Given $A$ a singular and symmetric matrix, and $B$ a symmetric matrix, I find that solving $(A+B)x=v$ by $$ x= (A+B)^+v $$ gives the correct solution to my physical system. The physical system
1) actually has one unique solution.
2) satisfies $Ax=0$ and $Bx=v$
3) $A$'s rank + $B$'s rank = $A$'s number of row (and column).
4) $A+B$ is singular.
example: $$ A= \begin{bmatrix} -2& 1 & -1 & 0 & -1 \\ 1 & -2 & -1 & -1 & 0 \\ -1& -1 & -2 & -1 & -1 \\ 0 & -1 & -1 & -2 & 1 \\ -1& 0 & -1 & 1& -2 \\ \end{bmatrix} $$
$$ B= \begin{bmatrix} 4& 0 & -2 & 0 & 0\\ 0& 0 & 0 & 0 & 0 \\ -2& 0 & 2 & 0 & -3 \\ 0 & 0 & 0 & 0 & 0 \\ 0& 0 & -3 & 0& 9 \\ \end{bmatrix} $$
where $A$ is rank 3 and $B$ is rank 2. If adding $B$'s rows to A, then the matrix rank is 5. So I guess this linear system has indeed one unique solution.
$v=(66,0,11,0,-132)$. The solution is known a prior $x=(16,16,-1,-15,-15)$.
But why pseudoinverse gives the correct solution ? I looked through the wiki page on Moore-Penrose inverse in vain.