linear lectures

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

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Matrix and Numerical Methods in Systems Engineering

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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

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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 = 0 3 or just A x = b

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Example 1 2 x - y = 0 - x + 2 y = 3 I Matrix form 2 - 1 - 1 2 x y = 0 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

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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 = 0 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 = 0 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 0

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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

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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 0 + y - 1 2 - 3 + z 0 - 1 4 = 0 - 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 0 + y - 1 2 - 3 + z 0 - 1 4 = 0 - 1 4 I Solution: x = y = 0, z = 1 I A x = b – Does a solution always exist?

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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

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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 0 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

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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 0 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 0 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

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Matrix Multiplication I matrix-by-column A b 1 b 2 b 3 = b 1 a 11 a 21 a 31 + b 2 a 12 a 22 a 32 + b 3 a 13 a 23 a 33
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