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brinethery

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

  • one year ago
  • one year ago

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

    • one year ago
  3. brinethery
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    hello?????

    • one year ago
  4. joemath314159
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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.

    • one year ago
  5. joemath314159
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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?

    • one year ago
  6. brinethery
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    For the first property, you're saying the the magnitude should just be 1?

    • one year ago
  7. brinethery
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    *that

    • one year ago
  8. joemath314159
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    yes that is correct, the magnitude of each column should be 1.

    • one year ago
  9. brinethery
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    I'm re-reading your paragraph, one sec :-)

    • one year ago
  10. joemath314159
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    sure thing, sry for rushing.

    • one year ago
  11. brinethery
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    It's no rush, I am just a slow learner lol

    • one year ago
  12. brinethery
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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)

    • one year ago
  13. brinethery
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    or wait... ugh

    • one year ago
  14. joemath314159
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    Does my explanation of why b = 0 and e = 0 make sense?

    • one year ago
  15. brinethery
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    yes

    • one year ago
  16. brinethery
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    sorry, OS froze up.

    • one year ago
  17. joemath314159
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    Good. Now maybe we can use the second property to figure out some stuff.

    • one year ago
  18. joemath314159
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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?

    • one year ago
  19. brinethery
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    I feel so stupid

    • one year ago
  20. brinethery
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    just a sec, let me go grab the white board

    • one year ago
  21. joemath314159
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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).

    • one year ago
  22. brinethery
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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)?

    • one year ago
  23. joemath314159
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    that is correct :)

    • one year ago
  24. brinethery
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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

    • one year ago
  25. joemath314159
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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,

    • one year ago
  26. brinethery
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    I tried doing the dot product but that doesn't look like it's getting me anywhere.

    • one year ago
  27. joemath314159
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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.

    • one year ago
  28. brinethery
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    Oh, I see what you're saying. I tried dotting all of the columns. hehe

    • one year ago
  29. brinethery
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    Okay, so f=0, but either a or d must equal 1.

    • one year ago
  30. brinethery
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    or some other number for both a and d?

    • one year ago
  31. joemath314159
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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?

    • one year ago
  32. brinethery
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    Oh my gosh, I did not even see that.

    • one year ago
  33. brinethery
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    Thank you very much for writing all of that up.

    • one year ago
  34. joemath314159
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    yw :)

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

    • one year ago
  36. joemath314159
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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.

    • one year ago
  37. brinethery
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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.

    • one year ago
  38. joemath314159
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    which book are you using?

    • one year ago
  39. brinethery
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    linear algebra with applications by Otto Bretscher

    • one year ago
  40. joemath314159
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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/

    • one year ago
  41. brinethery
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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 :-)

    • one year ago
  42. brinethery
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    If not, that's okay. I totally understand and you've helped me so much already.

    • one year ago
  43. joemath314159
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    My time is a little short, but what is the question?

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

    • one year ago
  45. brinethery
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    sec

    • one year ago
  46. Mathmuse
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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?

    • one year ago
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