Section 9.2: Matrices
Definition: A matrix is a rectangular array of numbers:


a11 a12 · · · a1n
 a21 a22 · · · a2n 


A =  ..
..
..  .
..
 .
.
.
. 
am1 am2 · · · amn
In general, aij denotes the (i, j) entry of A. That is, the entry in the ith row and jth column.
The matrix A can be denoted using the shorthand A = (aij ).
Definition: If A = (aij ) is an m × n matrix and c is a scalar, then cA is the m × n matrix
whose (i, j) entry is caij . This is known as scalar multiplication.
Definition: If A = (aij ) and B = (bij ) are both m × n matrices, then the sum A + B is the
m × n matrix whose (i, j) entry is aij + bij for each ordered pair (i, j).
Example: Consider the matrices
1 2
A=
0 3
0 1
B=
2 4
(a) Find A − B + 2C.
(b) Find a matrix D such that A + C = 2A − B + D
1
1 2
C=
.
1 −1
Definition: If A = (aij ) is an m × n matrix and B = (bij ) is an n × r matrix, then the product
AB = C = (cij ) is the m × r matrix with entries
cij =
n
X
aik bkj = ai1 b1j + ai2 b2j + · · · + ain bnj .
k=1
This is known as matrix multiplication.
Example: Consider the matrices
1 2
A=
3 4
and
2 2
B=
.
1 3
Compute the matrix products AB and BA.
Example: Consider the matrices


3 −2
A = 2 4 
1 −3
−2 1 3
and B =
.
4 1 6
Compute the matrix products AB and BA.
Note: In general, matrix multiplication is not commutative. That is, AB 6= BA.
2
Notation: If A is a square (n × n) matrix, it can be multiplied by itself a number of times.
If k is a positive integer, then
Ak = AA
· · · A} .
| {z
k factors
Example: Find an expression for An if
1 1
A=
.
1 1
Definition: The transpose of an m×n matrix A is the n×m matrix B with entries bij = aji .
That is, transposition interchanges the rows and columns of a matrix. The transpose of A
is denoted by AT or A0 .
Example: Consider the matrices
1 2
A=
3 4
and
Verify that (AB)T = B T AT .
3
2 1
B=
.
5 3
Just as the number 1 acts as an identity for multiplication of real numbers, there is a
special matrix I that acts as an identity for matrix multiplication. For any matrix, A,
IA = AI = A.
Definition: The identity matrix is a square (n × n) matrix In = (δij ), where
1, i = j
δij =
0, i 6= j
That is, In is the matrix with 1’s on the main diagonal and 0’s elsewhere. For example,


1 0 0
1 0
I1 = [1],
I2 =
,
I3 = 0 1 0
0 1
0 0 1
If the size of In is clear from the context, then we simply write I.
Definition: An n × n matrix A is called nonsingular if there exists a matrix B such that
AB = BA = I.
The matrix B is called the inverse of A and denoted by A−1 .
Example: Verify that the matrices


1 2 3
A = 0 1 4
0 0 1
and
are inverses of each other.
4


1 −2 5
B = 0 1 −4
0 0
1
Definition: An n × n matrix is called singular if it does not have an inverse.
1 0
Example: Verify that the matrix A =
is singular.
0 0
For every n × n matrix, we can define a scalar whose value will tell us whether the matrix
is singular or nonsingular.
Definition: The determinant of a 2 × 2 matrix
a11 a12
A=
a21 a22
is a scalar defined by
det A = a11 a22 − a12 a21 .
Theorem: If A is an n × n matrix, then A is nonsingular if and only if det A 6= 0. Moreover,
if A is a 2 × 2 matrix
a11 a12
A=
,
a21 a22
then its inverse is
−1
A
1
a22 −a12
=
.
det A −a21 a11
5
Example: Determine whether each matrix is nonsingular and find its inverse, if it exists.
3 5
(a) A =
2 4
2 4
(b) B =
3 6
2 −1
(c) C =
1 3
1 2
(d) D =
0 3
−2 6
(e) E =
3 −9
6
The inverse of a square matrix A can be calculated by augmenting A with the n × n
identity matrix and performing Gaussian elimination.
2 5
Example: Show that A =
is nonsingular and find its inverse.
1 3
Example: Find the inverse of


