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PhoenixFire

  • 2 years ago

Use elementary row operations to find the inverse of \left[\begin{matrix}-2 & -1 & 3 \\ 4 & -2 & 1 \\ 0 & 4 & -2\end{matrix}\right] I know the inverse as I've done it via cofactor and adjugate. And wolframalpha has confirmed the result. But no matter how many times I try with the elementary row operations I can't get it right.

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  1. UnkleRhaukus
    • 2 years ago
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    \[\qquad\qquad\quad\quad[\textbf A|\textbf I]\]\[\left[\begin{matrix}-2 & -1 & 3 \\ 4 & -2 & 1 \\ 0 & 4 & -2\end{matrix}\right|\left.\begin{matrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 &0 & 1\end{matrix}\right]\]

  2. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}-2 & -1 & 3 \\ 0 & -4 & 7 \\ 0 & 4 & -2\end{matrix}\right|\left.\begin{matrix}1 & 0 & 0\\ 2 & 1 & 0 \\ 0 &0 & 1\end{matrix}\right]\qquad R_2\rightarrow R_2+2R_1\]

  3. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}-2 & -1 & 3 \\ 0 & -4 & 7 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 0 & 0\\ 2 & 1 & 0 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_3\rightarrow R_3+R_2\]

  4. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}2 & 1 & 2 \\ 0 & -4 & 7 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 1\\ 2 & 1 & 0 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_1\rightarrow -(R_1-R_3)\]

  5. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}2 & 1 & 2 \\ 0 & 4 & -2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 1\\ 0 & 0 & 1 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_2\rightarrow -(R_2-R_3)\]

  6. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}2 & 1 & 2 \\ 0 & 1 & -1/2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 1\\ 0 & 0 & 1/4 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_2\rightarrow R_2/4\]

  7. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}2 & 0& 5/2 \\ 0 & 1 & -1/2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1 & 1 & 3/4\\ 0 & 0 & 1/4 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_1\rightarrow R_1-R_2\]

  8. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}1 & 0& 5/4 \\ 0 & 1 & -1/2 \\ 0 & 0& 5 \end{matrix}\right|\left.\begin{matrix}1/2 & 1/2 & 3/8\\ 0 & 0 & 1/4 \\ 2 &1 & 1 \end{matrix}\right]\qquad R_1\rightarrow R_1/2\]

  9. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}1 & 0& 5/4 \\ 0 & 1 & -1/2 \\ 0 & 0& 1 \end{matrix}\right|\left.\begin{matrix}1/2 & 1/2 & 3/8\\ 0 & 0 & 1/4 \\ 2/5 &1/5 & 1/5 \end{matrix}\right]\qquad R_3\rightarrow R_3/5\]

  10. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}1 & 0& 5/4 \\ 0 & 1 & 0 \\ 0 & 0& 1 \end{matrix}\right|\left.\begin{matrix}1/2 & 1/2 & 3/8\\ 1/5 & 1/10 & 7/20 \\ 2/5 &1/5 & 1/5 \end{matrix}\right]\qquad R_2\rightarrow R_2+R_3/2\]

  11. UnkleRhaukus
    • 2 years ago
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    \[\left[\begin{matrix}1 & 0& 0 \\ 0 & 1 & 0 \\ 0 & 0& 1 \end{matrix}\right|\left.\begin{matrix}0 & 1/4 & 1 /8\\ 1/5 & 1/10 & 7/20 \\ 2/5 &1/5 & 1/5 \end{matrix}\right]\qquad R_1\rightarrow R_1-5 R_3/4\]\[\qquad\qquad[\textbf I |\textbf A^{-1}]\]

  12. rajathsbhat
    • 2 years ago
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    wow that is a lot of work^

  13. UnkleRhaukus
    • 2 years ago
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    \[\color{gray} {\begin{matrix}0 & & 0 \\ & & \\ & \smile & \end{matrix}} \]

  14. PhoenixFire
    • 2 years ago
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    Wow thank you @UnkleRhaukus !! I understand why you make below the diagonal 0, but then you started working on the top triangle. Does the order matter? For example can I make the bottom triangle 0 then work down the diagonal making them 1? Is making the bottom triangle 0 the really only required step to be done first?

  15. UnkleRhaukus
    • 2 years ago
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    you can perform any of the allowed row operations at any time and in any order, some ways are certainly quicker than others , im not sure if the steps i used were the simplest, but they did work

  16. PhoenixFire
    • 2 years ago
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    Okay, thank you again.

  17. UnkleRhaukus
    • 2 years ago
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    http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi

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