Does there exist a 4 vectors are linearly independent modulo 2 but not linearly indepdent integrally? Does there exist a $4\times 4$ matrix with 0's and 1's as entries, which is row equivalent to the identity matrix modulo 2, but is not row equivalent to the identity matrix when working with the integers?
Equivalently, does there exist four vectors $a,b,c,d$, each with four components with 0's and 1's as entries, such that $a,b,c,d$ are linearly independent over ${\mathbb{Z}}/2$, but not linearly indepdent over ${\mathbb{Z}}?$ 
 A: Regarding your second question:
Let $A=(a_{ij})\in M_n(\mathbb{Z})$. Denote by $\overline{A}=(a_{ij}\ mod_2)\in M_n(\mathbb{Z}_2)$.
Notice that $det(\overline{A})=det(A)\ mod_2$.
If the rows of $A$ are linear dependent over $\mathbb{Z}$ then $det(A)=0$ then $det(\overline{A})=0$. Thus, the rows of $\overline{A}$ are linear dependent over $\mathbb{Z}_2$.
Regarding your first question:
If $A=(a_{ij})\in M_n(\mathbb{Z})$ is row equivalent to the identity(using only operations with integers) then  $A$ has an inverse in $M_n(\mathbb{Z})$.
Thus, $det(A)det(A^{-1})=1$. Since $det(A),det(A^{-1})\in \mathbb{Z} $ then $det(A),det(A^{-1})$ are 1 or -1.
Thus, you need more than only $det(A)\neq 0$ for $A$ to be row equivalent(using only operations with integers) to the identity. It is necessary that $det(A)=1$ or $-1$.
This site, http://mathworld.wolfram.com/HadamardsMaximumDeterminantProblem.html , says that the maximum determinant for binary matrices $(a_{ij}=0$ or $1)$ of order 4 is 3. Thus, exists a binary matrix  $A$ of order 4, with determinat equals to $3$, that is not row equivalent to the identity in $M_4(\mathbb{Z})$ but $det(\overline{A})\neq 0\ mod_2$. So, $\overline{A}$ is row equivalent to the identity in $M_4(\mathbb{Z}_2)$ using operations with coefficients in $\mathbb{Z}_2$.
Your first question is not equivalent to the second.  
Now, I am not sure if a matrix $A\in M_n(\mathbb{Z})$ with $det(A)=1$ is row equivalent to the identity. 
