A question about invertible matrices A square matrix $A$ over the reals  is said to be  invertible in practice if there exists a matrix $B$ of the same size s. t. all the  entries  of $AB$ differ from  the corresponding entries of the identity matrix $E$ less than or equal to $10^{-10}$. Does there exist invertible in practice marix which is not invertible? 
 A: The set of $n \times n$ matrices that are "invertible in practice" is exactly the set of $n \times n$ matrices that are invertible, as long as $n < 10^{10}$.
For $n \geq 10^{10}$, invertible matrices are invertible in practice, but not the other way around. For a counterexample, consider the matrix given by
$$
A_{ij} = 
\begin{cases}
1 - 1/n & i=j\\
-1/n & i \neq j
\end{cases}
$$
Noting that the row sums of $A$ are all zero, we may conclude that $A$ is not invertible.  Nevertheless, it is "invertible in practice".
A: Technically, yes, but practically, no.
Technical answer:
Consider the $n\times n$ matrix
$$M = \left[\begin{array}{ccccc}1 & x & x & \ldots & x\\x & 1 & 0 & \ldots & 0\\x & 0 & 1 & \ldots & 0\\ \vdots &  & & \ddots & \\ x & 0 & 0 &\ldots & 1\end{array}\right].$$
By row-reduction the determinant of this matrix is $1 - (n-1)x^2$ and in particular, $M$ is singular when $x = \sqrt{1/(n-1)}$. You can make $x$ arbitrary small (in particular, less than $10^{-10}$) by making $n$ arbitrary large; and then if $AB = M$, at least one of $A$ or $B^T$ is singular while being "practically invertible."

Practical answer: suppose your matrix is $n\times n$, where $n < 10^{10}$. Then any matrix $M$ that is close to the identity matrix, in the sense you describe, is strictly diagonally dominant and hence nonsingular. Therefore any $A, B$ with $AB=M$ are both nonsingular as well.
