Matrix inverse exists even determinate is zero. We know matrix inverse does not exist if det(matrix)=0. 
Now a 2*2 matrix with all entries $x$ has inverse as 2*2 matrix all entries $1/(4*x)$ .
So what is the gap of understanding?
 A: $\begin{bmatrix}x&x\\x&x\end{bmatrix}\begin{bmatrix}\frac{1}{4x}&\frac{1}{4x}\\\frac{1}{4x}&\frac{1}{4x}\end{bmatrix}=\begin{bmatrix}\frac12&\frac12\\\frac12&\frac12\end{bmatrix}$, and this is not the identity matrix of the full matrix ring $M_2(F)$, which is $\begin{bmatrix}1&0\\0&1\end{bmatrix}$.
However, $\begin{bmatrix}\frac12&\frac12\\\frac12&\frac12\end{bmatrix}$ is the identity matrix of the ring of matrices of the form $\left\{\begin{bmatrix}x&x\\x&x\end{bmatrix}\mid x\in F\right\}$.
The determinant-invertibility criterion does not hold for this subring. It is meant to work for the full matrix ring $M_2(F)$.
In the full matrix ring, you can easily see that $\begin{bmatrix}x&x\\x&x\end{bmatrix}$ isn't invertible since it is a zero divisor: $\begin{bmatrix}x&x\\x&x\end{bmatrix}\begin{bmatrix}1&0\\-1&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$.
If we were able to invert on the left, then we would wind up with $\begin{bmatrix}1&0\\-1&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$, a contradiction.
A: Assuming that $x\ne 0,$ we have $$\begin{bmatrix}x & x\\x & x\end{bmatrix}\begin{bmatrix}1/(4x) & 1/(4x)\\1/(4x) & 1/(4x)\end{bmatrix}=\begin{bmatrix}1/2 & 1/2\\1/2 & 1/2\end{bmatrix},$$ but this is not the identity matrix, as (for an easy example) $$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\begin{bmatrix}1/2 & 1/2\\1/2 & 1/2\end{bmatrix}=\begin{bmatrix}1/2 & 1/2\\1/2 & 1/2\end{bmatrix}\neq\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}.$$
