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?

  • 1
    $\begingroup$ "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 $\det(\operatorname{adj}(A))=0$. $\endgroup$ – user1551 Nov 24 '14 at 9:28

We have

$$A \operatorname{adj}(A)=\det (A)I_n$$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.