Math 2270, Fall 2015
Instructor: Thomas Goller
3 November 2015
Quiz #6 Solutions
(1) A dragon hunts down and eats either a horse or a linear algebra student each day. If the
dragon eats a horse on a given day, it is 70% likely to eat a horse on the next day and
30% likely to eat a student. If the dragon eats a student on a given day, it is 90% likely
to eat another student on the next day and only 10% likely to eat a horse instead.
(a) Write down a stochastic matrix P that models the situation. Label the columns and
rows. (1 point)
Solution: P =

.7 .1
(label the columns H and S and the rows H and S).
.3 .9
(b) Given that the dragon is 80% likely to eat a student today and only 20% likely to
eat a horse, how likely is the dragon to eat a student tomorrow? (1 point)
Solution:

.7 .1
.3 .9


.22
.2
=
, so 78%.
.8
.78
(c) Compute the steady-state vector ~q by computing Nul(P
Solution:
P
I=

.3
.3
.1
!
.1


.3 .1
1
!
0 0
0
I). (2 points)
1
3
0
,
1
x
3 2
1
so the parametric vector form of a solution to (P I)~x = ~0 is ~x =
= x2 3 .
x
1
2
⇢ 1
3
Thus Nul(P I) = Span
. The probability vector in this span is
1




1 13
3 13
1 1
.25
~q =
=
=
=
.
.75
4/3 1
4 1
4 3
(d) Check that P ~q = ~q . (1 point)
Solution: P ~q =
1
4

.7 .1
.3 .9

1
=
3
1
4

.7 + .3
= ~q
.3 + 2.7
(e) Given that the dragon has eaten a student on November 3, 2015, how likely is the
dragon to continue the tradition and eat a student on November 3, 2016? (1 point)
Solution: Exact answer: the second entry of P
entry of ~q , namely 75%.
365

0
. Rough answer: the second
1
Math 2270, Fall 2015
Instructor: Thomas Goller
3 November 2015
(2) In P2 , write down the change-of-coordinates matrix P from the basis B = {1 + t, 2t
t2 , 3 + 4t
E
B
5t2 } to the standard basis E = {1, t, t2 }. (1 point)
2
1
4
Solution: P = 1
E B
0
3
3
45
5
0
2
1
(3) Bonus problem: Consider the stochastic matrix P =

0 1
. (1 bonus point)
1 0
(a) Find a formula for P k .
8"
>
1
>
>
>
>
< 0
Solution: P k = "
>
>
0
>
>
>
: 1
0
1
1
0
#
if k is even
#
if k is odd
(b) Explain why no power of P has all positive entries. Thus P is not regular.
Solution: The formula in (a) shows that no power of P has all positive entries.
(c) Find the steady-state vector ~q for P . Check that P ~q = ~q .
Solution:
P
so Nul(P
I) = Span
I=
⇢
1
1

1
1
1
!
1


1 1
1
!
0 0
0
1
,
0
and the probability vector in this span is ~q =
1
2

1
.
1
(d) Invent an initial state vector ~x0 such that the Markov chain ~xk = P k ~x0 does not
converge to ~q as k ! 1. Explain why convergence fails.
Possible solution: ~x0 =

.2
. (Any probability vector ~x0 will work except ~q .)
.8