Solve a linear equation with no unique solution I want to calculate a unique equation for the following linear equation. The background is the page rank algorithm where I want to find a vector solution for the Google matrix:
The matrix on which I've started is:
$$\left(\begin{array}{rrrr}
1&13&1&1\\
1&1&7&7\\
13&1&1&1\\
1&1&13&1\end{array}\right).$$
Which is multiplied by $1/16$ in the end, but I leave this out, since its easier to do calculations on the matrix without this scalar. So yes, I am looking for the left eigenvector with eigenvalue 1, however I want to find a matching solution with linear equation. I used the Gauss algorithm on the matrix and came to the following solution:
$a = a/16$
$b = b/16 - a/16$
$c = c/16 + a/16 - 14b/16$
$d = 17a/16/16 - 15b/8/16 + c/16/16 + d/16 $
(This equation is derived from a matrix by using the Gauss algorithm)
For me it is obivous, that there is no unique solution to this equation. I was, hoever, told, that there is one, when adding the following constraints:
$a + b + c + d = 1$ and $a,b,c,d \geq 0$
This what I actually want, but I don't see how this could help. Any hints or suggestions?
 A: First version of the problem: Your first equation $a=a$ is tautological and so gives no information.  So you are left with three equations and four variables, so you should not expect a unique solution.  Add another equation and you may have the possibility of a unique solution which may satisfy your constraints; though you might have many solutions which satisfy your constraints, solution(s) which do not satisfy your constraints, or no solutions at all.  
Your second equation implies $a=0$, which combined with your third equation implies $b=0$, which combined with your fourth equation implies $c=0$, but $d$ can take any value.  Combining this with $a + b + c + d = 1$ implies $d=1$ and the solution $(0,0,0,1)$ satisfies your constraints so you are done.
Second version of the problem: Your first equation implies $a=0$, which combined with your second equation implies $b=0$, which combined with your third equation implies $c=0$, which combined with your fourth equation implies $d=0$.  So the unique solution is $(0,0,0,0)$. If you then combine this with $a + b + c + d = 1$ then you have no solution.
Third version of the problem: I think your matrix form results in $a=b=c=d$ so there is not a unique solution.  Combining this with $a + b + c + d = 1$ implies the unique solution $(0.25,0.25,0.25,0.25)$, which satisfies your constraints so you are done.  
