Markov Equations Balance equations and Normalising equations

Question

I am looking at a question involving three equations:

1. $A=0.6667A+0.2222B+0.1667C$
2. $B= 0.2A+0.3333B+0.5C$
3. $C=0.1333A+0.4444P+0.3333C$

The solution then goes on to say, that these equations can be re-written as:

4. $0.33A+0.22B+0.17C=0$
5. $-0.2A+0.67B+0.5C=0$
6. $-0.13A-0.44B+0.67C=0$

Can anyone explain how you get the second set of equations from the first? Pulling my hair out!!! If you take $A,B$ & $C$ from LHS over to RHS, then the signs change etc, but I don't get the second set at all.

One other equation that goes along with these are that $A+B+C=1$

Any help would be appreciated

Answer 1

Serval things happened. I will only consider equation 1.

$$A=0.6667A+0.2222B+0.1667C$$

First I subtract $0.6667A$ from both sides so that on the LHS I get $$A-0.6667A = (1-0.6667)A = 0.3333A$$ which was rounded to $0.33A$. Moving the other two terms to the LHS and more rounding gives $$0.33A+0.22B+0.17C=0$$

which is equation 4. Similar rounding occurs in the other equations.