangent line. So if I took the slope of the tangent line at this point, and I’m justapproximating it really, and then I took the slope of this, the slope at about this stage is 0.2,and you can get this by taking 2 points on a curve and calculating the slope. And then if Ilook at the function value, right there and I multiply it by this rate of growth, this k, well thefunction value right there is one so sure enough these things are equal and that’s what I saidthey should be. (Figure 2)

If I take the tangent line a little bit higher up, let’s say at when t = 6, the slope of that lineis almost 2

3, 0.664 or so, and if I look at the function values right there and I say 0.2 times

that function value, sure enough I get 0.664. So they are matching up, that’s what’s supposedto happen. Now, that’s a good financial model, we’ve used this when talking about moneyand, boy, you like to think your money can grow without bounds. But it is a bad populationmodel as illustrated in the previous example. You can’t just assume that things can grow,grow, grow, grow, grow, especially living things because they have to eat, they have to breath,they have to drink, they have to have water, and they have to have medicine sometimes.Things like this. So we need to look at a model where the resources are capped. You havesome kind of capacity for things living in an environment. (Figure 3)

So that’s the next example. That is, restricted exponential growth, that is where therate of growth of an amount is proportional to the product, you take them and multiply, its

19

Figure 2, Segment 3

Figure 3, Segment 3

20

proportional to the product of the amount and the amount less then the carrying capacity,and I’ll use a capital K for that, because I didn’t want to use C, I guess. But there’s somehigh number and I’ll say that’s the most you can have, that’s it. And after that things startto starve to death and business like that, or die off in someway, die of disease or whateverit is. Now the model for doing this is messy. Actually deriving that requires some calculus,so I’m not going to get into the derivation of it, but it’s this kind of, there is this patternto what’s going on here. This is called the logistic equation, and this is something you’d begiven to use as a model in some type of situation. We just have to fill in the blanks. A0 isour initial population, K is the carrying capacity, and this k is not the growth rate anymorebut it is some type of constant that we would need to solve for. (Figure 4)

Figure 4, Segment 3

The model in Figure 4, which is stated without proof in many algebra texts, is actuallythe solution of the differential equation mentioned in Figure 4:

dA

dt= kA(K − A), A(0) = A0,

with K > 0 and 0 ≤ A ≤ K. The differential equation is separable, so we may integratewith respect to both variables: ∫

dA

A(K − A)=

∫k dt + C∫

dA

A+

∫dA

K − A= kt + C

21

ln A− ln(K − A) = kt + C

lnA

K − A= kt + C

A

K − A= ekt+C = eCekt.

By letting M = eC , we may solve for A to get

A(t) =KMekt

Mekt + 1=

KM

M + e−kt.

By letting t = 0 and using our initial condition A(0) = A0, we see that M = A0

K−A0and so

A(t) =K

(A0

K−A0

)A0

K−A0+ e−kt

=KA0

A0 + (K − A0)e−kt.

So let me just convince you quickly that it does what I say it does. I’ll let the initialamount be 1, k be 0.5, and the highest amount be 10. If you graph this, here’s 10, you’llnotice that as it gets closer to this 10, it’s leveling off. I’ve filled in the blanks here, andsure enough that curve is doing what I said it would do. In fact if I check the tangent lines,the slopes here is about .45, this is when time is zero, and if I take the amounts, here’s theamount, and here’s 10 minus the amount, that would be 1 times 9, solve for n is about .05.(Figure 5) Now if I check at a different location, this is when time equals 6, the slope of thisline is a little over 1, if I say .05 times the amount of that curve, the y value of that curve,and 10 minus the y value of that curve when time is equal to 6, sure enough I get the sameamount. So its doing what I said it would do, but what I’d really like it to do is show youhow that can give you a realistic model of populations. (Figure 6)

So let me give you an example. Let’s say you have 40 cows like in our previous example,I’m not going to worry about tracking them by age group any more. I’m simply going to say,alright I’ve got 40 cows, I have the capacity to feed and sustain 100, okay? That’s all thepasture I’ve got, that’s all the feed I’ve got. And then I’d like to know, okay, I do want it togrow, I do want my numbers to grow, so say if at the end of the year I have 44 cows, that’sgoing to give me an indication of my growth rate, when can I expect that herd to grow to 80in number? So we’re going to use this logistic equation to model the situation, and what I’mgoing to do is I’m going to plug in the values, my initial population is 40, so 40 would gothere and there, etc. My carrying capacity is 100, so I’ll put 100 there and 100 there, andI don’t know my growth rate, k, yet. So this actually works out to this, I’ve simplified it alittle bit. (Figure 7)

What you have to do is use that bit of information, okay, I had 44 at the end of the year,to solve for k. And this is quickly just some solving of exponential equations, we do this incollege algebra, to kind of clean it up a bit. You can actually switch those out. Clean thatup and subtract the two from both sides. Go through then and divide by the 3. And then, tosolve for the k, you take the natural log of both sides, so you get − k equals this, and thenk would equal a negative logarithm, which just turns it upside down. That’s about 0.164303.

22

Figure 5, Segment 3

Figure 6, Segment 3

23

Figure 7, Segment 3

(Figure 8) And then you use that in your model and set you model equal to 80, and I knowI’m doing this very quickly, but you’ve seen this before. It’s the same step, you just takethings, swap them out, subtract 2 from both sides. you get this down to an exponential thing,take the log of both sides, and you get time, t, to be about, this kind of messy thing, but itsabout 10.9 years. And I’m relying on my calculator to get that. (Figure 9) And very quickly,I’ll just show you the graph of things to convince you very quickly that that is the correctanswer. Here’s the curve that I graphed, here’s 80 and it does hit about at 10.9, right there.(Figure 10)

Okay, a good example of how we can get a realistic model using exponential models. We’vegot a few summary pages and well come back and wrap things up.

Closing

I hope you’ve enjoyed the example that I’ve shown you today. I have to admit, I wentoff the deep end on algebra, and I realize that. Maybe I can make it up to the general mathstudents next time. But I’m really impressed by the power that you can access when you’reusing matrices and matrix multiplication. There’s a lot of powerful stuff that you can dowith algebra in general, and hopefully I’ve illustrated a few of those things. I did things veryquickly so if you want to download episodes of this, you can go to our web page and we’llflash that up in a minute. I’m done, thanks, and I’ll see you next week.

24

Figure 8, Segment 3

Figure 9, Segment 3

25

Figure 10, Segment 3

Summary page 1, Segment 3

26

Summary page 2, Segment 3

27