I'm not sure if this is possible at all - sorry if it isn't and I would delete my question if you told me so. But let's say there is a transition matrix of a Markov chain $$P=\begin{pmatrix} 0.3 & 0.3 & 0.4\\ 0.2 & 0.7 & 0.1\\ 0.2 & 0.3 & 0.5 \end{pmatrix}$$
And the question is something like this: "How many steps do you need to reach state $2$ from state $1$ with a probability of $0.42$?"
I know the solution is $n=2$ (because if you take $P^2$ then row $1$ , column $2$ of this matrix, the element is element $0.42$).
But is it possible to find this solution like that? I couldn't think of another formula than this one, but I don't know how to solve it for $n$ ? :c
$$0.42 = P^n$$
The characteristic polynomial of an $n \times n$ matrix gives you an $n$-step recurrence for the powers of the matrix. In this case the characteristic polynomial is $\lambda^3 - 1.5\; \lambda^2 + 0.54 \; \lambda - 0.04 $ so $$P^k - 1.5\; P^{k-1} + 0.54\; P^{k-2} - 0.04 \; P^{k-3} = 0$$ Solving the recurrence $a(k) - 1.5\; a(k-1) + 0.54 \; a(k-2) - 0.04\; a(k-3) = 0$ with initial condition $a(0) = 0$, $a(1) = P_{12} = 0.3$, $a(2) = P^2_{12} = 0.42$ gives you $$ a(k) = \frac{1 - (2/5)^k}{2}$$ and you can solve $a(k) = t$ for $k$ to find what power of $P$ has $P^k_{12}=t$.
However, in most cases there will be several terms with different constants to the power $k$, and also possibly powers of $k$, so in general it may not be possible to solve the equation for $a(k)$ in closed form.