{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture15wn_NucE309FA11

# Lecture15wn_NucE309FA11 - NucE 309 Fall 2011 Lecture 14...

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

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

View Full Document
NucE 309, Fall 2011 Lecture 14, September 23, 2011 Matrix Inverse Chapter 7, Section 7 The original matrix is: 1 2 -5 1 0 0 A= 2 -1 3 I= 0 1 0 -3 -1 0 0 0 1 Combine these to form an augmented matrix 1 2 -5 1 0 0 A + = 2 -1 3 0 1 0 -3 -1 0 0 0 1 Add a multiple of the first row to the second and third rows to eliminate the first element 1 2 -5 1 0 0 A + = 0 -5 13 -2 1 0 0 5 -15 3 0 1 Add a multiple of the second row to the third row to eliminate the second element 1 2 -5 1 0 0 A + = 0 -5 13 -2 1 0 0 0 -2 1 1 1 Now divide the last row by a(3,3) so that a(3,3)=1 1 2 -5 1 0 0 A + = 0 -5 13 -2 1 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

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

View Full Document
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
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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}