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

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
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 " # $ %
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
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:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 " # $ %
Background image of page 4
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.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon