Lecture Slides 1

# Lecture Slides 1 - Math20F Slides Chapter 1 of[Lay P Oswald...

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Math20F Slides: Chapter 1 of [Lay] P. Oswald Mathematics UCSD Fall 2015

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Outline 1 Linear Systems: Elimination 1.1-3 2 Linear Systems: Vector Language 1.3-5 3 Linear Independence 1.7 4 Linear Transformations 1.8-9
Outline 1 Linear Systems: Elimination 1.1-3 2 Linear Systems: Vector Language 1.3-5 3 Linear Independence 1.7 4 Linear Transformations 1.8-9

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m × n Linear System m linear equations in n unknowns form a m × n linear system: 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 Notation : a ij a i , j coefficients, b i right-hand side values. i = 1 , 2 , . . . , m index of equation, j = 1 , 2 , . . . , n index of unknown. Three cases : m = n ”square case” (appears most of the time), m > n overdetermined system (typical for noisy data fitting), m < n underdetermined system.
Augmented Matrix A short-hand notation for the above system is Ax = b , where A = a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . a m 1 a m 2 . . . a mn , b = b 1 b 2 . . . b m , x = x 1 x 2 . . . x m are called coefficient matrix , right-hand side (vector) , and solution (vector) , respectively. [ A | b ] = a 11 a 12 . . . a 1 n | b 1 a 21 a 22 . . . a 2 n | b 2 . . . . . . . . . . . . | . . . a m 1 a m 2 . . . a mn | b m is called augmented matrix associated with the linear system Ax = b .

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Elimination Method: Example 2 from [L,1.2] Solve ( = describe the set of all solution vectors x ) - 3 x 2 - 6 x 3 +4 x 4 = 9 - x 1 - 2 x 2 - x 3 +3 x 4 = 1 - 2 x 1 - 3 x 2 +3 x 4 = - 1 x 1 +4 x 2 +5 x 3 - 9 x 4 = - 7 ”Challenge”: Write down m , n , A , b , x , a 32 , b 2 , and [ A | b ] for this system! Elimination method : Main idea is to reduce the size of the system by eliminating unknowns in the order x 1 , x 2 , . . . using reversible transformations (that do not change the solution set of Ax = b ) in a recursive way until the reduced system becomes trivial. Will do this first with the system as is, then with the augmented matrix.
Step 1: Eliminate x 1 Choose an equation containing x 1 (call it pivot equation for x 1 ), make it the first equation, and add/subtract its multiples from the other equations such that x 1 disappears! E.g, choose the 4th equation as pivot equation, move it up, x 1 +4 x 2 +5 x 3 - 9 x 4 = - 7 - 3 x 2 - 6 x 3 +4 x 4 = 9 - x 1 - 2 x 2 - x 3 +3 x 4 = 1 - 2 x 1 - 3 x 2 +3 x 4 = - 1 and add it to the 3rd, and twice it to the 4th equation. This gives the smaller 3 × 3 system - 3 x 2 - 6 x 3 +4 x 4 = 9 2 x 2 +4 x 3 - 6 x 4 = - 6 5 x 2 +10 x 3 - 15 x 4 = - 15

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Step 2: Eliminate x 2 Before doing a similar step for eliminating x 2 , we simplify by removing common factors from the 2nd and 3rd equation (evidently, this leads to two identical equations, one of which could be dropped but since we act like machines, we will just do our routine). Take the 2nd equation as pivot equation for x 2 , move it up, x 2 +2 x 3 - 3 x 4 = - 3 - 3 x 2 - 6 x 3 +4 x 4 = 9 x 2 +2 x 3 - 3 x 4 = - 3 and add 3 times it to the 2nd, and subtract it from the 3rd equation. This gives a smaller 2 × 2 system for x 3 , x 4 which in this particular case degenerates into a trivial 2 × 1 system for x 4 : 0 · x 3 - 5 x 4 = 0 0 · x 3 +0 · x 4 = 0 ⇐⇒ - 5 x 4 = 0 0 · x 4 = 0 ⇐⇒ x 4 = 0.
Final Step Formally, we would continue with the next ”available” variable. Since

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