Homework Summary – Feb.23, 2011
Class began with a brief example of multiplying two matrices. Arrange the second matrix of the intended
product elevated and to the right of the first matrix so that the matrix containing the product can be
placed under the second and to the right of the first. This configuration allows a better relationship for
arranging the elements of the product.
We also revisited the methods of finding the determinant of a matrix. Beginning with the 2 x 2 and 3 x 3,
we reviewed finding the product of the diagonals and went into more detail on using permutations to
determine even (positive) or odd (negative). Also we found that the number of products calculated
from the diagonals to find the determinant of a matrix is equal to the order of the matrix factorial. For a
1 x 1 matrix, the determinant is the element itself, in a 2 x 2 matrix there are two diagonals for which to
find the product of its elements (2!), for a 3 x 3 there are 6 diagonals (3!), etc.
The determinant of a matrix can be found using expansion by minors, but a more efficient way –
especially if the determinant is 4 x 4 or larger, is to use row operations to convert the matrix to an upper
or lower diagonal matrix. Once the matrix is in this form, the determinant is found by multiplying the
elements of the diagonal.
After a brief review of the first to items in the talking points we went into more depth of the 3-birth
Fibonacci model. We used an Excel spreadsheet to calculate the growth rate and the stable population
profile. Then we calculated these without linear algebra. Beginning with the population a, b, c, d, the
age distribution for the next month is represented by the system of equations:
𝑏 + 𝑐 + 𝑑 = 𝜆𝑎
𝑎 = 𝜆𝑏
𝑏 = 𝜆𝑐
𝑐 = 𝜆𝑑
Using substitution: 𝑏 = 𝜆2 𝑑, 𝑎 = 𝜆3 𝑑 so, 𝑏 + 𝑐 + 𝑑 = 𝜆𝑎 becomes 𝜆2 𝑑 + 𝜆𝑑 + 𝑑 = 𝜆4 𝑑
Setting equal to zero and factoring we have: (𝜆4 − 𝜆2 − 𝜆 − 1)𝑑 = 0
Since 𝑑 ≠ 0, 𝑝(𝜆) = 𝜆4 − 𝜆2 − 𝜆 − 1. Graphing, we found the zeros to be -1 and 1.465571. To give this
1
profile as a percent, 𝑎 + 𝑏 + 𝑐 + 𝑑 = 1, 𝑜𝑟 (𝜆4 + 𝜆2 + 𝜆 + 1)𝑑 = 1 so 𝑑 = 𝜆4 +𝜆2 +𝜆+1 .
Therefore, d = .1288, a = .4056, b = .764, and c = .1888. These were the values that we observed using
the Excel spreadsheet.
Again with the power method, where 𝜆 = 𝜆1 is the dominant eigenvalue, we used the matrix 𝐴 =
0 1 1 1
1 0 0 0
[
], where the first row represents births and the next three rows represent the age
0 1 0 0
0 0 1 0
progression. Next we found the eigenvector, 𝑥⃗1 of the dominant eigenvalue, to get the stable age
𝜆 −1 −1 −1
−1 𝜆
0
0
distribution. 𝑝(𝜆)=𝑑𝑒𝑡 [
]
0 −1 𝜆
0
0
0 −1 𝜆
4
2
𝑝(𝜆) = 𝜆 − 𝜆 − 𝜆 − 1 . This being the same polynomial as our earlier computations will yield the
. 4056
. 2764
eigenvector 𝑥⃗1 = [
].
. 1888
. 1288
Our discussion then led to Leslie Models, named for Patrick Leslie. I remembered studying these several
years ago when taking a course entitled “Mathematical Modeling in Biology.” This course was entirely
on-line, therefore difficult for me because there was very little interaction from the instructor. Just from
the little bit we discussed in class Wednesday, I think I will be able to understand it much better. The
variables are defined as: age groups 𝑥0 , 𝑥1 , … , 𝑥𝑛 , birth rates 𝑏𝑘 , survival rates 𝑠𝑘 , and “t” is the
population in the next cycle.
𝑥0 𝑡 = 𝑏1 𝑥1 + 𝑏2 𝑥2 + ⋯ + 𝑏𝑛 𝑥𝑛
𝑥1 𝑡 = 𝑠1 𝑥0
𝑥2 𝑡 = 𝑠2 𝑥1
.
.
.
𝑥𝑛 𝑡 = 𝑠𝑛 𝑥𝑛−1
0 𝑏1 𝑏2 … 𝑏𝑛
𝑠
0
0
In matrix form, a Leslie Matrix is 𝐴 = 1
0 𝑠2⋱
0
[0
0
𝑠𝑛
0]
Example: Model 1
For 𝑏1 = .8, 𝑏2 = .9, 𝑏3 = .4, 𝑠1 = .6, 𝑠2 = .9, 𝑠3 = .8 , we have the Leslie Matrix
0 .8 .9 .4
.6 0 0 0
𝐴=[
]
0 .9 0 0
0 0 .8 0
Will the population survive or will it go extinct?
The growth rate is the dominant eigenvalue and the population profile is the eigenvector.
If 𝑝(𝜆) > 1, the population will survive. If 𝑝(𝜆) < 1, the population will die out.
Eigenvalue:
Use expansion by minors to find the determinant:
𝜆
−.8 −.9 −.4
−.8 −.9 −.4
𝜆
0 0
−.6 𝜆
0
0
𝑑𝑒𝑡 [
] = 𝜆 |−.9 𝜆 0| − (−.6) |−.9 𝜆
0 |
0 −.9 𝜆
0
0 −.8 𝜆
0 −.8 𝜆
0
0 −.8 𝜆
= 𝜆(𝜆3 ) + .6(−.8𝜆2 − .288 − .81𝜆)
= 𝜆4 − .48𝜆2 − .486𝜆 − .1728
𝑝(𝜆) = 𝜆4 − .48𝜆2 − .486𝜆 − .1728
Graph to find zeros: 𝜆 = −.4735645 𝑎𝑛𝑑 𝜆 = 1.0489757
The dominant eigenvalue is 1.0489757 indicating a growth rate of .0489757. Since this is >1, the
population will survive.
. 8𝑏 + .9𝑐 + .4𝑑 = 𝜆𝑎
. 6𝑎 = 𝜆𝑏
. 9𝑏 = 𝜆𝑐
. 8𝑐 = 𝜆𝑑
Solving, I find that 𝑐 =
𝜆𝑑
,𝑏
.8
=
𝜆2 𝑑
,𝑎
.72
𝜆3 𝑑
= .432
The profile in percent is found using 𝑎 + 𝑏 + 𝑐 + 𝑑 = 1
𝜆3 𝑑
This becomes .432 +
𝜆3
𝜆2
𝜆2 𝑑
.72
+
𝜆𝑑
.8
+𝑑 =1
𝜆
(.432 + .72 + .8 + 1) 𝑑 = 1
𝑑=
1
𝜆3 𝜆2 𝜆
+ + +1
.432 .72 .8
𝑑 = .0976047835
𝑐 = .1599766345
𝑏 = .2330716697
𝑎 = .5093469123
The eigenvector, which represents the population profile is:
. 5093469123
. 2330716697
[
]
. 1599766345
. 0976047835