We have $\mathbf{b} \in \operatorname{img} A$, which means there exists at least one vector $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{b}$. Also, $\operatorname{ker} A = \{0\}$ implies that the only solution to $A\mathbf{x} = 0$ is $\mathbf{x} = 0$. This tells
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