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Math Help - Prove that an nxn matrix is invertible

  1. #1
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    Prove that an nxn matrix is invertible

    Afternoon all,

    The question:

    Let A=\begin{bmatrix}1&x_1&x_1^2&...&x_1^{n-1}\\1&x_2&x_2^2&...&x_2^{n-1}\\1&.&.&...&.\\1&x_n&x_n^2&...&x_n^{n-1}\\\end{bmatrix} where x_1,x_2,...,x_n are distinct real numbers. Show that A is invertible.

    How far I have got:

    So there are several ways to do this from what I gather. The way I have tried is by trying to prove that AB=BA=I where B is a unique nxn matrix.

    So looking at my old uni notes I have:

    AB=I\implies C(AB)=CI\implies (CA)B=C

    The next part is where I get confused because I go on to say that CA=I \implies B=C \therefore A is invertible

    Can anyone explain to me if this is correct and why it is so?

    Thanks

    Chris
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  2. #2
    MHF Contributor

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    Tejas
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    Re: Prove that an nxn matrix is invertible

    if CA= I, and AB = I, then B = C:

    C =C I = C(AB) = (CA)B = IB = B.

    thus B ( = C) is "one inverse" for A. suppose B' is another, so that B'A = AB' = I:

    B' = B'I = B'(AB) = (B'A)B = IB = B.

    so a matrix that is invertible can only have one inverse, and if a matrix has a left-inverse and a right-inverse, both the left-inverse and the right-inverse

    are the same matrix, A^-1.

    so to find an inverse for A, you can just find a right-inverse or a left-inverse, which saves time checking by matrix multiplication.

    personally, i would check det(A) for this particular problem, because A^-1 exists iff det(A) ≠ 0.
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