Question

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

Answer 1

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

Answer 2

hello?????

Answer 3

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.

Answer 4

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?

Answer 5

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

Answer 6

*that

Answer 7

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

Answer 8

I'm re-reading your paragraph, one sec :-)

Answer 9

sure thing, sry for rushing.

Answer 10

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

Answer 11

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

Answer 12

or wait... ugh

Answer 13

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

Answer 14

yes

Answer 15

sorry, OS froze up.

Answer 16

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

Answer 17

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

Answer 18

I feel so stupid

Answer 19

just a sec, let me go grab the white board

Answer 20

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

Answer 21

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)?

Answer 22

that is correct :)

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

or some other number for both a and d?

Answer 30

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?

Answer 31

Oh my gosh, I did not even see that.

Answer 32

Thank you very much for writing all of that up.

Answer 33

yw :)

Answer 34

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

Answer 35

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.

Answer 36

Thank you for taking the time to really explain this to me. I can barely decipher what is written in my linear algebra book.

Answer 37

which book are you using?

Answer 38

linear algebra with applications by Otto Bretscher

Answer 39

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/

Answer 40

I am lost on another question as well. Would you have time to help me with one more? That will be all, I promise :-)

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

sec

Answer 45

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?