This question comes from a section before inverse matrices are introduced.
Suppose $AD=I_m$. Show that for any b in $R^m$, the equation $A$x$=$b has a solution. [Hint: Think about the equation $AD$b$=$b] Explain why $A$ cannot have more rows than columns.
My attempt:
Let b be any element of $R^m$, and define x as follows:
x$=D$b
Left multiply both sides by $A$
$A$x$=A(D$b$)$
$A$x$=I_m $b
$A$x$=$b
We have mapped every b in $R^m$ to some x in x$=D$b, so in $A$x=b we just map back from some x to any b in $R^m$
$A$ cannot have more rows than columns because if it did, the row echelon form of $A$ would necessarily have at least one row of zeroes, which would restrict b to having at least one $0$ entry.
IS my proof correct? Is there a better proof or a proof that uses the hint given?