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 March 13th, 2014, 12:50 PM #1 Newbie   Joined: Oct 2013 Posts: 29 Thanks: 1 Non-linear systems of equations Hello, I'm having a really tough time solving the following non-linear systems of equations. They are the equations that arose from the search for constraned extrema of multivariable functions. Any help would be greatly appreciated: 1) $z-2\lambda x=0$ $z-2\lambda y=0$ $x+y-2\lambda z=0$ $x^{2}+y^{2}+z^{2}=1$ 2) $-2xsin(x^{2}+y^{2})+2\lambda x=0$ $2ysin(x^{2}+y^{2})+2\lambda y=0$ $x^{2}+y^{2}=1$
 March 13th, 2014, 07:58 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra Re: Non-linear systems of equations In 1) you can rearrange each of the first two equations to give you expressions for x and y in terms of z and $\lambda$ only, You can then substitute those into the third equation. You then have an equation that expresses z in terms of $\lambda$ only. Substituting that back into the first two equations will give you x and y, each in terms of $\lambda$ only. And you can then substitute into the fourth equation to work out a value for $\lambda$. For 2) you can use the third equation directly in the first two, and from the results calculate $\lambda$. Except that, on first sight, that seems to give $\lambda= \sin{1}$ and $\lambda= -\sin{1}$. Are you sure that question is correct? If it is, then I conclude that either x or y is equal to zero, the other is therefore 1.
 March 13th, 2014, 10:17 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1127 Math Focus: Elementary mathematics and beyond Re: Non-linear systems of equations 1) $\,z\,-\,2\lambda x\,=\,0\,[1] \\ z\,-\,2\lambda y\,=\,0\,[2] \\ x\,+\,y\,-\,2\lambda z\,=\,0\,[3] \\ x^2\,+\,y^2\,+\,z^2\,=\,1\,[4]$ $\text{From [1]: }z\,=\,2\lambda x \\ \text{From [1] and [2]: }x\,=\,y \\ \text{From [3]: }2x\,-\,4\lambda^2x\,=\,0\,\Rightarrow\,\lambda\,=\,\pm\ frac{1}{\sqrt2} \\ \text{Now, from [1]: }z\,=\,\pm\frac{2}{\sqrt2}x \\ \text{Into [4]: }4x^2\,=\,1\,\Rightarrow\,x\,=\,\pm\frac12\text{ (recall }x\,=\,y\text{)}$ $(x,\,y,\,z,\,\lambda)\,=\,$$\frac12,\,\frac12,\,\f rac{1}{\sqrt2},\,\frac{1}{\sqrt2}$$,\,$$-\frac12,\,-\frac12,\,-\frac{1}{\sqrt2},\,\frac{1}{\sqrt2}$$,\,$$-\frac12,\,-\frac12,\,\frac{1}{\sqrt2},\,-\frac{1}{\sqrt2}$$,\,$$\frac12,\,\frac12,\,-\frac{1}{\sqrt2},\,-\frac{1}{\sqrt2}$$$ I think that's all but it's getting late here . . . goodnight!
March 14th, 2014, 01:49 AM   #4
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Re: Non-linear systems of equations

Quote:
 Originally Posted by Akcope 2) $-2xsin(x^{2}+y^{2})+2\lambda x=0$ $2ysin(x^{2}+y^{2})+2\lambda y=0$ $x^{2}+y^{2}=1$
If $x\ne0$ then $\lambda=\sin1$. If $y\ne0$ then $\lambda=-\sin1$.

Suppose $x,y\ne0$. Then

$\begin{pmatrix}-x=&x \\ y=&y\end{pmatrix}\begin{pmatrix}2\sin1 \\ 2\lambda\end{pmatrix}\=\ \begin{pmatrix}0 \\ 0\end{pmatrix}$ $\Rightarrow$ $\begin{pmatrix}2\sin1 \\ 2\lambda\end{pmatrix}\=\ \begin{pmatrix}-x &x \\ y=&y\end{pmatrix}^{-1}\begin{pmatrix}0 \\ 0\end{pmatrix}=$

If $x,y\ne0$, $\begin{pmatrix}-x & x \\ y & y\end{pmatrix}^{-1}$ exists, giving $2\sin1=0$, an absurdity. Hence one of $x,y$ must be 0. (Note though that $x^2+y^2=1$ implies they can’t both be 0.)

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