Solutions to Assignment 4 - March 16, 2006 Physics 681-481;...

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March 16, 2006 Physics 681-481; CS 483: Discussion of #4 I. (a) According to Eq. (1) in Assignment #4, the probability of Fnding 2 linearly inde- pendent vectors (doing arithmetic modulo 2) among a random set of four 3-vectors of 0’s and 1’s orthogonal to the vector a = 111, is q = ( 1 - 1 8 )( 1 - 1 16 ) = 105 128 . ( n = 3 , x = 1) (1) To check this note that the four vectors orthogonal to 111 are 000, 110, 101, and 011. Any two distinct nonzero vectors from this set are linearly independent, so the number of the 4 4 = 256 di±erent quartets without two linearly independent vectors, is just the number of quartets containing at most one kind of nonzero vector. To enumerate these note that: (a) There is just 1 way to have four zero vectors (000). (b) There are 12 ways to have one nonzero vector and three zero vectors: the nonzero vector can be either the Frst, second, third, or fourth vector in the quartet, and it can have one of the three forms (110), (101), or (011)). (c) There are 18 ways to have two identical nonzero vectors and two zero vectors:
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Solutions to Assignment 4 - March 16, 2006 Physics 681-481;...

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