$A^{-1}x \pmod{26}$ and coprime requirement in Hill cipher

I am reading Hill cipher from wiki page and I have been stuck on this thought for a while.
Why is there a requirement for $\det(A)$ and $26$ to be coprime in Hill cipher ?
Anybody familiar with Hill cipher and modular arithmetic help me understand this please ? 

The short answer is that for $A$ to be invertible over the integers mod 26, $det(A)$ must be a unit in that ring, i.e. coprime to 26.  Note that because $det(A)det(A^{-1}) = det(I) = 1$, the $det(A)$ must be invertible (as a "scalar").