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Matrix determinants and inverses

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    If A and S are n x n matrices with S invertible, show that det(S-1AS)=det(A). [HINT: Since S-1S=In, how are det(S-1) and det(S) related?]


    2. Relevant equations



    3. The attempt at a solution

    Not sure. The only thing I can think of doing is substituting S-1S=In into det(S-1AS), so you have det(InA)=det(A), but I really don't know. Can somebody get me started so we can work through it?
     
  2. jcsd
  3. Mar 2, 2009 #2

    Dick

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    Sure. You should have proved det(AB)=det(A)*det(B). det(I)=1. Start from there.
     
  4. Mar 3, 2009 #3
  5. Mar 3, 2009 #4

    HallsofIvy

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    You don't seem to have payed any attention to the hint! "S-1S=In, how are det(S-1) and det(S) related?" I presume you know that det(AB)= det(A)det(B).
     
  6. Mar 3, 2009 #5
    Is it:

    det(S-1AS) = det(S-1S)det(A)

    =det(In)det(A)

    =1det(A)

    =det(A)


    ????
     
  7. Mar 3, 2009 #6

    Dick

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    That's true. But don't make it look like you think the matrices commute. S^(-1)AS is NOT necessarily equal to S^(-1)SA. But det(S^(-1)AS)=det(S^(-1))*det(A)*det(S). Now you can rearrange.
     
  8. Mar 3, 2009 #7
    So first put them in 3 seperate determinants like you did, then move the S-1 next to S, then put it back together? Go backwards from two determinants multiplying each other to a single determinant, which would be equal to det(S-1S) ?
     
  9. Mar 3, 2009 #8

    Dick

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    That's it exactly.
     
  10. Mar 3, 2009 #9
    Thanks for the help :smile:

    Just on a curious note, are there examples of when doing what I did in post #5 would be wrong, or is it just an incorrect way of solving?
     
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