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3x3 matrix inverse unit vector

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi!

    I have the 3x3 matrix for L below, which I calculated. But now I need to figure out how the equation below actually means! Is it just the inverse of L (L^-1)? I cannot proceed if I don't know this step.

    2. Relevant equations
    See image

    3. The attempt at a solution
    I put in the values in the calculator, but still get a wrong answer. You can see the 3x3 matrix and the values.
     

    Attached Files:

  2. jcsd
  3. Oct 16, 2015 #2
    Sorry, I uploaded the images twice...
     
  4. Oct 16, 2015 #3

    Mark44

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    It looks to me like they are using Cramer's Rule to get the inverse.
    Fixed
     
  5. Oct 16, 2015 #4

    Ray Vickson

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    What numerical values do you get, and why do you think they are wrong?

    Anyway, your post is confusing: it does not make clear what has been given to you (by the person that set the question) and what you have developed yourself. Is the fordmula you gave for ##L^{-1}## one that somebody gave you, or is it your own? If it IS something you developed on your own, you should know that such formulas are just about the worst possible way to invert a (numerical) matrix, because of roundoff-error issues, and because they take more work than alternative methods.
     
  6. Oct 16, 2015 #5
    Sorry if I wasn't clear enough. It's an example from a book and I try to understand how they got the results. Note that I just want to understand how they solved this part and they don't show how.
    I put in the values in my calculator and did the inverse btw., still wrong. The first image is the one with their L values
     

    Attached Files:

    • ONE.png
      ONE.png
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    • TWO.png
      TWO.png
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  7. Oct 16, 2015 #6

    Ray Vickson

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    Still not clear. In the second attachment, is the numerical matrix (below the general formula) the one given by the book, or is it the one you calculated? If it is the one given by the book, show us the numerical results of your calculation, so we can compare.

    BTW: you should just type out the matrices, not attach images. It is probably less trouble to just go ahead and type the results than it is to scan, upload, etc. Certainly, it is a lot less trouble for the READER to read typed work than attached work. Usually I will not bother to respond to messages that do not type the results.
     
  8. Oct 16, 2015 #7
    No, both are from the book, which is in PDF form . They did some calculations and got the L matrix, which is the first attached image And then, I have no idea how, they performed a matrix operation and received the values in the second image. I haven't calculated anything yet, I just need to understand how they did this this.

    i.e. , the matrix is (from the beginning):

    L= [0.9473 0.5742 0.3007
    0.0155 0.5747 0.7399
    0.3201 0.5830 0.6018 ]

    the operation (inverse of the unit vector L? The formula can is in figure two) gives:

    L^-1=[28.593 56.908 –84.259
    –76.035 –158.366 232.712
    58.460 123.169 –178.99 ]

    How exactly? :S When I type [L]^-1 in the calculator, I get:

    [L]^-1 =[109.126 226.537 -330.791
    -300.197 -630.505 918.900
    232.764 490.312 -712.583]

    completely wrong
     
  9. Oct 16, 2015 #8

    SteamKing

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    Check to see if your calculator expects to see a matrix entered by rows or by columns.

    When I put L into Wolfram Alpha, the L-1 returned is pretty close to the printed version. The differences can be attributed to round off.

    If you multiply a matrix by its inverse, you should get the identity matrix as the result. This is how you check your calculations.
     
  10. Oct 16, 2015 #9

    vela

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    That doesn't explain the difference. The inverse of the transpose is the transpose of the inverse of the original matrix.

    Did you actually try to calculate the inverse using the formula in the book?
     
  11. Oct 16, 2015 #10

    Ray Vickson

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    When I use Maple to compute the inverse, I get something close to the book's answer:

    [tex] L^{-1} = \pmatrix{ 28.8646 & 57.4696 & -85.0803 \\
    -76.8017 & -159.951 & 235.031 \\
    59.0493 & 124.386 & -180.773} [/tex]

    Rounding errors are not an issue with Maple's result, because I input the elements of the matrix ##L## as rational numbers (instead of decimals), then had Maple do the exact rational inverse, printing out the results as decimal numbers after all computations were completed. I also checked that ##L^{-1} L = I##, as it should.

    I truly do not understand how your calculator's answer was obtained, but I can suggest you try to verify at least some of the entries directly, using the algebraic formula given in the book.That might help you decide if your calculator's inversion routine is reliable or not----valuable information for the future.
     

    Attached Files:

  12. Oct 17, 2015 #11
    Oh, that's nice. Yeah, I thin that I need to check that out. It's strange that it gives me the wrong answer. But about the Maple. I tried this with Matlab by inserting:
    A=sym([0.9473 0.5742 0.3007;0.0155 0.5747 0.7399; ;0.3201 0.5830 0.6018])
    inv(A), AND got the same answer as I did with my calculator. What did you write on Maple, because I'm clearly thinking wrong here.
     
  13. Oct 17, 2015 #12
    Never mind, I'll calculate it online instead. Thanks for the help, now I know that the calculator is the issue. Sat with this in an hour or so without knowing it. Cheers!
     
  14. Oct 17, 2015 #13

    Ray Vickson

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    If I take C = your calculated Inverse(L), and test it, it fails: neither C.A or A.C is near equal to the unit matrix I.
    In fact, C.A = [[1.0001, -0.00019, 1.3589],[-0.00956,0.99436,-3.7859],[-0.00006,-0.0005, 3.9415]] = [row1,row2,row3]. The final column is [1.3589,-3.7859, 3.9415], which is nowhere near [0,0,1]. I doubt that Matlab could have given you such an incorrect inverse matrix like C, so I can only conclude that somehow your inputs must be incorrect.

    The Maple commands I used were:
    with(LinearAlgebra): <----- loads the linear algebra package
    A:= upload_2015-10-17_0-59-49.png <---- entered as <column1|column2|column3>
    (This prints out the correct matrix A exactly as you have presented it). To get the inverse is a snap:
    B:=MatrixInverse(A);
    Here are the results, printed out to 6-digit accuracy:
    [[28.8646, 57.4696, -85.0803],[ -76.8017, -159.951, 235.031], [59.0493, 124.386, -180.773]] = [row1,row2,row3].
     

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