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102_solutions3

# 102_solutions3 - MATH 102 SOLUTIONS TO HW#3 Section 2.4...

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MATH 102 SOLUTIONS TO HW #3 Section 2.4, problem 8. In this case the rank of the matrix is r = n . Recall that in general the rank r is { # of pivot variables } , while the dimension of the null space is n - r = { # of free variables } . In this case the dimension of the null-space is 0. So the number of free variables must also be zero. Thus we have that n - r = 0. Let us write the matrix A in terms of column vectors as follows A = [ v 1 . . . v n ]. Then the null-space of A consists of all n × 1 vectors of the from c T = ( c 1 , . . . , c n ), which give the sum: c 1 v 1 + . . . + c n v n = 0 . Since the null-space is only the zero vector, then the only possible sum of this form has all c i = 0. That is the same as saying that the collection of vectors { v 1 , . . . , v n } are linearly independent. Section 2.4, problem 18. TO BE POSTED SOON. Section 2.4, problem 6. In this case the incidence matrix is: A = - 1 1 0 0 - 1 0 1 0 0 - 1 1 0 0 - 1 0 1 - 1 0 0 1 0 0 - 1 1 . In this case, the three elementary loop vectors are associated with the loops: (1) Loop #1: y 5 y 4

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102_solutions3 - MATH 102 SOLUTIONS TO HW#3 Section 2.4...

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