2 3 1 x1 1 setting x3 0 we have to solve doing this

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Unformatted text preview: ge 3, then edge 2 backwards. This corresponds to ￿ Row 1 + Row 3 -￿ Row ￿ being the 0 Row. 2￿ ￿￿ 3 −1 x1 1 Setting x3 = 0, we have to solve = . Doing this by hand, x2 0 −1 3 3/8 c x1 = 3/8, x2 = 1/8, so ￿ = 1/8 + c gives the potentials. x 0 c Finally, the currents are given by 100 −1 10 3/8 −1 10 3/8 1/4 0 1 1/8 = − −2 0 2 1/8 = 3/4 . −CA￿ = 0 2 0 −1 x 002 0 −1 1 0 0 −2 2 0 1/4 10) A = −1 1 0 −1 0 1 0 −1 1 0 −1 0 0 0 −1 0 0 0 1 1 → 1 −1 0 −1 0 1 0 −1 1 0 −1 0 0 0 −1 0 0 0 1 1 → 1 0 0 0 0 −1 0 −1 1 −1 1 −1 0 0 −1 0 0 0 1 1 1 −1 00 1 −1 00 1 −1 00 1 −1 00 0 0 0 0 1 −1 0 1 −1 0 1 −1 0 1 −1 0 0 0 0 → 0 0 0 0 → 0 0 −1 1 → 0 −1 −1 0 → 0 0 −1 0 0 0 01 0 −1 1 0 −1 1 0 00 0 0 −1 1 0 0 −1 1 0 0 00 0 0 00 an echelon form. From the second to last matrix (before the row swap to move zero rows to the bottom) we have that edges 1, 2 and 4 are kept. The reason is from edges 1 and 2 we can get edge 3. and from 3 and 4 we can get edge 5. The other spanning trees correspond to the following triples of edges: {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, , 5}, {2, 3, 4}, {2, 3, 5} and {2, 4, 5}. 1 0 1 0 1 0 1 0 12) (a) It’s almost a definition that A = so certainly AT A = . 1 0 1 0 1 0 1 0 T As A A and A have the same rank and we know from exercise 10 that A hass rank 3, then AT A has rank 3. (b) For any matrix B , B T B is always positive semidefinite as ￿ T B T B ￿ = (B ￿ )T B ￿ = x x x x 2 ||B ￿ || ≥ 0. x 1 0 1 0 Choosing B = A and noting A = , we see AT A is not positive definite. 1 0 1 0 ,...
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This note was uploaded on 01/30/2014 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell.

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