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brinethery Group Title

Find scalars a, b, c, d, e, f, and g so that the matrix A is orthogonal.

  • 2 years ago
  • 2 years ago

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  1. brinethery Group Title
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    • 2 years ago
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  2. brinethery Group Title
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    It would be really nice to get an explanation of how to go about doing this problem :-)

    • 2 years ago
  3. brinethery Group Title
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    hello?????

    • 2 years ago
  4. joemath314159 Group Title
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    There are two properties that orthogonal matrices have. When looking at the columns of the matrix, in this case the vectors v1, v2 and v3, if you take the inner product (or dot product) of a column with itself, you should get 1, and if you take the inner (dot) product of a column with any other column, you should get 0. Some how we need to use these two properties to figure out what the correct numbers are.

    • 2 years ago
  5. joemath314159 Group Title
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    Using the first property, we can conclude that b = e = 0, since:\[v_2\cdot v_2 = 1 \Longrightarrow b^2 +e^2 +1^2 =1\] which only has the solution b = 0 and e = 0. Does that make sense?

    • 2 years ago
  6. brinethery Group Title
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    For the first property, you're saying the the magnitude should just be 1?

    • 2 years ago
  7. brinethery Group Title
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    *that

    • 2 years ago
  8. joemath314159 Group Title
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    yes that is correct, the magnitude of each column should be 1.

    • 2 years ago
  9. brinethery Group Title
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    I'm re-reading your paragraph, one sec :-)

    • 2 years ago
  10. joemath314159 Group Title
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    sure thing, sry for rushing.

    • 2 years ago
  11. brinethery Group Title
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    It's no rush, I am just a slow learner lol

    • 2 years ago
  12. brinethery Group Title
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    So it looks like "a" has to equal a number and "d" also has to equal a number (both nonzero, of course)

    • 2 years ago
  13. brinethery Group Title
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    or wait... ugh

    • 2 years ago
  14. joemath314159 Group Title
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    Does my explanation of why b = 0 and e = 0 make sense?

    • 2 years ago
  15. brinethery Group Title
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    yes

    • 2 years ago
  16. brinethery Group Title
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    sorry, OS froze up.

    • 2 years ago
  17. joemath314159 Group Title
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    Good. Now maybe we can use the second property to figure out some stuff.

    • 2 years ago
  18. joemath314159 Group Title
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    Maybe taking the dot product of the second column with any of the other two columns will shed some light on this?

    • 2 years ago
  19. brinethery Group Title
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    I feel so stupid

    • 2 years ago
  20. brinethery Group Title
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    just a sec, let me go grab the white board

    • 2 years ago
  21. joemath314159 Group Title
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    just one step at a time. Remember, we now know the entire second column, from top to bottom its (0, 0, 1).

    • 2 years ago
  22. brinethery Group Title
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    So I wrote it up and it looks like g= 0. And then to find what the entire third column is, just take the magnitude. so sqrt(c^2 +(3/5)^2 +0^2)?

    • 2 years ago
  23. joemath314159 Group Title
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    that is correct :)

    • 2 years ago
  24. brinethery Group Title
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    haha nevermind again! I got confused for a sec with what I was doing. Okay, so I took the magnitude and set it equal to 1. I got c=sqrt(16/25), or 4/5

    • 2 years ago
  25. joemath314159 Group Title
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    ok, so now we know the second and third columns completely. Lets see what damage can be done to the first column,

    • 2 years ago
  26. brinethery Group Title
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    I tried doing the dot product but that doesn't look like it's getting me anywhere.

    • 2 years ago
  27. joemath314159 Group Title
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    what happens when you dot the second and the first columns? Think similarly to how you got that g = 0, but look at f this time.

    • 2 years ago
  28. brinethery Group Title
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    Oh, I see what you're saying. I tried dotting all of the columns. hehe

    • 2 years ago
  29. brinethery Group Title
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    Okay, so f=0, but either a or d must equal 1.

    • 2 years ago
  30. brinethery Group Title
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    or some other number for both a and d?

    • 2 years ago
  31. joemath314159 Group Title
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    hmm...from the first property, we have that:\[v_1\cdot v_1=1\Longrightarrow a^2 +d^2+0^2=1\Longrightarrow a^2+d^2=1\]and from taking the dot product of the first and tthird column, property two tells us:\[v_1\cdot v_3 = 0\Longrightarrow \frac{4}{5}a+\frac{3}{5}d+0\cdot 0=0\Longrightarrow \frac{4}{5}a+\frac{3}{5}d=0\]Maybe we can use these equations together as a system of equations to solve for a and d?

    • 2 years ago
  32. brinethery Group Title
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    Oh my gosh, I did not even see that.

    • 2 years ago
  33. brinethery Group Title
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    Thank you very much for writing all of that up.

    • 2 years ago
  34. joemath314159 Group Title
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    yw :)

    • 2 years ago
  35. brinethery Group Title
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    So a= sqrt(1-d^2) and sub that into the second eqn and then solve for d?

    • 2 years ago
  36. joemath314159 Group Title
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    While that will work, it might be a little easier to say:\[\frac{4}{5}a+\frac{3}{5}d=0\iff \frac{4}{5}a=-\frac{3}{5}d\iff a=-\frac{3}{4}d\]and substitute that into the first equation. Again, its not what you said is wrong, Im just trying to avoid square roots since they make things a little complicated.

    • 2 years ago
  37. brinethery Group Title
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    Thank you for taking the time to really explain this to me. I can barely decipher what is written in my linear algebra book.

    • 2 years ago
  38. joemath314159 Group Title
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    which book are you using?

    • 2 years ago
  39. brinethery Group Title
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    linear algebra with applications by Otto Bretscher

    • 2 years ago
  40. joemath314159 Group Title
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    If you need more resources, other than your book, MIT's OCW videos are decent. Here's their video on orthogonal matries: http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-17-orthogonal-matrices-and-gram-schmidt/

    • 2 years ago
  41. brinethery Group Title
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    I am lost on another question as well. Would you have time to help me with one more? That will be all, I promise :-)

    • 2 years ago
  42. brinethery Group Title
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    If not, that's okay. I totally understand and you've helped me so much already.

    • 2 years ago
  43. joemath314159 Group Title
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    My time is a little short, but what is the question?

    • 2 years ago
  44. brinethery Group Title
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    I'll just make a new post and give you the link. If you're busy, I understand completely.

    • 2 years ago
  45. brinethery Group Title
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    sec

    • 2 years ago
  46. Mathmuse Group Title
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    Thanks @Joemath314159 I've learned something. Just to add now that @brinethery has got it, since the transpose of a orthogonal matrix is also orthogonal, you can perform the same operations with dot product on the row vectors no?

    • 2 years ago
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