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I have 2 matrices, I don't believe that they are transitive but my friend is insisting that they are both transitive, is he correct?

a) $ \begin{bmatrix} 1& 1 & 1 & 1 & 1 &1 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $

b) $ \begin{bmatrix} 1& 0 & 0 & 0 & 0 &1 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} $

I'm trying to order them by pairs of relation (using positions as letters, e.g. (a,b)) and failing for both. Thank you.

K.defaoite
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Jonathan Grand
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  • The comments on this question explain how to answer this question by squaring the matrix. – saulspatz May 17 '21 at 17:23
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    Please avoid "do my homework for me"-style questions. For example, simply telling us where you got the problem from or explaining what you tried would be a big step forward! (For further feedback/help with asking questions, you can ask here.) – user1729 May 18 '21 at 14:20

1 Answers1

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As explained in the comments to this question, one way to check on transitivity is to square the matrix. If the square has a nonzero entry is a position where the original matrix has a zero, then the relation is non-transitive. if there is no such entry, then the relation is transitive.

Rather than going to the work of squaring the matrix, let's see if we can spot such an entry in the first matrix. Since the first row is all $1$s, multiplying any row with a $1$ in the first column will give a non-zero entry. The first row has no zeros, so that won't help, but the sixth row has a $1$ in the first column, and a $0$ in the second. That is, $(6,1)$ is in the relation, and $(1,2)$ is in the relation, but $(6,2)$ is not. The relation is not transitive.

Can you do the second part now?

saulspatz
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