anonymous
  • anonymous
Quick question...I'm given the matrix of the inverse, how do I directly calculate det(A) from this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
det(inverse of A) = 1/det(A)
anonymous
  • anonymous
seriously? cool, I'll try it out thanks!
anonymous
  • anonymous
yes, you're welcome

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anonymous
  • anonymous
its a 4x4 matrix if thatchanges things
anonymous
  • anonymous
it can be used on 4x4, 3x3, etc
anonymous
  • anonymous
k cool
anonymous
  • anonymous
but i dont think finding the determinant of 4x4 matrix easy. in my school the problem is usually 2x2 or 3x3
anonymous
  • anonymous
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
anonymous
  • anonymous
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)
anonymous
  • anonymous
yeah that one I knew, I'll have to look over my notes I think
anonymous
  • anonymous
MathTy, I worked it out last night. The determinant of this 4x4 is 1! :D
anonymous
  • anonymous
@MathTy: \[A*A^{-1}=I \implies \det(A*A^{-1})=\det(I) \implies \det(A)*\det(A^{-1})=1\]\[\implies \det(A^{-1})=1/\det(A).\] :)
anonymous
  • anonymous
awesome I found it to be 1, but not nicely put like this. Does that mean that all invetible matrices have det=1?
anonymous
  • anonymous
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
anonymous
  • anonymous
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

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