1 −1 −1
A =  2 −1 1  .
−1 1 −1
7
Solving Linear Systems
Consider a system of m linear equations in n variables
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
.
am1 x1 + am2 x2 + · · · + amn xn = bm .
This system can be expressed in the

a11 a12 · · ·
 a21 a22 · · ·

A =  ..
..
..
 .
.
.
am1 am2 · · ·
matrix form AX = B, where

 
 
a1n
x1
b1





a2n 
 x2 
 b2 
,
X
=
,
B
=




..
.
. .
 .. 
 .. 
. 
amn
xn
bm
If A is a square (n × n), nonsingular matrix, then the system has a unique solution given by
X = A−1 B.
Example: Consider the linear system
2x + y = 2
4x + 6y = 8.
(a) Write this system in matrix form.
(b) Solve this system using the inverse matrix.
8
Consider the linear system defined by
AX = ~0.
Clearly X = ~0 is a solution of this system. It is called a trivial solution. If A is nonsingular,
then the trivial solution is the only solution. Indeed,
X = A−1~0 = ~0.
However, if A is singular, then there exists a nontrivial solution X 6= ~0.
Theorem: Suppose that A is an n × n matrix, and X and ~0 are n × 1 matrices. Then the
equation
AX = ~0
has a nontrivial solution if and only if A is singular.
Example: Determine whether
1 1
A=
2 1
is nonsingular. If so, compute its inverse. In either case, solve AX = ~0.
Example: Determine whether
1 3
B=
1 3
is nonsingular. If so, compute its inverse. In either case, solve BX = ~0.
9
The Leslie Matrix
Suppose that we would like to model the population size of a species with discrete breeding
seasons and age-dependent reproductive fitness (e.g. perennial plants, cicadas, salmon).
Let N (t) denote the population size at generation t = 0, 1, 2, . . .. Then the population size
at generation t + 1 is a function of the population size at the previous generation,
N (t + 1) = f (N (t)).
An equation of this form is called a difference equation or recursion.
Recall that the difference equation modeling discrete exponential growth is
N (t + 1) = RN (t),
t = 0, 1, 2, . . .
where N0 ≥ 0 is the initial population size and R > 0 is a growth parameter (average number
of offspring per individual per breeding season). The solution of this difference equation is
N (t) = N0 · Rt
If R > 1, then the population experiences exponential growth over time, and if 0 < R < 1,
the population decays exponentially over time.
300
5
4
Population Size
Population Size
250
200
150
100
2
1
50
0
0
3
2
4
6
8
0
0
10
Time
2
4
6
8
10
Time
The discrete exponential growth model (as well as the Beverton-Holt, discrete logistic,
and Ricker logistic models) assume that all individuals have the same reproductive fitness.
However, in many species, reproduction is age dependent. Thus, we will consider an agestructured, discrete-time model introduced by Patrick Leslie in 1965.
Since only females produce offspring, we will model only the female population. The
number of individuals is determined at the end of each breeding season.
10
Suppose that the population is divided into m+1 distinct age classes and let Nx (t) denote
the number of females of age x = 0, 1, 2, . . . , m at time t = 0, 1, 2, . . .. Then the vector


N0 (t)
 N1 (t) 


 N2 (t) 
N (t) = 

 .. 
 . 
Nm (t)
describes the number of females in each age class at time t = 0, 1, 2, . . ..
Let Pj ∈ [0, 1] denote the fraction of females of age j at time t that survive to time t + 1.
Then
N1 (t + 1) = P0 N0 (t)
N2 (t + 1) = P1 N1 (t)
..
.
Nm (t + 1) = Pm−1 Nm−1 (t)
Let Fj ≥ 0 denote the average number of female offspring per female of age j which
survive to the end of the next breeding season. Then
N0 (t + 1) = F0 N0 (t) + F1 N1 (t) + F2 N2 (t) + · · · + Fm Nm (t).
These equations can be written in matrix form as
N (t + 1) = LN (t),
where

F0 F1 F2
 P0 0 0

 0 P1 0

L=0 0 P
2

 ..
..
..
.
.
.
0 0 0

· · · Fm−1 Fm
···
0
0 

···
0
0 

···
0
0 


..
...

