# Please explain how to multiply this matrices: Find X2 (the probability distribution of the system after two observations) for the distribution vector X0 and the transition matrix T. X0 = 0.35 0.20 0.45 , T = 0.1 0.1 0.2 0.8 0.7 0.4 0.1 0.2 0.4

Quick answer:

The probability distribution after two iterations using the given distribution and transition matrix is [.1255, .638, .2365]

## Expert Answers

We are asked to find the probability distribution of the given vector `X_0=([.35,.2,.45])` and the transition matrix `T=([.1,.8,.1],[.1,.7,.2],[.2,.4,.4])` after two observations (iterations).

This is equivalent to asking for the product `X_0T^2` .

I. We can compute `X_0*T*T`, always multiplying from left to right.

When multiplying the vector by the transition matrix, we take a 1x3 vector times a 3x3 matrix to get a 1x3 resultant. (1x3:3x3 the "inner" numbers match so the product is defined, the outer numbers give the "dimensions" of the product.)

There are three entries in the result, call them r(1), r(2) and r(3).

To get r(1) we take an element of the vector times the corresponding element of the 1st column of T, and then sum these. Thus:

r(1)=.35(.1)+.2(.1)+.45(.2)=.145
r(2)=.35(.8)+.2(.7)+.45(.4)=.6 *
r(3)=.35(.1)+.2(.2)+.45(.4)=.255 **

*Notice that we took the elements of the vector times the corresponding entry in the 2nd column.

**Notice that we took the elements of the vector times the corresponding entry in the 3rd column.

So `X_0*T=[.145, .6,.255])`

Repeat this process with `X_1=X_0*T`

r(1)=.145(.1)+.6(.1)+.255(.2)=.1255
r(2)=.145(.8)+.6(.7)+.255(.4)=.638
r(3)=.145(.1)+.6(.2)+.255(.4)=.2365

So `X_0*T^2=([.1255,.638,.2365])`

II. Since we know we want `X_0*T^2`, we could compute `T^2` . When multiplying a 3x3 matrix by a 3x3 matrix, the result will be a 3x3 matrix. (3x3:3x3 ==> 3x3). The entries in the product are typically labelled `a_(i,j)`, where i is the row number and j is the column number. So `a_(3,2)` is the entry in row three column 2 (the middle entry of the bottom row.)

`a_(1,1)` (the top, leftmost entry) is found by taking the sum of the products of the entries in the first row in the first matrix by the entries in the first column of the second matrix.

Here `a_(1,1)=.1(.1)+.8(.1)+.1(.2)=.11`

`a_(3,2)=.2(.8)+.4(.7)+.4(.4)=.6`

Doing all of the arithmetic yields: `T^2=([.11,.68,.21],[.12,.65,.23],[.14,.6,.26])`

So `X_0*T^2` again has 1 row and three columns:

r(1)=.35(.11)+.2(.12)+.45(.14)=.1255
r(2)=.35(.68)+.2(.65)+.45(.6)=.638
r(3)=.35(.21)+.2(.23)+.45(.26)=.2365 as before.

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