# Find Lambda for the linear system

• Apr 28th 2011, 12:56 PM
userod
Find Lambda for the linear system
Hi guys I am trying to figure out the following question can anyone help:
Find $\displaystyle \lambda$ for the Linear System:
$\displaystyle \lambda$x+y+z=1
x+$\displaystyle \lambda$y+z=1
x+y+$\displaystyle \lambda$z=1

For infinite solutions.
A unique solution.

And it's not asked but no solutions I wouldn't mind knowing either.

I understand how to do these with numbers
• Apr 28th 2011, 01:11 PM
TheEmptySet
Quote:

Originally Posted by userod
Hi guys I am trying to figure out the following question can anyone help:
Find $\displaystyle \lambda$ for the Linear System:
$\displaystyle \lambda$x+y+z=1
x+$\displaystyle \lambda$y+z=1
x+y+$\displaystyle \lambda$z=1

For infinite solutions.
A unique solution.

And it's not asked but no solutions I wouldn't mind knowing either.

I understand how to do these with numbers

by inspection is should be obvious that is lambda=1 all three equations are the same so you will get an infinite number of solutions, but there could be other values as well.

Now use equation 3 to eliminate the x variable from equations one and two to get

$\displaystyle -\lambda E_3+E_1=(1-\lambda)y+(1-\lambda^2)z=0 =E_3$

$\displaystyle -E_3+E_2=(\lambda-1)y+(1-\lambda)z=0=E_4$

Now use these two equations to eliminate y Can you finish from here?
• Apr 28th 2011, 01:39 PM
userod
Great thanks very much - when I add E3 and E4 I get a quadratic equation in lambda which gives lambda =1 or lambda = -2. So ruling out lambda=1 which gives an infinite number of solutions then lambda = -2. Would I be right? Thanks again
• Apr 28th 2011, 01:45 PM
TheEmptySet
Quote:

Originally Posted by userod
Great thanks very much - when I add E3 and E4 I get a quadratic equation in lambda which gives lambda =1 or lambda = -2. So ruling out lambda=1 which gives an infinite number of solutions then lambda = -2. Would I be right? Thanks again

We need to be vary careful right now.

Both $\displaystyle \lambda=1\quad \lambda =-2$ both give an infinite number of solutions.

Back in my first post I recommended that you muliply one of the equations by

$\displaystyle \lambda$ right?

What value can't lambda be? Asked another way when we multiply an equation by a number can we use any number or are there number(s) we can use?
• Apr 28th 2011, 02:07 PM
userod
I think we can use any number except 0? I'm not sure I'm kind of lost :(
• Apr 28th 2011, 02:50 PM
TheEmptySet
Quote:

Originally Posted by userod
I think we can use any number except 0? I'm not sure I'm kind of lost :(

You are correct. What happens when you put $\displaystyle \lambda =0$ into the original system does it have a solution then?
If so is it unique?
• Apr 28th 2011, 09:44 PM
userod
Ah yes lambda=0 seems to give a unique solution .5,.5,.5.

But it was just trial and error to give this? The simultaneous equation part didn't seem to help this atall?
• Apr 29th 2011, 02:37 PM
HallsofIvy
If you are lost, it might be best to go back and think about what you are doing. You originally said, "Find lambda for the linear system", and then gave a system of equations involving x, y, z and parameter lambda. Find lambda so that what happens? I suspect you want to find lambda so that the system has a unique solution. Or, the contrary- find lambda so that there is more than one solution. Why would that be of interest? I can't imagine why you would say "The simultaneous equation part didn't seem to help this atall?" Without the system of equations, you would only be saying "Find lambda" without any conditions on it. And that would make no sense at all!