Lecture15wn_NucE309FA11

Lecture15wn_NucE309FA11 - NucE 309 Fall 2011 Lecture 14...

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NucE 309, Fall 2011 Lecture 14, September 23, 2011 Matrix Inverse Chapter 7, Section 7 The original matrix is: 12- 5 100 A= 2 -1 3 I= 0 1 0 -3 -1 0 0 0 1 Combine these to form an augmented matrix 5100 A + = 2- 13010 - 3 - 10001 Add a multiple of the first row to the second and third rows to eliminate the first element A + = 0- 5 1 3 - 21 0 05 - 1 5 301 Add a multiple of the second row to the third row to eliminate the second element A + = 5 1 3 - 0 00- 2111 Now divide the last row by a(3,3) so that a(3,3)=1 A + = 5 1 3 - 0 0 0 1 -0.5 -0.5 -0.5 Now use a backward elimination to change upper triangular values in the last column to zero 1 2 0 -1.5 -2.5 -2.5 A + = 0 -5 0 4.5 7.5 6.5 0 0 1 -0.5 -0.5 -0.5 Now normalize the second row. 1 2 0 -1.5 -2.5 -2.5 A + = 0 1 0 -0.9 -1.5 -1.3 0 0 1 -0.5 -0.5 -0.5
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Use backward substitution to change the upper row. 1 0 0 0.3 0.5 0.1 A + = 0 1 0 -0.9 -1.5 -1.3 0 0 1 -0.5 -0.5 -0.5 The inverse matrix is the right side. 0.3 0.5 0.1 A -1 = -0.9 -1.5 -1.3 -0.5 -0.5 -0.5 To check, we can multiply matrix A and A -1 1 2 -5 0.3 0.5 0.1 1 0 0 2 -1 3 -0.9 -1.5 -1.3 = 0 1 0 -3 -1 0 -0.5 -0.5 -0.5 0.00 0 1
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NucE 309, Fall 2011 Lecture 14, September 23, 2011 L-U Factorization (using Doolittle's Method) 20.2, pp. 840-842 was 18.2, pp. 894-896
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This document was uploaded on 10/27/2011 for the course NUC E 309 at Penn State.

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Lecture15wn_NucE309FA11 - NucE 309 Fall 2011 Lecture 14...

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