At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
det(inverse of A) = 1/det(A)
seriously? cool, I'll try it out thanks!
yes, you're welcome
its a 4x4 matrix if thatchanges things
it can be used on 4x4, 3x3, etc
but i dont think finding the determinant of 4x4 matrix easy. in my school the problem is usually 2x2 or 3x3
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
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)
yeah that one I knew, I'll have to look over my notes I think
MathTy, I worked it out last night. The determinant of this 4x4 is 1! :D
awesome I found it to be 1, but not nicely put like this. Does that mean that all invetible matrices have det=1?
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
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