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  "body": "\n## Proof\n\n1. Since $A$ is [[invertible|0TWTS88TWN]], there is an $n \\times n$ matrix $A^{-1}$ with $A A^{-1} = A^{-1} A = I_n$.\n2. By the [[product rule|04CPJ6AF00]], $\\det A \\cdot \\det(A^{-1}) = \\det(A A^{-1}) = \\det I_n$.\n3. By the [[determinant of the identity|0Z4BWARC38]], $\\det I_n = 1$. So $\\det A \\cdot \\det(A^{-1}) = 1$.\n4. Since $K$ is commutative, also $\\det(A^{-1}) \\cdot \\det A = 1$.\n5. By the definition of a [[unit|019NFGFPR5]], steps 3 and 4 show that $\\det A$ is a unit of $K$ with inverse $\\det(A^{-1})$.\n6. So $\\det(A^{-1}) = (\\det A)^{-1}$.\n\n## Idea\n\nThe determinant turns products of matrices into products of ring elements and sends $I_n$ to $1$. So the equation $A A^{-1} = I_n$ becomes $\\det A \\cdot \\det(A^{-1}) = 1$.\n",
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