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Linear dependency matrix problem

  1. May 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I am having an issue understanding what to do with this question. The question is displayed below:

    upload_2017-5-9_1-19-11.png

    I have hand wirtten my working, as I dont now how to do matrices fully on latext.

    I used the definition to get this far for part a, but not sure about the second part.
    I have attche my working with what I have so far
    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. May 8, 2017 #2

    FactChecker

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    You should start by thinking about properties of det(A) that you might use. Do you know anything about the effect of row manipulations on determinants?
     
  4. May 9, 2017 #3
    I know that is u have a triangular matrix then the det would be 0. The propble is beacuse its an arbitary matrix its throwing me off. If r2 and r1 ard linearly independent the r3 would have to = 0 which means that one of the rows could equal each other
     
  5. May 9, 2017 #4

    FactChecker

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    Since this is a homework problem, there must be some facts about the determinant that you can use. What formulas do you know about the determinant? What do you know about the effect of row manipulations on the determinant?
     
  6. May 9, 2017 #5
    I know the basic where the determniant is: ##D=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{31}-a_{23}a_{32})+a_{31}(a_{21}a_{32}-a_{22}a_{31})## I also know after reading that if you use the ro echol matrix which is done by row mainpulation then it will not effect the determinant of the matrix
     
  7. May 9, 2017 #6

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    Ok. Use the linear dependence of the first two rows to guide you in row manipulations that zero out one of them without changing the determinant. Then calculate the determinant.
     
    Last edited: May 9, 2017
  8. May 9, 2017 #7

    Ray Vickson

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    No, if you have a triangular matrix there is no reason to have zero determinant. The 3x3 identity matrix is certainly upper triangular, but has determinant = 1. Also: just because rows 1 and 2 are dependent, that would have no effect whatsoever on row 3. I don't know why you think row 3 would have to be 0. Furthermore, there is no need for any two of the rows to equal each other, as you seem to think.
     
  9. May 10, 2017 #8
    Ok I have finally realised what I have to do. So linear dependency means in this case you have two vectors one a scalar multiple of the other so the matrix of (1,2,3 | 2,4,6|,5,10,11) will do, to show this condition. I was under the impression I had to prove this from the question. It obvious to me that is any row in a matrix is propotional to each other would mean that the det will equal 0. Once again I thank you for your help.
     
  10. May 10, 2017 #9

    Ray Vickson

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    Still, no. What you say is OK in part (b), but not in part (a). For example, the matrix
    $$\begin{bmatrix}
    2 & 0 & -1 \\ 0 & 2 & 2 \\2 & 2 & 1
    \end{bmatrix}
    $$
    has any two rows linearly independent, but the three rows taken together are linearly dependent.
     
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