Let $A$ be a square matrix. Prove that $A$ is non-singular if and only if $AX=b$ has a solution for each $b$.

Assuming $A$ is non-singular I can show that the system $AX=b$ has a solution for each $b$. Now I assume that the system $AX=b$ has a solution for each $b$. To prove $A$ is non-singular that is $A$ is invertible. I know that $A$ is invertible if and only if $A$ is row-equivalent to identity matrix. Now how to show that $A$ is equivalent to a identity matrix of same size?

• Questions like this highly depend on the theorems allowed. – Kenny Lau Aug 11 '18 at 14:46
• Using the dimension of the image should be the most elegant way to finish the proof. – Peter Aug 11 '18 at 15:14

Since the system has a solution for every $b$, therefore we can choose specific $b$’s. Namely the standard basis vectors $e_i=(0,0, \ldots , 1, \ldots ,0)^T$. So there is a solution $X_1$ for $AX=e_1$, solution $X_2$ for $AX=e_2$, and so on.
Thus you have a matrix $M$ with $X_i$’s as column vectors such that $AM=I$. Hopefully you can take it from here.
A quick one: suppose $\;A\;$ is an $\;n\times n\;$ matrix over some field $\;\Bbb F\;$ , then
$\;AX=b\;$ has a unique solution for each $\;b\in\Bbb F^n\;$ iff $\;A\;$ is onto (as a linear operator) iff it is one-to-one iff it is an isomorphism iff it is regular iff it is non-singular.