Triangular Markov chain question A triathlon consists of $3$ disciplines: swimming, cycling and running. A  triathlete does a training session every day.
However he doesn’t want to pay for professional coaching advice so instead his
strategy for training is to choose the next day’s training discipline uniformly
likely from the two disciplines that he didn’t do on the current day.
1)Supposing that on day zero the training session undertaken is swimming;
what is the probability that the first day’s session is swimming?
2)Supposing that on day zero the training session undertaken is swimming;
what is the probability that the $n$th day’s session is swimming?
So I constructed a stochastic matrix $P$ in the following way:
$$P=\begin{pmatrix}
0 & 1\over 2 & 1\over 2 \\
1\over 2 & 0 & 1\over 2 \\
1\over 2 & 1\over 2 & 0 \\
\end{pmatrix}$$ since going from swimming or any other sport to a different sport has probability $1\over 2$.
So the answer to the first question is $0$ since if we start at swimming then the next day is supposed to be a different sport day.
and then for the second question I tried finding the $p^n_{s_0s_n}$ -th value- where $p_{i_0,j_n}$ means probability of starting at day $i$ sat day $0$ and finishing at day $j$ on day $n$ , and $p^n$ is the value of the specific entry on with the matrix to the $n$-th power. Also $s$ is denoted as swimming day.
But in pursuit of finding this value I got stuck. Any help would be appreciated
 A: Because this kind of matrix appears frequently in Markov chain theory, let's generalize a little.  The matrix $\mathbb P$ is an instance of the family
$$\mathbb{P}_d = \pmatrix{0 & \frac{1}{d-1} & \frac{1}{d-1} & \cdots & \frac{1}{d-1} \\
                          \frac{1}{d-1} & 0 & \frac{1}{d-1} & \cdots & \frac{1}{d-1} \\
                          \vdots & \frac{1}{d-1} & \ddots & \cdots & \frac{1}{d-1} \\
                          \frac{1}{d-1} & \frac{1}{d-1} & \cdots & \frac{1}{d-1} & 0}
$$
for $d = 2, 3, \ldots.$  It can be expressed as a "perturbation" of the $d$-dimensional identity matrix $\mathbb{I}_d$ via
$$\mathbb{P}_d = \frac{1}{d-1} \left(\pmatrix{1 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \ddots & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1} - \pmatrix{1 & 0 &  \cdots & 0 \\ 0 & 1  & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 \\0   & \cdots & 0 & 1}\right) = \frac{1}{d-1} \left(\mathbf{1\, 1^\prime} - \mathbb{I}_d\right)$$
where $\mathbf{1} = (1, 1, \ldots, 1)^\prime$ is the column $d$-vector of ones.
This is a huge simplification because the powers of the rank-1 matrix $\mathbf{1\, 1^\prime}$ are simply found by successive multiplications by $d:$
$$\left(\mathbf{1\, 1^\prime}\right)^2 = \left(\mathbf{1\, 1^\prime}\right)\left(\mathbf{1\, 1^\prime}\right) = \mathbf{1}\, \left(\mathbf{1^\prime 1} \right)\, \mathbf{ 1^\prime} = \mathbf{1}\, d\, \mathbf{1^\prime} = d\left(\mathbf{1\,1^\prime}\right),$$
whence (recursively) for $n \ge 1$
$$\left(\mathbf{1\, 1^\prime}\right)^n = d^{-1}\,d^{n}\mathbf{1\, 1^\prime}.$$
Because $\mathbb{I}_d$ commutes with $\mathbf{1\,1^\prime},$ the Binomial Theorem applies to give
$$(d-1)^n\mathbb{P}_d ^ n = \sum_{k=0}^n \binom{n}{k} (-\mathbb{I}_d)^{n-k} \left(\mathbf{1\,1^\prime}\right)^k = (-1)^n\mathbb{I}_d + d^{-1}\sum_{k=1}^n \binom{n}{k}(-1)^{n-k} d^{k}\,\left(\mathbf{1\,1^\prime}\right) $$
If we were to include a $k=0$ term, equal to $(-1)^k,$ in the last sum, it would be the Binomial expansion of $d^{-1}$ times the value of $(-1+d)^n,$ whence

$$\mathbb{P}_d^n = \frac{1}{(d-1)^n} \left((-1)^n \mathbb{I}_d + \frac{(d-1)^n - (-1)^n}{d}\mathbf{1\,1^\prime}\right).$$

This formula actually works for $n=0,$ too.
The question concerns the case $d=3.$
Let's translate this answer into a formula for the components $\left(\mathbb{P}_d^n\right)_{ij}.$  It says first to compute $\frac{(d-1)^n - (-1)^n}{d\,(d-1)^n}.$ When $i\ne j,$ this is the value to use.  When $i=j,$ add $\frac{(-1)^n}{(d-1)^n}$ to it.

The same approach works for finding powers of any matrix proportional to $\mathbb{I}_d + \mathbf{u\, v^\prime}$ where  $\mathbf u$ and $\mathbf v$ are any column vectors with $d$ components: recursively find the powers
$$\left(\mathbf{u\, v^\prime}\right)^n = \left(\mathbf{v^\prime\, u}\right)^{n-1}\left(\mathbf{u\, v^\prime}\right)$$
($\left(\mathbf{v^\prime\, u}\right)$ is just a number) and apply the Binomial Theorem to obtain

$$ \left(\mathbb{I}_d + \mathbf{u\, v^\prime}\right)^n = \mathbb{I}_d + \frac{\left(\mathbf{v^\prime\, u} + 1\right)^n - 1}{\mathbf{v^\prime\, u}}\, \mathbf{u\, v^\prime};\quad n=0,1,2,\ldots$$

Letting $d=3,$ $\mathbf{u} = \mathbf{1},$ and $\mathbf{v} = -\mathbf{1}$  gives a formula for powers of twice the matrix in the question. 
