What Values of ##\lambda## Allow Non-Trivial Solutions in This Linear System?

[nightingale123]

Homework Statement

3.For which values of $\lambda$ does the following system of equations also have non trivial solutions

Homework Equations

The Attempt at a Solution

What I tried doing first is to put all variables on the same side and got
$v+y-\lambda*x=0\\ x+z-\lambda*y=0\\ y+u-\lambda*z=0\\ z+v-\lambda*u=0\\ u+x-\lambda*v=0$
and when I wrote the coefficient into the matrix i got
$\begin{bmatrix} -\lambda& 1 &0&0&1\\ 1&-\lambda&1&0&0\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 1&0&0&1&-\lambda\\ \end{bmatrix}$

here I noticed that all the columns sum to the same number $2-\lambda$ there I summed everything into the first row and got
$\begin{bmatrix} 2-\lambda & 2-\lambda&2-\lambda&2-\lambda&2-\lambda\\ 1&-\lambda&1&0&0\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 1&0&0&1&-\lambda\\ \end{bmatrix}$
here I looked into 2 different possibilities if a) $\lambda=2$ and b) $\lambda\neq2$.
However a) is pretty simple and it's mostly b) that I'm having trouble with.
Here I thought if $\lambda\neq2$ then I can divide the first row by $2-\lambda$
When I did this my matrix looked like this
$\begin{bmatrix} 1 & 1&1&1&1\\ 1&-\lambda&1&0&0\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 1&0&0&1&-\lambda\\ \end{bmatrix}$
Then I subtracted the first row from the second and last one and got
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&-\lambda-1&0&-1&-1\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 0&-1&-1&0&-\lambda-1\\ \end{bmatrix}$
then I just rearranged some rows so that it would be easier for me to read
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&1&-\lambda&1&0\\ 0&-1&-1&0&-\lambda-1\\ 0&-\lambda-1&0&-1&-1\\ 0&0&1&-\lambda&1\\ \end{bmatrix}$
then I added the second row to the third and forth one and switched the third and forth row
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&1&-\lambda&1&0\\ 0&-\lambda&-\lambda&0&-1\\ 0&0&-1-\lambda&1&-\lambda-1\\ 0&0&1&-\lambda&1\\ \end{bmatrix}$
Lastly I added the last row to the forth one and switched them
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&1&-\lambda&1&0\\ 0&-\lambda&-\lambda&0&-1\\ 0&0&1&-\lambda&1\\ 0&0&-\lambda&1-\lambda&-\lambda\\ \end{bmatrix}$
Here is where I get stuck. I don't know how to continue from here on out. Maybe I made a mistake somewhere in my addition however I went through it at least a few times and I was not able to find it:
Any help / tips are greatly appreciated
Thanks

 

[Mark44]

Homework Statement

3.For which values of $\lambda$ does the following system of equations also have non trivial solutions

View attachment 195591

Homework Equations

The Attempt at a Solution

What I tried doing first is to put all variables on the same side and got
$v+y-\lambda*x=0\\ x+z-\lambda*y=0\\ y+u-\lambda*z=0\\ z+v-\lambda*u=0\\ u+x-\lambda*v=0$
and when I wrote the coefficient into the matrix i got
$\begin{bmatrix} -\lambda& 1 &0&0&1\\ 1&-\lambda&1&0&0\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 1&0&0&1&-\lambda\\ \end{bmatrix}$

here I noticed that all the columns sum to the same number $2-\lambda$ there I summed everything into the first row and got
$\begin{bmatrix} 2-\lambda & 2-\lambda&2-\lambda&2-\lambda&2-\lambda\\ 1&-\lambda&1&0&0\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 1&0&0&1&-\lambda\\ \end{bmatrix}$
here I looked into 2 different possibilities if a) $\lambda=2$ and b) $\lambda\neq2$.
However a) is pretty simple and it's mostly b) that I'm having trouble with.
Here I thought if $\lambda\neq2$ then I can divide the first row by $2-\lambda$
When I did this my matrix looked like this
$\begin{bmatrix} 1 & 1&1&1&1\\ 1&-\lambda&1&0&0\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 1&0&0&1&-\lambda\\ \end{bmatrix}$
Then I subtracted the first row from the second and last one and got
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&-\lambda-1&0&-1&-1\\ 0&1&-\lambda&1&0\\ 0&0&1&-\lambda&1\\ 0&-1&-1&0&-\lambda-1\\ \end{bmatrix}$
then I just rearranged some rows so that it would be easier for me to read

After your first step (when all the entries in row 1 were 1), you used the first entry in row 1 to eliminate all the entries below it. Continue this process by using the 2nd entry in row 2 (the pivot) to eliminate all entries above and below it. Continue this process until you have a diagonal matrix or until the matrix is as reduced as possible.

$\begin{bmatrix} 1 & 1&1&1&1\\ 0&1&-\lambda&1&0\\ 0&-1&-1&0&-\lambda-1\\ 0&-\lambda-1&0&-1&-1\\ 0&0&1&-\lambda&1\\ \end{bmatrix}$
then I added the second row to the third and forth one and switched the third and forth row
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&1&-\lambda&1&0\\ 0&-\lambda&-\lambda&0&-1\\ 0&0&-1-\lambda&1&-\lambda-1\\ 0&0&1&-\lambda&1\\ \end{bmatrix}$
Lastly I added the last row to the forth one and switched them
$\begin{bmatrix} 1 & 1&1&1&1\\ 0&1&-\lambda&1&0\\ 0&-\lambda&-\lambda&0&-1\\ 0&0&1&-\lambda&1\\ 0&0&-\lambda&1-\lambda&-\lambda\\ \end{bmatrix}$
Here is where I get stuck. I don't know how to continue from here on out. Maybe I made a mistake somewhere in my addition however I went through it at least a few times and I was not able to find it:
Any help / tips are greatly appreciated
Thanks

 

[nightingale123]

Thank you for the reply.
I though about doing that however does that mean that I have to multiply some rows with $\lambda$ to get the other ones to cancel out ?
If so do I also need to check what happens when $\lambda=0$ ?

 

[Mark44]

Thank you for the reply.
I though about doing that however does that mean that I have to multiply some rows with $\lambda$ to get the other ones to cancel out ?
If so do I also need to check what happens when $\lambda=0$ ?

Yes to the first question. For the second, you could go back to your first matrix and replace $\lambda$ with 0, and see what you get from that.