# Adjacency Matrix / Multiplication Table

• Mar 1st 2009, 04:19 AM
h2o pete
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!

:)

http://i176.photobucket.com/albums/w...rieff/math.jpg
• Mar 1st 2009, 05:16 AM
HallsofIvy
Quote:

Originally Posted by h2o pete
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?

Quote:

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

:)

http://i176.photobucket.com/albums/w...rieff/math.jpg
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?
• Mar 1st 2009, 05:50 AM
h2o pete
Quote:

Originally Posted by HallsofIvy
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
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? :(