The fact that we can solve for p_11^{(n)} in the stated form follows from the fact that p_{11}^{(n)} satisfies recurrence relations given by

q(P)_{11} = 0

where q(x)=det(xI−P) is the characteristic polynomial. By the Caley-Hamilton theorem we know q(P)=0 (see

**IB Linear Algebra**)). You know how to solve such recurrence relations (from**IA Differential Equations**). I have already said something about this is my earlier blog entry for Lecture 2. In particular, although some matrices cannot be diagonalized, every square matrix P can be written in Jordan normal form, P=UJU^{−1} (see**IB Linear Algebra**), and P^n=UJ^nU^{−1} is then of the form that takes account of repeated eigenvalues.
I believe that a good mathematician these days should not only understand theory and proofs, but also be able to use modern computing tools to quickly work out examples.

I highly recommend that you install Mathematica and use it while you study and work on examples sheets (in many courses). I personally use it on almost a daily basis. You can download a free copy:

http://www.damtp.cam.ac.uk/internal/computing/software/mathematica/

http://www.damtp.cam.ac.uk/internal/computing/software/mathematica/

It is very easy install and learn to use. The time you spend learning to use it (and similarly MATLAB) is a very good investment.

Below is a short Mathematica program that does Example Sheet 1, #7. I think that a well-motivated student might like do this question by hand, and then subsequently check the answer using Mathematica. By the way, I would not expect an examiner to set a question in tripos that is as difficult to do by hand as doing all of (a) (b) and (c) in #7. However, it would be fair to ask the answer for just one value of p.

`Clear[P,p,p11]`

`P={{0,1,0},{0,2/3,1/3},{p,1−p,0}};`

`(* Solution to (a) *)`

mu=Eigenvalues[P /. p→1/16]

p11[n_]=a mu[[1]]^n+ b mu[[2]]^n+ c mu[[3]]^n;

Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]

p11[n] /. %[[1]] //Expand

mu=Eigenvalues[P /. p→1/16]

p11[n_]=a mu[[1]]^n+ b mu[[2]]^n+ c mu[[3]]^n;

Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]

p11[n] /. %[[1]] //Expand

`Out[1]= {1,-(1/4),-(1/12)}`

Out[2]= {{a->1/65,b->-(2/5),c->18/13}}

Out[3]= 1/65-1/5 (-1)^n 2^(1-2 n)+1/13 (-1)^n 2^(1-2 n) 3^(2-n)

Out[2]= {{a->1/65,b->-(2/5),c->18/13}}

Out[3]= 1/65-1/5 (-1)^n 2^(1-2 n)+1/13 (-1)^n 2^(1-2 n) 3^(2-n)

`(* Solution to (b) *)`

mu=Eigenvalues[P /. p->1/6]

p11[n_]=a mu[[1]]^n+ b mu[[2]]^n+ c mu[[3]]^n;

Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]

p11[n] /. %[[1]] //ComplexExpand

mu=Eigenvalues[P /. p->1/6]

p11[n_]=a mu[[1]]^n+ b mu[[2]]^n+ c mu[[3]]^n;

Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]

p11[n] /. %[[1]] //ComplexExpand

`Out[4]= {1,-(1/6)+I/6,-(1/6)-I/6}`

Out[5]= {{a->1/25,b->12/25-(9 I)/25,c->12/25+(9 I)/25}}

Out[6]= 1/25+1/25 2^(3-n/2) 3^(1-n) Cos[(3 n \[Pi])/4]+1/25 2^(1-n/2) 3^(2-n) Sin[(3 n \[Pi])/4]

Out[5]= {{a->1/25,b->12/25-(9 I)/25,c->12/25+(9 I)/25}}

Out[6]= 1/25+1/25 2^(3-n/2) 3^(1-n) Cos[(3 n \[Pi])/4]+1/25 2^(1-n/2) 3^(2-n) Sin[(3 n \[Pi])/4]

`(* Solution to (c) *)`

mu=Eigenvalues[P /. p->1/12]

p11[n_]=a +(b+c n)mu[[2]]^n;

Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]

p11[n] /. %[[1]]

Out[7]= {1,-(1/6),-(1/6)}

Out[8]= {{a->1/49,b->48/49,c->-(6/7)}}

Out[9]= 1/49+(-(1/6))^n (48/49-(6 n)/7)

mu=Eigenvalues[P /. p->1/12]

p11[n_]=a +(b+c n)mu[[2]]^n;

Solve[{p11[0]==1,p11[1]==0,p11[2]==0},{a,b,c}]

p11[n] /. %[[1]]

Out[7]= {1,-(1/6),-(1/6)}

Out[8]= {{a->1/49,b->48/49,c->-(6/7)}}

Out[9]= 1/49+(-(1/6))^n (48/49-(6 n)/7)