Solving System of Equations Issue

Question

$$ \begin{aligned} x^2+4y^2+9z^2 &= 1 \\ 8yz+2ax &=0 \\ 8zx+8ay &= 0\\ 8xy+18az &= 0 \end{aligned}$$

How would I solve these equations? I tried making $a$ the subject of the final $3$ but struggling to solve it.

Answer 1

Assuming that $x\neq0$, $y\neq0$ and $z\neq0$ you can re-formulate the equations as

$$\begin{aligned} x^2+4y^2+9z^2 &= 1 \\ (8)xyz+(2a)x^2 &=0 \\ (8)xyz+(8a)y^2 &= 0\\ (8)xyz+(18a)z^2 &= 0 \end{aligned}$$

by multiplying the 2nd, 3rd and 4th equations with $x$, $y$ and $z$ respectively

and again by normalizing the quadratic terms in the 2nd-4th equations

$$\begin{aligned} x^2+4y^2+9z^2 &= 1 \\ \tfrac{8}{2a} xyz+x^2 &=0 \\ \tfrac{8}{8a} xyz+y^2 &= 0\\ \tfrac{8}{18a}xyz+z^2 &= 0 \end{aligned}$$

Now use the 2nd-4th equations in the 1st equation

$$ -\tfrac{4}{a} x y z - \tfrac{4}{a} x y z - \tfrac{4}{a} x y z = 1 $$

$$ - \tfrac{12}{a} x y z = 1 $$

$$ \boxed{ x y z = -\tfrac{a}{12} } $$

Use the product $xyz$ in the equations above to get

$$\begin{aligned} -(8)\tfrac{a}{12}+(2a)x^2 &=0 \\ -(8)\tfrac{a}{12}+(8a)y^2 &= 0\\ -(8)\tfrac{a}{12}+(18a)z^2 &= 0 \end{aligned}$$

which is used to solve for $(x,y,z)$ as $a$ cancels out.

$$\boxed{\begin{aligned} x^2 &= \tfrac{1}{3} \\ y^2 &= \tfrac{1}{12} \\ z^2 &= \tfrac{1}{27} \\ \end{aligned}}$$

Finally, the 1st boxed equation above is used to find $a$

$$ \boxed{ a = \pm \tfrac{2 \sqrt{3}}{9} }$$

Answer 2

Let us use these three equations: \begin{align*} ax&=-4yz\\ ay&=-xz\\ az&=-\frac49xy \end{align*}

From the first two equations you get: \begin{align*} (ax)\cdot(xz)&=(4yz)\cdot(ay)\\ azx^2&=4azy^2\\ x^2&=4y^2 \tag{1} \end{align*} The last step is valid under the assumption that $a\ne0$ and $z\ne0$.

Using similar arguments you can get: $x^2=9z^2$.

Once you know that $x^2=4y^2=9z^2$, you immediately get $$x^2=4y^2=9z^2=\frac13,$$ i.e., $x=\pm\frac1{\sqrt3}$, $y=\pm\frac1{2\sqrt3}$ and $z=\pm\frac1{3\sqrt3}$.

You should be able to use this to calculate $a$ and to check which combinations of the signs are actually possible.

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All of the above was done under the assumption that $a\ne0$, $x\ne0$, $y\ne0$, $z\ne0$. So it still remains to check what happens if one of those variables is equal to zero. (I would hazard a guess that there are no such solution - but there's still a bit work to check this thoroughly.)

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Another possibility to get $(1)$ is to view the system \begin{align*} 4yz+xa&=0\\ xz+ya&=0 \end{align*} as a system of two linear equations with the unknowns $z$ and $a$. If the determinant $\det\begin{pmatrix} 4y & x \\ x & y \\ \end{pmatrix}$ is non-zero, then we get $a=z=0$ as the only solution. Using the remaining equations we find out that there is no such solution of the original system.

So the determinant must be zero and we get $$\det\begin{pmatrix} 4y & x \\ x & y \\ \end{pmatrix}=4y^2-x^2=0.$$