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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  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 $Adj(A) = A^{n1}\!$. 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. 
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. 