# SolSec 4_2 - Problems and Solutions for Section 4.2 (4.19...

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Problems and Solutions for Section 4.2 (4.19 through 4.33) 4.19 Calculate the square root of the matrix M = 13 ! 10 ! 10 8 " # \$ % Hint: Let M 1/2 = a ! b ! b c " # \$ % ; calculate M 1/2 ( ) 2 and compare to M . " # \$ % Solution: Given: M = 13 ! 10 ! 10 8 " # \$ % If M 1/2 = a ! b ! b c " # \$ % , then M = M 1/2 M 1/2 = a ! b ! b c " # \$ % a ! b ! b c " # \$ % = a 2 + b 2 ! ab ! bc ! ab ! bc b 2 + c 2 " # \$ % = 13 ! 10 ! 10 8 " # \$ % This yields the 3 nonlinear algebraic equations: a 2 + b 2 = 13 ab + bc = 10 b 2 + c 2 = 8 There are several possible solutions but only one that makes M 1/2 positive definite which is a = 3, b = c = 2 as determined below in Mathcad. Choosing these values results in M 1/2 = 3 ! 2 ! 2 2 " # \$ %

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4.20 Normalize the vectors 1 ! 2 " # \$ % , 0 5 " # \$ % , ! 0.1 0.1 " # \$ % first with respect to unity (i.e., x T x = 1 ) and then again with respect to the matrix M (i.e., x T Mx = 1 ), where M = 3 ! 0.1 ! 0.1 2 " # \$ % Solution: (a) Normalize the vectors x 1 = 1 ! 2 " # \$ % ( 1 = 1 x T x = 1 5 Normalized: x 1 = 1 5 1 ! 2 " # \$ % = 0.4472 ! 0.8944 " # \$ % x 2 = 0 5 ! " # \$ % 2 = 1 x T x = 1 5 Normalized: x 2 = 0 1 ! " # \$ % x 3 = ! 0.1 0.1 " # \$ % 3 = 1 x T x = 1 0.02 Normalized:

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x 3 = 50 ! 0.1 0.1 " # \$ % = 1 2 ! 1 1 " # \$ % = ! 0.7071 0.7071 " # \$ % (b) Mass normalize the vectors x 1 = 1 ! 2 " # \$ % ( 1 = 1 x T M x = 1 11.4 Mass normalized: x 1 = 1 11.4 1 ! 2 " # \$ % = 0.2962 ! .5923 " # \$ % x 2 = 0 5 ! " # \$ % ! 2 = 1 x T M x = 1 50 x 2 = 1 50 0 5 ! " # \$ % = 1 2 0 1 ! " # \$ % = 0 0.7071 ! " # \$ % x 3 = ! 0.1 0.1 " # \$ % 3 = 1 x T Mx = 1 0.052 Mass normalized: x 3 = 1 0.052 ! 0.1 0.1 " # \$ % = ! 0.4385 0.4385 " # \$ %
4.21 For the example illustrated in Figure P4.1 with c 1 = c 2 = c 3 = 0 , calculate the matrix ! K . Solution: From Figure 4.1, m 1 0 0 m 2 ! " # # \$ % !! x + k 1 + k 2 k 2 k 2 k 2 + k 3 ! " # # \$ % x = 0 ! K = M ! 1/2 KM ! 1/2 = m 1 ! 1/2 0 0 m 2 ! 1/2 " # \$ \$ % k 1 + k 2 ! k 2 ! k 2 k 2 + k 3 " # \$ \$ % m 1 ! 1/2 0 0 m 2 ! 1/2 " # \$ \$ % ! K = m 1 ! 1 k 1 + k 2 ( ) ! m 1 ! 1/2 m 2 ! 1/2 k 2 ! m 1 ! 1/2 m 2 ! 1/2 k 2 m 1 ! 1 k 2 + k 3 ( ) " # \$ \$ % Since ! K T = ! K , ! K is symmetric. Using the numbers given in problem 4.2 yields ! K = 3 ! 1 ! 1 6 " # \$ % This is obviously symmetric.

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