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.


1 Answer 1


If $A$ is invertible then $\exists \, M \in \mathcal{R}^{n \times n}$ such that $AM=MA=I_n$. Then determinant being a homomorphism gives $$\det(AM)=\det(A)\det(M)=\det(I)=1_{\mathcal{R}}.$$ Thus $\det(A)$ must be a unit in $\mathcal{R}$.

For the other way:

Assume $\det(A)$ is invertible. Since $$\det(A)I_n = A\operatorname{adj}(A) = \operatorname{adj}(A)A$$ therefore determinant being invertible implies the inverse exists, since the adjoint always exists.

  • $\begingroup$ Am sorry I didnt quote this..this way i have proved already..but I have to prove the forward implication,$ i.e.,det(A) $ is invertible $\implies A$ has an inverse. $\endgroup$
    – David
    Nov 16, 2015 at 7:14
  • $\begingroup$ I think you have proved only one way implication.@anurag $\endgroup$
    – David
    Nov 16, 2015 at 7:19
  • $\begingroup$ @gloom I have added the proof for the converse as well. $\endgroup$
    – Anurag A
    Nov 16, 2015 at 7:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .