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Math Help - transition matrix

  1. #1
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    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
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  2. #2
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    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] <br />
\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)

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