I am getting det(INV(A)) = det(A)

Apr 2019
4
0
india
Sorry for dumb question :D
\(\displaystyle

INV(A) = (Adj(A))/|A|

\)

now apply det on both sides

as we know


\(\displaystyle
|adj(A)| = |A|^2
\)
 
Last edited:

romsek

Math Team
Sep 2015
2,753
1,534
USA
I'm not seeing a question here.
 
Apr 2019
4
0
india
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 a moderator:

skipjack

Forum Staff
Dec 2006
21,301
2,377
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.
 
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Apr 2019
4
0
india
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.
 
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