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Refer to Exercise $2 .$ Which food will the animal prefer after many trials?

Equilibrium vector: $\left( \begin{array} { l } { 1 / 3 } \\ { 1 / 3 } \\ { 1 / 3 } \end{array} \right)$

Calculus 3

Chapter 4

Vector Spaces

Section 9

Applications to Markov Chains

Vectors

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In this example, we have a three by three stochastic matrix, and it's a scraping different brands of dog food. So, for example, if we look at, say this entry here, it tells us that there is a 25% chance that a dog will transition from brand to to brand one, and so on would teach with each of them entries. So for this type of transition matrix or stochastic matrix, we'd like to determine which pet food will eventually be favored after many, many trials. To do this, we need to solve for the steady state vector Que, and the first step in that process is to solve the Matrix equation. P X equals X. This can be rearranged as p minus I times X equals zero vector. And that's where we'll start by using augmented matrix and roll, reducing first p minus. I is going to subtract one a lock, the main diagonal, which affects thes highlighted entries. So will have negative 10.5, then copy 0.25 and a zero, and this forms are first row. Then we'll have a 00.25 Another negative 0.5, then copy 0.25 and zero Finally, we have 0.25 points to five and another negative 0.5 a zero. And this is our matrix that's augmented from the Matrix equation. Now we can do some real operations as follows. Let's make the old Row three be the new row one, and also divide by 0.25 will obtain one one negative to zero. Let's divide row to buy 0.25 and keep it as a row. Two. We'll have one negative 210 And finally, let's make Row one our new Row three and divide by 2.25 again to obtain native to 11 and zero, then after row, reducing this into row reduced echelon form the Matrix becomes 10 negative. 10 01 negative 10 and the final row contains all zeros. So this tells us the solution for X will involve X one equals x three x two is going to be equal to x three and the third column has no pivot. So x three is a free variable. All right, X three equals x three. So with this set up, we have finally a solution X to the matrix equation if we pick X three to be one than X is 111 And from this we confined to the steady state Vector que He was going to be equal to the following first take the some of the entries in the Vector X and then we'll form 1/3 times X. And this produces the steady state vector containing 1/3 for each of its entries. But this tells us then that and the very, very long run each one of the brands of dog food are going to be equally preferred, since all the probabilities are exactly equal.

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