A matrix is invertible iff its determinant is a unit

Statement

Let $K$ be a commutative ring and let $A$ be an $n \times n$ matrix over $K$. Then $A$ is invertible if and only if its determinant $\det A$ is a unit of $K$. In particular, over a field, $A$ is invertible if and only if $\det A \ne 0$.

Remarks

If $\det A$ is a unit, then the adjugate gives the inverse $A^{-1} = (\det A)^{-1} \operatorname{adj}(A)$. Over $\Z$, the matrix $\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ has determinant $2 \ne 0$, but it has no inverse with integer entries. Over a field, this is one of many equivalent conditions for invertibility, with $\rank A = n$ and $\ker A = 0$.

Explanations

L1 human

A real square matrix has an inverse exactly when its determinant is not zero. A zero determinant means that the matrix squeezes space into a lower dimension. Such a squeeze cannot be undone.