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You cannot solve what has no solutions That matrix is singular, so the system has either zero or infinite solutions. In the case of your system, I think some Perron-Frobenius theory can be used to prove that there are zero: there is a non-negative vector such that $Sv=v$, hence $0=x(I-S)v=bv > 0$, contradiction. (Those $x$ and $b$ are row vectors, right?)...