1. ## transition matrix

steady state vector [7/13 6/13] [.4 .6]
[m n]

Whenever I do it I get .4 and .6 for n what am I doing wrong

2. Hello, nafiro!

What method are you using?

Steady state vector: . $\begin{pmatrix} \frac{7}{13}& \frac{6}{13} \\ \\[-4mm] \frac{2}{5} & \frac{3}{5} \end{pmatrix}$

We want the vector $(m,n)$, where $m+n \:=\:1$ .[1]

. . and: . $(m,n)\begin{pmatrix}\frac{7}{13} & \frac{6}{13} \\ \\[-3mm]\frac{2}{5} & \frac{3}{5}\end{pmatrix} \;=\;(m,n)$

Multiplying, we get: . $\begin{array}{cccccccc}\frac{7}{13}m + \frac{2}{5}n \:=\:m & \Rightarrow & 30m - 26n \:=\:0 \\ \\[-3mm]
\frac{6}{13}m + \frac{3}{5}n \:=\:n & \Rightarrow & 30m - 26n \:=\:0 \end{array}$
. $\Rightarrow\quad 15m - 13n \:=\:0$ .[2]

Solve the system: . $\begin{array}{cccc}m + n &=& 1 & [1] \\ 15m - 13n &=& 0 & [2] \end{array}$

$\begin{array}{cccccc}\text{Multiply [1] by 13:} & 13m + 13n &=& 13 \\ \text{Add [2]:} & 15m - 13n &=& 0 \end{array}$

And we have: . $28m \:=\:13 \quad\Rightarrow\quad m \:=\:\tfrac{13}{28} \quad\Rightarrow\quad n \:=\:\tfrac{15}{28}$

Therefore: . $(m,n) \;=\;\left(\frac{13}{28},\:\frac{15}{28}\right)$