A grasshopper jumps about at random between the corners of a triangle. If he is at one corner, he is equally likely to jump to either of the other two corners (but he never jumps straight up in the air and lands on the same corner where he was before). For each positive integer $t$, find the probability that, after $t$ jumps, the grasshopper is back on the corner it started from.
The state space $S$, is the corners of the triangle, label them $A,B,C$, with the Markov chain having transition matrix $$\begin{pmatrix} 0&\frac12&\frac12\\ \frac12&0&\frac12\\ \frac12&\frac12&0 \end{pmatrix}.$$ Without loss of generality, suppose the grasshopper starts at corner $A$. Clearly, when $t=1$, we have that it is impossible for the grasshopper to return back to corner $A$, so the probability of which is zero.
Now suppose $t=2$. Well we can either take the path $(BA)$ or $(CA)$, so the probability here is $$2\Big(\frac{1}{2}\cdot\frac{1}{2}\Big)=\frac{1}{2}.$$ Now suppose $t=3$. We can take paths $(BCA)$ or $(CBA)$, so the probability here is $$2\Big(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\Big)=\frac{1}{4}.$$
This is the "standard" matrix approach to finding the probability. The transition matrix diagonalises as $$A=\begin{bmatrix} 0&\frac12&\frac12\\ \frac12&0&\frac12\\ \frac12&\frac12&0 \end{bmatrix}=PDP^{-1}$$ $$=\begin{bmatrix} -1&-1&1\\ 0&1&1\\ 1&0&1 \end{bmatrix} \begin{bmatrix} -\frac12&0&0\\ 0&-\frac12&0\\ 0&0&1 \end{bmatrix} \begin{bmatrix} -\frac13&-\frac13&\frac23\\ -\frac13&\frac23&-\frac13\\ \frac13&\frac13&\frac13 \end{bmatrix}$$ The required answer is the first entry of $$A^t\begin{bmatrix}1\\0\\0\end{bmatrix}=PD^tP^{-1}\begin{bmatrix}1\\0\\0\end{bmatrix}$$ which turns out to be $$\frac13+\frac23\left(-\frac12\right)^t$$