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Unformatted text preview: AAE 250 Homework 7 Solutions http://nsgsun.aae.uiuc.edu/AAE250 1. Solution: (a) We calculate K q = 3 1 0 1 3 1 0 1 2 1 2 1 = 1 4  q T K = h 1 2 1 i 3 1 0 1 3 1 0 1 2 = h 1 4 0 i q T K T = h 1 2 1 i 3 1 0 1 3 1 0 1 2 = h 1 4 0 i 1 2 q T K q = 1 2 h 1 2 1 i 3 1 0 1 3 1 0 1 2 1 2 1 = 1 2 h 1 4 0 i 1 2 1 = 9 2 (b) The matrix K is positive definite if all of its principal minors are positive definite. These minors are given by 1 =  k 11  =  3  > 2 = 3 1 1 3 = 8 > 3 = 3 1 0 1 3 1 0 1 2 = 13 > So, since all of the principal leading minors are positive, the matrix K is positive definite. (c) The matrix K 2 I is given by K 2 I = 1 1 0 1 1 1 0 1 0 1 The principal leading minors are given by 1 =  k 11  = 1 > 2 = 1 1 1 1 = 0 3 = 1 1 0...
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 Spring '09
 SRI

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