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I have been learning markov chains for a while now and understand how to produce the steady state given a 2x2 matrix.

For example given the matrix,

[.5 .5] 
[.8 .2]

To find the steady state it will be

[x,y] * [.5 .5] = [x,y]
        [.8 .2]

.5x + .8y = x
.5(1-y) + .8y = 1-y
........

Knowing that x + y = 1, I can do substitution and elimination to get the values of x and y.

However for a 3x3 matrix, I am confused how I could compute the steady state. I will like to have an example with steps given this sample matrix :

 [.5 .4 .1] 
 [.6 .2 .2]
 [.5 .3 .2] 
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1 Answer 1

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Solve the 3x3 linear system

          [.5 .4 .1]
[x,y,z] * [.6 .2 .2] = [x,y,z]   with  x+y+z = 1
          [.5 .3 .2]
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  • $\begingroup$ The same way than for a 2x2 system: rewrite the first equation as x=ay+bz for some (a,b) and plug this into the second equation. This yields y=cz for some c. Use x=ay+bz again to deduce that x=(ac+b)z. Use the normalization x+y+z=1 to deduce that dz=1 with d=(a+1)c+b+1, hence z=1/d. Deduce that y=c/d and that x=(ac+b)/d. Done. $\endgroup$
    – Did
    Dec 8, 2014 at 13:59

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