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Thread: Compute p(A) where p(x) = ...?

  1. #1
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    Compute p(A) where p(x) = ...?

    Hi! Can someone please explain this problem to me? I'm not sure what it's asking for or what I'm supposed to do...

    Compute p(A) where p(x) = x^2 - (a+d)x + (ad - bc)

    and A =

    [ a b
    c d ]

    ^ That's a matrix.


    I really don't understand the question.
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  2. #2
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    Re: Compute p(A) where p(x) = ...?

    Quote Originally Posted by csgirl504 View Post
    Hi! Can someone please explain this problem to me? I'm not sure what it's asking for or what I'm supposed to do...

    Compute p(A) where p(x) = x^2 - (a+d)x + (ad - bc)

    and A =

    [ a b
    c d ]

    ^ That's a matrix.


    I really don't understand the question.
    This doesn't really make much sense. If you were to substitute the matrix, you would end up with a matrix plus a scalar, which obviously does not have matching dimensions...

    Maybe you need to substitute $\displaystyle \displaystyle |\mathbf{A}|$ instead?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Compute p(A) where p(x) = ...?

    Quote Originally Posted by csgirl504 View Post
    Compute p(A) where p(x) = x^2 - (a+d)x + (ad - bc)
    and A =
    [ a b
    c d ]
    If $\displaystyle p(x)=a_nx^n+\ldots +a_1x+a_0$ then, by definition $\displaystyle p(A)=a_nA^n+\ldots +a_1A+a_0I$ .

    P.S. If you have covered the Cayley-Hamilton theorem you can immediately find that in our case the solution is the zero matrix
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Compute p(A) where p(x) = ...?

    Quote Originally Posted by Prove It View Post
    This doesn't really make much sense. If you were to substitute the matrix, you would end up with a matrix plus a scalar, which obviously does not have matching dimensions...
    See my previous post.
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