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Math Help - what will be the determinant of the matrix A

  1. #1
    Junior Member sorv1986's Avatar
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    Lightbulb what will be the determinant of the matrix A

    Let A be 3 by 3 matrix over real numbers satisfying A⁻= I -2A. Then what will be the

    determinant of the matrix A?

    thanks in advance. regards.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by sorv1986 View Post
    Let A be 3 by 3 matrix over real numbers satisfying A⁻= I -2A. Then what will be the determinant of the matrix A?
    Perhaps (perhaps) that is not the exact formulation of the problem. Could you review it?
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    Re: what will be the determinant of the matrix A

    There are two possible answers. First, the determinant of a matrix is the same as the product of all of its eigenvalues. Second, if A satisfies that equation, its eigenvalues must also. You can manipulate the equation to a quadratic that has two solutions. Since A is 3 by 3, it has three eigenvalues. Since only two numbers sastisfy that equation, one of them must be a duplicate eigenvalue. The determinant depends upon which of them is the duplicate.
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  4. #4
    Junior Member sorv1986's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by FernandoRevilla View Post
    Perhaps (perhaps) that is not the exact formulation of the problem. Could you review it?
    i m sry .but that was the only information given about the matriX A.
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    Re: what will be the determinant of the matrix A

    det(AB) = det(A)det(B)

    Determinant is non-zero if it is invertible

    det(A) = 1/(det(A^-1))
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    MHF Contributor FernandoRevilla's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by HallsofIvy View Post
    The determinant depends upon which of them is the duplicate.
    But that is not possible. We have A^{-1}=I-2A\Leftrightarrow 2A^2-A+I=0 . That is , p(\lambda)=2(\lambda-\lambda_1)(\lambda-\lambda_2) is an annihilator polynomial of A being \lambda_1=(1+\sqrt{7}i)/4 and \lambda_2=(1-\sqrt{7}i)/4 . As A\in\mathbb{R}^{3\times 3} , A-\lambda_1I\neq 0 and A-\lambda_2I\neq 0 . This means that the minimal polynomial of A is just (1/2)p(\lambda) and \lambda_1,\lambda_2 are eigenvalues of A as you said. The third one \lambda_3 should be necessarily real and A is diagonalizable in \mathbb{C} . So, A is similar (in \mathbb{C}) to D=\textrm{diag}\;(\lambda_1,\lambda_2,\lambda_3) and as a consequence:

    2D^2-D+I\Rightarrow 2\lambda_i^2-\lambda_i+1=0\;(i=1,2,3)

    Again, as you said \lambda_3=\lambda_1 or \lambda_3=\lambda_2 which implies \det A=\det D=\lambda_1\lambda_2\lambda_3=(1/2)\lambda_3\not\in\mathbb{R} (contradiction).
    Last edited by FernandoRevilla; November 24th 2011 at 11:54 PM.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by CaramelCardinal View Post
    Determinant is non-zero if it is invertible det(A) = 1/(det(A^-1))
    Rigorously true.
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  8. #8
    Junior Member sorv1986's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by FernandoRevilla View Post
    But that is not possible. We have A^{-1}=I-2A\Leftrightarrow 2A^2-A+I=0 . That is , p(\lambda)=2(\lambda-\lambda_1)(\lambda-\lambda_2) is an annihilator polynomial of A being \lambda_1=(1+\sqrt{7}i)/4 and \lambda_2=(1-\sqrt{7}i)/4 . As A\in\mathbb{R}^{3\times 3} , A-\lambda_1I\neq 0 and A-\lambda_2I\neq 0 . This means that the minimal polynomial of A is just (1/2)p(\lambda) and \lambda_1,\lambda_2 are eigenvalues of A as you said. The third one \lambda_3 should be necessarily real and A is diagonalizable in \mathbb{C} . So, A is similar (in \mathbb{C}) to D=\textrm{diag}\;(\lambda_1,\lambda_2,\lambda_3) and as a consequence:

    2D^2-D+I\Rightarrow 2\lambda_i^2-\lambda_i+1=0\;(i=1,2,3)

    Again, as you said \lambda_3=\lambda_1 or \lambda_3=\lambda_2 which implies \det A=\det D=\lambda_1\lambda_2\lambda_3=(1/2)\lambda_3\not\in\mathbb{R} (contradiction).
    how [COLOR=rgb(0, 0, 0)]lambda3=lambda1 or lambda3= lambda2[/COLOR] is possible? lambda_3 must be real as the matrix is 3 by 3.Is that problem is wrong or something? Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me
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  9. #9
    Junior Member sorv1986's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by CaramelCardinal View Post
    det(AB) = det(A)det(B)

    Determinant is non-zero if it is invertible

    det(A) = 1/(det(A^-1))
    how the det of A^-1 i.e. det (I-2A) would be calculated?
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: what will be the determinant of the matrix A

    Quote Originally Posted by sorv1986 View Post
    Is that problem is wrong or something?
    Plainly: A does not exist.

    Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me
    The minimal polynomial \mu (\lambda) of A divides to any annihilating polynomial of A so \mu(\lambda)=\lambda-\lambda_1 or \mu(\lambda)=\lambda-\lambda_2 or \mu(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2) . As A-\lambda_1I\neq 0 and A-\lambda_2I\neq 0 we conclude that \mu(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2) is the minimal polynomial of A .
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