Linear Algebra with Applications (3rd Edition)

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 2 SSM: Linear Algebra Chapter 2 2.1 1. Not a linear transformation, since y 2 = x 2 + 2 is not linear in our sense. 3. Not linear, since y 2 = x 1 x 3 is nonlinear. 5. By Fact 2.1.2, the three columns of the 2 × 3 matrix A are T ( ~ e 1 ) , T ( ~ e 2 ), and T ( ~ e 3 ), so that A = 7 6 - 13 11 9 17 . 7. Note that x 1 ~v 1 + · · · + x m ~v m = [ ~v 1 . . .~v m ] x 1 · · · x m , so that T is indeed linear, with matrix [ ~v 1 ~v 2 · · · ~v m ]. 9. We have to attempt to solve the equation y 1 y 2 = 2 3 6 9 x 1 x 2 for x 1 and x 2 . Reducing the system 2 x 1 + 3 x 2 = y 1 6 x 1 + 9 x 2 = y 2 we obtain x 1 + 1 . 5 x 2 = 0 . 5 y 1 0 = - 3 y 1 + y 2 . No unique solution ( x 1 , x 2 ) can be found for a given ( y 1 , y 2 ); the matrix is noninvertible. 11. We have to attempt to solve the equation y 1 y 2 = 1 2 3 9 x 1 x 2 for x 1 and x 2 . Reducing the system x 1 + 2 x 2 = y 1 3 x 1 + 9 x 2 = y 2 we find that x 1 = 3 y 1 - 2 3 y 2 x 2 = - y 1 + 1 3 y 2 . The inverse matrix is 3 - 2 3 - 1 1 3 . 13. a. First suppose that a 6 = 0. We have to attempt to solve the equation y 1 y 2 = a b c d x 1 x 2 for x 1 and x 2 . ax 1 + bx 2 = y 1 cx 1 + dx 2 = y 2 ÷ a x 1 + b a x 2 = 1 a y 1 cx 1 + dx 2 = y 2 - c ( I ) x 1 + b a x 2 = 1 a y 1 ( d - bc a ) x 2 = - c a y 1 + y 2 28
Image of page 1

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

View Full Document Right Arrow Icon
SSM: Linear Algebra Section 2.1 x 1 + b a x 2 = 1 a y 1 ( ad - bc a ) x 2 = - c a y 1 + y 2 We can solve this system for x 1 and x 2 if (and only if) ad - bc 6 = 0, as claimed. If a = 0, then we have to consider the system bx 2 = y 1 cx 1 + dx 2 = y 2 swap : I II cx 1 + dx 2 = y 2 bx 2 = y 1 We can solve for x 1 and x 2 provided that both b and c are nonzero, that is if bc 6 = 0. Since a = 0, this means that ad - bc 6 = 0, as claimed. b. First suppose that ad - bc 6 = 0 and a 6 = 0. Let D = ad - bc for simplicity. We continue our work in part (a): x 1 + b a x 2 = 1 a y 1 D a x 2 = - c a y 1 + y 2 · a D x 1 + b a x 2 = 1 a y 1 x 2 = - c D y 1 + a D y 2 - b a ( II ) x 1 = ( 1 a + bc aD ) y 1 - b D y 2 x 2 = - c D y 1 + a D y 2 x 1 = d D y 1 - b D y 2 x 2 = - c D y 1 + a D y 2 ( Note that 1 a + bc aD = D + bc aD = ad aD = d D . ) It follows that a b c d - 1 = 1 ad - bc d - b - c a , as claimed. If ad - bc 6 = 0 and a = 0, then we have to solve the system cx 1 + dx 2 = y 2 bx 2 = y 1 ÷ c ÷ b x 1 + d c x 2 = 1 c y 2 x 2 = 1 b y 1 - d c ( II ) x 1 = - d bc y 1 + 1 c y 2 x 2 = 1 b y 1 It follows that a b c d - 1 = - d bc 1 c 1 b 0 = 1 ad - bc d - b - c a (recall that a = 0), as claimed. 29
Image of page 2
Chapter 2 SSM: Linear Algebra 15. By Exercise 13a, the matrix a - b b a is invertible if (and only if) a 2 + b 2 6 = 0, which is the case unless a = b = 0. If a - b b a is invertible, then its inverse is 1 a 2 + b 2 a b - b a , by Exercise 13b. 17. If A = - 1 0 0 - 1 , then A~x = - ~x for all ~x in R 2 , so that A represents a reflection about the origin. This transformation is its own inverse: A - 1 = A . 19. If A = 1 0 0 0 , then A x 1 x 2 = x 1 0 , so that A represents the orthogonal projection onto the ~ e 1 axis. (See Figure 2.1.) This transformation is not invertible, since the equation A~x = 1 0 has infinitely many solutions ~x . Figure 2.1: for Problem 2.1.19 . 21. Compare with Example 5.
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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