hwsol2 - Homework 2 Solutions Course Reader Exercises 5.9...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 2 Solutions Course Reader Exercises 5.9 The goal is to find scalars a and b such that a 2 1 + b 3 2 = 1 2 This leads to the system of equations 2 a + 3 b = 1 1 a + 2 b = 2 The second equation implies a = 2- 2 b , so inserting this into the first equation gives 4- 4 b + 3 b = 1, or b = 3. Plugging back into either equation gives a =- 4, so- 4 2 1 + 3 3 2 = 1 2 5.10 We want to know if there exist scalars c 1 ,c 2 ,c 3 such that c 1 2 1 1 + c 2 5 8 + c 3 1 6- 2 = 8- 5- 11 This leads to the system of equations 2 c 1 + 5 c 2 + c 3 = 8 c 1 + 8 c 2 + 6 c 3 =- 5 c 1- 2 c 3 =- 11 which has augmented matrix A = 2 5 1 1 8 6 1 0- 2 8- 5- 11 Since rref( A ) = 1 0- 2 0 1 1 0 0 1 the last equation of the reduced system is 0 = 1, so there are no solutions. Thus v is not in the span of the given set of vectors. 5.11 We need to find a solution to the system of equations that corresponds to the following matrix of coefficients, which we row reduce : 1 2 5 11 8 1 8 2- 5 1 0- 12- 11 1 0- 12- 11 2 5 11 8 1 8 2- 5 1 0- 12- 11 0 5 35 30 0 8 14 6 1 0- 12- 11 0 1 7 6 0 4 7 3 1 0- 12- 11 0 1 7 6 0 0- 21- 21 1 0- 12- 11 0 1 7 6 0 0 1 1 1 0 0 1 0 1 0- 1 0 0 1 1 So we conclude that 1 2 1 1 - 1 5 8 + 1 11 2- 12 = 8- 5- 11 5.12 We want to solve c 1 1 1 1 1 + c 2 1 2 3 4 + c 3 4 3 2 1 = 1 5 9 13 The augmented matrix corresponding to this system is A = 1 1 4 1 2 4 1 3 2 1 4 1 1 5 9 13 Since...
View Full Document

Page1 / 8

hwsol2 - Homework 2 Solutions Course Reader Exercises 5.9...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online