Question: Let u = (2, 3, 1), v = (1, 3, 0), and w = (2, -3, 3). Since (1/2)u - (2/3)v - (1/6)w = (0,

Let u = (2, 3, 1), v = (1, 3, 0), and w = (2, -3, 3). Since (1/2)u - (2/3)v - (1/6)w = (0, 0, 0), can we conclude that the set {u, v, w} is linearly dependent over Z7?

This is what I have so far

Over the field of Z7

The inverse of 2 is 4

The inverse of 3 is 5

The inverse of 6 is 6

As such 4u - 2(5v) - 6w = (0, 0, 0)

The additive inverse of 2 in Z7 = 5

The multiplicative inverse of 3 in Z7 = 5

5 *5 = 25 = 4 mod 7

Converting -1 / 6 in Z7 would be -1 mod 7 = 6 so converting 1 / 6 mod 7 would leave us with 6

This leaves us with 4u - 4v - 6w = (0, 0, 0)

As such the set {u, v, w} is linearly dependent over Z7 according to the definition of linear dependency since there is nontrivial combinations of the sets over Z7 that result in (0, 0, 0).

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