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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It would be really nice to get an explanation of how to go about doing this problem :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0There 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Using 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For the first property, you're saying the the magnitude should just be 1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes that is correct, the magnitude of each column should be 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm rereading your paragraph, one sec :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure thing, sry for rushing.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's no rush, I am just a slow learner lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So it looks like "a" has to equal a number and "d" also has to equal a number (both nonzero, of course)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does my explanation of why b = 0 and e = 0 make sense?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Good. Now maybe we can use the second property to figure out some stuff.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Maybe taking the dot product of the second column with any of the other two columns will shed some light on this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just a sec, let me go grab the white board

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just one step at a time. Remember, we now know the entire second column, from top to bottom its (0, 0, 1).

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, so now we know the second and third columns completely. Lets see what damage can be done to the first column,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I tried doing the dot product but that doesn't look like it's getting me anywhere.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, I see what you're saying. I tried dotting all of the columns. hehe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, so f=0, but either a or d must equal 1.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or some other number for both a and d?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hmm...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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh my gosh, I did not even see that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you very much for writing all of that up.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So a= sqrt(1d^2) and sub that into the second eqn and then solve for d?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0While 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you for taking the time to really explain this to me. I can barely decipher what is written in my linear algebra book.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which book are you using?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0linear algebra with applications by Otto Bretscher

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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/1806linearalgebraspring2010/videolectures/lecture17orthogonalmatricesandgramschmidt/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am lost on another question as well. Would you have time to help me with one more? That will be all, I promise :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If not, that's okay. I totally understand and you've helped me so much already.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0My time is a little short, but what is the question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll just make a new post and give you the link. If you're busy, I understand completely.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks @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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