Solving a linear equation to find a stationary matrix I'm trying to solve the following system of linear equations derived from a transitional matrix for a regular Markov chain. I can't use matrix methods since that would involve finding the inverse of a non-square matrix, and I'm not clear on how the elimination method would work with this. Any ideas?
$$-0.25x + 0.15y + 0.05z = 0$$
$$0.05x - 0.25y + 0.1z = 0$$
$$0.2x + 0.1y - 0.15z = 0$$ 
$$x + y + z = 1$$
The solution is $x = 0.25, y = 0.25, z = 0.5$.
 A: You give four equations for three unknowns. Fortunately, there is a point of major interest :
substracting the first equation from the third equation gives the second equation. Then, either the first or the third equation must be discarded. So we have now three linear  equations for three unknowns.
Suppose that we discard the first equation. Extract  $x = 5 y - 2 z$ from the second equation and plug into the third equation from which now you extract $y = z /2$ (so $x= z /2$). Now the fourth equation gives the result.   
In fact all the problem was to detect the collinearities.
A: Firstly, to make things easier, I would multiply each of the three equations by $20$ to get rid of the decimals:
$$-5x + 3y + z = 0$$
$$x - 5y + 2z = 0$$
$$4x + 2y - 3z = 0$$
Now, add $5$ times of the equation $4$ to equation $1$ to get a new equation, equation $5$:
$$8y + 6z = 5$$
Similarly, subtract $1$ time(s) of the equation $4$ to equation $2$ to get another new equation, equation $6$:
$$-6y + z = -1$$
Now, after having eliminated $x$, go ahead and solve the equations $5$ and $6$ simultaneously to find the values of $y, z$. (Hint : Subtract $6$ times of the equation $6$ from equation $5$).
Then, substitute back into either of the first four equations to get $x$.
It's possible to use the same approach to eliminate $y$ or $z$ as well, depending on whichever feels easier.
