Matrix question

[ptex]

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

[HallsofIvy]

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}"?

[ptex]

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]X=Y=Z={1,2,3,4,5,6} in that order and,
R1 = {(x,y)| x+y < or = to 7}
R2 = {(y,z)| y-z > or = to 3}

a) The matrix A1 of the relation R1 (relative to the given orderings)

b) The matrix A2 of the relation R1 (relative to the given orderings)

c) The matrix product A1 A1

d) Use the result of part (c) to find the matrix of the relation R20 R1.

e) Use the result of part (d) to find the relation R20 R1. (as a set of ordered pairs).

For R1 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

[gnome]

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)

[ptex]

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

[gnome]

Looks correct.

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