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Matrix question

  1. Apr 28, 2004 #1
    X=Y=Z={1,2,3,4,5,6}
    Is this the same as
    X={1,2,3,4,5,6}
    Y={1,2,3,4,5,6}
    Z={1,2,3,4,5,6}

    If so {(y,z)|y-z> or = 3}
    is the y the vertical (or side)and z the Horizontal (or top)?

    Like this;
    ? 1 2 3 4 5 6
    1 0 0 0 0 0 0
    2 0 0 0 0 0 0
    3 0 0 0 0 0 0
    4 1 0 0 0 0 0
    5 1 1 0 0 0 0
    6 1 1 1 0 0 0
     
    Last edited: Apr 28, 2004
  2. jcsd
  3. Apr 28, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm afraid you will have to give more detail as to exactly what your question is.

    Matrices in general do NOT have X, Y, Z parts- AND do not have 3 dimensions.

    Perhaps the problem you are working on specifies in some way how X, Y, Z affects the matrix but you will have to tell us. Where did you get "X=Y=Z={1,2,3,4,5,6}"?
     
  4. Apr 28, 2004 #3
    You :rolleyes: helped me with a question like this before and I got it. Now I am studing for a final.
    Ok the whole question;
    Code (Text):
    X=Y=Z={1,2,3,4,5,6} in that order and,
    R[sub]1[/sub] = {(x,y)| x+y < or = to 7}
    R[sub]2[/sub] = {(y,z)| y-z  > or = to 3}

    a) The matrix A[sub]1[/sub] of the relation R[sub]1[/sub] (relative to the given orderings)

    b) The matrix A[sub]2[/sub] of the relation R[sub]1[/sub] (relative to the given orderings)

    c) The matrix product A[sub]1[/sub] A[sub]1[/sub]

    d) Use the result of part (c) to find the matrix of the relation R[sub]2[/sub]0 R[sub]1[/sub].

    e) Use the result of part (d) to find the relation R[sub]2[/sub]0 R[sub]1[/sub]. (as a set of ordered pairs).

    For R[sub]1[/sub] I have;
    ? 1 2 3 4 5 6
    1 1 1 1 1 1 1
    2 1 1 1 1 1 0
    3 1 1 1 1 0 0
    4 1 1 1 0 0 0
    5 1 1 0 0 0 0
    6 1 0 0 0 0 0
     
    Last edited: Apr 29, 2004
  5. Apr 28, 2004 #4
    Your A1 looks fine to me. Of course, R1 would come out the same no matter which way you look at it. For R2, orientation matters.

    You want each A(i,j) (where i is the row and j is the column) to have a 1 when Yi is related to Zj according to the relation R2. So, for example, A(6,1) = 1 because the value in Y6 minus the value in Z1 ≥ 3 (6-1=5)
     
  6. Apr 29, 2004 #5
    So are these correct?

    For R1 or A1 I have;
    ? 1 2 3 4 5 6
    1 1 1 1 1 1 1
    2 1 1 1 1 1 0
    3 1 1 1 1 0 0
    4 1 1 1 0 0 0
    5 1 1 0 0 0 0
    6 1 0 0 0 0 0

    For R2 or A2 I have;
    ? 1 2 3 4 5 6
    1 0 0 0 0 0 0
    2 0 0 0 0 0 0
    3 0 0 0 0 0 0
    4 1 0 0 0 0 0
    5 1 1 0 0 0 0
    6 1 1 1 0 0 0
     
    Last edited: Apr 29, 2004
  7. Apr 29, 2004 #6
    Looks correct.

    PS: those code windows are annoying -- too much scrolling.
     
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