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  "body": "\n## Proof\n\n1. Suppose $A$ is [[invertible|0TWTS88TWN]]. By the result on the [[determinant of the inverse|0783S9D0ZH]], $\\det A$ is a unit of $K$.\n2. 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.\n3. Let $n = 1$ and $A = (a)$. Then $A$ is $a$ times the $1 \\times 1$ identity matrix $I_1 = (1)$.\n4. By [[multilinearity|0PPG793642]] in the only row and the [[determinant of the identity|0Z4BWARC38]], $\\det A = a \\det I_1 = a$.\n5. So $a$ is a unit. The matrix $(a^{-1})$ satisfies $(a)(a^{-1}) = (a^{-1})(a) = (1) = I_1$, so $A$ is invertible.\n6. By steps 1 to 5, $A$ is invertible if and only if $\\det A$ is a unit.\n7. Let $K$ be a field. By the definition of a field, every nonzero element has a multiplicative inverse, and $0$ has none since $0 \\cdot x = 0 \\ne 1$.\n8. So the [[units|019NFGFPR5]] of a field are its nonzero elements. Hence over a field, $A$ is invertible if and only if $\\det A \\ne 0$.\n\n## Idea\n\nOne 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.\n",
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