The answer is A=B=C= 1/3 I can't seem to finish the logic below and I wonder if there is a faster way that anyone can suggest please. This is related to steady state probabilities and I know I could use "Doubly Stochastic", but I really need to be able to solve these without shortcuts.
I used;
A=7A+2B+1C
B=2A+6B+2C
C=1A+2B+7C
A+B+C=10A+10B+10C
A=B+C
Then
A=7A+2B+1C
-(C=1A+2B+7C) =
A-C=6A+0B-6C
-5A=-6C
A=C
Now I have these A+B+C=1 and A=B+C and A=C, but keep going around in circles not able to actually arrive at the solution to the answer "A=B=C= 1/3".
Rewrite \begin{cases} A = 7A + 2B + C \\ B = 2A + 6B + 2C \\ C = A + 2B + 7C \\ 1 = A + B + C \end{cases} as \begin{cases} 0 = 6A + 2B + C \\ 0 = 2A + 5B + 2C \\ 0 = A + 2B + 6C \\ 1 = A + B + C \tag{1}\label{1} \end{cases} Subtract the first from the third equation to get $5C-5A=0$, which gives $A = C$. \begin{cases} 0 = 7A + 2B \\ 0 = 4A + 5B \\ 1 = 2A + B \tag2\label2 \end{cases} From the first two equations in \eqref{2}, it's clear that $A = B = 0$, which doesn't satisfy \eqref{2}, so your system is inconsistent.
Remarks:
We're getting something wrong not because of the steps in the question body. The true reason is that OP has asked the wrong question. More precisely, he has set up a wrong linear system of simultaneous equations.
The stochastic matrix is in fact $S=\begin{pmatrix}7/10&2/10&1/10\\2/10&6/10&2/10\\1/10&2/10&7/10\end{pmatrix}$, and he's trying to find an invariant measure $p=(p_1,p_2,p_3)$ so that $pS = p$. (i.e. a left-eigenvector $p$ of $S$ corresponding to the eigenvalue $1$.) The "shortcut" is to add up the rows of $S$ and observe that it becomes $(1,1,1)$. (i.e. $(1,1,1)S = (1,1,1)$) This gives $p_1 = p_2 = p_3 = 1/3$, which OP wants.
I appreciate OP's effort to translate the problem of "solving for an invariant measure from a doubly stochastic matrix" into "solving for $p$ from $pS = p$ with a given $S$". The reason is that not everyone is familiar with the jargons in stochastic-processes. From the actually numeric value of $p$, we can see that OP's mistake is a minor one: when he's stripping off the denominator $10$ from the doubly stochastic matrix $S$, he'd forgotten to multiply another side by $10$. It should be $p(10S)=10p$ instead of $p(10S) = p$.
i.e. We should have \begin{cases} 10A = 7A + 2B + C \\ 10B = 2A + 6B + 2C \\ 10C = A + 2B + 7C \\ 1 = A + B + C \end{cases} instead. This should give $(A,B,C) = (1/3,1/3,1/3)$.
Making careless mistakes can help understanding things.