finding steady-state vectors for a matrix

[buckaroobill]

My book gives an example for solving for a steady state vector for a matrix, but I'm a little confused.

Okay, so it gives you the matrix:

M =

[ 0 .5 0
.5 0 1
.5 .5 0]

Then, it tells you that in order to find the steady state vector for the matrix, you have to multiply

[-1 .5 0
.5 -1 1
.5 .5 -1]

by

[x1
x2
x3]

to get

[0
0
0]

I understand that they got the:
[-1 .5 0
.5 -1 1
.5 .5 -1]

by doing M - the identity matrix.

However, the book came up with these steady state vectors without an explanation of how they got there:
x1 = .222
x2 = .444
and x3 = .333

I'm a little confused about how you come up with those. Do you just row reduce
[-1 .5 0
.5 -1 1
.5 .5 -1]
to get them? Or is another method involved here?

 

[soroban]

Re: Question

Hello, buckaroobill!

My book gives an example for solving for a steady state vector for a matrix.

It gives you the matrix: \(\displaystyle \:M \:=\:\begin{bmatrix} 0 & \,0.5\, & 0 \\ 0.5 & 0 & 1 \\ 0.5 & 0.5 & 0\end{bmatrix}\)

Then, it tells you that in order to find the steady state vector, you have to solve:

. . \(\displaystyle \begin{bmatrix}-1 & \,0.5\, & 0 \\ 0.5 &-1 & 1 \\ 0.5 & 0.5 & -1\end{bmatrix}\,\cdot\,\begin{bmatrix}x \\ y \\ z\end{bmatrix} \;=\;\begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}\;\) [A]

I understand that they got: \(\displaystyle \begin{bmatrix}-1 & \,0.5\, & 0 \\ 0.5 & -1 & 1 \\ 0.5 & 0.5 & -1 \end{bmatrix}\;\) by doing \(\displaystyle M\,-\,I.\)

However, the book came up with these steady state vectors
without an explanation of how they got there:
. . \(\displaystyle \begin{bmatrix}x \\ y \\ z\end{bmatrix}\:=\:\begin{bmatrix}\frac{2}{9} \\ \\ \frac{4}{9} \\ \\ \frac{1}{3}\end{bmatrix}\)

[A] gives us a system of equations: \(\displaystyle \:\begin{array}{ccc}-x\,+\,0.5y & \,=\, & 0 \\ 0.5x\,-\,y\,+\,z & = & 0 \\ 0.5x\,+\,0.5y\,-\,z & = & 0\end{array}\;\begin{array}{cc}(1) \\ (2) \\ (3)\end{array}\)

Add (2) and (3): \(\displaystyle \:x\,-\,0.5y \:=\:0\) . . . which is equivalent to (1).

But there is a third equation available: \(\displaystyle \:x\,+\,y\,+\,z\:=\:1\)


So we have this system of equations: \(\displaystyle \:\begin{array}{cccc}-x\,+\,\frac{1}{2}y & \,=\, & 0 & \1) \\ \frac{1}{2}x\,-\,y\,+\,z & \,=\, &0 & \2) \\ x\,+\,y\,+\,z &\,=\,&1 & \3)\end{array}\)

Subtract (2) from (3): \(\displaystyle \:\frac{1}{2}x\,+\,2y\:=\:1\)
Multiply (1) by \(\displaystyle \frac{1}{2}:\;\;\;\;\;-\frac{1}{2}x\,+\,\frac{1}{4}y\:=\:0\)

Add: \(\displaystyle \:\frac{9}{4}y \,=\,1\;\;\Rightarrow\;\;\fbox{y\,=\,\frac{4}{9}}\)

Substitute into (1): \(\displaystyle \:-x\,+\,\frac{1}{2}\left(\frac{4}{9}\right)\:=\:0\;\;\Rightarrow\;\;\fbox{x\,=\,\frac{2}{9}}\)

Substitute into [3]: \(\displaystyle \:\frac{2}{9}\,+\,\frac{4}{9}\,+\,z\:=\:1\;\;\Rightarrow\;\;\fbox{z\,=\,\frac{1}{3}}\)

. . . . . There!

 

[buckaroobill]

Thanks! That's clear now.

I'm confused about how to do the following problem, though, because there are so many variables:

I was given a transition matrix R and I have to find the steady state vector of that. I did the first step, which is to do R - the identity matrix. This is what I got (it's in table form just so you can tell the numbers apart more easily...it's a thumbnail so you can click it to make it larger and more readable).


Now, I know that in order to find a steady state vector I have to do this matrix multiplied by column vector [x1...x9] to get the column vector [0, 0, 0, 0, 0, 0, 0, 0, 0].

I'm just confused as to how to find the x1 through x9 because there are so many equations and variables. Any help would be greatly appreciated!

This is what I was thinking...

Would I just get 0 for all of x1 through x9 because...

Each column contains fractions that add up to one plus -1. This gives you 0.