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Math Help - Adjacency Matrix / Multiplication Table

  1. #1
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    Adjacency Matrix / Multiplication Table

    Hi guys

    I'm stuck on these questions.

    I've done the first part of question 1 but I don't get the x^2 = 3 bit.

    I've done 2. a) and c) but don't get b).

    Can someone help me understand it?

    I don't want someone to do the questions lol I just don't get how to do it!



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  2. #2
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    Quote Originally Posted by h2o pete View Post
    Hi guys

    I'm stuck on these questions.

    I've done the first part of question 1 but I don't get the x^2 = 3 bit.
    x^2 just means x*x. Look at the multiplication table you wrote for part (a). Are there any numbers in it such that x*x= 3? Another way you could do this, without using the table is just go through the numbers. What is 0*0 (mod 6)? What is 1*1 (mod 6)? What is 2*2 (mod 6)? What is 3*3 (mod 6)? What is 4*4 (mod 6)? What is 5*5 (mod 6)? Are any of those equal to 3?

    [quote]I've done 2. a) and c) but don't get b).

    Can someone help me understand it?

    I don't want someone to do the questions lol I just don't get how to do it!



    Have you found A^2? That is, have you multiplied the matrix A by itself?
    One thing I notice immediately is that "first row times first column" is 1+1= 2 so the "11" component of A^2 is 2. I don't know how you have labled the vertices in your graph but if we label them "1", "2", etc. The fact that the first row is "0 1 1 0 0 0" means there is a "bridge", line, from 1 to 2 and from 1 to 3 but no others. Of course that means that there is also a bridge from 2 to 1 and from 3 to 1 so those same numbers appear in the first column. I could go from 1 to 2 and then back again or I could go from 1 to 3 and then back again: there are 2 paths of length 2 from 1 to itself. Now, where should you look to find how many paths of length 2 from 1 to 3?
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    x^2 just means x*x. Look at the multiplication table you wrote for part (a). Are there any numbers in it such that x*x= 3? Another way you could do this, without using the table is just go through the numbers. What is 0*0 (mod 6)? What is 1*1 (mod 6)? What is 2*2 (mod 6)? What is 3*3 (mod 6)? What is 4*4 (mod 6)? What is 5*5 (mod 6)? Are any of those equal to 3?
    Thanks! If I've done the multiplication table correctly then there is a number 3 in row 3 column 1, row 1 column 3, row 3 column 3, row 5 column 3 and row 3 column 5. Do you know if that's right?

    Quote Originally Posted by HallsofIvy View Post
    Have you found A^2? That is, have you multiplied the matrix A by itself?
    One thing I notice immediately is that "first row times first column" is 1+1= 2 so the "11" component of A^2 is 2. I don't know how you have labled the vertices in your graph but if we label them "1", "2", etc. The fact that the first row is "0 1 1 0 0 0" means there is a "bridge", line, from 1 to 2 and from 1 to 3 but no others. Of course that means that there is also a bridge from 2 to 1 and from 3 to 1 so those same numbers appear in the first column. I could go from 1 to 2 and then back again or I could go from 1 to 3 and then back again: there are 2 paths of length 2 from 1 to itself. Now, where should you look to find how many paths of length 2 from 1 to 3?
    Ah okay.

    How exactly do I do that?

    If the matrix is:

    011000
    101000
    110100
    001011
    000101
    000110

    Then wouldn't A^2 just be the same?

    0 x 0 = 0
    1 x 1 = 1

    Am I doing it wrong?
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