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# MATH-M303 Section 1.5 Notes MATH-M 303

IU

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This 3 page Class Notes was uploaded by Kathryn Brinser on Tuesday September 27, 2016. The Class Notes belongs to MATH-M 303 at Indiana University taught by Keenan Kidwell in Summer 2016. Since its upload, it has received 3 views. For similar materials see Linear Algebra for Undergraduate in Mathematics at Indiana University.

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Date Created: 09/27/16

M303 Section 1.5 Notes- Solution Sets of Linear Systems 9-12-16 equations would have much more variancee much variety; general system of algebraic Homogeneous system- can be written in matrix form as A x=0 , ie. right sides of all equations are 0 o Can simply row reduce A - no operations will change augmented column when it is0 o Always consistent- no possibilit[00…0∨b] ; also always have trivial solution 0 o Will focus on nontrivial solutionset of vectors If there is one, system has infinitely many solutions To find, row reducA to see if free variables appear; if all columns have pivots, zero solution is the only one o Ex.Determine if the following homogeneous system has nontrivial solutions and give the solution set. 3x1+5x2−4x 30 −3 x −2 x +4 x =0 1 2 3 6x1+x2−8x 30 5 −4 3 5 −4 0 3 5 −4 0 1 3 3 0 −3 −2 4 0 EF 0 3 0 0 REF 0 [6 1 −8 0] [ ]0 0 0 0 [ ] 1 0 |0 0 0 0 System consistent;x3 free, so system does have nontrivial solutions x1= x 3 Solution set: 3 { x2=0 x3= free General solution vector(1,x2,x3=(4 x3,0,3) ( )4,0,1 =x3v 3 3 Set of all scalar multiples of vevt- its span In general, solution set Ax=0 is Span v} o Ex.Give the solution set forx1−3x2−2x 30 . [0 −3 −2 0 ]EF 1 −3 −1|0 [ 10 5 ] 3 1 x1=10 2 +5x3 x = free { 2 x3= free 3 1 (10x2+3x 3x 2x3) 3 1 ¿(10 x2,2 ,)(+3x3,0,x3) ¿x2 3 ,1,0 +3 1,0,1 (10 ) ( )3 ¿x2v1+x3 2 ¿ all linear combinations of1,2 ¿Span { ,v} 1 2 Non-homogeneous- A x=b where b≠ 0 o Solution set looks like a “translate” of solution set of corresponding homogeneous system A x=0 3 5 −4 7 o Ex.Give the solution set foA x=b where A= −3 −2 4 , b= −1 [ 6 1 8 ] [−4 −4 1 0 3 −1 REF 0 1 0 2 [ ] | 0 0 0 0 Solution set: 4 x1= x 31 Parametric expression: 3 { x2=2 x3= free Parametric vector form (general solution vector) (x1,2 ,3)= 4 x3−1,2,3 (3 ) o When expanding, one vector will not have free variables in it 4 ¿(3 x3,0,) ( −1,2,) ¿x3 4 ,0,0 ( −1,2,) (3) ¿ 3 v+p Homogeneous solutions are multiples ofv , translate by adding constant vector p Takes form of “span + vector” v here is same as we found by solving homogeneous A x=0 (above); p is a solution toA x=b Check by multiplying originaA by p to hopefully getb Structure of Solution Set Theorem- Given equatioA x=b , if p is a solution, then any other solution has formh+p where vh is a solution to homogeneous equation A x=0 o Proof: Ifp and q are solutions toAx=b , then we define vh=q−p ; then q=v hp . Check that vh satisfiesA x=0 by plugging in: Avh=A (q−p ) ¿Aq−A p ¿b−b ¿0 ∴v h is a solution to homogeneous equation Parametric vector form- expressing solution set to Ax=b by writing out x1 general solution vector ⋮ =s1v1+…+s k kp where si are parameters (free [x n variables) o To write solution of consistent system: Row reduce augmented matrix to REF Express each pivot variable in terms of any free variables Write a typical solution vectox in terms of free variables Decompose x into non-homogeneous part (if system non-homogeneous) followed by linear combination of vectors with free variables as parameters o Ex.Give the parametric vector form of the solution set tA x=b , where 1 4 −5 0 A= 2 −1 8 ] and b=[9 . [ | ] 1 4 −5 0 REF 1 0 3 4 [2 −1 8 9 ] [ 0 1 −2 −1 ] Solution set: x1=−3x 34 x2=2x 31 { x3= free (1,x2,x3) (3x 34,2 x 31,x 3) ¿ 3−3,2,1 +(4,−1,0) ¿x3v+p

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