# If det(A) is zero, what is det(adj(A))?

I wanted to prove that det(adj(A))=det(A)^n-1 for an nxn matrix A.

I separate the proof in two cases: singular and non-singular matrix A.

For the non-invertible A, det(A)=0. In my head, I know that adj(A)=0, but I cannot prove it. How does one proceed?

• "I know that $\operatorname{adj}(A)=0$, but I cannot prove it." Hmm ... of course you cannot prove it, because your assertion is wrong. What is true is that $\det(\operatorname{adj}(A))=0$. – user1551 Nov 24 '14 at 9:28

$$A \operatorname{adj}(A)=\det (A)I_n$$ so if $\det(\operatorname{adj}(A))\ne0$ which means that it's invertible then we get $A=0$ but in this case $\operatorname{adj}(A)=0$ which's a contradiction. Conclude.