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June 28th, 2018, 01:11 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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Last edited by skipjack; June 28th, 2018 at 06:05 AM.
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June 28th, 2018, 06:46 AM   #2
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No, because "equating an equation to zero" doesn't make sense. What did you equate to zero?
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June 28th, 2018, 07:59 AM   #3
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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 01:22 PM.
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June 28th, 2018, 01:25 PM   #4
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Calculate the determinant: $\lambda^3+4\lambda-36=0$. This has only 1 real root, which is the desired solution.
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June 28th, 2018, 01:27 PM   #5
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Quote:
Originally Posted by MathsLearner123 View Post
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.
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June 28th, 2018, 04:27 PM   #6
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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?
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June 28th, 2018, 07:49 PM   #7
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It's incorrect.
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June 29th, 2018, 07:50 AM   #8
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Quote:
Originally Posted by MathsLearner123 View Post
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 07:52 AM.
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June 29th, 2018, 09:54 AM   #9
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The correct equation doesn't factorize that simply.
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June 29th, 2018, 11:01 AM   #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.
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File Type: jpg page2.jpg (8.5 KB, 3 views)
File Type: jpg Page1.jpg (70.2 KB, 6 views)

Last edited by skipjack; June 29th, 2018 at 09:05 PM.
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