linear lectures - Matrix and Numerical Methods in Systems...

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Unformatted text preview: Matrix and Numerical Methods in Systems Engineering Linear Algebra Systems of Linear Equations I Linear equation: ax + by = c I variables: x , y I constants: a , b , c I System of linear equations a 11 x 1 + a 12 x 2 + ··· + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + ··· + a 2 n x n = b 2 . . . . . . a m 1 x 1 + a m 2 x 2 + ··· + a mn x n = b m I n variables I m equations I nm constants Example 1 2 x- y = 0- x + 2 y = 3 Example 1 2 x- y = 0- x + 2 y = 3 I Matrix form 2- 1- 1 2 x y = 3 or just A x = b Example 1 2 x- y = 0- x + 2 y = 3 I Matrix form 2- 1- 1 2 x y = 3 or just A x = b I Three interpretations I row I column I matrix Example 1 I Row I 2 x- y = 0 and- x + 2 y = 3 I solution: x = 1, y = 2 I number of solutions Example 1 I Row I 2 x- y = 0 and- x + 2 y = 3 I solution: x = 1, y = 2 I number of solutions I Column I linear combination x 2- 1 + y- 1 2 = 3 I solution: x = 1, y = 2 I number of solutions Example 1 I Row I 2 x- y = 0 and- x + 2 y = 3 I solution: x = 1, y = 2 I number of solutions I Column I linear combination x 2- 1 + y- 1 2 = 3 I solution: x = 1, y = 2 I number of solutions I Matrix 2- 1- 1 2 2 1 = 2 2- 1 + 1- 1 2 = 3 Example 2 2 x- y =0- x +2 y- z =- 1- 3 y +4 z =4 Example 2 2 x- y =0- x +2 y- z =- 1- 3 y +4 z =4 I Row: planes instead of lines Example 2 2 x- y =0- x +2 y- z =- 1- 3 y +4 z =4 I Row: planes instead of lines I Column x 2- 1 + y - 1 2- 3 + z - 1 4 = - 1 4 I Solution: x = y = 0, z = 1 Example 2 2 x- y =0- x +2 y- z =- 1- 3 y +4 z =4 I Row: planes instead of lines I Column x 2- 1 + y - 1 2- 3 + z - 1 4 = - 1 4 I Solution: x = y = 0, z = 1 I A x = b – Does a solution always exist? Elimination I Natural I Two stages I forward elimination I backward substitution Elimination I Natural I Two stages I forward elimination I backward substitution I Example x +2 y + z =2 3 x +8 y + z =12 4 y + z =2 Elimination I Natural I Two stages I forward elimination I backward substitution I Example x +2 y + z =2 3 x +8 y + z =12 4 y + z =2 I Forward step: pivots A = 1 2 1 3 8 1 0 4 1 → 1 2 1 2- 2 0 4 1 → 1 2 1 0 2- 2 0 0 5 = U I U – upper triangular matrix Elimination I What can go wrong I pivots 1 & 2 equal to 0: interchange rows I pivot 3 equals to 0: no way out Elimination I What can go wrong I pivots 1 & 2 equal to 0: interchange rows I pivot 3 equals to 0: no way out I Augmented matrix 1 2 1 2 3 8 1 12 0 4 1 2 → 1 2 1 2 2- 2 6 0 4 1 2 → 1 2 1 2 0 2- 2 6 0 0 5- 10 = [ U c ] x +2 y + z =2 2 y- 2 z =6 5 z =- 10 Elimination I What can go wrong I pivots 1 & 2 equal to 0: interchange rows I pivot 3 equals to 0: no way out I Augmented matrix 1 2 1 2 3 8 1 12 0 4 1 2 → 1 2 1 2 2- 2 6 0 4 1 2 → 1 2 1 2 0 2- 2 6 0 0 5- 10 = [ U c ] x +2 y + z =2 2 y- 2 z =6 5 z =- 10 I Backward substitution: z =- 2, y = 1, x = 2 Matrix Multiplication...
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This note was uploaded on 03/26/2012 for the course ESI 4567c taught by Professor Momcilovic during the Spring '12 term at University of Florida.

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linear lectures - Matrix and Numerical Methods in Systems...

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