Computing limiting transition probabilities with absorbing states A bank classifies loans as paid in full (F), in good standing (G), in arrears (A), or as a bad debt (B). Loans move between the categories according to the following transition probability:
$$B = \begin{pmatrix}1.0000 & 0 & 0 & 0 \\\ 0.1000 & 0.8000 & 0.1000 & 0 \\\ 0.1000 & 0.4000 & 0.4000 & 0.1000 \\\ 0 & 0 & 0 & 1.0000 \end{pmatrix} $$
What fraction of loans in good standing are eventually paid in full? What is the answer for those in arrears?
My Work
I am not sure how to go about this. I know how to compute transition probabilities for irreducible, aperiodic chains, but when I raise this transition matrix to a high power on MATLAB, I get 
$$ ans = \begin{pmatrix} 1.0000 & 0 & 0 & 0 \\\ 0.8750 & 0 & 0 & 0.1250 \\\ 0.7500 & 0 & 0 & 0.2500 \\\ 0 & 0 & 0 & 1.0000\end{pmatrix} .$$
So I can answer this question, I'm just not sure how to symbolically find these limiting probabilities. 
 A: To answer that question you need to know a bit of linear algebra, and in particular about matrix diagonalization.
You can decompose matrix $B$ as the following matrix product.
$$B = P \Lambda P^{-1}, $$
where the columns of P are eigenvectors and $\Lambda$ is a diagonal matrix, with eigenvalues on the diagonal. Note that
$$B^n = P \Lambda P^{-1} \ldots P \Lambda P^{-1} = P \Lambda^n P^{-1},$$
and that $\Lambda^n$ is very easy to compute, because it consists of the eigenvalues raised to the power $n$. In addition, all we need to know in order to compute the limit of $\Lambda^n$ is whether the values qre larger or smaller than 1. 
When all values in $B$ are positive, the Perron-Frobenius guarantees that 1 is the unique largest (in modulus) eigenvalue. In general you can prove that 1 is always an eigenvalue of $B$ and in your case it is an eigenvalue of order 2 because you have two absorbing states. The other two eigenvalues are smaller than 1, and quickly go to 0 when raised to the power $n$. So the answer is the result of
$$ P \begin{pmatrix}1 & 0 & 0 & 0 \\\ 0 & 1 & 0 & 0 \\\ 0 & 0 & 0 & 0 \\\ 0 & 0 & 0 & 0 \end{pmatrix} P^{-1}.$$
Here is how you can do it with R
B = matrix(c(1, .1, .1, 0, 0, .8, .4, 0, 0, .1, .4, 0, 0, 0, .1, 1), ncol=4)
eig = eigen(B)
# Show the eigen values.
eig[['values']]
# [1] 1.0000000 1.0000000 0.8828427 0.3171573
# Notice that 1 is there two times and the other two are
# less than 1.
P = eig[['vectors']]
# Show the matrix P.
P
# Show the answer.
P %*% diag(c(1,1,0,0)) %*% solve(P)

A: For a system this small, you can do everything by hand.
Let $x_i$ be the probability that you end up paid in full starting in state $i$. Use the transition probabilities to write one equation per state, saying that $x_i$ equals the weighted average of $x_j$, where the weight is the transition probability from $i$ to $j$ in one step. In other words, the chance to end up paid in full equals the expected value of the chance to be paid in full after the next step. For example, $x_G = 0.1 x_F + 0.8 x_G + 0.1 x_A$.
This gives you $4$ equations in $4$ unknowns, which should sound promising. However, $2$ of those equations are $x_F = x_F$ and $x_B = x_B$. You should replace those with $x_F = 1$ and $x_B = 0$.
