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Math Help - Matrix Multiplication in Z9 and Z101

  1. #1
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    Matrix Multiplication in Z9 and Z101

    Hey, I present to you two problems:

    First one:

    To find the Inverse of the matrix A = \begin{matrix} 4 & 2\\7 & 4\end{matrix} in  \mathbb{Z}9.

    I have worked out the determinent which is equal to 12, but in  \mathbb{Z}9 that is equal to 3.
    I know I have to inverse the determinent in \mathbb{Z}9 but I am a little unsure about how to do that. But I'm stuck at the line:
    A.^-1 = \frac{1}{3}\begin{matrix} 4 & -2\\-7 & 4\end{matrix} in  \mathbb{Z}9

    Second one:

    Given a matrix:
    M = \begin{matrix} 6 & 31 \\ 37 & 95\end{matrix} in  \mathbb{Z}101. work out M^2 and find M^{2*m*n} where m = 34 and n = 84.

    I have worked out M^2 but I'm not sure how to proceed with the other part, I think I must use Euler's formula, but that specifies for two numbers pq that are prime, clearly 34 and 84 are not.

    Regards
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  2. #2
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    Quote Originally Posted by ramdrop View Post
    Hey, I present to you two problems:

    First one:

    To find the Inverse of the matrix A = \begin{matrix} 4 & 2\\7 & 4\end{matrix} in  \mathbb{Z}9.

    I have worked out the determinent which is equal to 12, but in  \mathbb{Z}9 that is equal to 3.


    Nop. That matrix's determinant is 4\cdot 4-7\cdot 2=2=2\!\!\pmod 9 ...., and thus:

    \displaystyle{A^{-1}=\frac{1}{2}\begin{pmatrix}4&-2\\-7&4\end{pmatrix}=5\,\begin{pmatrix}4&-2\\-7&4\end{pmatrix}=\begin{pmatrix}2&-1\\1&2\end{pmatrix}

    because 2^{-1}=5\!\!\pmod 9 .

    Tonio


    I know I have to inverse the determinent in \mathbb{Z}9 but I am a little unsure about how to do that. But I'm stuck at the line:
    A.^-1 = \frac{1}{3}\begin{matrix} 4 & -2\\-7 & 4\end{matrix} in  \mathbb{Z}9

    Second one:

    Given a matrix:
    M = \begin{matrix} 6 & 31 \\ 37 & 95\end{matrix} in  \mathbb{Z}101. work out M^2 and find M^{2*m*n} where m = 34 and n = 84.

    I have worked out M^2 but I'm not sure how to proceed with the other part, I think I must use Euler's formula, but that specifies for two numbers pq that are prime, clearly 34 and 84 are not.

    Regards
    .
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  3. #3
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    I don't know why I have written 12 here, but I have 2 on my piece of paper, many thanks!
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  4. #4
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    Hello, ramdrop!

    \text{Find the Inverse of the matrix: }\:A \:=\:\begin{pmatrix} 4 & 2\\7 & 4\end{pmatrix}\:\text{ in }\mathbb{Z}9.

    I used the Augmented Matrix approach.

    \text{We have: }\;\left[\begin{array}{cc|cc} 4 & 2 & 1 & 0 \\ 7 & 4 & 0 & 1 \end{array}\right]\;\text{(mod 9)}


    \begin{array}2R_1 - R_2 \\  \\ \end{array} \left[\begin{array}{cc|cc}<br />
1 & 0 & 2 & \text{-}1 \\ 7 & 4 & 0 & 1 \end{array}{\right] \text{(mod 9)} \;\equiv \;\left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 7 & 4 & 1 & 2\end{array}\right]\text{(mod 9)}


    \begin{array}{c} \\ R_2-7R_1\end{array} \left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 0 & 4 & \text{-}14 & \text{-}55 \end{array}\right] \text{(mod 9)} \;\equiv \;\left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 0 & 4 & 4 & 8 \end{array}\right]\text{(mod 9)}


    . . \begin{array}{c} \\ \frac{1}{4}R_2 \end{array} \left[\begin{array}{cc|cc}1 & 0 & 2 & 8 \\ 0 & 1 & 1 & 2\end{array}\right] \text{(mod 9)}


    \text{Therefore: }\:A^{-1} \;\equiv\;\begin{bmatrix}2 & 8 \\ 1 & 2 \end{bmatrix}\text{(mod 9)}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Check

    \begin{bmatrix}4 & 2 \\ 7 & 4 \end{bmatrix}\begin{bmatrix}2 & 8 \\ 1 & 2 \end{bmatrix} \;=\; \begin{bmatrix}10 & 36 \\ 18 & 64\end{bmatrix} \;\equiv\;\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\text{(mod 9)}

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  5. #5
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    That works very nicely, but I'm still stuck on the second question.
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