hwsol1 - Homework 1 Solutions Course Reader Exercises 0 4 2...

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

View Full Document Right Arrow Icon
Homework 1 Solutions Course Reader Exercises 1.7 (a) 2 1 1 + 2 - 4 0 - 4 2 - 2 = 0 - 5 3 (b) 2 1 1 + 2 - 3 - 4 2 = 2 1 1 + - 6 - 8 4 = - 4 - 7 5 (c) 4 6 - - 9 6 = 13 0 (d) 6 9 + - 6 4 = 0 13 (e) 2(1) + 1( - 2) + 1(0) = 0 (f) 2(5) + 1(0) + 1( - 2) = 8 (g) 2( - 3) + 3(2) = 0 (h) p (2 - ( - 3)) 2 + (3 - 2) 2 = 26 2.3 The first two vectors are nonzero and neither is a scalar multiple of the other, so they span a plane. To see if the third vector is a linear combination of the first two, consider the equation c 1 2 1 1 + c 2 5 8 0 = 1 6 - 2 which is equivalent to the system 2 c 1 + 5 c 2 = 1 c 1 + 8 c 2 = 6 c 1 = - 2 The solution is c 1 = - 2, c 2 = 1. Thus - 2 2 1 1 + 5 8 0 = 1 6 - 2 and therefore the set spans a plane. 2.4 We know the span of these vectors is not a point, because the vectors are not all zero; and it is not a line, because clearly the second vector is not a multiple of the first. So we know the span is either a plane, or all of R 3 . 1
Image of page 1

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

View Full Document Right Arrow Icon
To distinguish between these possibilities, it is enough to determine if the given vectors are dependent or independent. We therefore attempt to find solutions to the system of equations c 1 2 1 1 + c 2 5 8 0 + c 3 2 6 - 2 = 0 0 0 We row reduce the corresponding matrix: 2 5 2 0 1 8 6 0 1 0 - 2 0 1 0 - 2 0 1 8 6 0 2 5 2 0 1 0 - 2 0 0 8 8 0 0 5 6 0 1 0 - 2 0 0 1 1 0 0 5 6 0 1 0 - 2 0 0 1 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern