Question

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?

Answer 1

ab
cd

inverse, isnt the inverse divided by the det?

Answer 2

it has to be 1

Answer 3

hard to parse your postings ... but i see where you have it

Answer 4

remember det(kA)=k^2 det(A)

Answer 5

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...

Answer 6

no thats not how it works

Answer 7

u cannot multiply scalar like that look what mathmate said

Answer 8

think about exapanding the scalar into your matrix and then what happens when u take determinant

Answer 9

so dan Det(A^-1) does not equal 1/(det(A))?

Answer 10

it does

Answer 11

what is n/n^2 ?

Answer 12

*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)

Answer 13

let det(A) = n

nd -nb
-nc na

whats our determinant of the inverse?

Answer 14

http://prntscr.com/8zjj4h

Answer 15

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...

Answer 16

no thats what im trying to tell u

Answer 17

look at amistres post.. when u expand it over a 2by 2 matrix now u multiply the constant twice

Answer 18

slight mistype ... let 1/det(A) = n :)