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brinethery

  • 2 years ago

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

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

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

  4. joemath314159
    • 2 years ago
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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.

  5. joemath314159
    • 2 years ago
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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?

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

  7. brinethery
    • 2 years ago
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    *that

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

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

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

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

  12. brinethery
    • 2 years ago
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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)

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

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

  15. brinethery
    • 2 years ago
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    yes

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

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

  18. joemath314159
    • 2 years ago
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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?

  19. brinethery
    • 2 years ago
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    I feel so stupid

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

  21. joemath314159
    • 2 years ago
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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).

  22. brinethery
    • 2 years ago
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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)?

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

  24. brinethery
    • 2 years ago
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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

  25. joemath314159
    • 2 years ago
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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,

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

  27. joemath314159
    • 2 years ago
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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.

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

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

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

  31. joemath314159
    • 2 years ago
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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?

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

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

  34. joemath314159
    • 2 years ago
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    yw :)

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

  36. joemath314159
    • 2 years ago
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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.

  37. brinethery
    • 2 years ago
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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.

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

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

  40. joemath314159
    • 2 years ago
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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/

  41. brinethery
    • 2 years ago
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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 :-)

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

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

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

  45. brinethery
    • 2 years ago
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    sec

  46. Mathmuse
    • 2 years ago
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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?

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