A rectangular matrix of full rank can be multiplied by infinitely many matrices to form the identity

Question

Let $A$ be an $m \times n$ matrix with $m < n$ and $\operatorname{rank}(A) = m$. Prove that there exist infinitely many matrices $B$ such that $AB = I$.

Stumped. How do I begin to prove this?

Answer 1

Hint: the $j$'th column of $B$ must be a vector $\bf x$ such that $A{\bf x} = \ldots$. What do you know about solving equations $A {\bf x} = \bf b$ when $A$ is $m \times n$ with rank $m$ and $m < n$?

Answer 2

Hint-For a matrix $A$ of order $m×n$, $m<n$ ; the system $Ax_j$=$b$ has infinitely many solutions. Now, consider $x_j$ as $j^th$ column of matrix $B$.