Prove that det($A$) is non-zero iff $A$ is row equivalent to the $n\times n$ identity matrix $A$  is  an $n\times n$ matrix. Now  if  the  row-reduced  echelon  form  for  this  $A$  is   $E$  then  after  all  the  row  operations  we  have $\det(A)=M\det(E)$   where  $M$  is  a non-zero  scalar  from  the   field. 
 If  $E$  is  the  $n\times n$ identity  matrix  then  $\det(A) \neq 0$.
For  the  converse, $\det(A)$   is  non-zero, so  is  $M$. Then obviously $\det(E)\neq0$. How to  reach the  conclusion  that  $E$  is  the  identity matrix,from  here  just  need  a  little  help  with  that.
 A: You have reduced the proof to this: If $E$ is an $n\times n$ matrix in reduced row echelon form, and $\det(E) \neq0$, then $E$ is the identity matrix.
To show this, I think you can work directly from the definition of RREF.  
The leading coefficient in the first row must be in the first column.  Otherwise $e_{11} = 0$ and $\det(E)$ would be zero.  So $a_{11} = 1$ and there are only zeroes below it.
Suppose now that $e_{11} = e_{22} = \dots e_{kk} = 1$, and all entries above and below these are zero.  That is, $E$ has a block decomposition
$$
    E = \begin{pmatrix} I_k & * \\ 0 & E' \end{pmatrix}
$$
where $I_k$ is the $k\times k$ identity matrix, and $E'$ is a $(n-k)\times(n-k)$ matrix in RREF. By the same argument as in the previous paragraph, the leading coefficient of the first row of $E'$ must occur in the first position.  Therefore $e_{k+1,k+1} = 1$, and all entries above and below it are zero. 
Therefore by induction $e_{ii} = 1$ for all $i$ from $1$ to $n$, and all entries above and below these diagonal entries are zero.  So $E$ is the identity matrix.
