Answer
See explanation
Work Step by Step
Suppose$A=\left[\begin{array}{lll}.4 & .2 & .3 \\ .3 & .6 & .3 \\ .3 & .2 & .4\end{array}\right]$. Then $A^{2}=\left[\begin{array}{lll}.31 & .26 & .30 \\ .39 & .48 & .39 \\ .30 & .26 & .31\end{array}\right]$
Instead of computing $A^{3}$ next, speed up the calculations by computing
\[
\begin{array}{l}
A^{4}=A^{2} A^{2}=\left[\begin{array}{ccc}
.2875 & .2834 & .2874 \\
.4251 & .4332 & .4251 \\
.2874 & .2834 & .2875
\end{array}\right] \\
A^{8}=A^{4} A^{4}=\left[\begin{array}{ccc}
.2857 & .2857 & .2857 \\
.4285 & .4286 & .4285 \\
.2857 & .2857 & .2857
\end{array}\right] \\
& \\
& \\
&
\end{array}
\]
To four decimal places, as k increases, $A^{k}=\left[\begin{array}{ccc}.2857 & .2857 . & .28577 \\ .4286 & .4286 & .4286 \\ .2857 & .2857 & .2857\end{array}\right]$
rational format $A^{k}=\left[\begin{array}{ccc}2 / 7 & 2 / 7 & 2 / 7 \\ 3 / 7 & 3 / 7 & 3 / 7 \\ 2 / 7 & 2 / 7 & 2 / 7\end{array}\right]$
\[
\begin{array}{l}
\text { If } B=\left[\begin{array}{rrr}
0 & .2 & .3 \\
.1 & .6 & .3 \\
.9 & .2 & .4
\end{array}\right] \\
B^{2}=\left[\begin{array}{rrr}
.29 & .18 & .18 \\
.33 & .44 & .33 \\
.38 & .38 & .49
\end{array}\right]
\end{array}
\]
$B^{4}=\left[\begin{array}{ccc}.2119 & .1998 & .1998 \\ .3663 & .3764 & .3663 \\ .4218 & .4218 & .4339\end{array}\right]$
$B^{8}=\left[\begin{array}{ccc}.2024 & .2022 & .2022 \\ .3707 & .3709 & .3707 \\ .4269 & .4269 & .4271\end{array}\right]$
Or in rational format, $B^{k}=\left[\begin{array}{ccc}18 / 89 & 18 / 89 & 18 / 89 \\ 33 / 89 & 33 / 89 & 33 / 89 \\ 38 / 89 & 38 / 89 & 38 / 89\end{array}\right]$
