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Modelling Inbreeding With Matrices

In the brother system problem, a male and a female mate, and from their direct descendants, two individuals of opposite sex are selected at random. Direct offspring of these two are mated and the process continues.
The possible genotype for the parents 1st generation are AA, Aa and aa. The possible pairings are 1 - AA x AA, 2 - AA x Aa, 3 - Aa x Aa, 4 - Aa x aa, 5 - aa x aa, AA x aa. We can construct the table for the probabilities of the genotypes of Generation 2.
Generation 1\Generation 2 AA x AA AA x Aa Aa x Aa Aa x aa aa x aa AA x aa
AA x AA 1 0 0 0 0 0
AA x Aa 1/4 1/2 1/4 0 0 0
Aa x Aa 1/16 1/4 1/4 1/4 1/6 1/8
Aa x aa 0 0 1/4 1/2 1/4 0
aa x aa 0 0 0 0 1 0
AA x aa 0 0 1 0 0 0
The pairings AA x AA and aa x aa determine that all offspring will be AA and aa respectively. To find the expected number of generations needed reach any of these two pairings find the fundamental matrix. Rewrite the table as
Generation 1\Generation 2 AA x AA aa x aa AA x Aa Aa x Aa Aa x aa AA x aa
AA x AA 1 0 0 0 0 0
aa x aa 0 1 0 0 0 0
AA x Aa 1/4 0 1/2 1/4 0 0
Aa x Aa 1/16 1/16 1/4 1/4 1/4 1/8
Aa x aa 0 1/4 0 1/4 1/2 0
AA x aa 0 0 0 1 0 0
This is of the form  
\[\left( \begin{array}{cc} I & O \\ S & Q \end{array} \right)\]
  where
\[I=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right), \; O= \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right), \; S= \left( \begin{array}{cc} 1/4 & 0 \\ 1/16 & 1/16 \\ 0 & 1/4 \\ 0 & 0 \end{array} \right), \; Q= \left( \begin{array}{cccc} 1/2 & 1/4 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 1/8 \\ 0 & 1/4 & 1/2 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)\]

Pick out the 4 x 4 matrix in the lower right.
\[Q= \left( \begin{array}{cccc} 1/2 & 1/4 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 1/8 \\ 0 & 1/4 & 1/2 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right)\]

The fundamental matrix is  
\[F=(I-Q)^{-1}= \left( \begin{array}{cccc} 8/3 & 4/3 & 2/3 & 1/6 \\ 4/3 & 8/3 & 4/3 & 1/3 \\ 2/3 & 4/3 & 8/3 & 1/6 \\ 4/3 & 8/3 & 4/3 & 4/3 \end{array} \right)\]
.
\[(I-Q)^{-1}S= \left( \begin{array}{cc} 1/4 & 1/4 \\ 1/2 & 1/2 \\ 1/4 & 1/4 \\ 1/2 & 1/2 \end{array} \right) \]
.
These columns correspond to AA x AA and aa x aa, so after repeated inbreeding the genotype is either AA or aa.