Let 


* $X$ be compact Hausdorff topological space,
* $C(X)$ denote the algebra of complex-valued continuous functions on $X$,
* $b\in \mathbb{C}^m$,
* $\mathbf{A}\in C(X)^{m\times n}$,
* for all $x\in X$, $b\in \textrm{range}(\mathbf{A}(x))$.


Question: Does there exist an $\mathbf{x}\in C(X)^{n\times 1}$ such that for all $x\in X$, $\mathbf{A}(x) \mathbf{x}(x)=b$?



$Ax=b$ in a function space
Let $$A(x)=\pmatrix{x&1\cr 0&x},$$ for $x\in [-1,1]$ and $$b=\pmatrix{1\cr 0}.$$ Then $b$ is in the range of $A(x)$ for every $x$, but for $x\neq 0$, we have $$A^{-1}(x)b=\pmatrix{1/x\cr 0},$$ so there is no continuous solution.


