91Ó°ÊÓ

Referring to the exercise2,

The Matrix G and its inverse is,

\(G = \left( {\begin{array}{*{20}{c}}{ - 0.3}&1\\{0.6}&1\end{array}} \right)\)

\({G^{ - 1}} = \frac{{AdjG}}{{\left| G \right|}}\)

So,

\(AdjG = \left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 0.6}&{ - 0.3}\end{array}} \right)\)

And,

\(\begin{aligned}{}\left| G \right| &= - 0.3 - 0.6\\ &= - 0.9\\ &= - \frac{9}{{10}}\end{aligned}\)

So,

\({G^{ - 1}} = - \frac{{10}}{9}\left( {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 0.6}&{ - 0.3}\end{array}} \right)\)

The bottom row of \({G^{ - 1}}\) is \(\left( {\frac{2}{3},\frac{1}{3}} \right)\)

Therefore, the bottom row of \({G^{ - 1}}\) is \(\left( {\frac{2}{3},\frac{1}{3}} \right)\), is the unique stationary distribution for the Markov chain.