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June 28th, 2018, 01:11 AM  #1 
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2  Infinite solutions problem
I am trying to solve following problem. I am trying to find x,y,z in terms of $\displaystyle \lambda$. And then equating the resultant equation to 0. Am I correct?
Last edited by skipjack; June 28th, 2018 at 06:05 AM. 
June 28th, 2018, 06:46 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039 
No, because "equating an equation to zero" doesn't make sense. What did you equate to zero?

June 28th, 2018, 07:59 AM  #3 
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2 
After replacing y and z, finally I arrived at the following equation if I hopefully did not a mistake $\displaystyle x*\lambda*(\lambda^2+4*\lambda3)=0 $ So I find that there are 3 solutions, with one of them being 0. Am I correct? Please advise. Last edited by skipjack; June 28th, 2018 at 01:22 PM. 
June 28th, 2018, 01:25 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689 
Calculate the determinant: $\lambda^3+4\lambda36=0$. This has only 1 real root, which is the desired solution.

June 28th, 2018, 01:27 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039  
June 28th, 2018, 04:27 PM  #6 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140 
These three equations represent three planes in 3 dimensional space. To have an infinite number of solutions these three planes must either be coincident or must intersect in a common line. The planes cannot be coincident since the attitude numbers (coefficients of x,y,z) are not proportional. To have a common line, the three planes must have two points in common. (0,0,0) is common to all three so all you need is one more. I think that $\displaystyle \left( \dfrac{\lambda^2+4\lambda8}{2},\lambda+2,8\lambda \right)$ is also common to all three planes (although I'm not sure of my arithmetic). If I'm correct then it appears that any value of $\displaystyle \lambda$ would make a line with (0,0,0) that produces an infinite number of solutions. Can someone check this? 
June 28th, 2018, 07:49 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039 
It's incorrect.

June 29th, 2018, 07:50 AM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra  Quote:
Unfortunately, the quadratic has no real roots. Last edited by v8archie; June 29th, 2018 at 07:52 AM.  
June 29th, 2018, 09:54 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,475 Thanks: 2039 
The correct equation doesn't factorize that simply.

June 29th, 2018, 11:01 AM  #10 
Member Joined: Aug 2017 From: India Posts: 50 Thanks: 2 
This is the attempt I have done. The final equation is $\displaystyle x(\lambda^3+4\lambda40)=0; $ I am not sure why I got a different equation last time. Sorry if the attachments are not clear. Next time, I will try to use math editor. Please advise whether I am correct. Last edited by skipjack; June 29th, 2018 at 09:05 PM. 

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