Vaz_MAT_343_ONLINE_B_Fall_2019.hourogne.Section_1.5_Elementary_Matrices.pdf - Hassan Ouro-gneni Vaz MAT 343 ONLINE B Fall 2019 Assignment Section 1.5

Vaz_MAT_343_ONLINE_B_Fall_2019.hourogne.Section_1.5_Elementary_Matrices.pdf

This preview shows page 1 - 3 out of 9 pages.

Hassan Ouro-gneni Vaz MAT 343 ONLINE B Fall 2019 Assignment Section 1.5 Elementary Matrices due 10/21/2019 at 11:59pm MST 1. (1 point) Suppose that A is a 3 × 3 matrix. For each of the question (a)-(d), find the elementary matrix E such that EA gives the matrix resulting from A after the given row operation is performed. Then give E - 1 . a. R 2 1 5 R 2 E = , E - 1 = b. R 3 ← - 3 R 3 E = , E - 1 = c. R 1 R 3 E = , E - 1 = d. R 2 R 2 + 6 R 3 E = , E - 1 = Solution: a. E 1 is obtained by multiplying by 1 2 the first row of the identity matrix I = 1 0 0 1 . This gives E 1 = 1 2 0 0 1 The inverse of E 1 is obtained by multiplying the first row of the identity by 2. This gives E - 1 1 = 2 0 0 1 . b. E 2 is obtained by multiplying by - 6 the second row of the identity matrix I = 1 0 0 1 . This gives E 2 = 1 0 0 - 6 The inverse of E 2 is obtained by dividing the first row of the identity by - 6. This gives E - 1 2 = 1 0 0 1 - 6 . c. E 3 is obtained by switching the first and second rows of the identity matrix I = 1 0 0 1 . This gives E 3 = 0 1 1 0 The inverse of E 3 is obtained by switching the first and second rows of the identity . This gives E - 1 3 = 0 1 1 0 . Note that E 3 = E - 1 3 . This is always the case for permutation matrices, that is, matrices that are obtained by permuting the rows of the identity matrix. d. E 4 is obtained by performing the row operation r 2 r 2 + 7 r 1 on the identity matrix I = 1 0 0 1 . This gives E 4 = 1 0 7 1 The inverse of E 4 is obtained by performing the row op- eration r 2 r 2 - 7 r 1 on the identity matrix. This gives E - 1 4 = 1 0 - 7 1 . e. E 5 is obtained by multiplying by 1 5 the second row of the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 5 = 1 0 0 0 1 5 0 0 0 1 The inverse of E 5 is obtained by multiplying the second row of the identity by 5. This gives E - 1 5 = 1 0 0 0 5 0 0 0 1 . f. E 6 is obtained by multiplying by - 3 the third row of the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 6 = 1 0 0 0 1 0 0 0 - 3 The inverse of E 6 is obtained by dividing the third row of the identity by - 3. This gives E - 1 6 = 1 0 0 0 1 0 0 0 1 - 3 . g. E 7 is obtained by switching the fist and the third row of the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 7 = 0 0 1 0 1 0 1 0 0 The inverse of E 7 is obtained by switching the first and third row of the identity . This gives E - 1 7 = 0 0 1 0 1 0 1 0 0 . h. E 8 is obtained by performing the row operation r 2 r 2 + 6 r 3 on the identity matrix I = 1 0 0 0 1 0 0 0 1 . This gives E 8 = 1 0 0 0 1 6 0 0 1 The inverse of E 8 is obtained by performing the row operation r 2 r 2 - 6 r 3 the identity . This gives E - 1 8 = 1 0 0 0 1 - 6 0 0 1 . Answer(s) submitted: 1 0 0 0 1/5 1
Image of page 1

Subscribe to view the full document.

0 0 0 1 1 0 0 0 5 0 0 0 1 1 0 0 0 1 0 0 0 -3 1 0 0 0 1 0 0 0 -1/3 0 0 1 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 0 0 1 6 0 0 1 1 0 0 0 1 -6 0 0 1
Image of page 2
Image of page 3
  • Fall '08
  • ringhofer

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

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes