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Math Help - Proving adj(a^trace)=[adj(a)]^trace?

  1. #1
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    Proving adj(a^trace)=[adj(a)]^trace?

    Hello, I had 3 homework problems that I am having trouble with. I really have problems when it comes to proving things. Here goes

    Assume A is an n x n invertible matrix. Prove the following identities:

    a) adj( A^trace) = [adj(A)]^trace

    b) adj( A^-1) = [adj(A)]^-1

    c) adj(k*A) = k^(n-1) * adj(A), k E Real Number.


    So I've tried a bit, starting out with

    A^-1 = 1/det(a) * adj(a) then
    A^(-1) * det(a) = adj(a) then I do the trace of both sides and don't really know where to go from there...

    [det(a)*A^(-1)]^trace = [adj(A)]^trace

    CAn anyone help me out?
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  2. #2
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    Quote Originally Posted by forensic91 View Post
    Hello, I had 3 homework problems that I am having trouble with. I really have problems when it comes to proving things. Here goes

    Assume A is an n x n invertible matrix. Prove the following identities:

    a) adj( A^trace) = [adj(A)]^trace

    b) adj( A^-1) = [adj(A)]^-1

    c) adj(k*A) = k^(n-1) * adj(A), k E Real Number.


    So I've tried a bit, starting out with

    A^-1 = 1/det(a) * adj(a) then
    A^(-1) * det(a) = adj(a) then I do the trace of both sides and don't really know where to go from there...

    [det(a)*A^(-1)]^trace = [adj(A)]^trace

    CAn anyone help me out?
    for a) and c) use the definition of adjoint matrix (cofactors!). part b) is trivial because: I=\text{adj}(AA^{-1})=\text{adj}(A^{-1}) \text{adj}(A).
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