# 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.

• The answer posted is correct, so I won't post an answer, but my intuition here is: Firstly a bijective function has an inverse; it's easy to prove that the inverse is linear. Secondly non-singular means injective. It's an interesting property of finite-dimensional vector spaces that there's some duality between range and null space; this commonly encoded in the rank+nullity theorem. From which it follows that an injective map must be surjective too. Nov 13, 2016 at 0:46
• @Gaffney's comments are far better, in the sense of being explanatory, than the accepted answer. The accepted answer is literally correct, of course, but to my mind is not explanatory. (There is a frequent, unfortunate possibility in mathematics of proving that something is true ... without effectively explaining why it is true. A further unfortunateness is that there is little professional pressure to do better, and also there's the pop mythology that "it's all about logic"... ) Nov 13, 2016 at 0:54

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

$$\mathbf{A}^{-1}=\dfrac{\textbf{adj}\mathbf{(A)}}{\textbf{det}\mathbf{(A)}} \tag{1}$$

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 $(1)$.

$\mathbf A\cdot \operatorname{adj}(\mathbf A)=\det(\mathbf A)\cdot\mathbf I$
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 $\det\mathbf A=\pm 1)$