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# SolSec 4_9 - Problems and Solutions for Section 4.9(4.80...

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Problems and Solutions for Section 4.9 (4.80 through 4.90) 4.80 Consider the mass matrix M = 10 ! 1 ! 1 1 " # \$ % & and calculate M -1 , M -1/2 , and the Cholesky factor of M . Show that LL T = M M ! 1/2 M ! 1/2 = I M 1/2 M 1/2 = M Solution: Given M = 10 ! 1 ! 1 1 " # \$ % & The matrix, P , of eigenvectors is P = ! 0.1091 ! 0.9940 ! 0.9940 0.1091 " # \$ % & The eigenvalues of M are ! 1 = 0.8902 ! 2 = 10.1098 From Equation M ! 1 = P diag 1 " 1 , 1 " 2 # \$ % & ( P T , M ! 1 = 0.1111 0.1111 0.1111 1.1111 # \$ % & ( From Equation M ! 1/2 = Vdiag " 1 ! 1/2 , " 2 ! 1/2 # \$ % & V T M ! 1/2 = 0.3234 0.0808 0.0808 1.0510 # \$ % & ( The following Mathcad session computes the Cholesky decomposition.

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4.81 Consider the matrix and vector A = 1 ! " ! " " # \$ % & ( b = 10 10 # \$ % & ( use a code to solve Ax = b for ε = 0.1, 0.01, 0.001, 10 -6 , and 1. Solution: The equation is 1 ! " ! " " # \$ % & ( x = 10 10 # \$ % & ( The following Mathcad session illustrates the effect of ε on the solution, a entire integer difference. Note that no solution exists for the case ε = 1. So the solution to this problem is very sensitive, and ill conditioned, because of the inverse.
4.82 Calculate the natural frequencies and mode shapes of the system of Example 4.8.3. Use the undamped equation and the form given by equation (4.161). Solution: The following MATLAB program will calculate the natural frequencies and mode shapes for Example 4.8.3 using Equation (4.161).

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