My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum


Thanks Tree1Thanks
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
April 30th, 2019, 01:13 PM   #1
Newbie
 
Joined: Apr 2019
From: india

Posts: 4
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.
dcwl is offline  
 
April 30th, 2019, 01:42 PM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,549
Thanks: 1399

I'm not seeing a question here.
romsek is offline  
April 30th, 2019, 08:10 PM   #3
Newbie
 
Joined: Apr 2019
From: india

Posts: 4
Thanks: 0

Quote:
Originally Posted by romsek View Post
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.
dcwl is offline  
May 1st, 2019, 03:54 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 20,968
Thanks: 2216

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
skipjack is offline  
May 1st, 2019, 10:28 AM   #5
Newbie
 
Joined: Apr 2019
From: india

Posts: 4
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.
dcwl is offline  
Reply

  My Math Forum > College Math Forum > Linear Algebra

Tags
deta, deta1, detinva



Thread Tools
Display Modes






Copyright © 2019 My Math Forum. All rights reserved.