MATH 1700 Homework
Han-Bom Moon
Homework 8 Solution
Section 4.1 ∼ 4.3.
• For every problem, explain your answer. It is not acceptable to write the answer
only.
1. Construct the linear competition model that satisfies:
(a) The growth rate of the first species is 1.5, and of the second is 1.25; each is
diminished by 0.4 times the population of the other.
From the conditions, r1 = 1.5, r2 = 1.25, s1 = s2 = 0.4. Therefore
Pn+1 = 1.5Pn − 0.4Qn
Qn+1 = −0.4Pn + 1.25Qn .
(b) In addition to the characteristics of (a), the first species undergoes immigration at 2500 per step, and the second migrates at 1200 per step.
Pn+1 = 1.5Pn − 0.4Qn + 2500
Qn+1 = −0.4Pn + 1.25Qn − 1200.
2. Find the unique fixed point of
xn+1 = 0.75xn + 0.5yn + 20
yn+1 = 4xn − yn + 40.
Suppose that (x, y) is a fixed point. Then
x = 0.75x + 0.5y + 20
y = 4x − y + 40,
or equivalently,
0.25x − 0.5y = 20
−4x + 2y = 40.
From x − 2y = 4(0.25x − 0.5y) = 4 · 20 = 80,
−3x = x − 2y + (−4x + 2y) = 80 + 40 = 120 ⇒ x = −40
x − 2y = 80 ⇒ −40 − 2y = 80 ⇒ y = −60
So the fixed point is (−40, −60).
1
MATH 1700 Homework
Han-Bom Moon
3. Explain why a unique fixed point does not exist for
xn+1 = 2xn + yn + 2
yn+1 = 3xn + 4yn + 9.
A fixed point is a solution (x, y) of the system of linear equations
x = 2x + y + 2
y = 3x + 4y + 9,
or equivalently,
−x − y = 2
−3x − 3y = 9.
Then 0 = −3(−x − y) + (−3x − 3y) = −3 · 2 + 9 = 3, which is impossible. So
there is no solution of this system of linear equations, and it implies that there is
no fixed point.
4. Find the fixed point of
xn+1 = 2.5xn
yn+1 = 0.8yn + 2000
and determine whether it is a sink, source or saddle.
If (x, y) is the fixed point, x = 2.5x, y = 0.8y + 2000. So x = 0, y = 10000 and the
fixed point is (0, 10000).
The general solution of xn+1 = 2.5xn is xn = 2.5n x0 , and the general solution of
yn+1 = 0.8yn + 2000 is
yn = 0.8n y0 + 2000 ·
0.8n − 1
= 0.8n y0 + 10000(1 − 0.8n ).
0.8 − 1
For the initial condition (1, 10000), (xn , yn ) = (2.5n , 0.8n ·10000+10000(1−0.8n )) =
(2.5n , 10000) and it diverges from the fixed point as n → ∞. On the other hand,
for another initial condition (0, 0), (xn , yn ) = (0, 10000(1 − 0.8n )) and as n → ∞,
it approaches (0, 10000). Therefore the fixed point is a sink.
"
#
" #
−6
5
5. For u =
and v =
, compute:
0
3
(a) 3u;
"
3u = 3
−6
0
#
"
=
2
3 · (−6)
3·0
#
"
=
−18
0
#
MATH 1700 Homework
Han-Bom Moon
(b) u + v;
"
u+v =
#
−6
0
"
+
(c) − 12 v;
1
1
− v=−
2
2
"
#
5
3
5
3
#
"
5
3
"
=
"
#
−6 + 5
0+3
=
− 21 · 5
− 12 · 3
#
"
"
−1
3
=
#
"
=
#
#
− 52
− 32
(d) 2u − v;
"
−6
0
2u − v = 2
"
6. For A =
#
−
#
2.75 1.5
2.5 −1
"
and B =
=
−4 −3
1.5 2
#
2 · (−6) − 5
2·0−3
"
=
−17
−3
#
#
, compute:
(a) 5A;
"
2.75 1.5
2.5 −1
5A = 5
#
"
5 · 2.75 5 · 1.5
5 · 2.5 5 · (−1)
=
#
"
=
13.75 7.5
12.5 −5
#
(b) A + B;
"
−4 −3
1.5 2
#
2.75 + (−4) 1.5 + (−3)
2.5 + 1.5
−1 + 2
#
2.75 1.5
2.5 −1
A+B =
"
=
#
"
+
"
=
−1.25 −1.5
4
1
#
(c) 4A − 2B.
"
4A − 2B = 4
"
7. For A =
2 0
5 1
#
#
"
−2
−4 −3
1.5 2
#
4 · 2.75 − 2 · (−4) 4 · 1.5 − 2 · (−3)
4 · 2.5 − 2 · 1.5
4 · (−1) − 2 · 2
=
"
2.75 1.5
2.5 −1
"
,u=
3
7
#
"
, and v =
0
−2
#"
"
#
"
=
#
, compute:
(a) Au;
"
Au =
2 0
5 1
3
7
3
#
=
2·3+0·7
5·3+1·7
#
"
=
6
22
#
19 12
7 −8
#
MATH 1700 Homework
Han-Bom Moon
(b) Av.
"
# "
# "
#
0
2 · 0 + 0 · (−2)
0
Av =
=
=
−2
5 · 0 + 1 · (−2)
−2
"
#
"
#
2.75 1.5
−4 −3
8. For A =
and B =
, compute:
2.5 −1
1.5 2
2 0
5 1
#"
(a) AB;
"
AB =
2.75 1.5
2.5 −1
#"
−4 −3
1.5 2
#
"
2.75 · (−4) + 1.5 · 1.5 2.75 · (−3) + 1.5 · 2
=
2.5 · (−4) + (−1) · 1.5 2.5 · (−3) + (−1) · 2
"
#
−8.75 −5.25
=
−11.5 −9.5
#
(b) BA;
"
BA =
−4 −3
1.5 2
#"
2.75 1.5
2.5 −1
#
"
−4 · 2.75 + (−3) · 2.5 −4 · 1.5 + (−3) · (−1)
=
1.5 · 2.75 + 2 · 2.5
1.5 · 1.5 + 2 · (−1)
"
#
−18.5 −3
=
9.125 0.25
#
(c) A2 ;
"
A2 =
2.75 1.5
2.5 −1
#"
2.75 1.5
2.5 −1
#
"
2.752 + 1.5 · 2.5
2.75 · 1.5 + 1.5 · (−1)
=
2.5 · 2.75 + (−1) · 2.5
2.5 · 1.5 + (−1)2
"
#
11.3125 2.625
=
4.375
4.75
#
(d) A3 .
"
A3 = A2 · A =
11.3125 2.625
4.375
4.75
"
#"
2.75 1.5
2.5 −1
#
11.3125 · 2.75 + 2.625 · 2.5 11.3125 · 1.5 + 2.625 · (−1)
=
4.375 · 2.75 + 4.75 · 2.5
4.375 · 1.5 + 4.75 · (−1)
"
#
37.671875 14.34375
=
23.90625
1.8125
4
#
MATH 1700 Homework
Han-Bom Moon
9. Write the system
3x − 6y = 0
5x − 9y = 1
in vector/matrix form and then use the inverse to solve.
"
#" # " #
3 −6
x
0
=
5 −9
y
1
"
x
y
#
"
=
3 −6
5 −9
#−1 "
#
1
=
3 · (−9) − (−6) · 5
"
# "
#
1
6
2
=
=
3 −3
−1
0
1
So x = 2, y = −1.
5
"
−9 6
−5 −3
#"
0
1
#