Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 151
This Octave session produced the output given here.
octave:1> v0=[0;0;0;1;0;0]
v0 =
0
0
0
1
0
0
octave:2> p=.5
p = 0.50000
octave:3> v1=coin(p,v0)
v1 =
0.00000
0.00000
0.50000
0.00000
0.50000
0.00000
octave:4> v2=coin(p,v1)
v2 =
0.00000
0.25000
0.00000
0.50000
0.00000
0.25000
This continued for too many steps to list here.
octave:26> v24=coin(p,v23)
v24 =
0.39600
0.00276
0.00000
0.00447
0.00000
0.59676
(b) Using these formulas
p 1 (n + 1) = 0.5 · p 2 (n)
p 2 (n + 1) = 0.5 · p 1 (n) + 0.5 · p 3 (n)
p 3 (n + 1) = 0.5 · p 2 (n) + 0.5 · p 4 (n) p 5 (n + 1) = 0.5 · p 4 (n)
and these initial conditions
⎛ ⎞ ⎛ ⎞
p 0 (0) 0
p 1 (0)
0
p 2 (0)
⎜p 3 (0)
=
0
⎟ ⎜1
⎟
⎝p 4 (0) ⎠ ⎝0⎠
p 5 (0) 0
we will prove by induction that when n is odd then p 1 (n) = p 3 (n) = 0 and when n is even then
p 2 (n) = p 4 (n) = 0. Note first that this is true in the n = 0 base case by the initial conditions. For
the inductive step, suppose that it is true in the n = 0, n = 1, . . . , n = k cases and consider the
n = k + 1 case. If k + 1 is odd then the two
p 1 (k + 1) = 0.5 · p 2 (k) = 0.5 · 0 = 0
p 3 (k + 1) = 0.5 · p 2 (k) + 0.5 · p 4 (k) = 0.5 · 0 + 0.5 · 0 = 0
follow from the inductive hypothesis that p 2 (k) = p 4 (k) = 0 since k is even. The case where k + 1
is even is similar.
(c) We can use, say, n = 100. This Octave session
octave:1> B=[1,.5,0,0,0,0;
> 0,0,.5,0,0,0;
> 0,.5,0,.5,0,0;
> 0,0,.5,0,.5,0;