Math 148 Exam II Practice Problems
This review should not be used as your sole source for preparation for the exam. You
should also re-work all examples given in lecture, all homework problems, all lab assignment
problems, and all quiz problems.
1. Assume that the size of a population evolves according to the logistic equation with
intrinsic growth rate r = 1.25 and carrying capacity K = 500.
(a) Find the differential equation that describes the growth rate of this population.
(b) Find all equilibria and discuss the stability of the equilibria using the graphical
approach.
(c) Find the eigenvalues associated with the equilibria and use the stability criterion
to determine stability of the equilibria.
2. Consider a single-species population model with an Allee effect
N
dN
= 2N (N − 150) 1 −
,
dt
400
with initial condition N (0) = N0 ≥ 0.
(a) Find all equilibria of this differential equation.
(b) Determine the stability of the equilibria using the graphical approach.
(c) Find the eigenvalues associated with the equilibria and use the stability criterion
to determine stability of the equilibria.
(d) If N0 = 100, find lim N (t).
t→∞
(e) If N0 = 200, find lim N (t).
t→∞
(f) If N0 = 500, find lim N (t).
t→∞
3. Solve the linear system
x+y+z = 1
3x − y − z = 4
x + 5y + 5z = −1
by forming the augmented matrix and performing Gaussian elimination.
4. Solve the linear system
x + 2y − 3z = −2
3x − y − 2z = 1
2x + 3y − 5z = −3
by forming the augmented matrix and performing Gaussian elimination.
1
5. Solve the linear system
2x + y − z = 5
x+y−z = 4
−x − 2y + 3z = −8
by using an inverse matrix.
6. Three non-profit organizations are engaged in alternate-energy research. Organization
1 spends 60% of their funds on solar energy research, 30% on wind energy research, and
10% on the motion of ocean tides research. Organization 2 spends 40% of their funds
on solar energy research, 30% on wind energy research, and 30% on the motion of ocean
tides research. Organization 3 spends 20% of their funds on solar energy research, 60%
on wind energy research, and 20% on the motion of ocean tides research. How much
was awarded to each organization if total solar research was $9.2 million, total wind
research was $9.6 million, and total motion of ocean tides research was $5.2 million?
7. Determine whether each matrix is nonsingular. If so, compute its inverse. In either
case, solve AX = ~0.
1 1
(a) A =
2 1
4 1
(b) A =
4 1
8. Assume that a population is divided into three age classes and that 70% of the females
of age 0 and 40% of the females of age 1 survive until the end of the next breeding
season. Assume further that females of age 1 have an average of 1.5 female offspring
and females of age 2 have an average of 2.4 female offspring. If the initial population
consists of 1000 females of age 0, 400 females of age 1, and 50 females of age 2, find
the Leslie matrix and the age distribution at time t = 1.
−→
9. Let A = (0, 1, −3) and B = (1, 3, −2). Find a unit vector in the direction of AB.
10. Find a vector of magnitude r = 6 that forms an angle of 120◦ measured counterclockwise from the positive x-axis.
11. Consider the triangle with vertices A = (2, 1, 5), B = (−1, −3, 7), and C = (3, 2, 0).
Find the angle at the vertex B.
12. Find the equation of the line through (2, 0) and orthogonal (perpendicular) to h1, −2i.
13. Find the equation of the plane through (3, −1, 2) and orthogonal (perpendicular) to
h4, 1, 2i.
14. Find parametric equations of the line through A = (2, 0, 4) and B = (4, 1, 0).
2
15. Find the eigenvalues and eigenvectors of
−4 2
A=
−3 1
16. Without explicitly computing the eigenvalues, determine whether the real parts of both
eigenvalues are negative.
2
4
(a) A =
−3 −4
−2 1
(b) B =
4 1
17. Consider the matrix
5
7
A=
−2 −4
If ~x = h3, 2i, find A5~x without using a calculator.
18. Suppose that a population is divided into two
2
L=
0.3
age classes with Leslie matrix
4
0
Determine the growth parameter and stable age distribution.
3
Solutions
1. Assume that the size of a population evolves according to the logistic equation with
intrinsic growth rate r = 1.25 and carrying capacity K = 500.
