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

- anonymous

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

- schrodinger

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- anonymous

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- anonymous

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

- anonymous

hello?????

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## More answers

- anonymous

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.

- anonymous

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?

- anonymous

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

- anonymous

*that

- anonymous

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

- anonymous

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

- anonymous

sure thing, sry for rushing.

- anonymous

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

- anonymous

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

- anonymous

or wait... ugh

- anonymous

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

- anonymous

yes

- anonymous

sorry, OS froze up.

- anonymous

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

- anonymous

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

- anonymous

I feel so stupid

- anonymous

just a sec, let me go grab the white board

- anonymous

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

- anonymous

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)?

- anonymous

that is correct :)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

or some other number for both a and d?

- anonymous

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?

- anonymous

Oh my gosh, I did not even see that.

- anonymous

Thank you very much for writing all of that up.

- anonymous

yw :)

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

which book are you using?

- anonymous

linear algebra with applications by Otto Bretscher

- anonymous

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/

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

sec

- anonymous

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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