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In Exercises $5-8,$ find the steady-state vector.$$\left[\begin{array}{ccc}{.7} & {.1} & {.1} \\ {.2} & {.8} & {.2} \\ {.1} & {.1} & {.7}\end{array}\right]$$

$\underline { x } = \left[ \begin{array} { c } { \frac { 1 } { 4 } } \\ { \frac { 1 } { 2 } } \\ { \frac { 1 } { 4 } } \end{array} \right]$

Calculus 3

Chapter 4

Vector Spaces

Section 9

Applications to Markov Chains

Vectors

Oregon State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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in this example, we have a three by three probability Matrix P that's provided and Arkle here is to find the steady state vector. The first step in this process is going to be to solve the equation that p of X is equal to X. We could rearrange this equation. So vets of the form P minus the identity matrix Times X equals the zero vector. Now, from this particular matrix equation, we can augment as follows First P minus. I is going to cause a subtraction by one just along the main diagonal. So I'm going to be copying the entries otherwise So for the first entry, take 10.7 minus one for a negative 10.3, then copy 0.1 and 0.1 for row two will copy point to then take the entry 0.8 minus one so that we have a negative point to here. Now back to copying points to 0.1 point one. Then take 0.7 minus one so that we have another negative 10.3 that will go here. So this is P minus I and what we would like to do is augment this with zero vector. So let me place this zero vector here and now Finally closed this matrix to reduce this matrix weaken taken initial step where we do the following I'm going to multiply row three by 10 So that will have integers and make it a row one So now we have a 11 negative 30 in row one, I'll make the old row one here Row three and also multiplied by 10. So we have a negative 100.3 110 Then multiply row to buy 10 So that will have to negative 22 and zero. Now with this matrix weaken, begin road reduction as usual and the arithmetic Matic operations won't be as bad Now that we have integers will find that this is row equivalent to 10 negative one zero in row one row to is 01 negative to zero and the third row contains only zeros. This that implies that the solution for the vector X indicated here will have first next one equals x three, which came from the first row From the second row we find that X two equals two x three and x three is the free variable soldiers right X three equals X three. Now we have infinitely many solutions, but we just need one. So it's set X three equal toe one. Then the solution X is going to be equal to 1 to 1. We have to be careful, though, since this is not the steady state vector, since it's not even a probability vector. But it's very close to being on the right track. If we take the some of the entries in this vector x one plus two plus one, we have all together four. And now we can claim that our answer is Cue the steady state vector for in spite, dividing this vector X by four So we'll have 1/4 2 4th or 1/2 if we reduce and then 1/4. So this vector Q is the steady state vector for the Matrix P.

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