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Math Help - Matrix Algebra - fairly basic

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    Matrix Algebra - fairly basic

    Using the definition of the inverse of a matrix and the principles of matrix algebra, prove that if A and B are invertible matrices of the same order, then AB is invertible and (AB)^{-1} =  B^{-1} A^{-1} .

    Please help.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by WWTL@WHL View Post
    Using the definition of the inverse of a matrix and the principles of matrix algebra, prove that if A and B are invertible matrices of the same order, then AB is invertible and (AB)^{-1} =  B^{-1} A^{-1} .

    Please help.
    Hint: recall that if a matrix A is invertible, and we denote A^{-1} as its inverse matrix, then AA^{-1} = I and A^{-1}A = I

    here, we simply need to show that (AB) \left( B^{-1}A^{-1}\right) = \left( B^{-1}A^{-1} \right)(AB) = I, and the desired result follows immediately. use the fact that if A,B,C,D are matrices, then (AB)(CD) = A(BC)D
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    Yes, that makes sense. But I don't understand why you'd need to use this fact . How does that relate to the ?

    Could you please expand on this a bit more, or give a more detailed explanation please? I am really, really poor at maths.

    Thanks.
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    Quote Originally Posted by WWTL@WHL View Post


    Yes, that makes sense. But I don't understand why you'd need to use this fact . How does that relate to the ?

    Could you please expand on this a bit more, or give a more detailed explanation please?
    i give you one more hint.

    Consider (AB) \left( B^{-1}A^{-1} \right)

    in light of the fact that i told you to note (it would also do you well to recall the properties of the identity matrix), we will have that:

    (AB) \left( B^{-1}A^{-1} \right) = A \left(B B^{-1} \right)A^{-1}

    I am really, really poor at maths.
    you and me both
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     = A (I) A^{-1}

     = (I) A A^{-1}

     = I^2

    It looks like I've committed a mathematical sin there, and I don't think I've got anywhere.

    Just realised I^2 = I!! Thanks, I think I've got it.

    Thanks so much Jhevon!!!
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    Quote Originally Posted by WWTL@WHL View Post


     = A (I) A^{-1}

     = (I) A A^{-1}

     = I^2

    It looks like I've committed a mathematical sin there, and I don't think I've got anywhere.
    i told you to recall the properties of the identity matrix.

    Definition: If we have a matrix, I, such that IA = AI = A for any matrix A, then we call I the identity matrix. It is a matrix of the same order of A with a diagonal of 1's on it's main diagonal and 0's everywhere else. Multiplying any matrix by the identity matrix gives the original matrix itself

    in other words: AIA^{-1} = AA^{-1}
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    Quote Originally Posted by Jhevon View Post
    i told you to recall the properties of the identity matrix.

    Definition: If we have a matrix, I, such that IA = AI = A for any matrix A, then we call I the identity matrix. It is a matrix of the same order of A with a diagonal of 1's on it's main diagonal and 0's everywhere else. Multiplying any matrix by the identity matrix gives the original matrix itself

    in other words: AIA^{-1} = AA^{-1}
    Yes - of course.

    Thank you so much for your patience, and I'm sorry you had to walk me through this one. I really should've understood it. I'm only 2 weeks in to my maths course at university, and I'm already stuggling. Ho-hum.....
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by WWTL@WHL View Post
    Yes - of course.

    Thank you so much for your patience, and I'm sorry you had to walk me through this one. I really should've understood it. I'm only 2 weeks in to my maths course at university, and I'm already stuggling. Ho-hum.....
    that's fine. i understand what it is to struggle with math, really. now can you complete the proof?
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    Quote Originally Posted by Jhevon View Post
    that's fine. i understand what it is to struggle with math, really. now can you complete the proof?
    Yep.
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    Quote Originally Posted by WWTL@WHL View Post
    Yep.
    good!

    (you can post it if you want to make sure everything is in order)
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