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Math Help - Matrix Inverse

  1. #1
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    Matrix Inverse

    Using Row Operations, find the inverse of  \begin{pmatrix}2&3&4\\0&4&1\\1&0&3 \end{pmatrix} in  M_3 (\mathbb{F}_7) .
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  2. #2
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    Hello, funnyinga!

    Using Row Operations, find the inverse of: . A \;=\; \begin{bmatrix}2&3&4\\0&4&1\\1&0&3 \end{bmatrix}

    We have: . \left[ \begin{array}{ccc|ccc}2 &3&4&1&0&0 \\ 0&4&1&0&1&0\\ 1&0&3&0&0&1\end{array}\right]


    . . \begin{array}{c}\text{Switch} \\ R_1 \& R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&4&1&0&1&0\\2&3&4&1&0&0 \end{array}\right]


    \begin{array}{c}\\ \\ R_3-2R_1 \end{array} \left[\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&4&1&0&1&0\\ 0&3&\text{-}2&1&0&\text{-}2 \end{array}\right]


    \begin{array}{c} \\ R_2-R_3 \\ \\ \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1\\ 0&1&3&\text{-}1&1&2\\ 0&3&\text{-}2&1&0&\text{-}2 \end{array}\right]


    \begin{array}{c}\\ \\ R_3-3R_2 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\ 0&1&3&\text{-}1&1&2 \\ 0&0&\text{-}11 &4&\text{-}3 & \text{-}8 \end{array}\right]


    . . \begin{array}{c} \\ \\ \text{-}\frac{1}{11}R_3 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\0&1&3&\text{-}1&1&2 \\ 0&0&1&\text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11} \end{array}\right]


    \begin{array}{c}R_2-3R_3 \\ R_2-3R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&0 & \frac{12}{11} & \text{-}\frac{9}{11} & \text{-}\frac{13}{11} \\ \\[-4mm] 0&1&0 & \frac{1}{11} & \frac{2}{11} & \text{-}\frac{2}{11} \\ \\[-4mm]  0&0&1& \text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11}\end{array}\right]


    Therefore: . A^{-1} \;=\;\frac{1}{11}\begin{bmatrix}12 &\text{-}9 &\text{-}13 \\ 1&2&\text{-}2 \\ \text{-}4 &3&8 \end{bmatrix}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, funnyinga!


    We have: . \left[ \begin{array}{ccc|ccc}2 &3&4&1&0&0 \\ 0&4&1&0&1&0\\ 1&0&3&0&0&1\end{array}\right]


    . . \begin{array}{c}\text{Switch} \\ R_1 \& R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&4&1&0&1&0\\2&3&4&1&0&0 \end{array}\right]


    \begin{array}{c}\\ \\ R_3-2R_1 \end{array} \left[\begin{array}{ccc|ccc}1&0&3&0&0&1 \\ 0&4&1&0&1&0\\ 0&3&\text{-}2&1&0&\text{-}2 \end{array}\right]


    \begin{array}{c} \\ R_2-R_3 \\ \\ \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1\\ 0&1&3&\text{-}1&1&2\\ 0&3&\text{-}2&1&0&\text{-}2 \end{array}\right]


    \begin{array}{c}\\ \\ R_3-3R_2 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\ 0&1&3&\text{-}1&1&2 \\ 0&0&\text{-}11 &4&\text{-}3 & \text{-}8 \end{array}\right]


    . . \begin{array}{c} \\ \\ \text{-}\frac{1}{11}R_3 \end{array} \left[\begin{array}{ccc|ccc} 1&0&3&0&0&1 \\0&1&3&\text{-}1&1&2 \\ 0&0&1&\text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11} \end{array}\right]


    \begin{array}{c}R_2-3R_3 \\ R_2-3R_3 \\ \end{array} \left[\begin{array}{ccc|ccc}1&0&0 & \frac{12}{11} & \text{-}\frac{9}{11} & \text{-}\frac{13}{11} \\ \\[-4mm] 0&1&0 & \frac{1}{11} & \frac{2}{11} & \text{-}\frac{2}{11} \\ \\[-4mm] 0&0&1& \text{-}\frac{4}{11} & \frac{3}{11} & \frac{8}{11}\end{array}\right]


    Therefore: . A^{-1} \;=\;\frac{1}{11}\begin{bmatrix}12 &\text{-}9 &\text{-}13 \\ 1&2&\text{-}2 \\ \text{-}4 &3&8 \end{bmatrix}
    i hate to add anything to Soroban's beautiful work, but the last step needs a little care! since everything here is in \mathbb{F}_7, we have \frac{1}{11}=11^{-1}=2 and so: A^{-1}=\begin{pmatrix}3 & 3 & 2 \\ 2 & 4 & 3 \\ 6 & 6 & 2 \end{pmatrix}.
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  4. #4
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    Hello, NonCommAlg!

    I was so involved with the inverse, I forgot the modulo theme.

    Thanks for catching it!

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