Kenshin
  • Kenshin
Hi all, I have encountered a contradiction which I cannot resolve by myself, so someone please help me out here: we know that the formula for the inverse of a 2x2 A if A is [a b; c d] then the inverse of A is (1/(ad-bc)) * [d -b; -c a] but from my textbook it says: det (A^-1) should = 1/(det(A)); because of: det(A*A^-1) = det(I) = 1 = det(A) * det(A^-1) and that makes det(A^-1) = 1/det(A) now if i was to use the 2x2 matrix as an example and do det(A), i get ad-bc, and det(A^-1) = det((1/(ad-bc)*[d -b; -c a]) is just 1! since you get (da-bc)/(ad-bc). That sure isn't right?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
ab cd inverse, isnt the inverse divided by the det?
dan815
  • dan815
it has to be 1
amistre64
  • amistre64
hard to parse your postings ... but i see where you have it

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mathmate
  • mathmate
remember det(kA)=k^2 det(A)
Kenshin
  • Kenshin
the inverse from the 2x2 inverse formula makes it (1/(ad-bc)) * [d -b] [-c a] taking the det of the above inverse makes it 1. but det (A) is just ad-bc, so according to the book, 1/(ad-bc) should be the det of the above inverse...
dan815
  • dan815
no thats not how it works
dan815
  • dan815
u cannot multiply scalar like that look what mathmate said
dan815
  • dan815
think about exapanding the scalar into your matrix and then what happens when u take determinant
Kenshin
  • Kenshin
so dan Det(A^-1) does not equal 1/(det(A))?
dan815
  • dan815
it does
amistre64
  • amistre64
what is n/n^2 ?
mathmate
  • mathmate
*determinant(kA)=k^n determinant(A) In this case, n=2, so determinant((1/(ad-bc))*A) =(1/(ad-bc))^2 determinant (A) =(ad-bc)/(ad-bc)^2 =1/(ad-bc)
amistre64
  • amistre64
let det(A) = n nd -nb -nc na whats our determinant of the inverse?
dan815
  • dan815
http://prntscr.com/8zjj4h
Kenshin
  • Kenshin
so i can't take the 1/(ad-bc) out of the bracket as a scalar for the det of the inverse? i remember det(kA) = kdet(A) though...
dan815
  • dan815
no thats what im trying to tell u
dan815
  • dan815
look at amistres post.. when u expand it over a 2by 2 matrix now u multiply the constant twice
amistre64
  • amistre64
slight mistype ... let 1/det(A) = n :)
dan815
  • dan815
|dw:1446774702144:dw|
Kenshin
  • Kenshin
oh i just read mathmate's post now, ok and thanks all, so i can't do the step involving taking the 1/(ad-bc) out as a scalar?
amistre64
  • amistre64
how would you?
dan815
  • dan815
|dw:1446774792666:dw|
Kenshin
  • Kenshin
it would have to be taken out as k^n because of the way it applies to the n x n matrix in its arrays.
dan815
  • dan815
im not too sure, yet but i want to say for 3 by 3 matrix its k^3, and n by n matrix its k^n,
dan815
  • dan815
yes
dan815
  • dan815
just a quick way to think about it is in terms of sub divisions when u take determinants
dan815
  • dan815
|dw:1446774899681:dw|
dan815
  • dan815
u can see that if u expand k through now ull have k and k^2, from each of thsoe sub detereminants
dan815
  • dan815
|dw:1446775009413:dw|
dan815
  • dan815
thus by induction u can see it really has to be k^n factored out for n by n matrices
Kenshin
  • Kenshin
thanks i got it now
dan815
  • dan815
okay :)

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