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

#### dcwl

Sorry for dumb question
$$\displaystyle INV(A) = (Adj(A))/|A|$$

now apply det on both sides

as we know

$$\displaystyle |adj(A)| = |A|^2$$

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#### romsek

Math Team
I'm not seeing a question here.

#### dcwl

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?

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#### skipjack

Forum Staff
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.

1 person

#### dcwl

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 $$\displaystyle |Adj(A)|=|A|^{nâˆ’1}\!$$, the result remains valid.

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