TA: Geoff Williams
Tutorials: Thursday 1:30-2:20 in TSH-120
E-mail:williagg@math.mcmaster.ca
Office hours: Wednesday 1:30-3:30 at the Math Help Centre
Tutorial # 8
Some test review.
This is a very similar question to sample test Sample test # 5 (M1D03
Test #2) question #6.
We are looking at a population of fish, squid and seals. A state vector
for their pop. is:


ft
S t =  qt 
st
where ft is the fish pop. at time t, qt is the squid pop. at time t and st is the
seal pop. at time t. Fish reproduce but are eaten by squid. Squid are eaten
by seals. It is determined that the pop. obeys a linear dynamical system
where St+1 = ASt and


17 −128
0
A =  0.5 0.5 −3.5 
0
0.5
0.5
T
(a). X1 = 448 21 1
is an eigenvector of A. Find the eigenvalue λ1
associated with it.
AX1 = λ1 X1



17 −128
0
448
AX1 =  0.5 0.5 −3.5   21 
0
0.5
0.5
1


4928
=  231 
11
1

448
= 11  21 
1

So λ1 = 11.
We are also given that λ2 = 6, with X2 =
T
X3 = 8 1 1

448 128

21 11
P =
1
1
448 21 1
T
and λ3 = 1, with

8
1 
1
(b). We are given that the det of P is 2000, find P −1 .
1 adj.P
det P
P −1 =


10
−20
10
cij (A) =  −120 440 −320 
40 −280 2240


1 −12 4
1 
−2 44 −28 
P −1 =
200
1 −32 224
T
(c). The initial pop. is S0 = 1288 76 6 . Write S0 as a linear
combination of our eigenvectors.
Want S0 = b1 X1 + b2 X2 + b3 X3
b1 b2

1
1 
−1
−2
P =
200
1

T
= P −1 S0


−12 4
1288
44 −28   76 
−32 224
6
 
2

= 3 
1

b3
1288

76  = 2X1 + 3X2 + X3
So
6
2
Question Find a real irreducible quadratic equation with z = 2 + 5i as
a root.
Answer The other root will be z̄.
x2 + ax + b
−a = z + z̄ = 4
p
b = |z| = 22 + 52 = 29
x2 − 4x + 29
Question Show that u = 1 + i is a root of x2 − 3ix + (−3 + i) = 0 and
find the other root.
Answer let x = 1 + i.
(1 + i)2 − 3i(1 + i) + (−3 + i)
2i − 3i + 3 − 3 + i = 0 X
u + v = 3i
1 + i + v = 3i
v = −1 + 2i
so the other root is v = −1 + 2i.
Question Find all solutions to z 3 = 27
Answer Obviously 3 works; but we know there are two other solutions.
27 in polar form is 27ei(0+2πk)
z 3 = 27ei(0+2πk)
z = 3e
z0 = 3e0 = 3
√
2π
3
3 3
2π
= 3(cos
+ i sin ) = − + i
3
3
2
2
√
4π
4π
3
3 3
= 3(cos
+ i sin ) = − − i
3
3
2
2
k=0→
k=1→
z1 = 3e
k=2→
z2 = 3e
2πi
3
4πi
3
2πki
3
3