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?
Stacey Warren - Expert brainly.com
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inverse, isnt the inverse divided by the det?
it has to be 1
hard to parse your postings ... but i see where you have it
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remember det(kA)=k^2 det(A)
the inverse from the 2x2 inverse formula makes it (1/(ad-bc)) * [d -b]
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...
no thats not how it works
u cannot multiply scalar like that look what mathmate said
think about exapanding the scalar into your matrix and then what happens when u take determinant
so dan Det(A^-1) does not equal 1/(det(A))?
what is n/n^2 ?
In this case, n=2, so
=(1/(ad-bc))^2 determinant (A)
let det(A) = n
whats our determinant of the inverse?
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...
no thats what im trying to tell u
look at amistres post.. when u expand it over a 2by 2 matrix now u multiply the constant twice
slight mistype ... let 1/det(A) = n :)
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?
how would you?
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.
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,
just a quick way to think about it is in terms of sub divisions when u take determinants
u can see that if u expand k through now ull have k and k^2, from each of thsoe sub detereminants
thus by induction u can see it really has to be k^n factored out for n by n matrices