0
$\begingroup$

Let $A$ be an invertible $n\times n$ matrix. Prove that $\det(\operatorname{adj}(A^{-1})) = (\det(A))^{1-n}$

I tried starting with $A^{-1} = 1/\det(A) \cdot \operatorname{adj}(A)$

I tried everything to reduce it to the proof..

the closest I could get was substituting $A^{-1}$ for $B$ and $A$ for $B^{-1}$ and wind up with the proof with $A$ replaced by $B$... but that can't be right...

$\endgroup$
1

2 Answers 2

1
$\begingroup$

If you substitute $A^{-1}$ for $A$ you get $\text{adj}(A^{-1})=(\det(A^{-1}))A$, so now we have

$\det(\text{adj}(A^{-1}))=(\det(A^{-1}))^{n}\det(A)=(\det(A))^{-n}\det(A)=(\det(A))^{1-n}$

$\;\;\;\;\;$since $\det(cA)=c^{n}\det(A)$ and $\det(A^{-1})=(\det(A))^{-1}$.

$\endgroup$
4
  • $\begingroup$ I'm sorry, but I think you skipped too many steps, can you state that more thoroughly? $\endgroup$
    – J L
    Oct 11, 2014 at 21:00
  • $\begingroup$ Yes, I will add some explanation. $\endgroup$
    – user84413
    Oct 11, 2014 at 21:01
  • $\begingroup$ now I see what you did. I thought however that if you substitute A for A^-1 you'd have to substitute it back at the end of the proof? thats pretty much how I did it at first but I didn't think it was acceptable.. $\endgroup$
    – J L
    Oct 11, 2014 at 21:18
  • $\begingroup$ You can see that you had the right idea; the formula is valid for any invertible matrix, so it's true for $A^{-1}$ as well as for A. $\endgroup$
    – user84413
    Oct 11, 2014 at 21:20
0
$\begingroup$

For any square matrix $\;B\;$ , we have that

$$B\cdot\text{adj}\,B=|B|I\;,\;\;|B|=\det B$$

The above, for $\;B=A^{-1}\;$ , means

$$A^{-1}\cdot\text{adj}\,A^{-1}=|A|^{-1}I\;,\;\;\text{since we know}\;\;\det A^{-1}=(\det A)^{-1}\implies$$

using the product theorem for determinants, we get:

$$|A|^{-1}|\text{adj}\,A^{-1}|=\det(|A|^{-1}I)=|A|^{-n}\det I=|A|^{-n}$$

and from here the wanted equality follows at once.

$\endgroup$

You must log in to answer this question.

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