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June 28th, 2018, 02:11 AM   #1
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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?
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 solutions.jpg (13.1 KB, 8 views)

Last edited by skipjack; June 28th, 2018 at 07:05 AM.

 June 28th, 2018, 07:46 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 No, because "equating an equation to zero" doesn't make sense. What did you equate to zero?
 June 28th, 2018, 08:59 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 $\displaystyle x*\lambda*(\lambda^2+4*\lambda-3)=0$ 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 02:22 PM.
 June 28th, 2018, 02:25 PM #4 Global Moderator   Joined: May 2007 Posts: 6,680 Thanks: 658 Calculate the determinant: $\lambda^3+4\lambda-36=0$. This has only 1 real root, which is the desired solution.
June 28th, 2018, 02:27 PM   #5
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Quote:
 Originally Posted by MathsLearner123 I hopefully did not a mistake . . .
Can you post all your work, so that the mistake(s) in it can be identified?

The three equations are satisfied by $x = y = z = 0$. In the right circumstances, there may be infinitely many other solutions.

 June 28th, 2018, 05:27 PM #6 Senior Member     Joined: Feb 2010 Posts: 702 Thanks: 137 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 $\displaystyle \left( \dfrac{-\lambda^2+4\lambda-8}{2},\lambda+2,8-\lambda \right)$ 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, 08:49 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 It's incorrect.
June 29th, 2018, 08:50 AM   #8
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Quote:
 Originally Posted by MathsLearner123 After replacing y and z, finally I arrived at the following equation if I hopefully did not a mistake $\displaystyle x*\lambda*(\lambda^2+4*\lambda-3)=0$ So I find that there are 3 solutions, with one of them being 0. Am I correct? Please advise.
This is the correct conclusion to draw from your equation if the quadratic has two roots, yes. To satisfy the equation $x$, $\lambda$ or $(\lambda^2+4*\lambda-3)$ must be equal to zero. If neither $\lambda \ne 0$ and $(\lambda^2+4*\lambda-3)\ne 0$ you must have $x=0$ which leads to a single solution. On the other hand, if either $\lambda = 0$ or $(\lambda^2+4*\lambda-3) = 0$, the equation is satisfied by any value of $x$ and you thus have an infinite number of solutions.

Unfortunately, the quadratic has no real roots.

Last edited by v8archie; June 29th, 2018 at 08:52 AM.

 June 29th, 2018, 10:54 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,262 Thanks: 1958 The correct equation doesn't factorize that simply.
June 29th, 2018, 12:01 PM   #10
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This is the attempt I have done. The final equation is
$\displaystyle x(\lambda^3+4\lambda-40)=0;$
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.
Attached Images
 page2.jpg (8.5 KB, 3 views) Page1.jpg (70.2 KB, 6 views)

Last edited by skipjack; June 29th, 2018 at 10:05 PM.

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