Let $X$ be a compact Hausdorff topological space, and $C(X)$ denote the ring of complex-valued, continuous functions on $X$. Let $A$ be a matrix with entries from $C(X)$ of size $m\times n$ and $b\in \mathbb{C}^{m\times 1}$. Suppose that for each $x\in X$, the equation $A(x) v=b$ has a solution $v\in \mathbb{C}^n$. Then does there exists an $V\in C(X)^{n\times 1}$ such that $AV=b$? 



Continuity of solutions to $Av=b$
No. Take $n=m=2$, $A(x)=\pmatrix{1&0\\x&x^2}$, $x\in \mathbb{R}$, $b=\pmatrix{1\\0}$. For $x\ne 0$ the only solution of $AV=b$ is $V(x)=\pmatrix{1\\-1/x}$, it has discontinuity at 0 for any choice of $V(0)$. 


