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How do we get the second matrix from the first one in this example

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data
    I have a matrix 1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
    0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
    0, 12, 143/12, 11/12 |-13/12,0,1,0
    0, -12, -143/12, -11/12|-11/12,0,0,1
    ---------------------------------------------------
    1 , 0, 0 , 0 |
    0, 1, 0 , 0 |
    0, 0, 1, 0 |
    0, 0, 0 , 1 |
    And I can't understand how can one get the following block matrix from the previous one with elementary row operations:
    1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
    0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
    0, 12, 143/12, 11/12 |-13/12,0,1,0
    0, -12, -143/12, -11/12|-11/12,0,0,1
    ---------------------------------------------------
    1 , 0, -1/12 ,-1/12 |
    0, 1, 0 , 0 |
    0, 0, 1, 0 |
    0, 0, 0 , 1 |

    And ideas?



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 3, 2008 #2
    I have a matrix:
    1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
    0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
    0, 12, 143/12, 11/12 |-13/12,0,1,0
    0, -12, -143/12, -11/12|-11/12,0,0,1
    ---------------------------------------------------
    1 , 0, 0 , 0 |
    0, 1, 0 , 0 |
    0, 0, 1, 0 |
    0, 0, 0 , 1 |
    And I can't understand how can one get the following block matrix from the previous one with elementary row operations:
    1, 0, 1/12, 1/12 | 1/12, 0, 0, 0
    0, 12, 143/12, 11/12 |-1/12, 1, 0, 0
    0, 12, 143/12, 11/12 |-13/12,0,1,0
    0, -12, -143/12, -11/12|-11/12,0,0,1
    ---------------------------------------------------
    1 , 0, -1/12 ,-1/12 |
    0, 1, 0 , 0 |
    0, 0, 1, 0 |
    0, 0, 0 , 1 |

    And ideas?
     
  4. Oct 3, 2008 #3
    That's a bit confusing. I see what looks like three matrices. Can you rewrite it using the latex array format? Example: \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array} gives:

    [tex]\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}[/tex]
     
  5. Oct 3, 2008 #4
    And how do I put this ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}) into a message (reply to thread)?
     
  6. Oct 3, 2008 #5
    Use the tex delimiters. Just click on the matrix in my post, you should get a popup with the source code.
     
  7. Oct 3, 2008 #6
    Could you please say where are the TEX delimeters?
     
  8. Oct 3, 2008 #7
    You type them [x] \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array} [/x] where x=tex. Use cccc for a 4x4 matrix instead of ccc.
     
  9. Oct 3, 2008 #8
    ok i think i see now how to type it
     
  10. Oct 3, 2008 #9
    [tex] Matrix A: \begin{array}{cccc} 1 & 0 & 1/12 & 1/12 \\ 0 & 12 & 143/12 & 11/12\\ 0 & 12& 143/12 & 11/12 \\ 0 & -12 & -143/12 & -11/12\end{array}
    Matrix B : \begin{array}{cccc} 1/12 & 0 & 0 & 0 \\ -1/12 & 1 & 0 & 0\\ -13/12 & 0& 1 & 0 \\ -11/12 & 0 & 0 & 1\end{array}
    Matrix C : \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0& 1 & 0 \\ 0 & 0 & 0 & 1\end{array}


    [/tex]
    They are located this way in a block matrix :
    AB
    C


    And how can I get this with elementary row operations?

    [tex] Matrix A: \begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 12 & 143/12 & 11/12\\ 0 & 12& 143/12 & 11/12 \\ 0 & -12 & -143/12 & -11/12\end{array}
    Matrix B : \begin{array}{cccc} 1/12 & 0 & 0 & 0 \\ -1/12 & 1 & 0 & 0\\ -13/12 & 0& 1 & 0 \\ -11/12 & 0 & 0 & 1\end{array}
    Matrix C : \begin{array}{cccc} 1 & 0 &-1/12 & -1/12 \\ 0 & 1 & 0 & 0\\ 0 & 0& 1 & 0 \\ 0 & 0 & 0 & 1\end{array}


    [/tex]
    So I have a new matrix

    AB
    C
     
  11. Oct 3, 2008 #10
    Forget about A and B. You can do the operations on C alone. Look at rows 7 and 8.
     
  12. Oct 3, 2008 #11
    hmm... I see what you mean. But the thing is I thought in this example I was doing things with matrix A and matrices B and C were the 'reflections' of what I was doing with matrix A.
    And what about the first row in matrix A? You see it has changed also.
     
    Last edited: Oct 3, 2008
  13. Oct 3, 2008 #12
    ohh, I think I see now the answer to my question.

    I can now add the 7th row multiplyed by -1/12 and the 8th multiplyed by -1/12 to the first one and I GET WHAT I SEE! ;)
     
  14. Oct 3, 2008 #13
    but this way C is not a 'reflection'...
     
  15. Oct 4, 2008 #14
    I see. In your original question, A's first row was unchanged. In that case, just use the new C row to get the result.
     
  16. Oct 4, 2008 #15
    thank you very much, cellotim
     
  17. Oct 9, 2008 #16

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Madou! Welcome to PF! :smile:

    Hi cellotim! (are we related?:wink:)

    There's an easier way to do matrices: use the CODE tag (next to the QUOTE tag)…

    it's really intended for writing computer code, but it does matrices nicely, because it prints with equal spacing, as on a typewriter …

    Code (Text):

    1    0    1/12   1/12
    0   12  143/12  11/12
    0   12  143/12  11/12
    0   12 -143/12 -11/12
     
     
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