Linear Algebra: Extra Practice Worksheet 3

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  • Опубликовано: 16 ноя 2024

Комментарии • 13

  • @jan-willemreens9010
    @jan-willemreens9010 2 года назад +1

    Good evening Dr. Valerie, As an interested outsider I have also been busy, in addition to viewing your easy-to-follow Linear Algebra lectures playlist, with the first chapters of Otto Bretscher's Linear Algebra textbook, so I also looked at your video, if you don't mind. What I like about a question like number 5 is that it gives you a good indication of whether you have really understood the theory correctly! I have become much wiser the past few days thanks to you, and I'm still enjoying the process of learning more. Dr. Valerie, many thanks for your unwavering commitment to education!

    • @DrValerieHower
      @DrValerieHower  2 года назад +1

      You are welcome :). I appreciate your kind words.

  • @shubhreetbhullar4845
    @shubhreetbhullar4845 Год назад

    For the projection of the matrix, would we not square root U1 squared plus U2 squared. I mean to say, for problem 1a, shouldn't it be 1/sqrt(17) and not 1/17?

    • @DrValerieHower
      @DrValerieHower  Год назад

      1/17 is the correct multiplier here. I am using a formula to calculate the matrix. You can also project e1. We get 1/17 * (1,4) then project e2. We get 1/17 * (4,16). Then put these as columns in a matrix. The result is the same. I hope this helps.

  • @javierarturorizzosudario.8505
    @javierarturorizzosudario.8505 Год назад +1

    Dear Dr. Valerie, I have a question regarding our reflection formula: 2 projL(vector x) - vector x . I was wondering why in the problem 1b (video 6:45), we could do 2 projL(vector x) - (the identity) ? shouldn't we minus the vector x?

    • @DrValerieHower
      @DrValerieHower  Год назад +1

      Correct that reflection is 2projection(x)-x. If we think about this as a difference of functions, We have 2projection function minus the identity function (the identity maps every vector to itself). Thinking this way (as a difference of functions), we can translate to the corresponding matrices. The 2x2 matrix of reflection is therefore 2projection matrix minus identity matrix. I hope this helps.

    • @javierarturorizzosudario.8505
      @javierarturorizzosudario.8505 Год назад +1

      It's very helpful. Thank you very much for your time!

  • @mayrajloredo87
    @mayrajloredo87 2 года назад

    Can you upload the practice problems please.

    • @DrValerieHower
      @DrValerieHower  2 года назад +1

      Probably this summer I want to make a website with all my worksheets/reviews/etc. Until then, here is this worksheet: drive.google.com/file/d/1fLCl3ue76TPHzFBGjzcC0xEJwBqT3_Or/view?usp=sharing

  • @mayrajloredo87
    @mayrajloredo87 2 года назад

    Is there another way to find the inverse without using rref?

    • @DrValerieHower
      @DrValerieHower  2 года назад

      Generally one can solve a linear system at REF. However to find the inverse of a 3x3 matrix, we are really solving 3 linear systems: Ax=e1, Ax=e2, and Ax=e3. I think it is more work to stop at REF back substitute three times.

    • @mayrajloredo87
      @mayrajloredo87 2 года назад

      @@DrValerieHower ok its because I saw a different video where they solve a 3x3 matrix using matrix of minors and cofactors, adjugate and determinant.

    • @DrValerieHower
      @DrValerieHower  2 года назад

      In our textbook, determinant comes in chapter 6. Inverse of a matrix is in chapter 2. The determinant itself can be a complicated calculation. Is there a formula for the inverse that uses the determinant? Probably. There is certainly a shortcut for 2x2 matrices using it. But, I don't see how that method could be generally shorter than the standard one used here.