Let $\mathbb K$ be a commutative ring with unity. Let $det:\mathbb K^{n\times n}\to \mathbb K$ be determinant function.
Prove that,
$det A$ is invertible $\iff$ $A$ has an inverse.
I proved this for $\mathbb K= a \, Field$. Because, in that case it reduces to $det(A)\neq 0$. But, how we will prove if $\mathbb K$ is s commutative ring with unity.
Work already done:
If $A$ has an inverse, then $AA^{-1}=I=A^{-1}A$ where $A^{-1}$ being the inverse of $A$. Hence, $$det(AA^{-1})=det(I)=det(A^{-1}A)$$ $$det(A)det(A^{-1})=1=det(A^{-1})det(A)$$ which proves that $det(A)$ is invertible.
Now, we have to prove that $det(A)$ is invertible $\implies A$ has an inverse.