(a) Find the differential equation that describes the growth rate of this population.
The general form of the logistic equation is
dN
N
= rN 1 −
,
dt
K
where N (t) represents the population size at time t ≥ 0. For the parameter values
r = 1.25 and K = 500, we have
dN
N
5
N
= 1.25N 1 −
= N 1−
.
dt
500
4
500
(b) Find all equilibria, and discuss the stability of the equilibria using the graphical
approach.
To find the equilibria, let
5
dN
= N
dt
4
N
1−
= 0.
500
It follows that the equilibria are N̄ = 0 and N̄ = 500.
The graph of the function f (N ) = 1.25N (1 − N/500) is a parabola which opens
downward (because of the negative in front of the N 2 term) that crosses the
horizontal axis at N = 0 and N = 500. The graph of f (N ) is shown below.
(Note: Since N denotes a population size, we only sketch the graph for N ≥ 0.)
For values of N between 0 and 500, dN/dt > 0 (since the graph of f (N ) = dN/dt
lies above the horizontal axis). Thus, the population size N increases on (0, 500).
For values of N larger than 500, dN/dt < 0, so the population size N decreases
for N > 500. Therefore, we see that N̄ = 0 is unstable and N̄ = 500 is locally
stable.
4
(c) Find the eigenvalues associated with the equilibria and use the stability criterion
to determine stability of the equilibria.
Consider the function
5
f (N ) = N
4
N
1−
500
5
1 2
= N−
N .
4
400
The derivative of this function is
f 0 (N ) =
1
5
−
N.
4 200
The eigenvalues associated with the equilibria N̄ = 0 and N̄ = 500 are
5
>0
4
5
f 0 (500) = − < 0
4
f 0 (0) =
According to the stability criterion, N̄ = 0 is unstable and N̄ = 500 is locally
stable.
2. Consider a single-species population model with an Allee effect
dN
N
= 2N (N − 150) 1 −
,
dt
400
with initial condition N (0) = N0 ≥ 0.
(a) Find all equilibria of this differential equation.
To find the equilibria, let
N
dN
= 0.
= 2N (N − 150) 1 −
dt
400
It follows that the equilibria are N̄ = 0, N̄ = 150, and N̄ = 400.
5
(b) Determine the stability of the equilibria using the graphical approach.
The graph of the function f (N ) = 2N (N −150)(1−N/400) is a cubic that crosses
the horizontal axis at N = 0, N = 150, and N = 400. The graph of f (N ) is shown
below.
(Note: Since N denotes a population size, we only sketch the graph for N ≥ 0.)
For values of N between 0 and 150, dN/dt < 0 (since the graph of f (N ) = dN/dt
lies below the horizontal axis. Thus, the population size N decreases on (0, 150).
For values of N between 150 and 400, dN/dt > 0, so the population size N
increases on (150, 400). For values of N larger than 400, dN/dt < 0, so the
population size N decreases for N > 400. Therefore, we see that N̄ = 0 is locally
stable, N̄ = 150 is unstable, and N̄ = 400 is locally stable.
(c) Find the eigenvalues associated with the equilibria and use the stability criterion
to determine stability of the equilibria.
Consider the function
1 3 11 2
N
N + N − 300N.
=−
f (N ) = 2N (N − 150) 1 −
400
200
4
The derivative of this function is
f 0 (N ) = −
3 2 11
N + N − 300.
200
2
The eigenvalues associated with the equilibria N̄ = 0, N̄ = 150, and N̄ = 400 are
f 0 (0) = −300 < 0
375
f 0 (150) =
>0
2
f 0 (400) = −500 < 0
According to the stability criterion, N̄ = 0 is locally stable, N̄ = 150 is unstable,
and N̄ = 400 is locally stable.
6
(d) If N0 = 100, find lim N (t).
t→∞
Since N0 lies between the equilibria N̄ = 0 and N̄ = 150, the population size N
decreases (see part (b)) until it reaches the locally stable equilibrium N̄ = 0 from
which there is no change. Thus,
lim N (t) = 0.
t→∞
(e) If N0 = 200, find lim N (t).
t→∞
Since N0 lies between the equilibria N̄ = 150 and N̄ = 400, the population size
N increases (see part (b)) until it reaches the locally stable equilibrium N̄ = 400
from which there is no change. Thus,
lim N (t) = 400.
t→∞
(f) If N0 = 500, find lim N (t).
t→∞
Since N0 is greater than the equilibrium N̄ = 400, the population size N decreases
(see part (b)) until it reaches the locally stable equilibrium N̄ = 400 from which
there is no change. Thus,
lim N (t) = 400.
t→∞
7
3. Solve the linear system
x+y+z = 1
3x − y − z = 4
x + 5y + 5z = −1
by forming the augmented matrix and performing Gaussian elimination.
The augmented matrix for this linear

