My Math Forum I am getting det(INV(A)) = det(A)

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 April 30th, 2019, 01:13 PM #1 Newbie   Joined: Apr 2019 From: india Posts: 3 Thanks: 0 I am getting det(INV(A)) = det(A) Sorry for dumb question $\displaystyle INV(A) = (Adj(A))/|A|$ now apply det on both sides as we know $\displaystyle |adj(A)| = |A|^2$ Last edited by dcwl; April 30th, 2019 at 01:30 PM.
 April 30th, 2019, 01:42 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1371 I'm not seeing a question here.
April 30th, 2019, 08:10 PM   #3
Newbie

Joined: Apr 2019
From: india

Posts: 3
Thanks: 0

Quote:
 Originally Posted by romsek I'm not seeing a question here.
On solving

$\displaystyle |Inv(A)| = |Adj(A)|/|A|$

$\displaystyle |Inv(A)| = |A|^2 / |A| = |A|$

resulting in

$\displaystyle |Inv(A)| = |A|$

but

$\displaystyle A(Inv(A)) = I$

$\displaystyle |A|*|Inv(A)| = 1$

$\displaystyle |Inv(A)| = 1 / |A|$

First and second result are not the same; where am I going wrong?

Last edited by skipjack; May 1st, 2019 at 03:55 AM.

 May 1st, 2019, 03:54 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,820 Thanks: 2159 I'll assume that $A$ has order $n$ and is invertible, so that $|Adj(A)| = |A|^{n-1}\!$. That's a consequence of $1/|A| = |Inv(A)| = |Adj(A)/|A|| = |Adj(A)|/|A|^n$, which is a corrected version of your initial thinking. Thanks from dcwl
 May 1st, 2019, 10:28 AM #5 Newbie   Joined: Apr 2019 From: india Posts: 3 Thanks: 0 Thanks mate I was just looking at the n=3 picture till now. Got it now by seeing your answer. As $\displaystyle 1/|A|$ is scalar, it gets a power of n when Det is applied. And as $\displaystyle |Adj(A)|=|A|^{n−1}\!$, the result remains valid. Last edited by skipjack; May 1st, 2019 at 06:25 PM.

 Tags deta, deta1, detinva

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