Absorbing Markov Chains: An efficient algorithmic approach

Question

Following this procedure I have successfully written a program to calculate the probability of ending in a given absorbing state given the initial state. The procedure is as follows:

1. Given the transition matrix (P), row and column swap until the identity matrix is in the bottom right corner.

$$ P = \begin{bmatrix} Q & R \\ 0 & I \end{bmatrix} $$

2. Calculate the matrix N, and multiply by R to give the final probability matrix (B).

$$N = (I-Q)^{-1}$$ $$B = N*R$$

Where the values in B represent the probability of moving from an initial non-absorbing state (rows) to a final absorbing state.

My question is if there is a more efficient method for solving this problem if I know what my initial state is. It seems wasteful to calculate the entire B matrix, given that I will only ever use one row in it.

I am writing a program to do this, so the matrix inversion step is particularly inefficient. Can I avoid this altogether?