Show that if $A$ is a 2-by-3 matrix and B is a 3-by-2 matrix, then $BA\not=$Id. Claim: Suppose $A$ is a 2-by-3 matrix and $B$ is a 3-by-2 matrix. Then $BA$$\not=$$Id$.
My attempt
Proof. Let $T$:$F^3$$\rightarrow$$F^2$ and $S$:$F^2$$\rightarrow$$F^3$ whereby their matrix representations are given by $A$ and $B$, respectively. Thus $ST$ is a linear map from $F^3$ to $F^3$. Suppose $ST=Id$. Then $ST$ is injective, which imples that $T$ must also be injective. By the Fundamental Theorem of Linear Maps, we have that $\dim(F^3)=\dim\newcommand{\null}{\operatorname{null}}\newcommand{\range}{\operatorname{range}}\null T+\dim\range T$. Now because $T$ is injective and $\dim\range T\le\dim(F^2)$, it follows that $\dim(F^3)=\dim\range T\le \dim(F^2)$. However, this is impossible since $3$ $\not\le$ $2$. So if $ST=Id$, then $T$ is not a linear map from $F^3$ to $F^2$ and hence $A$ $\notin$${F}^{2,3}$. Therefore, $BA$$\not=$$Id$ whenever $A$$\in$ ${F}^{2,3}$ and $B$$\in$ ${F}^{3,2}$, as claimed.

$r(A)\leq{2},r(B)\leq{2}$
So,
$r(BA)\leq{min(r(B),r(A))}\leq{2}<3$
Therefore,it is impossible $BA=I$
