How does the rank of matrix affects solvability of $Ax = b$?

Question

I was going through lecture 8 of Linear algebra by MIT (at 36:48)i didn't understand the concept where when rank of a matrix is equal to the number of rows ($r = m$, for $m\times n$ matrix), the solution for $Ax = b$ exists for every b.Can someone please elaborate on that. Thanks

Answer 1

The rank is the dimension of image of the map $v\mapsto A.v$. So, asserting that $\operatorname{rank}A=m$ means that that map is surjective, which is the same thing as asserting that every equation $A.x=b$ has a solution.

Answer 2

Since the rank of the given matrix $A$ is equal to the number of rows, this means each row of $A$ has a pivot. Said differently, when you consider the augmented matrix $[A \, | \, b]$ of the system $Ax=b$, there will be NO pivots that will appear in the right hand side column (after row operations). This means the system will always be consistent.