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Math Help - Determinant of a matrix

  1. #1
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    Determinant of a matrix

    Hello everyone,
    Please help me with finding the determinant of
    \begin{vmatrix}x&-1&0&\cdots &0&0\\0&x&-1&\codts &0&0\\0&0&x&\ddots&\vdots&\vdots\\ \vdots&\vdot&\vdots &\ddots&-1&0\\0&0&0&\cdots&x&-1\\a_n&a_{n-1}&a_{n-2}&\cdots&a_1&a_0\end{vmatrix}

    I've tried to solve by myself, but I coudln't get anywhere. I do wish to improve my skill in mathematics; I am not asking for the complete answer. Just give me what to start, then I will try again. I do appreciate your help. Thank you.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Determinant of a matrix

    D_2=\begin{vmatrix}{x}&{-1}\\{a_1}&{a_0}\end{vmatrix}=a_0x+a_1

    D_3=\begin{vmatrix}{x}&{-1}&{0}\\{0}&{x}&{-1}\\{a_2}&{a_1}&{a_0}\end{vmatrix}=xD_2+a_2\begin{  vmatrix}{-1}&{\;\;0}\\{\;\;x}&{-1}\end{vmatrix}=a_0x^2+a_1x+a_2

    Conjecture D_{n+1}=a_0x^n+a_1x^{n-1}+\ldots +a_n and use induction. The way we have found D_3 will help you.
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  3. #3
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    Re: Determinant of a matrix

    Thank you very much!
    Assuming D_{k+1} = a_0x^k+a_1x^{k-1}+\cdots +a_k,
    D_{k+2}=xD_{k+1}+(-1)^{k+3}a_{k+1}\begin{vmatrix}-1&0&\cdots&\cdots&0\\x&-1&0&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&0\\0&\cdots&\cdots&x&-1\end{vmatrix}.

    and as \begin{vmatrix}-1&0&\cdots&\cdots&0\\x&-1&0&\cdots&0\\ \vdots&\vdots&\ddots&\ddots&0\\0&\cdots&\cdots&x&-1\end{vmatrix}=(-1)^{k+1}, we get

    D_{k+2}=xD_{k+1}+a_{k+1}.

    Therefore

    D_{k+2}=a_0^{k+1}+a_1x^{k}\cdots+a_{k+1}

    D_2 satisfies the first assumption, therefore via mathematical induction, D_{n+1}=a_0x^n+a_1^{n-1}+\cdots+a_n\ \ \ ^\forall n\in \mathbb{Z},n\geq 0.

    Am I correct? Thank you very much for your help, and I will work even harder.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Determinant of a matrix

    Quote Originally Posted by joll View Post
    Am I correct?
    Yes, you are. Only one thing: it should be n\geq 1 instead of n\geq 0 . If n=0, we have D_{n+1}=D_1=\det [x]=x i.e. we have no a_i and the rule defining the given determinant has no sense.


    Thank you very much for your help,
    You are welcome!

    and I will work even harder.
    Good. I am pretty sure.
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  5. #5
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    Re: Determinant of a matrix

    Thank you very much.
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  6. #6
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    Re: Determinant of a matrix

    Quote Originally Posted by joll View Post
    Hello everyone,
    Please help me with finding the determinant of
    \begin{vmatrix}x&-1&0&\cdots &0&0\\0&x&-1&\codts &0&0\\0&0&x&\ddots&\vdots&\vdots\\ \vdots&\vdot&\vdots &\ddots&-1&0\\0&0&0&\cdots&x&-1\\a_n&a_{n-1}&a_{n-2}&\cdots&a_1&a_0\end{vmatrix}

    I've tried to solve by myself, but I coudln't get anywhere. I do wish to improve my skill in mathematics; I am not asking for the complete answer. Just give me what to start, then I will try again. I do appreciate your help. Thank you.
    it can also be done without using induction. just expand the determinant along the last row and you'll quickly get the answer.
    Last edited by NonCommAlg; July 10th 2011 at 02:14 AM.
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  7. #7
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    Re: Determinant of a matrix

    I see... that way, the determinant can be written

    (-1)^{n+1}a_n\begin{vmatrix}-1&0&\cdots&0&0\\x&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots&&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}+ (-1)^{n+2}a_{n-1}\begin{vmatrix} x&0&\cdots&0&0\\0&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots &&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}\dots.

    The determinent term in (n,i)cofactor (I mean, the \vmatrix part of the equation above) of the matrix is given by

    (-1)^{n-i}x^{i-1}

    Therefore

     D_{n+1}=a_0x^n+\dots a_n(am I correct?).


    I haven't been learning linear algebra for long yet, so knowing many ways to solve one problem really helps me. Thank you very much.
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  8. #8
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    Re: Determinant of a matrix

    Quote Originally Posted by joll View Post
    I see... that way, the determinant can be written

    (-1)^{n+1}a_n\begin{vmatrix}-1&0&\cdots&0&0\\x&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots&&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}+ (-1)^{n+2}a_{n-1}\begin{vmatrix} x&0&\cdots&0&0\\0&-1&\cdots&0&0\\0&x&\ddots&\vdots&\vdots\\ \vdots &&\ddots&-1&0\\0&0&\cdots&x&-1\end{vmatrix}\dots.

    The determinent term in (n,i)cofactor (I mean, the \vmatrix part of the equation above) of the matrix is given by

    (-1)^{n-i}x^{i-1}

    Therefore

     D_{n+1}=a_0x^n+\dots a_n(am I correct?).


    I haven't been learning linear algebra for long yet, so knowing many ways to solve one problem really helps me. Thank you very much.
    yes, you've got the idea but you forgot that the original determinant is (n+1) \times (n+1) not n \times n.

    so, after expanding along the last row, you'll get that the determinant is equal to

    \sum_{i=1}^{n+1}(-1)^{n+1+i}a_{n+1-i}(-1)^{n+1-i}x^{i-1}=\sum_{i=1}^{n+1}a_{n+1-i}x^{i-1}=a_n + \ldots + a_0x^n.
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  9. #9
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    Re: Determinant of a matrix

    Ah..yes, I forgot that. Thank you.
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