Lecture 19 - ENGR-1100 Introduction to Engineering Analysis...

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Lecture 19 Notes courtesy of: Prof. Yoav Peles ENGR-1100 Introduction to Engineering Analysis
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Homework Assignment (due next class) LAS Section 1.6 Problems 4, 7 LAS Section 2.1 Problems 7, 9
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Lecture outline Procedure (algorithm) for finding the inverse of invertible matrix. Investigate the system of linear equation and invertibility of matrices. Determinants: Finding determinants by duplicate column method. Finding determinants by cofactor expansion along the first row. Finding determinants by cofactor expansion along any row or column.
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Definition: row equivalent A B Finite sequence of row operation Matrices that can be obtained from one another by a finite sequence of elementary row operations are said to be row equivalent . A B Inverse of the row operation in reverse order A is row equivalent to B
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Theorem 1 If A is an n x n matrix, then the following statements are equivalent, that is, if one is true then all are: (a) A is invertible (b) AX=0 has only the trivial solution (the only solution is x 1 =0, x 2 =0…x n =0 ) (c) A is row equivalent to I n . If A -1 exists then A n Finite sequence of row operation
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(b) AX=0 has only the trivial solution (the only solution is x 1 =0, x 2 =0…x n =0 ) Finite sequence of row operation x 1 =0, x 2 =0…x n =0 a 11 a 12….… a 1n 0 a 21 a 22….… a 2n 0 : : : : : : : : a n1 a n2….… a nn 0 A= 1 0 0 ….… 0 0 0 1 0 ….… 0 0 0 0 1 ….… 0 0 : : : 0 0 0 ….… 1 0
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Theorem 2 If A is an n x n invertible matrix, then the sequence of row operations that reduces A to I n reduces I n to A -1 . A I n The same sequence of row operation A I n -1
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The procedure A | I n I n | A -1 The same sequence of row operation
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Example 1: Find the inverse of 1 2 3 2 5 3 1 0 8 A= 1 2 3 2 5 3 1 0 8 1 0 0 0 1 0 0 0 1 * –2 * –1 1 2 3 0 1 -3 0 -2 5 1 0 0 -2 1 0 -1 0 1 * 2
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1 2 3 0 1 -3 0 0 -1 1 0 0 -2 1 0 -5 2 1 * -1 1 2 3 0 1 -3 0 0 1 1 0 0 -2 1 0 5 -2 -1 * 3 1 2 3 0 1 0 0 0 1 1 0 0 13 -5 -3 5 -2 -1 * -3
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1 2 0 0 1 0 0 0 1 -14 6 3 13 -5 -3 5 -2 -1 * -2 1 0 0 0 1 0 0 0 1 -40 16 9 13 -5 -3 5 -2 -1 Thus: A -1 = -40 16 9 13 -5 -3 5 -2 -1
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This note was uploaded on 04/04/2011 for the course ENGR 1100 taught by Professor Anderson during the Fall '06 term at Rensselaer Polytechnic Institute.

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Lecture 19 - ENGR-1100 Introduction to Engineering Analysis...

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