$A$ has more columns than rows and has full row rank, show there exist infinitely many $B$ s.t. $AB=I$ If A $\in M_{m\times n}(R)$ such that $n>m$. 
Prove that if $\text{rank} (A) = m$ then there are infinitely many matrices $B \in \ M_{n\times m} (R)$ such that $ AB = I_m$
So the question is defining a matrix $A$, in a set of matrices where the number of columns is always greater than the number of rows, and if $A$ has full row rank, then there are infinitely many matrices $B$ such that $AB$ creates the identity matrix. 
I'm not sure where to go with this question, could anyone please help?
ETA
Thank you to everyone for their answers, the question makes sense now. 
 A: Full row rank implies existence of at least one right inverse, because the linear system $Ax=b$ has a solution for every $b$; if $b_i$ is a solution of $Ax=e_i$ (where $e_i$ is the $i$-th column of the $m\times m$ identity) and $B=[b_1\ b_2\ \dots\ b_m]$, then $AB=I$ by construction.
Next we show that we can modify $B$ without changing the result. Let $v\ne0$ be any vector such that $Av=0$. Such a vector surely exists because of the rank-nullity theorem (alternatively, because the consistent system $Ax=0$ has free variables). Consider the matrix
$$
C=[\,\underbrace{v\ v\ \dots\ v}_{m}\,]
$$
so that
$$
A(B+C)=AB+AC=AB+0=I_m
$$
Since there are infinitely many choices of a vector $v$, you're done.
A: The matrix $A$ is a surjective linear transformation $\mathbb{R}^n \to \mathbb{R}^m$ and its kernel, say $K$, has dimension $k = n - m > 0$.
Take a standard basis vector $e_i \in \mathbb{R}^m$. Since $A$ is surjective and not injective (kernel has positive dimension) you have infinitely many preimages w.r.t $A$. Hence take some $b_i \in \mathbb{R}^n$ such that $A b_i = e_i$.
Now define $B \colon \mathbb{R}^m \to \mathbb{R}^n$ by simply declaring that the standard basis vector $e_i \in \mathbb{R}^m$ is mapped to $b_i$, i.e. $B(e_i) = b_i$. This amounts to say that the $i$-th column of $B$ is precisely $b_i$.
By construction $A B (e_i) = A b_i = e_i$ for any $i = 1, \dots, m$. Hence $A B = I_m$. 
A: The objective of this answer is to give a quite general form for matrices $B$.
The full rank condition implies that there exist a $m \times m$ invertible submatrix of $A$.
Let us assume at first that the left-most block $C$ of $A$ is such a submatrix. 
Then, with the following notations:
$A=\begin{pmatrix}
C & D
\end{pmatrix}
\ \ \ (1)$
$B=\begin{pmatrix}
C^{-1}(I_m-DE)\\
E
\end{pmatrix}
\ \ \ (2)
$
we have $AB=I_m$ for any $E$.
Remark: 
The set of such matrices $B$ is an affine space $R=v+V$, where $V$ is a vector space with dimension at most $m \times (n-m)$.
If the leftmost block is not invertible, consider the columns $A_{k_1}, A_{k_2}, \cdots A_{k_m}$ of $A$, which assembled, give an invertible matrix. We consider the $n \times n$ permutation matrix that sends $k_1, k_2, \cdots k_m$ to $1,2,\cdots, m$, and leaves unchanged the other indices of $1,2...n$. It is rather easy to see that if the lines of $B$ (as described by (2)) are "scrambled" by the same permutation, the result is preserved. 
(one can give a more rigorous treatment for this last part).
