# Thread: Stuck solving set of linear equations [not solved]

1. ## Stuck solving set of linear equations [not solved]

Hi all, first post.

I'm stuck trying to solve this set of linear equations. I need to find the coefficients a, b and c and express them in terms of x, y, z.

I've tried setting c=1 and finding expressions for a and b but when I check these with arbitrary numbers I just keep getting it wrong.
Somebody has suggested reducing the 3 eigenvectors to row echelon form but I don't understand why. If possible, I would rather solve this algebraically.

The vectors being multiplied by a, b and c are normalised eigenvectors which I am going to assume are correct!

a*(1/sqrt2,0,-1/sqrt2) + b*(1/sqrt2,-sqrt2/2,1/2) + c*(1/2,sqrt2/2,1/2) =(x,y,z)

Any help or advice would be appreciated!
Thanks Although this is university work, it was set as a revison exercise so apologies if this in the wrong section!

2. I'll try to help until someone else gets to it.

I get that you have three vectors that are unit vectors. As your final answer though you want to have something like x=?, y=?, z=? in terms of a,b and c?

Are x,y and z just unit vectors themselves? I'm not quite following what you want to solve. It sounds like you are just stuck with the algebra, which I might be able to help with, but I'm lost conceptually to what you meant.

Welcome to MHF by the way!

3. Thanks for the fast response!
Sorry for being unclear. (x,y,z) is defined as an arbitrary column vector.

I need to find expressions for a,b,c. ie. a=... b=... c=... in terms of x,y,z for any vector (x,y,z).

That is how I interpreted the question anyway, but just incase I'll write this part of the question down exactly as it is written:

(...after having found normalised eigenvectors p, q, r of a matrix...)

"Show that an arbitrary column vector x=(x,y,z) can be written as a linear combination of the three eigevectors, i.e.

x=ap+bq+cr

and find expressions for the coefficients a,b and c"

Where p,q and r are the unit vectors in my first post.

Hope I've explained that well enough now

4. That helps a lot. One more thing though...

Your three unit vectors are written in terms of (x,y,z) components so the three equations you get from the problem are like:

$\displaystyle x=ax_1 + bx_2+cx_3$, where x_n is the x component from each vector. y and z are formed the same way.

Am I understanding this right?

5. Yes, that's the form I'd been trying solve them from. Thanks

6. Ok so the three equations we get are:

$\displaystyle x= a( \frac{\sqrt{2}}{2})+b(\frac{\sqrt{2}}{2})+(\frac{1 }{2})c$

$\displaystyle y=a(0)-b(\frac{\sqrt{2}}{2})+c(\frac{\sqrt{2}}{2})$

$\displaystyle z= -a(\frac{\sqrt{2}}{2})+b(\frac{1}{2})+c(\frac{1}{2} )$

Wow this is gross. RREF would definitely be quicker.

I'll just post this and think some more...

EDIT: I would definitely multiply everything by 2 to start.

$\displaystyle 2x= a(\sqrt{2})+b(\sqrt{2})+c$

$\displaystyle 2y=a(0)-b(\sqrt{2})+c(\sqrt{2})$

$\displaystyle 2z= -a(\sqrt{2})+b+c$

7. Wow this is gross. RREF would definitely be quicker.
Lol. I'll try the RREF method then if you think it's quicker. I can reduce it to the identity matrix but how does this help?

I could post each row operation if that's any good?

8. What about using the $\displaystyle A^{-1}B$ method? A is a 3x3 matrix of coefficients plus variables a,b and c. B is a 3x1 matrix of [x,y,z]. The resulting matrix will be a 3x1 matrix and I think it should solve for a,b and c.

Do you have to do this by hand? Try plugging this into a calculator to see if it works. I think it should.