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

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

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 0 + 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 0 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

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

View Full Document
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 0 + 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 rref( A ) = 1 0 5 0 1 - 1 0 0 0 0 0 0 - 3 4 0 0 the reduced system is c 1 + 5 c 3 = - 3 c 2 - c 3 = 4 Choosing c 3 = 0 we get c 1 = - 3 and c 2 = 4. Thus 1 5 9 13 = - 3 1 1 1 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern