Solving for 0 in a system of equations

[Adamantine Chains]

How does one find a combination of these columns that equals zero:

(1) u + v + w = 0
(2) 2u + 3w = 0
(3) 3u + v + 4w = 0

I'm looking for the algorithm.

I tried solving algebraically, but that does not work when I only have
coefficients. I end up producing messier equations that lead me back to
where I started.

To try to understand deeper principles at work, I came up with a smaller
system of equations:

4x - 6y = 0
-2x + 3y = 0

I noticed that x = 3/2y and y = 2/3x. The answer is in that relation: the
dividend is the correct coefficient for the left hand side variable and the
divisor is the correct coefficient for the right hand side variable. This
relation does not appear (at least to me) when n = 3 with n equations and n
unknowns.

I also tried to solve for u, v, and w separately and combine them into one
equation, but that did not produce the right values despite several
attempts.

My approaches are clearly wrong.

Any suggestions would be appreciated.

 

[Ishuda]

How does one find a combination of these columns that equals zero:

(1) u + v + w = 0
(2) 2u + 3w = 0
(3) 3u + v + 4w = 0

I'm looking for the algorithm.

I tried solving algebraically, but that does not work when I only have
coefficients. I end up producing messier equations that lead me back to
where I started.

To try to understand deeper principles at work, I came up with a smaller
system of equations:

4x - 6y = 0
-2x + 3y = 0

I noticed that x = 3/2y and y = 2/3x. The answer is in that relation: the
dividend is the correct coefficient for the left hand side variable and the
divisor is the correct coefficient for the right hand side variable. This
relation does not appear (at least to me) when n = 3 with n equations and n
unknowns.

I also tried to solve for u, v, and w separately and combine them into one
equation, but that did not produce the right values despite several
attempts.

My approaches are clearly wrong.

Any suggestions would be appreciated.

You still have the same sort of thing but for larger matrices [maybe beyond 3X3's] there is an easier way to go about it. For methods, Google
cramer's rule gauss jordan

For a 3X3, here is a simple example
(1) u + v + w = 2
(2) 5u + v + w = 6
(3) u + v - 3 w = 2
from (1) we have
(1') u = 2 - v - w
Now use that in (2) to get
5 (2-v-w) + v + w = 6
or
(2') v + w = 1
and in (3) to get
5 (2-v-w) + v - 3 w = 2
or
(3') v + 2 w = 1

Now do your usual thing for a 2X2 [(2') and (3')] to get v=1 and w=0. Using (1') we have u=1.