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Lecture 23

# Lecture 23 - Chapter 10 Matrix Inverse and Condition Matrix...

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Chapter 10 Matrix Inverse and Condition

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How do we get inverse [ A ] - 1 ? Consider solving [ A ]{ x i } = { b i } with Put together x’s to get inverse of [A] as {x 1, x 2 ,x 3 ,x 4 } { } { } { } { } 1 2 3 4 1 0 0 0 0 1 0 0 , , , 0 0 1 0 0 0 0 1 b b b b                         = = = =                                 Matrix Inverse Sequentially! Sequentially!
Matrix Inverse - 2 0 1 1 0 3 2 1 1 1 0 0 0 1 0 0 0 1 1 1 2 3 0 1 1 0 2 - Inverse matrix can be obtained using Gaussian-Jordan method * Consider: A= A -1 =? * Add unit vectors to A: 3 0 1 1 1 2 1 0 2 0 1 0 1 0 0 0 0 1 - * Pivoting : R2 R 1 I

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Matrix Inverse * Gaussian elimination 0 1 3 0 0 1 1 0 1 0 1.66 .333 7 0 0 1.667 0 0 0.333 1 - - - * Continue with elimination to reduce the upper triangular matrix to obtain identity matrix on the left of the augmented matrix 0 0.4 0 1 0 0 .2 1 0 1 0 0.2 0 0 1 0 0 0 .6 1 - - - U 1 0 0.4 0.2 1 0 1 0 0.2 0.6 A - - - - = * No need to worry about row interchanges, no matter how often I
Matrix Inverse More efficient approach: LU decomposition [ A ] [ A ] - 1 = [ I ] Use LU factorization to find [ X ] = [ A ] - 1 LU decomposition: [ A ] = [ L ][ U ] Forward substitution: [ L ][ Y ] = [ I ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] X Y X U L I L I U Y = = =

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Lecture 23 - Chapter 10 Matrix Inverse and Condition Matrix...

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