## 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

1. brinethery Group Title

2. brinethery Group Title

It would be really nice to get an explanation of how to go about doing this problem :-)

3. brinethery Group Title

hello?????

4. joemath314159 Group Title

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 Group Title

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 Group Title

For the first property, you're saying the the magnitude should just be 1?

7. brinethery Group Title

*that

8. joemath314159 Group Title

yes that is correct, the magnitude of each column should be 1.

9. brinethery Group Title

10. joemath314159 Group Title

sure thing, sry for rushing.

11. brinethery Group Title

It's no rush, I am just a slow learner lol

12. brinethery Group Title

So it looks like "a" has to equal a number and "d" also has to equal a number (both nonzero, of course)

13. brinethery Group Title

or wait... ugh

14. joemath314159 Group Title

Does my explanation of why b = 0 and e = 0 make sense?

15. brinethery Group Title

yes

16. brinethery Group Title

sorry, OS froze up.

17. joemath314159 Group Title

Good. Now maybe we can use the second property to figure out some stuff.

18. joemath314159 Group Title

Maybe taking the dot product of the second column with any of the other two columns will shed some light on this?

19. brinethery Group Title

I feel so stupid

20. brinethery Group Title

just a sec, let me go grab the white board

21. joemath314159 Group Title

just one step at a time. Remember, we now know the entire second column, from top to bottom its (0, 0, 1).

22. brinethery Group Title

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 Group Title

that is correct :)

24. brinethery Group Title

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 Group Title

ok, so now we know the second and third columns completely. Lets see what damage can be done to the first column,

26. brinethery Group Title

I tried doing the dot product but that doesn't look like it's getting me anywhere.

27. joemath314159 Group Title

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 Group Title

Oh, I see what you're saying. I tried dotting all of the columns. hehe

29. brinethery Group Title

Okay, so f=0, but either a or d must equal 1.

30. brinethery Group Title

or some other number for both a and d?

31. joemath314159 Group Title

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 Group Title

Oh my gosh, I did not even see that.

33. brinethery Group Title

Thank you very much for writing all of that up.

34. joemath314159 Group Title

yw :)

35. brinethery Group Title

So a= sqrt(1-d^2) and sub that into the second eqn and then solve for d?

36. joemath314159 Group Title

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 Group Title

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 Group Title

which book are you using?

39. brinethery Group Title

linear algebra with applications by Otto Bretscher

40. joemath314159 Group Title

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 Group Title

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 Group Title

If not, that's okay. I totally understand and you've helped me so much already.

43. joemath314159 Group Title

My time is a little short, but what is the question?

44. brinethery Group Title

I'll just make a new post and give you the link. If you're busy, I understand completely.

45. brinethery Group Title

sec

46. Mathmuse Group Title

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?