Prove:$$det(incubadoradeartistas.combfA^-1) = frac1incubadoradeartistas.combfdet(A)$$

I understand that $(A)(A^-1) = I$, however I to be not certain what to perform with the knowledge.

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$egingroup$ The equation around determinants stop for all $A$ and $B$, so that holds if $B$ is miscellaneous particular. $endgroup$

$egingroup$ since $det(AB)=det Adet B$, letting $B=A^-1$ offers $det I=1=det Adet(A^-1)$, therefore $det(A^-1)=frac1det A$. $endgroup$

first of all we recognize that

$$det(A cdot B)=det(A) imesdet(B)$$

also we know that

$$A imes A^-1=I$$

we understand that

$$det(A cdot A^-1)=det(I)$$

or

$$det(A) imesdet(A^-1)=det(I)$$

Can you proceed from this? asking yourself: what is $det(I)$?) take the example of the $3 imes 3$ identification matrix:

$$ i = eginpmatrix1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 endpmatrix$$

$$ det(I) = 1$$

So,

$$det(A) imesdet(A^-1)=1$$

or

$$oxeddet(A^-1)=frac1det(A)$$

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exactly how to prove the determinant deserve to take any kind of real worth using only this definition of the determinant?

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