Find all real values of λ, if any, for which the system has a nontrivial solution. 2x1 + 3x2 = λx1 4x1 + 3x2 = λx2

Question

Find all real values of λ, if any, for which the system has a nontrivial solution.  2x1 + 3x2 = λx1 4x1 + 3x2 = λx2

fibonaccichick666 · Answer

What do your 'x's mean?

anonymous · Answer

they are the variables in the system

anonymous · Answer

$$2x_{1} + 3x _{2} = \lambda x _{1}$$

fibonaccichick666 · Answer

ah ok

anonymous · Answer

$$4x _{1} + 3x _{2} = \lambda x _{2}$$

fibonaccichick666 · Answer

What can you tell me about how many equations you need in relation to the number of variables you have?

anonymous · Answer

with two variables, don't you need two equations?

fibonaccichick666 · Answer

yup, how about 3?

anonymous · Answer

3... ?

fibonaccichick666 · Answer

and 4?

anonymous · Answer

equations?

fibonaccichick666 · Answer

4 variables?

anonymous · Answer

would need 4 equations

fibonaccichick666 · Answer

and n variables?

anonymous · Answer

n equations

fibonaccichick666 · Answer

good, so now, how many variables do you have?

anonymous · Answer

2

fibonaccichick666 · Answer

now, do you have enough equations?

anonymous · Answer

yes

fibonaccichick666 · Answer

(since we are saying $\lambda$ isn't a real variable)

anonymous · Answer

no, its acting as as scalar isn't it?

fibonaccichick666 · Answer

yes, sort of.

zepdrix · Answer

Hmm I haven't taken much Linear so I'm not sure how much help I can be.
But lemme share what I'm thinking.
$$\Large\rm 2x_{1} + 3x _{2} = \lambda x _{1}$$$$\Large\rm 4x _{1} + 3x _{2} = \lambda x _{2}$$Subtract the right hand sides so we have all of our x's on one side.
Then writing it in matrix form gives us:$$\Large\rm \left[\begin{matrix}2-\lambda & 3 \ 4 & 3-\lambda\end{matrix}\right]\vec X =\left(\begin{matrix}0 \ 0\end{matrix}\right)$$And then we can do something with that matrix yes?
Taking the determinate gives us some Lambda values...
I don't remember what that means though :P

Any ideas Kmendall?

anonymous · Answer

actually I think I may be able to figure something out from there...

fibonaccichick666 · Answer

Oh I was assuming no linear alg. that makes life easier, ok

anonymous · Answer

sorry, this is a linear algebra question!

fibonaccichick666 · Answer

make the modified variable matrix, then get it down to reduced row echelon and you should see some lambda values

anonymous · Answer

ok, I'm having a moment... how do I make the modified variable matrix?

fibonaccichick666 · Answer

it's essentially what zep gave ya,

fibonaccichick666 · Answer

just put it all together so it's a 2 by 3

anonymous · Answer

ok, that's what I was thinking, just doubting myself
Thanks

fibonaccichick666 · Answer

you got it