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.