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DLS Group Title

Find the inverse of the following Matrix.

  • one year ago
  • one year ago

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  1. e.mccormick Group Title
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    2 ways to invert any invertable matrix: Augment on the identity, solve the first for the identity and the identity you adjoined becomes the inverse. 2) Find the adjoing, multiply that by the inverse of the determinant of the original.

    • one year ago
  2. DLS Group Title
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    \[\LARGE \left[\begin{matrix}2 & 3 \\ 5 & 7\end{matrix}\right]\] Using elementary transformations only. I want to know how to think on this question or any shortcuts or such,establishing a relation between 2.3.5 and 7..

    • one year ago
  3. e.mccormick Group Title
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    Ah, using elementary transformations is method 1.

    • one year ago
  4. DLS Group Title
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    yes I know

    • one year ago
  5. e.mccormick Group Title
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    OK. There is only one shortcut, than that would be finding the determinant to see if it is invertable or not. Other than that, it is jsut elementary operations. The best ways to do those depend on preferences and if you like working with fractions or not.

    • one year ago
  6. e.mccormick Group Title
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    If you like fractions, do 5/2 times the first row and use the result to eliminate the first column from the second row. Otherwise, it is the typical back and forth.

    • one year ago
  7. DLS Group Title
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    \[\LARGE A^{-1}=\frac{A_{adjacent}}{|A|}\] I can find the A^-1 using this but I wnt toknow the method 1

    • one year ago
  8. waterineyes Group Title
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    For two by two matrix, there is a shortcut method to determine the inverse of the matrix..

    • one year ago
  9. DLS Group Title
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    anything else than these 2? and does it involve elementary operations?it is necessary to use that,i would like to know it though :)

    • one year ago
  10. e.mccormick Group Title
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    \[\LARGE \left[\begin{matrix}2 & 3 &1 &0\\ 5 & 7 &0 &1\end{matrix}\right]\] That is the only way to start it for if you are required to solve through elementary operations and not allowed to use the Adj.

    • one year ago
  11. waterineyes Group Title
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    Oh no, that is just a trick to find the inverse in a jiffy..

    • one year ago
  12. e.mccormick Group Title
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    This must always be written as: \[\large A^{-1}=\frac{1}{det(A)}adj(A)\] You can not write this as: \[\large A^{-1}=\frac{adj(A)}{det(A)}\] Because the adjoint is a matrix and there is no such thing as dividing a matrix by a number. In fact, there as no such thing as dividing a matrix by anything. You are multiplying through by a quotent.

    • one year ago
  13. DLS Group Title
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    \[\LARGE \left[\begin{matrix}2 & 3 \\ 5 &7\end{matrix}\right]=\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]A\] I would use.. R1->R1/2 R2->R2-5R1 Then, R2-->R2/7 am i correct till here?

    • one year ago
  14. e.mccormick Group Title
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    Why R2/7?

    • one year ago
  15. waterineyes Group Title
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    7-15/2 = -1/2

    • one year ago
  16. e.mccormick Group Title
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    Your first steps produce: \[\LARGE \left[\begin{matrix} 1 & 3/2 & 1/2 & 0\\ 0 & -1/2 & -5/2 & 1 \end{matrix}\right]\]

    • one year ago
  17. e.mccormick Group Title
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    OK. Want to see the rest with elementary row ops from that point, or work it and have us check? Either way, I can explain the why along the way.

    • one year ago
  18. DLS Group Title
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    i got it ! thanks! :)

    • one year ago
  19. e.mccormick Group Title
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    OK. So you saw what we were talking about? I finished with: R1+3R2-->R1 \[\LARGE \left[\begin{matrix} 1 & 0 & -14/2 & 3\\ 0 & -1/2 & -5/2 & 1 \end{matrix}\right] \] -2R2-->R2 \[ \LARGE \left[\begin{matrix} 1 & 0 & -14/2 & 3\\ 0 & 1 & 5 & -2 \end{matrix}\right] \]

    • one year ago
  20. DLS Group Title
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    yeah!

    • one year ago
  21. e.mccormick Group Title
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    Which obviusly simplifies a little... hehe.. -7

    • one year ago
  22. e.mccormick Group Title
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    Yah, once you get the basic principal, this is easy to do. Only problem is arithmetic errors. Those still kick me in the anatomy at times...

    • one year ago
  23. DLS Group Title
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    is there any shortcut in this elementary method?

    • one year ago
  24. e.mccormick Group Title
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    Not really. It is kind of the brute force thing. However, it works pretty easy no matter the size of the matrix. The problem with th adjoint method is very large matrixes where finding the cofactors becomes a recursive loop of determinants.

    • one year ago
  25. DLS Group Title
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    hmm..okay! its more or less like Rubik's cube. :/

    • one year ago
  26. e.mccormick Group Title
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    By recursive loop of determinants I mean: Start with a \(10\times 10\) matrix. To find the minor of the first row, first column, you need the determinant of a sub matrix that is everything remaining when you co0ver up the first row and first column. This means you need to get the determinant of a \(9\times 9\) matrix. Now you loop into another problem because to get this, guess what you need? The determinant of an \(8\times 8\) matrix... wheee. Where as by doing the augment with \(\mathrm{I}_n\) and solve, it becomes the shortcut for large matrices. For \(3\times 3\) and smaller, the adjoint method can be faster. It can even work faster for \(4\times 4\). But for really large ones, it just becomes too much work.

    • one year ago
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