.
· · · Pm−1 0
The matrix L is called the Leslie matrix. The Leslie matrix model is illustrated below.
11
Example: Assume that a population is divided into three age classes and that 80% of the
females of age zero and 10% of the females of age one survive until the end of the next
breeding season. Moreover, assume that females of age one have an average of 1.6 female
offspring and females of age 2 have an average of 3.9 female offspring. If the initial population
consists of 1000 females of age zero, 100 females of age one, and 20 females of age two, find
the Leslie matrix and age distribution at time t = 3.
12
Example: Assume that a population is divided into four age classes and that 65% of the
females of age zero, 40% of the females of age one, and 30% of the females of age two survive
until the end of the next breeding season. Moreover, assume that females of age one have an
average of 2.8 female offspring, females of age 2 have an average of 7.2 female offspring, and
females of age 3 have an average of 3.7 female offspring. If the initial population consists of
1500 females of age zero, 500 females of age one, 250 females of age two, and 50 females of
age three, find the Leslie matrix and age distribution at time t = 3.
13
Example: Consider the Leslie matrix


0
5 0
L = 0.8 0 0 .
0 0.3 0
Determine the number of age classes in the population, the fraction of one-year-olds that
survive until the end of the next breeding season, and the average number of female offspring
of a two-year-old female.
Example: Consider the Leslie matrix

2
3
2
0.4 0
0
L=
 0 0.6 0
0
0 0.8

1
0
.
0
0
Determine the number of age classes in the population, the fraction of two-year-olds that
survive until the end of the next breeding season, and the average number of female offspring
of a one-year-old female.
14
Stable Age Distribution
Example: Suppose that a population is divided into two age classes with Leslie matrix
1.5 2
L=
,
0.08 0
and suppose that N0 (0) = 100 and N1 (0) = 100.
(a) Find the age distributions for t = 0, 1, 2, . . . , 7.
The following table summarizes the age distributions for t = 0, 1, 2, . . . , 7.
Time, t
0
1
2
3
4
5
6
7
15
N0 (t)
100
350
541
868
1388
2221
3553
5685
N1 (t)
100
8
28
43
69
111
178
284
(b) Compute the successive ratios
q0 (t) =
N0 (t)
N0 (t − 1)
and q1 (t) =
N1 (t)
N1 (t − 1)
for t = 1, 2, . . . , 7, and determine lim q0 (t) and lim q1 (t).
t→∞
t→∞
The following table summarizes the successive ratios for t = 1, 2, . . . , 7.
Time, t
1
2
3
4
5
6
7
q0 (t)
3.5
1.55
1.6
1.5991
1.6001
1.5997
1.6001
q1 (t)
0.08
3.5
1.536
1.605
1.609
1.604
1.596
Therefore, we suspect that
lim q0 (t) = lim q1 (t) = 1.6.
t→∞
t→∞
In the long run, the population size is increasing by a factor of 1.6 each generation.
16
(c) Compute the fraction of females which are age 0,
p(t) =
N0 (t)
N0 (t) + N1 (t)
for t = 0, 1, 2, . . . , 7, and determine lim p(t).
t→∞
The following table summarizes the fraction of females which are age 0 for t = 0, 1, 2, . . . , 7.
Time, t
p(t)
0
0.5
1
0.9777
2
0.9508
3
0.9528
4
0.9526
5
0.9524
6
0.9523
7
0.9524
Therefore, we suspect that
lim p(t) = 0.9524.
t→∞
In the long run, approximately 95.24% of the population is age 0.
17
Although the population size is increasing (by a constant factor of 1.6 each generation),
the fraction of females in age class 0 is convergent (to 95.24%). This constant fraction is
called the stable age distribution.
If we start the population in the stable age distribution, the fraction of females in age
class 0 will remain the same, about 95.24%, and the population will increase by a constant
factor of 1.6 each generation. A stable age distribution for this population is
2000
N (0) =
.
100
It follows that
1.5 2
N (1) = LN (0) =
0.08 0
2000
3200
=
.
100
160
The fraction of females in age class 0 remains the same:
3200
2000
N0 (t)
=
=
= 0.9524.
N0 (t) + N1 (t)
3200 + 160
2000 + 100
Similarly,
2000
3200
1.6
=
100
160
yields the same result.
Therefore, if N = [N0 , N1 ]T denotes a stable age distribution, then
LN = λN,
where λ = 1.6, and
N0
≈ 95.24%.
N0 + N1
In Section 9.3, we obtain a much simpler method to compute the stable age distribution.
18