When will this system have: one solution, infinity solutions. zero solutions?

Sep 2012
837
87
Canada
Hello,

I have the linear system of presumably 3 variables ( the book does not say):

\(\displaystyle x+ay=0\)
\(\displaystyle y+bz=0\)
\(\displaystyle z+cx=0\)

We are asked:
Find (if possible) conditions a,b,c such that the system has infinitely many solutions, no solutions or one solution.
In the answer, they claim that for \(\displaystyle abc\neq-1\) there is a unique solution and so for \(\displaystyle abc=-1\) there is are infinitely many solutions.

My question is, how did they find the base case abc=-1?

Thank you,
-Sakon
 
Last edited:
Mar 2015
9
1
Sweden
Hmm, try and solve z=-cx input Z in R2 => y-bcx=0 => y=bcx input Y in R1 => x+abcx= 0 => -1=abc I think.
 
Last edited:
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Aug 2008
12,883
4,999
Hello,

I have the linear system of presumably 3 variables ( the book does not say):

\(\displaystyle x+ay=0\)
\(\displaystyle y+bz=0\)
\(\displaystyle z+cx=0\)

We are asked:


In the answer, they claim that for \(\displaystyle abc\neq-1\) there is a unique solution and so for \(\displaystyle abc=-1\) there is are infinitely many solutions.

My question is, how did they find the base case abc=-1?

Thank you,
-Sakon
You have this system of equations:

$\displaystyle \begin{align*} x + a\,y &= 0 \\ y + b\,z &= 0 \\ z + c\,x &= 0 \end{align*}$

which can be written in matrix form as

$\displaystyle \begin{align*} \left[ \begin{matrix} 1 & a & 0 \\ 0 & 1 & b \\ c & 0 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] = \left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \end{align*}$

This will have a nontrivial solution for $\displaystyle \begin{align*} \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \end{align*}$ as long as $\displaystyle \begin{align*} \left| \begin{matrix} 1 & a & 0 \\ 0 & 1 & b \\ c & 0 & 1 \end{matrix} \right| \neq 0 \end{align*}$, so evaluate this determinant and see if you can find when it is not going to be 0...
 
Sep 2012
837
87
Canada
Yes indeed, then if x(1+abc)=0 , one unique solution when abc is not -1 and infinitely many solutions when abc=-1 as x can be replaced by a parameter of any value?
 
Sep 2012
837
87
Canada
Prove it, I am kind of lost. How would you go about solving that matrix in your last sentence?