Is it possible for a non-singular matrix to not have an inverse? I know that a singular matrix cannot have an inverse, but I was curious as to whether or not a matrix with a non-zero determinant can have an inverse. And from what I understand the requirement for a matrix to be non-singular is that it has a determinant that is not zero.
 A: Every square matrix $\mathbf{A}$ has an adjugate matrix $\textbf{adj}(\mathbf{A})$ formed from the cofactors of $\mathbf{A}$, even if $\mathbf{A}$ is singular. The inverse of a square matrix is the matrix 
\begin{equation}
\mathbf{A}^{-1}=\dfrac{\textbf{adj}\mathbf{(A)}}{\textbf{det}\mathbf{(A)}} \tag{1}
\end{equation}
Since, if $\mathbf{A}$ is nonsingular, then its determinant is non-zero, every non-singular square matrix $\mathbf{A}$ has an inverse defined by equation $(1)$.
Wikipedia reference: https://en.wikipedia.org/wiki/Adjugate_matrix
A: I will add my two cents to the answer above:
One has:
$\mathbf A\cdot \operatorname{adj}(\mathbf A)=\det(\mathbf A)\cdot\mathbf I$
whre $\operatorname{adj}(\mathbf A)$ is the adjugate of matrix $\mathbf A$
So if the elements of $\mathbf A$ are in a field (like real or complex numbers), nonzero determinant implies $\mathbf A$ is invertible with $\mathbf A^{-1}=\det(\mathbf A)^{-1}\cdot\operatorname{adj}(\mathbf A)$
However, if the matrix is over a commutative ring which is not field (for example $\mathbb Z$), $\mathbf A$ is invertible only if $\det\mathbf A$ is invertible in the ring (in the case of $\mathbb Z$, we should have $\det\mathbf A=\pm 1)$
In other words, it is entirely possible for a matrix with a non-zero determinant to not have an inverse. Just not for real or complex matrices.
