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Inverse of Matrix

  1. Mar 4, 2016 #1
    • Member warned about posting homework with no effort shown
    1. The problem statement, all variables and given/known data

    I have to find the inverse for this generic matrix (the dimensions are not specified, but I assume its a square matrix, I don't know if that is necessary).

    ##A=\left [
    \begin{matrix}
    1 & -1 & -1 & -1 & \dots & -1 & -1 \\
    0 & 1 & -1 & -1 & \dots & -1 & -1 \\
    0 & 0 & 1 & -1 & \dots & -1 & -1 \\
    \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
    0 & 0 & 0 & 0 & \dots & 0 & 1 \\
    \end{matrix}
    \right]##

    I think there must be a clever and fast way for calculating ##A^{-1}##, but I don't know how to do it.

    Thanks in advance.
     
  2. jcsd
  3. Mar 4, 2016 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    It is necessary that matrix A be square for an inverse to exist.

    It's not clear why you need to calculate the inverse. Are you trying to solve a system of linear equations?

    In any event, the matrix A is a special kind of triangular matrix called an atomic triangular matrix, and there is a simple and easy method of calculating its inverse.
    See this article for the details:

    https://en.wikipedia.org/wiki/Triangular_matrix
     
  4. Mar 4, 2016 #3
    No, it's just an exercise. It asks explicitly to give the inverse Matrix.

    Thank you verymuch :)

    One detail, it's not an atomic matrix, is just upper triangular.
     
    Last edited: Mar 4, 2016
  5. Mar 4, 2016 #4

    SteamKing

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    I see that now. Sorry.
     
  6. Mar 4, 2016 #5
    No problem. I did it my way, I think its okey. I've assumed an ##n \times n## matrix, and found:

    ##A^{-1}=\left [
    \begin{matrix}
    1 & 1 & 2 & 3 & \dots & n-2 & n-1 \\
    0 & 1 & 1 & 2 & \dots & n-3 & n-2 \\
    0 & 0 & 1 & 1 & \dots & n-4 & n-3 \\
    \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & 0 & 0 & \dots & 0 & 1 \\
    \end{matrix}
    \right]##

    This is not right, I get trouble after the third column.
     
    Last edited: Mar 4, 2016
  7. Mar 4, 2016 #6

    Mark44

    Staff: Mentor

    Instead of jumping right in to n x n matrices, start with smaller matrices. I looked at the case with a 3 x 3 matrix, and then with a 4 x 4 matrix. There's a definite pattern that develops.
     
  8. Mar 4, 2016 #7
    Yes, I've tried that way, but wasn't that clear to me. The case for the ##3\times3## is contained in the first 3 rows and columns.

    ##A_{3\times3}^{-1}
    =\left [
    \begin{matrix}
    1 & 1 & 2 \\
    0 & 1 & 1 \\
    0 & 0 & 1 \\
    \end{matrix}
    \right]##

    Then for the ##4\times 4##:
    ##A_{4\times4}^{-1}=
    =\left [
    \begin{matrix}
    1 & 1 & 2 & 4\\
    0 & 1 & 1 & 2 \\
    0 & 0 & 1 & 1 \\
    0 & 0 & 0 & 1 \\
    \end{matrix}
    \right]##

    And for ##5\times5##

    ##A_{5\times5}^{-1}=\left [
    \begin{matrix}
    1 & 1 & 2 & 4 & 8\\
    0 & 1 & 1 & 2 & 4 \\
    0 & 0 & 1 & 1 & 2 \\
    0 & 0 & 0 & 1 & 1 \\
    0 & 0 & 0 & 0 & 1 \\
    \end{matrix}
    \right]##

    Then:

    ##A_{n\times n}^{-1}=\left [
    \begin{matrix}
    1 & 1 & 2 & 3 & \dots & 2^{n-3} & 2^{n-2} \\
    0 & 1 & 1 & 2 & \dots & 2^{n-4} & 2^{n-3} \\
    0 & 0 & 1 & 1 & \dots & 2^{n-5} & 2^{n-4} \\
    \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & 0 & 0 & \dots & 0 & 1 \\
    \end{matrix}
    \right]##
     
    Last edited: Mar 4, 2016
  9. Mar 4, 2016 #8

    Ray Vickson

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    Science Advisor
    Homework Helper

    So, was your original matrix
    [tex] \left[ \begin{matrix} 1 & -1 & -1 & -1 \\
    0 & 1 & -1 & -1 \\
    0 & 0 & 1 & -1 \\
    0 & 0 & 0 & 1
    \end{matrix} \right]
    [/tex]
    not what was intended? When did the switch occur?
     
  10. Mar 4, 2016 #9

    Mark44

    Staff: Mentor

    The original matrix was n X n, not 4 X 4.
     
  11. Mar 4, 2016 #10

    Mark44

    Staff: Mentor

    Your last matrix isn't following the pattern of the previous ones. Check the first row in the matrix just above.
     
  12. Mar 4, 2016 #11

    Ray Vickson

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    Science Advisor
    Homework Helper

    Yes, I know. But the OP did some 3x3 and 4x4 examples, and I was just looking at the 4x4 case.
     
  13. Mar 4, 2016 #12
    Its done. I did as Mark said, by trying some tractable dimensions, and then generalized it. Thanks.
     
  14. Mar 4, 2016 #13
    That 3 was just a typo. Sorry.
     
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