1
 3
1
system is

1
1
1
−1 −1 4 
5
5 −1
Using Gaussian elimination, we have




1
1
1 1
1
1 1
1
 3 −1 −1 4  ∼  0 −4 −4 1 
1 5
5 −1
0 4
4 −2


1 1
1
1
∼  0 −4 −4 1 
0 0
0 −1
Since 0 6= −1, the system is inconsistent. There is no solution.
8
R2 − 3R1
R3 − R1
(R3 + R2 )
4. Solve the linear system
x + 2y − 3z = −2
3x − y − 2z = 1
2x + 3y − 5z = −3
by forming the augmented matrix and performing Gaussian elimination.
The augmented matrix for this linear

1
 3
2
system is

2 −3 −2
−1 −2 1 
3 −5 −3
Using Gaussian elimination, we have




1 2 −3 −2
1 2 −3 −2
 3 −1 −2 1  ∼  0 −7 7
7 
1
2 3 −5 −3
0 −1 1


1 2 −3 −2
∼  0 1 −1 −1 
0 −1 1
1


1 2 −3 −2
∼  0 1 −1 −1 
0
0 0 0
R2 − 3R1
R3 − 2R1
1
− R2
7
(R3 + R2 )
Therefore, z = α is a free variable. Using the second equation, we have
y = z−1
y = α − 1.
Using the first equation, we have
x = −2y + 3z − 2
x = −2(α − 1) + 3α − 2
x = α
Thus, the solution set is {(α, α − 1, α)|α ∈ R}.
9
5. Solve the linear system
2x + y − z = 5
x+y−z = 4
−x − 2y + 3z = −8
by using an inverse matrix.
This linear system can be written in matrix form as

   
2
1 −1 x
5
1
1 −1 y  =  4 
−1 −2 3
z
−8
(1)
To solve for x, y, z, we must find the inverse of the matrix


2
1 −1
1 −1
A= 1
−1 −2 3
Augmenting the matrix with the identity matrix, and performing Gaussian elimination,
we have




1
1 −1 0 1 0
2
1 −1 1 0 0
 1
1 −1 0 1 0  ∼  2
1 −1 1 0 0 
(R1 ↔ R2 )
−1 −2 3 0 0 1
−1 −2 3 0 0 1


1 1 −1 0 1 0
R2 − 2R1


0 −1 1 1 −2 0
∼
R3 + R1
0 −1 2 0 1 1


1 0 0 1 −1 0
R
+
R
1
2
∼  0 −1 1 1 −2 0 
R3 − R2
0 0 1 −1 3 1


1 0 0 1 −1 0
∼  0 −1 0 2 −5 −1 
(R2 − R3 )
1
0 0 1 −1 3


1 0 0 1 −1 0
∼  0 1 0 −2 5 1 
(−R2 )
0 0 1 −1 3 1
Therefore, solving Equation (1) above, we
  
