---
tag: 91NHW201A1
kind: proof
name: Proof that a matrix is invertible iff its determinant is a unit
msc: [15A15]
proves: 03Q7RE2HXR
uses: [0783S9D0ZH, 0ENVM493VK, 0TWTS88TWN, 0PPG793642, 0Z4BWARC38, 019NFGFPR5]
sources:
  - url: https://en.wikipedia.org/wiki/Determinant
    locator: Invertibility
---

## Proof

1. Suppose $A$ is [[invertible|0TWTS88TWN]]. By the result on the [[determinant of the inverse|0783S9D0ZH]], $\det A$ is a unit of $K$.
2. Conversely, suppose $\det A$ is a unit of $K$. If $n \ge 2$, the [[adjugate formula for the inverse|0ENVM493VK]] shows that $A$ is invertible.
3. Let $n = 1$ and $A = (a)$. Then $A$ is $a$ times the $1 \times 1$ identity matrix $I_1 = (1)$.
4. By [[multilinearity|0PPG793642]] in the only row and the [[determinant of the identity|0Z4BWARC38]], $\det A = a \det I_1 = a$.
5. So $a$ is a unit. The matrix $(a^{-1})$ satisfies $(a)(a^{-1}) = (a^{-1})(a) = (1) = I_1$, so $A$ is invertible.
6. By steps 1 to 5, $A$ is invertible if and only if $\det A$ is a unit.
7. Let $K$ be a field. By the definition of a field, every nonzero element has a multiplicative inverse, and $0$ has none since $0 \cdot x = 0 \ne 1$.
8. So the [[units|019NFGFPR5]] of a field are its nonzero elements. Hence over a field, $A$ is invertible if and only if $\det A \ne 0$.

## Idea

One direction uses the product rule: an inverse matrix forces $\det A \cdot \det(A^{-1}) = 1$. For the other direction, the adjugate gives an explicit inverse, namely the adjugate divided by $\det A$, and this division needs $\det A$ to be a unit.