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Lecture

Lesson 2b - Types of Solutions to Linear Systems.pdf

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Department
Mathematics
Course
MATH 1107
Professor
All Professors
Semester
Fall

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05/09/2013 Different Solutions? Connect: Linear Systems in High School: One Unique Solution: No solution: Infinitely many solutions: Different Solutions With Higher Orders Solution to a System: A solution to a linear system is an ordered(x1,x 2x 3...,xn) such that when you substitute the values into every equation, equality will hold. Example: (3,2,1) is a solution to the following: 3x −2x + x = 6 1 2 3 Sub x1= 3, x2= 2, 6 = 6 − x1+ x 2 x =30 and x = 1 0 = 0 2 x + x + x =2 12 3 1 2 3 12 =12 Solution Set: A set of all possible solutions that will solve the system. We use the notation as follows:S ={list of solutions} 1 05/09/2013 Finding Solutions To find a solution to a linear system, we use elimination to continuously reduce the number of variables: x −2x +3x = 6 x −2x +3x = 6 1 2 3 E2▯(E2)-2(E1) 1 2 3 E2÷5 2x1+ x 2 x =37 5x 25x = −3 E3▯(E1)+(E3) − x1+ x 2 x =3−2 − x2+2x = 3 x −2x +3x = 6 x −2x +3x = 6 1 2 3 E3▯(E3)+(E2) 1 2 3 x2− x 3 −1 x2− x 3 −1 − x 22x = 3 x3= 3 Sub into E2 x3= 3 x 23= −1 Sub into E1 x1−2(2)+3(3) = 6 x = 2 x +5 = 6 2 1 x1=1 The solution isS ={(1,2,3)} No Solutions Consider the following system: x1−2x +2x = 6 3 E2▯(E2)-2(E1) x1−2x +32 = 6 3 2x + x + x = 7 5x −5x = −5 E2÷5 1 2 3 2 3 E3÷5 3x − x +4x = −2 E3▯-3(E1)+(E3) 5x −5x = −20 1 2 3 2 3 x −2x +3x = 6 1 2 3 E3▯(E3)-(E2) x1−2x +32 = 6 3 x − x = −1 2 3 x2− x 3 −1 x2− x 3 −4 0 = −3 There is no Solution! 2 05/09/2013 Infinite Solutions Consider the following system: x −2x +3x = 6 x −2x +3x = 6 1 2 3 E2▯(E2)-2(E1) 1 2 3 E2÷5 2x 1 x +2x = 3 5x 25x = −3 E3▯-3(E1)+(E3) E3÷5 3x 1 x +2x =133 5x 25x = 35 x1−2x +2x = 6 3 x −2x +3x = 6 E3▯(E3)-(E2) 1 2 3 x 2 x =3−1 x2− x 3 −1 x − x = −1 2 3 0 = 0 x3can be what ever we want! This means x =3t, where t can be any number you want it to be. Substituting this back into the equations we get: Sub into E2 x 3 t x2−t = −1 Sub into E1 x1−2(−1+t)+3(t) = 6 x +t +2 = 6 x2= −1+t 1 x = 4−t 1 The solution is S ={(4−t,−1+t,t):t∈R} Different Types of Solutions Consistent System: A system that has a solution (either one or many). Inconsistent System: A system that has no sol
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