This preview shows pages 1–20. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
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.
 Spring '12
 Momcilovic

Click to edit the document details