x
1 −1
y  = −2 5
z
−1 3
Thus, the solution is (1, 2, −1).
10
have
   
5
1
0




4 = 2
1
1
−8
−1
6. Three non-profit organizations are engaged in alternate-energy research. Organization
1 spends 60% of their funds on solar energy research, 30% on wind energy research,
and 10% on the motion of ocean tides research. Organization 2 spends 40% of their
funds on solar energy research, 30% on wind energy research, and 30% on the motion
of ocean tides research. Organization 3 spends 20% of their funds on solar energy research, 60% on wind energy research, and 20% on the motion of ocean tides research.
How much was awarded to each organization if total solar research was $9.2 million,
total wind research was $9.6 million, and total motion of ocean tides research was $5.2
million?
Let x, y, and z denote the amounts (in millions of dollars) awarded to Organization 1,
Organization 2, and Organization 3, respectively. Therefore,
Solar: 0.6x + 0.4y + 0.2z = 9.2
Wind: 0.3x + 0.3y + 0.6z = 9.6
Ocean: 0.1x + 0.3y + 0.2z = 5.2
To simplify this linear system, we multiply each equation by 10:
6x + 4y + 2z = 92
3x + 3y + 6z = 96
x + 3y + 2z = 52
Using an augmented matrix and Gaussian elimination, we have




1 3 2 1 46
6 4 2 92
R
2 1
 3 3 6 96  ∼  1 1 2 32 
1
R
3 2
1 3 2 52
1 3 2 52


1 1 2 32
∼  3 2 1 46 
(R1 ↔ R2 )
1 3 2 52


1 1
2
32
R
−
3R
2
1
∼  0 −1 −5 −50 
R3 − R1
0 2
0
20


1 1 2 32
−R2


0 1 5 50
∼
1
R
2 3
0 1 0 10
It follows that y = 10. Using the second equation, we have
5z = 50 − y
5z = 40
z = 8.
11
Using the first equation, we have
x = 32 − y − 2z
x = 32 − 10 − 16
x = 6.
Therefore, Organization 1 was awarded $6 million, Organization 2 was awarded $10
million, and Organization 3 was awarded $8 million.
7. Determine whether each matrix is nonsingular. If so, compute its inverse. In either
case, solve AX = ~0.
1 1
(a) A =
2 1
The determinant of A is
det(A) = 1 − 2 = −1 6= 0.
Therefore, A is nonsingular with inverse
1 −1
−1 1
−1
A = −1
=
−2 1
2 −1
Since A is nonsingular, the only solution of AX = ~0 is the trivial solution X = ~0.
4 1
(b) A =
4 1
The determinant of A is
det(A) = 4 − 4 = 0.
Therefore, A is singular. Consider the equation AX = ~0:
4 1 x
4x + y
0
=
=
4 1 y
4x + y
0
It follows that y = −4x. Let x = α be a free variable. The solution set is
{(α, −4α)|α ∈ R}.
12
8. Assume that a population is divided into three age classes and that 70% of the females
of age 0 and 40% of the females of age 1 survive until the end of the next breeding
season. Assume further that females of age 1 have an average of 1.5 female offspring
and females of age 2 have an average of 2.4 female offspring. If the initial population
consists of 1000 females of age 0, 400 females of age 1, and 50 females of age 2, find
the Leslie matrix and the age distribution at time t = 1.
The Leslie matrix is


