Question

solving simultaneous equations
1a+2b+1c=0
6a-1b+0c=0
-1a-2b-1c=0

Answer 1

Add the first equation to the third equation and tell me how you feel about the result.

Answer 2

Yeah I got that equation 2 remains. so from there:
6a-1b+0c=0
6a=1b
therefore a=1 and b=6? Im trying to solve Eigenvectors from these 3 equations. Basically need to find a, b and c

Answer 3

The point of the first exercise was to suggest that there are infinitely many solutions. You cannot solve explicitly for a, b, and c. You can solve for a and b in terms of c.

Answer 4

Cant I solve for a an b above, then just replace values back into other equation to find c?

Answer 5

Did you try that? It is an underdefined system. You can't do it. Pick ANY value you like for 'c', then solve for the corresponding values for a and b. Seriously, ANY value for c.

Answer 6

Hmmm I'll show what I did for a previous set of equations.
5a+2b+c=0 -eq1
6a+3b+0c=0 - eq2
-a-2b+3c=0 -eq3
so solving equation 1 and 3
4a+0+4c=0 -eq4
4a=-4c
a=-c
so a/1=c/-1, values of a is 1 and c=-1
Then I just substituted eq4 into eq1
a=-c -eq4
5c+2b+c=0 -eq1
5c+eb+(-c)=0
4c+2b=0
2b=-4a
c=-2a
therefore -b/2=a/1
so values are a=1, b=-2, and c=-1

Answer 7

For starters, youhave one equation that says "a = -c" and another that says "c = -2a". Unless a and c are both zero, that will be a problem.

Second, maybe I'm just not following your process. Now that you've done it with a = -c, and arbitrarily chosen c = -1, do it again with c = 2. This leads immediately to a = -2. Now find a value for b.

Answer 8

hmmm I see what you mean it doesnt look right does it. Im a little lost with this.

Answer 9

Can you find the Eigenvalues, anyway?

Answer 10

Thats what Im trying to do I thought I had with the set above, but you've made me question what Ive done. Hmm back to the drawing board I guess.

Answer 11

I had found the Eigenvalues for the given matrix, now Im just trying to find the Eigenvectors from the Eigenvalues and given matrix but Im stumped.

Answer 12

What did you get for Eigenvalues?

Answer 13

0,3 & -4

Answer 14

and the given matrix was
1 2 1
6 -1 0
-1 -2 -1

Answer 15

That's right. And now you're trying to calcualte the eigenvector associated with 0. That's good. You are on the right track. And, frankly, the system so generated SHOULD be underdefined. Sorry, I hadn't yet seen that one of the eigenvalues was zero (0).

Answer 16

so what are the eigenvectors for 0?

Answer 17

We say earlier that adding Eq1 and Eq3 makes things disappear. We can then modify the system to this:

a + 2b = -c
6a - b = 0

And proceed to solve this for a and b.

Twice the bottom, added to the top gives 13*a = -c and a = -c/13
Finally, from the bottom, given 'a', we have b = -6c/13

Finally, Finally, we pick something easy for 'c', probably c = 1 is best, and we get:

(-1/13, -6/13, 1)^T

Answer 18

Note: It's probably considered bad form to keep 'c' in this process after we have elimiated it and decided that we can pick anyting we like. Traditionally, one pick the value "t" when it moves to the right-hand side - just to emphasize that we have to pick something.

Answer 19

Thanks for that, helped alot!!
for my eigenvalues of -4 I get eigenvectors of 1, -2, -1
and for 3 i get 1, 0, 2. Do these look correct?

Answer 20

I'm not thrilled with (1,-2,-1)^T, but only because you chose -1 rather than 1. No bid deal.

(1,0,2)^T though, I'm not seeing. You should get (-1, -3/2, 1)^T

Answer 21

Hmm really? Now Im lost I thought I had this. I followed the process to the letter from my text. I think Im getting lost with the solving of the equations. The process is correct just not getting the right answers.