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.
You are watching: Det(a^-1)
$egingroup$ 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
$$det(A cdot A^-1)=det(I)$$
Can you proceed from this? asking yourself: what is $det(I)$?) take the example of the $3 imes 3$ identification matrix:
$$ i = eginpmatrix1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 endpmatrix$$
$$ det(I) = 1$$
Thanks because that contributing solution to incubadoradeartistas.comematics Stack Exchange!Please be sure to answer the question. Carry out details and also share your research!
But avoid …Asking because that help, clarification, or responding to other answers.Making statements based upon opinion; ago them up with referrals or an individual experience.
Use incubadoradeartistas.comJax to style equations. incubadoradeartistas.comJax reference.
To discover more, watch our tips on writing great answers.
article Your price Discard
Not the price you're looking for? Browse various other questions tagged linear-algebra matrices determinant inverse or asking your very own question.
exactly how to prove the determinant deserve to take any kind of real worth using only this definition of the determinant?
See more: Why Dont You Have A Seat - Why Don'T You Take A Seat Right Over Here
site design / logo design © 2021 stack Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.10.29.40598