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What is the inverse of the 3x3 matrix mod 26

  1. Jul 13, 2014 #1
    1. The problem statement, all variables and given/known data
    What is the inverse of the 3x3 matrix mod 26?
    K = [tex]
    \begin{pmatrix}
    17 & 17 & 5\\
    21 & 18 & 21\\
    2 & 2 & 19
    \end{pmatrix}
    [/tex]




    2. Relevant equations



    3. The attempt at a solution
    So I found all the cofactors and then took the transpose of the matrix. I then divided new matrix, by the determinate -939. After which I would multiply this by 17 because 23-1 mod 26 = 17 to get the inverse. I found 17 by using the euclidean algorithm. This was all UPLOADED. However I am confused because even if I do this I do not get the answer in the book. They get:

    [tex]
    \begin{pmatrix}
    4 & 9 & 15\\
    15 & 17 & 6\\
    24 & 0 & 17
    \end{pmatrix}
    [/tex]

    I have so far without multiplying it by 17:

    [tex]
    \begin{pmatrix}
    300/-939 & -313/-939 & 267/-939\\
    -357/-939 & 313/-939 & -252/-939\\
    6/-939 & 0 & -51/-939
    \end{pmatrix}
    [/tex]

    I realize that even if I go ahead I will not reach what the book has, what have I done wrong? All of my work has been UPLOADED.
     

    Attached Files:

  2. jcsd
  3. Jul 14, 2014 #2
    I'll do the center column - because it's easy:
    [tex]
    \begin{pmatrix}
    k_{11} & k_{12} & k_{13}\\
    k_{21} & k_{22} & k_{23}\\
    k_{31} & k_{32} & k_{33}
    \end{pmatrix}
    \begin{pmatrix}
    17 & 17 & 5\\
    21 & 18 & 21\\
    2 & 2 & 19
    \end{pmatrix} =
    \begin{pmatrix}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 1
    \end{pmatrix}
    \ \\
    \ \\
    \ \\
    A)\ 17K_{11} + 21K_{12} + 2K_{13} = 1 \\
    B)\ 17K_{11} + 18K_{12} + 2K_{13} = 0 \\
    C)\ 5K_{11} + 21K_{12} + 19K_{13} = 0 \\
    D)\ 17K_{21} + 21K_{22} + 2K_{23} = 0 \\
    E)\ 17K_{21} + 18K_{22} + 2K_{23} = 1 \\
    F)\ 5K_{21} + 21K_{22} + 19K_{23} = 0 \\
    G)\ 17K_{31} + 21K_{32} + 2K_{33} = 0 \\
    H)\ 17K_{31} + 18K_{32} + 2K_{33} = 0 \\
    I)\ 5K_{31} + 21K_{32} + 19K_{33} = 1 \\
    \ \\
    \ \\
    Modulo 26\ table\ for\ Y=3X: 0, 3, 6, 9, 12, 15, 18, 21, 24, 1, 4, 7, 10, 13, 16, 19, 22, 25, ...\\
    A-B)\ \ \ 3K_{12} = 1 \\
    A-B/3)\ K_{12} = 9 \\
    D-E)\ \ \ 3K_{22} = 25 \\
    D-E/3)\ K_{22} = 17 \\
    G-H)\ \ \ 3K_{32} = 0 \\
    G-H/3)\ 3K_{32} = 0 \\
    \ \\
    \ \\
    \begin{pmatrix}
    k_{11} & 9 & k_{13}\\
    k_{21} & 17 & k_{23}\\
    k_{31} & 0 & k_{33}
    \end{pmatrix}
    \ \\
    [/tex]
    That book answer is looking good to me.
     
  4. Jul 14, 2014 #3
    Mod 26 (-939) is 23.
    Mod 26 (1/23) is 17.
    So 17 is the right number to use.

    [tex]
    \begin{pmatrix}
    17*300 & 17*-313 & 17*267\\
    17*-357 & 17*313 & 17*-252\\
    17*6 & 0 & 17*-51
    \end{pmatrix} =
    \begin{pmatrix}
    17*14 & 17*25 & 17*7\\
    17*7 & 17*1 & 17*8\\
    17*6 & 0 & 17*1
    \end{pmatrix} =
    \begin{pmatrix}
    4 & 9 & 15\\
    15 & 17 & 6\\
    24 & 0 & 17
    \end{pmatrix}
    [/tex]
    So you already had the solution.
    If you're using a recent Windows operating system, you have a calculator with the Mod function.
     
  5. Jul 14, 2014 #4
    Ok cool so I did get the answer.
     
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