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Solve this system of eq: $$\begin{cases} yz &= 2 \lambda x &\,\,\,(a)\\ xz &= 2 \lambda y &\,\,\,(b)\\ xy &= 2 \lambda z &\,\,\,(c) \\ x^2+y^2+z^2−3&=0 &\,\,\,(d) \end{cases}$$

I wanted to do the following: from we dive whats on the right to get a $1$, for a and b and c, so that $$1=\frac{yz}{2 \lambda x}=\frac{xz}{2 \lambda y}=\frac{xy}{2 \lambda z}$$ and now conclude $x^2=y^2$ and $y^2=z^2$ so that $$x^2=y^2=z^2=1$$

(Note, -366 was the one Umberto solved).

Have I done anything wrong? I miss out on some of the solutions – is it from the divding?

Edit 1 It was -3 not -366, but no reason to change your whole calculation Umberto, I get it.

Edit 2 any general tips on how to solve these algebraically demanding lagrange-equations in the future? I often miss out on solutions due to not paying attention to details.

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I don't know why GitGud wanted it to be $-366$, it is actually $3$. – jacob Apr 16 '14 at 19:13
up vote 2 down vote accepted

You can separate into cases whether $\lambda = 0$ or not.

If $\lambda = 0$ you have $xy = xz = yz = 0$ so that either two or three of $x$, $y$, $z$ are equal to zero. Since $x^2 + y^2 + z^2 = 3$ they can't all be zero, so exactly two of $x$, $y$, $z$ equal zero and the third equals $\pm \sqrt{3}$. This gives six solutions of the triple $(x,y,z)$: $$ (\sqrt{3},0,0),\, (-\sqrt{3},0,0),\, (0,\sqrt{3},0),\, (0,-\sqrt{3},0),\, (0,0,\sqrt{3}),\, (0,0,-\sqrt{3}).$$

If $\lambda \not = 0$ consider $x = 0$. By (b) you must also have $y = 0$, and by (c) you must also have $z = 0$. Since $(0,0,0)$ is not a valid solution you get $x \not= 0$. A similar argument shows $y \not= 0$ and $z \not= 0$.

Consider (b) and (c). You get $x^2 z = 2 \lambda x y = 4\lambda^2z$. Since $z \not= 0$ you get $x^2 = 4 \lambda^2$. Likewise, $y^2 = 4 \lambda^2$ and $z^2 = 4 \lambda^2$. Put these into (d) to obtain $12\lambda^2 = 3$, that is, $\lambda = \pm \dfrac 12$. Thus $x = \pm 1$, $y = \pm 1$, and $z = \pm 1$.

We're almost there. You can use equation (a), (b), and (c) to determine which signs to use. For instance, if $x = y = z = 2\lambda$ then (a), (b), and (c) all hold. You can check that if one of $x,y,z$ equals $-2\lambda$, then exactly one other value is also equal to $-2\lambda$ and the third equals $+2\lambda$. This gives you the solutions $$ (1,1,1),\,(-1,-1,1),\,(-1,1,-1),\,(1,-1,-1)$$when $\lambda = \dfrac 12$ and $$(-1,-1,-1),\,(1,1,-1),\,(1,-1,1),\,(-1,1,1)$$when $\lambda = -\dfrac 12$.

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How come you attacked the 0 cases so much? What made you "see" or "test" 0 on every place? Since it is often easy, or because something in this problem made you think of it? – jacob Apr 16 '14 at 19:15
"and the third equals $-2 \lambda$" I think you mean "$+2 \lambda$" – jacob Apr 16 '14 at 19:21
I changed the 366 to 3 to make it easier to read. It often helps to consider the case when one variable is zero separately since 1) it reduces the number of variables to solve for, and 2) once the zero case is finished you are free to divide by the variable. For instance a simple example is $x^2 = x$. A common error is to divide by $x$ and get $x=1$. This is only a partial solution because $x=0$ was neglected at the division. – Umberto P. Apr 17 '14 at 12:45
I usually do these system of equations wrong. I forget solutions, most often, and sometimes I even get false solutions. Have an exam tomorrow, what can I do to not miss these "not thery heavy" parts? Any tips? – jacob Apr 21 '14 at 15:05

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