0 1.5 2.4
0
L = 0.7 0
0 0.4 0
The initial population distribution is


1000
N (0) =  400 
50
Therefore, at time t = 1, we have


  
0 1.5 2.4 1000
720
0   400  = 700
N (1) = LN (0) = 0.7 0
0 0.4 0
50
160
The population at time t = 1 consists of 720 females of age 0, 700 females of age 1,
and 160 females of age 2.
−→
9. Let A = (0, 1, −3) and B = (1, 3, −2). Find a unit vector in the direction of AB.
−→
The vector AB is given by
−→
AB = h1, 2, 1i .
The magnitude of this vector is
−→
||AB|| =
√
√
1 + 4 + 1 = 6.
−→
Therefore, a unit vector in the direction of AB is
1 −→
1 2 1
√ ,√ ,√
.
−→ AB =
6 6 6
||AB||
13
10. Find a vector of magnitude r = 6 that forms an angle of 120◦ measured counterclockwise from the positive x-axis.
Using polar coordinates, the vector has components
hr cos θ, r sin θi = h6 cos 120◦ , 6 sin 120◦ i
* √ !+
1
3
=
6 −
,6
2
2
√
= h−3, 3 3i.
11. Consider the triangle with vertices A = (2, 1, 5), B = (−1, −3, 7), and C = (3, 2, 0).
Find the angle at the vertex B.
Consider the vectors
−→
~v = BA = h3, 4, −2i
−→
w
~ = BC = h4, 5, −7i
It follows that
~v · w
~ = 12 + 20 + 14 = 46
√
√
9 + 16 + 4 = 29
||~v || =
√
√
√
||w||
~
=
16 + 25 + 49 = 90 = 3 10
If β denotes the angle at vertex B, then
cos β =
46
46
~v · w
~
= √ √ = √
.
||~v ||||w||
~
3 29 10
3 290
Therefore,
β = arccos
46
√
3 290
14
radians.
12. Find the equation of the line through (2, 0) and orthogonal (perpendicular) to h1, −2i.
If (x, y) is another point on the line, then the vector ~v = hx − 2, y − 0i lies on the line.
Therefore,
1(x − 2) − 2(y − 0) = 0
x − 2 − 2y = 0
2y = x − 2
1
y =
x − 1.
2
13. Find the equation of the plane through (3, −1, 2) and orthogonal (perpendicular) to
h4, 1, 2i.
If (x, y, z) is another point on the plane, then the vector ~v = hx − 3, y + 1, z − 2i lies
in the plane. Therefore,
4(x − 3) + 1(y + 1) + 2(z − 2) = 0
4x − 12 + y + 1 + 2z − 4 = 0
4x + y + 2z = 15
14. Find parametric equations of the line through A = (2, 0, 4) and B = (4, 1, 0).
−→
A vector which defines the direction of the line is AB = h2, 1, −4i. Therefore, the
parametric equaitons of the line are
x = 2 + 2t
y = t
z = 4 − 4t
where t ∈ R is a parameter.
15
15. Find the eigenvalues and eigenvectors of
−4 2
A=
−3 1
To find the eigenvalues of A, we set
det(A − λI) = 0.
That is,
−4 − λ
2
−3
1−λ
= 0
λ2 + 3λ + 2 = 0
(λ + 1)(λ + 2) = 0.
Thus, the eigenvalues of A are λ1 = −1 and λ2 = −2.
To find the eigenvectors associated with these eigenvalues, we set
(A − λI)~v = 0.
If λ1 = −1, then we have
−3 2
(A − λI)~v1 =
−3 2
−3v1 + 2v2
v1
= ~0.
=
−3v1 + 2v2
v2
It follows that 3v1 = 2v2 and so ~v1 = h2, 3i is an eigenvector associated with λ1 = −1.
If λ2 = −2, then we have
−2 2
(A − λI)~v2 =
−3 3
v1
−2v1 + 2v2
=
= ~0.
v2
−3v1 + 3v2
It follows that v1 = v2 and so ~v2 = h1, 1i is an eigenvector associated with λ2 = −2.
