Can a nonsingular square matrix be made singular by changing exactly one element or vice versa? 
  
*
  
*Given a nonsingular square  matrix $A$, can changing just one element  make it singular? 
  
*Given a singular square  matrix $A$, can changing just one element  make it nonsingular? 
  

For  $1$)  I  was  thinking  that  the  reduced  row  echelon  form  of  $A$,say $E$ ($E$=$E_{1}E_{2}E_{3}.....E_{n}A$), must  be  identity  so  changing  just  one $1$  to  $0$  would make  it  singular $\bar E$. But  then , does  that  mean  alteration  in  just one  element  in  the  original  matrix  when the  same  row  operations  are  inverted  on  the  new  matrix  i.e ${E_{n}}^{-1}.....{E_{1}}^{-1}\bar E = \bar A$  is singular  but  do $A$  and $\bar A$  differ  by  only  one  element?
For  $2$)  May  be  wrong  because  in this  case  one  or  more  than  one  rows  are zero  rows so  putting  one  pivotal  $1$  in  each  row  will  make it non-singular and  reversing  the  row  operations  will  give a non-singular  matrix.But  the  same  problem  here .  How many  of  original  elements  are  altered? If  the  RRE  form  had more  than  $1$  zero  rows  then  I  guess  it  is not  just  $1$  element.
Am  I  going  in   the  right  direction  or   totally  messed  up?
Thanks   for  the  answers @Servaes  and  @Dustan Lavenstein. Those  were  really  helpful. 
Can  anybody  please  throw  some  light  on  the  process  I  was trying ? Can  the  answer  be  obtained  in  that  way, especially  for  the  first  question?
 Thanks.
 A: *

*Your method to get a singular matrix in the first question is not going to work. For example, you have got identity matrix as $I=EA$ where $E$ is the product of all elementary row operations. It means that $E^{-1}=A$ and after pre-multiplying with it, you basically have the identity $A=A\cdot I$. Then you switch one unity in the identity matrix to zero, but this turns the whole column in the new $A$ to zero.
$$
A=\left[\matrix{a & b\\ c & d}\right], \qquad \left[\matrix{a & b\\ c & d}\right]\left[\matrix{1 & 0\\ 0 & 0}\right]=\left[\matrix{a & 0\\ c & 0}\right].
$$
A correct way to make the matrix singular by changing one element only is to look at the effect of this one element, say $x$, on the determinant. Using the cofactor expansion of the determinant along the row that contains the element $x$ we get $n$ terms, of which one is "$x$ times the cofactor of $x$" and all others do not depend on $x$, i.e.
$$
\det A=ax+b
$$
where $a$ is the cofactor and $b$ is a constant. If $a\ne 0$ the determinant can be set to zero by setting $x=-\frac{b}{a}$. If $a=0$ then we pick another element in the row. Clearly some element in the row must have non-zero cofactor. If all cofactors in the row are zeros then we conclude from the same expansion that $\det A=0$, which contradicts the assumption that $A$ is non-singular.

*In your second question you may take a zero matrix as a counterexample.


P.S. As to the first question, we have proved a stronger claim. We have shown that a non-singular matrix can be made singular by changing only one element in a pre-defined row (or column).
A: For 2 ($A$ singular), the answer is  YES iff $rank(A)=k-1$.
Proof. Let $A=[a_{k,l}]$; if we change $a_{i,j}$ with $a_{i,j}+x$, then $A$ becomes $A_{x,i,j}$. Then $\det(A_{x,i,j})=cofactor(a_{i,j})x+\det(A)$.
$(\Rightarrow)$ Here $\det(A)=0$ and there is $(i,j),x$ s.t. $\det(A_{x,i,j})\not= 0$. Thus $cofactor(a_{i,j})\not= 0$ and $rank(A)=k-1$.
$(\Leftarrow)$ There is $(i,j)$ s.t. $cofactor(a_{i,j})\not= 0$; thus $\det(A_{x,i,j})=cofactor(a_{i,j})x$ and we choose any $x\not= 0$.
For 1 ($A$ is invertible) The answer is YES for every $A\in GL_k$.
Proof. There is $(i,j)$ s.t. $cofactor(a_{i,j})\not= 0$. Then take $x=\dfrac{-\det(A)}{cofactor(a_{i,j})}$.
A: If your base field is algebraically closed (e.g. $\Bbb{C}$) then the answer is yes:
A $k\times k$-matrix is singular if and only if its determinant is zero. The determinant is a (degree-$k$) polynomial over the base field in terms of the $k^2$ entries. Leaving one entry of the matrix variable and all others fixed therefore yields a polymial in one variable over the base field, which certainly has a root if the base field is algebraically closed. This means that for any non-singular $k\times k$-matrix, any entry can be changed to make the matrix singular. Moreover, in general there will be $k$ different values of this entry making the matrix singular.
The converse, however, is not true. If the matrix has two columns or rows that are all zeroes, then changing only one entry will leave the matrix singular. So your idea for the second question is correct. Take for example
$$A=\begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}.$$
