By transposition, your equation is equivalent to Q'y=1y where y:=x' (a column vector) where Q' is the transpose of Q (matlab notation...) which means that y is an eigenvector associated with eigenvalue 1 for matrix Q'. Such an eigenvector always exists for a Markov matrix. Let s be the sum of the entries of column vector y. Two cases can occur :
either s is not 0 ; then it suffices to divide all coordinates of y by s : we obtain a vector that is still an eigenvector, with a coordinate sum equal to 1.
or s=0 and there is no solution to your problem.
Here is a Matlab program that does the work for a 3 x 3 matrix :
M=[.2 .3 .5
.1 .8 .1
.4 .4 .2]
[P,D]=eig(M')
Y=P(:,3)
M'*Y - Y,% should be 0
Z=Y/sum(Y),%the sum of Z's coordinates is 1
M'*Z-Z,% should be 0