Question

Quick question...I'm given the matrix of the inverse, how do I directly calculate det(A) from this?

Answer 1

det(inverse of A) = 1/det(A)

Answer 2

seriously? cool, I'll try it out thanks!

Answer 3

yes, you're welcome

Answer 4

its a 4x4 matrix if thatchanges things

Answer 5

it can be used on 4x4, 3x3, etc

Answer 6

k cool

Answer 7

but i dont think finding the determinant of 4x4 matrix easy. in my school the problem is usually 2x2 or 3x3

Answer 8

Where does this equation come from? Is it unique, or does it derive from something else? It seems that it follows from A^-1=1/A..and the answer was right using the equation

Answer 9

i forgot where it came from. it was on my notebook. my teacher gave it to me some months ago, along with : det(transpose of A) = det (A)

Answer 10

yeah that one I knew, I'll have to look over my notes I think

Answer 11

MathTy, I worked it out last night. The determinant of this 4x4 is 1! :D

Answer 12

@MathTy: \[A*A^{-1}=I \implies \det(A*A^{-1})=\det(I) \implies \det(A)*\det(A^{-1})=1\]\[\implies \det(A^{-1})=1/\det(A).\] :)

Answer 13

awesome I found it to be 1, but not nicely put like this. Does that mean that all invetible matrices have det=1?

Answer 14

What do you mean "not nicely put like this"? xD That little proof just means that the product of the determinants of the regular matrix and the inverse matrix is always equal to 1. Not necessarily that the determinant of all invertible matrices is 1 -- this is just a clean exception. :P

Answer 15

oh ok, by simple like this I just mean that to find the determinant I used cofactors to break it down into a3x3 matrix, then used rule of Sarrus to get the det of the 3x3, which came to 1