16
16. Without explicitly computing the eigenvalues, determine whether the real parts of both
eigenvalues are negative.
2
4
(a) A =
−3 −4
The trace and determinant of A are
tr(A) = 2 − 4 = −2
and
det(A) = −8 + 12 = 4.
Since tr(A) < 0 and det(A) > 0, both eigenvalues of A have negative real parts
by the Routh-Hurwitz Criterion.
−2 1
(b) B =
4 1
The trace and determinant of B are
tr(B) = −2 + 1 = −1
and
det(B) = −2 − 4 = −6.
Since det(A) < 0, at least one of the eigenvalues of B has a positive real part by
the Routh-Hurwitz Criterion.
17
17. Consider the matrix
5
7
A=
−2 −4
If ~x = h3, 2i, find A5~x without using a calculator.
To calculuate A5~x, we will make use of the eigenvalues and (linearly independent)
eigenvectors of A. To find the eigenvalues of A, we set
det(A − λI) = 0.
That is,
5−λ
7
−2 −4 − λ
= 0
λ2 − λ − 6 = 0
(λ + 2)(λ − 3) = 0.
Thus, the eigenvalues of A are λ1 = −2 and λ2 = 3.
If λ1 = −2, then
7
7
v1
7v1 + 7v2
(A − λI)~v1 =
=
= ~0
−2 −2 v2
−2v1 − 2v2
It follows that v1 = −v2 and so ~v1 = h1, −1i is an eigenvector associated with λ1 = −2.
If λ2 = 3, then
2
7
v1
2v1 + 7v2
(A − λI)~v1 =
=
= ~0
−2 −7 v2
−2v1 − 7v2
It follows that 2v1 = −7v2 and so ~v2 = h7, −2i is an eigenvector associated with λ2 = 3.
Since λ1 6= λ2 , the eigenvalues are linearly independent. Therefore, the vector ~x = h3, 2i
can be written uniquely as a linear combination of the eigenvectors ~v1 and ~v2 . That is,
c1
c ~v + c ~v = ~x
1 1 2 2
1
7
3
+ c2
=
−1
−2
2
c1 + 7c2
3
=
−c1 − 2c2
2
It follows that c1 = −4 and c2 = 1. Therefore,
A5~x =
=
=
=
=
=
A5 (−4v~1 + v~2 )
−4A5 v~1 + A5 v~2
−4λ51 v~1 + λ52 v~2
−4(−2)5 h1, −1i + (3)5 h7, −2i
h128, −128i + h1701, −486i
h1829, −614i.
18
18. Suppose that a population is divided into two
2
L=
0.3
age classes with Leslie matrix
4
0
Determine the growth parameter and stable age distribution.
Since the population consists of two age classes, the growth parameter is the largest
eigenvalue of the Leslie matrix. Moreover, the eigenvector corresponding to the largest
eigenvalue can be used to calculate the stable age distribution.
To find the eigenvalues of L, we set
det(L − λI) = 0.
That is,
2−λ 4
0.3 −λ
= 0
λ2 − 2λ − 1.2 = 0.
Using the quadratic formula, we have
√
1
2 ± 4 + 4.8
2
√ 1
λ =
2 ± 8.8
2
√ 1
λ =
2 ± 2 2.2
2 √
λ = 1 ± 2.2
λ =
Therefore, the growth parameter is given by the larger eigenvalue, λ = 1 +
√
2.2.
To find the eigenvector associated with this eigenvalue, we set
(L − λI)~v = 0.
√
If λ = 1 + 2.2, then
√
√
v1
(1 − 2.2)v1√+ 4v2
1 − 2.2
4√
=
(A − λI)~v =
= ~0
0.3
−1 − 2.2 v2
0.3v1 − (1 + 2.2)v2
It follows that
√
( 2.2 − 1)v1 = 4v2
Thus, a stable age distribution is
or equivalently
√
10(1 + 2.2)
v1 =
v2 .
3
√
~v = h4, 2.2 − 1i.
Therefore, the stable age distribution is
N0 (t)
4
√
=
≈ 89.22%.
N0 (t) + N1 (t)
4 + 2.2 − 1
19