Matrix $BA\neq$$I_{3}$ If $\text{A}$ is a $2\times3$ matrix and $\text{B}$ is a $3\times2$ matrix, prove that $\text{BA}=I_{3}$ is impossible.
So I've been thinking about this, and so far I'm thinking that a homogenous system is going to be involved in this proof. Maybe something about one of the later steps being that the last row of the matrix would be $0\neq \text{a}$, where a is any real number. I've also been thinking that for a $2\times3$ matrix, there is a (non-zero) vector $[x,y,z]$ such that $\text{A}[x,y,z]=[0,0]$ because the dot product could possibly yield $0$. I'm not sure if that's helpful at all though.
Trouble is I'm not really too sure how to continue, or even begin.
Any help would be appreciated. 

Consider the possible dimension of the columnspace of the matrix $BA$. In particular, since $A$ has at most a two-dimensional columnspace, $BA$ has at most a two-dimensional columnspace. Stated more formally, if $A$ has rank $r_a$ and $B$ has rank $r_b$, then $BA$ has rank at most $\min\{ r_a, r_b \}$.
