System of equations on $\mathbb R$

Question

$\textbf{Question : }$ Find all $x,y,z \in \mathbb{R}$ that satisfy the following system of equations. $$3x-4y+\frac{1}{xy}=12$$ $$4z-12x+\frac{1}{zx}=3$$ $$12y-3z+\frac{1}{yz}=4$$

$\textbf{My Attempt :}$ Let $t = \frac{1}{xyz}$ then each of our equation transforms to

$$3x-4y+tz=12$$ $$-12x+ty+4z=3$$ $$tx+12y-3z=4$$

I thought I would zero out the coefficient matrix which led to nothing :

$$\Delta = -t^3-169t=0 \implies t = 0 ,\pm13i$$ none of which is in $t$'s domain as $t \in \mathbb{R} - \{0\}$

Next I thought to write it in a form where all the constant terms are zero so I could surely zero out the coefficient matrix and find out the value of $t$. So here are those equations $$51x-4y(t+1)+z(t-16)=0$$ $$3(t+16)x+4(9-t)y-25z=0$$ $$3(1-t)x-40y+z(t+9)=0$$

as it turns out the coefficient matrix is zero for all $t \in \mathbb{R} - \{0\}$. So what to do now?

Answer 1

Solve the set of equations you have $$3x-4y+tz=12$$ $$-12x+ty+4z=3$$ $$tx+12y-3z=4$$for $x$, $y$ and $z$ in terms of $t$. It looks as if it will be horrible, but in fact you can eliminate $y$, for example, to get (multiply first equation by t and second by 4, then add) $$(3t-48)x+(t^2+16)z=12t+12$$ and (then multiply first equation by 3 and add to last) $$(t+9)x+(3t-3)z=40.$$ Now multiply the first of these equations by $(3t-3)$ and the second by $(t^2+16)$ and subtract, to get $$(9(t-1)(t-16)-(t+9)(t^2+16))x=36(t^2-1)-40(t^2+16)$$ or $$(-t^3-169t)x=-4(t^2+169),$$ so $x= \dfrac{4}{t}$ and back substitution gives $y= \dfrac{3}{t}$ and $z= \dfrac{12}{t}$.

Then $t= \dfrac{1}{xyz} = \dfrac{t^3}{144}$, so $t=\pm 12$, which completes the solution.

Answer 2

I don’t know why you equate $\Delta$ to zero. Instead, at that juncture, you could have easily apply Cramer's rule as shown below.

$$x=\dfrac{ \begin{bmatrix} 12 & -4& t \\ 3 & t & 4 \\ 4 & 12 & -3 \\ \end{bmatrix}}{\Delta} =\dfrac{-36t-576-100+36t-4t^2}{-t\left(t^2+169\right)}=\dfrac{-676-4t^2}{-t\left(t^2+169\right)}=\dfrac{t}{4}. $$

Similarly, the other two unknowns $y$ and $z$ can be expressed in terms of $t$ by simplifying the two equation given below.

$$y=\dfrac{ \begin{bmatrix} 3 & 12& t \\ -12 & 3 & 4 \\ t & 4 & -3 \\ \end{bmatrix}}{\Delta} \qquad\text{and}\qquad z=\dfrac{ \begin{bmatrix} 3 & -4 & 12 \\ -12 & t & 3 \\ t & 12 & 4 \\ \end{bmatrix}}{\Delta} $$

Answer 3

If you write your system of equations in a reverse order :

$$\begin{cases}tx+12y-3z&=&4\\-12x+ty+4z&=&3\\3x-4y+tz&=&12\end{cases} \iff $$

$$\begin{cases}12y-3z&=&4-tx\\-12x+4z&=&3-ty\\3x-4y&=&12-tz\end{cases},$$

you recognize in the LHS a cross-product :

$$\begin{pmatrix}x\\y\\z\end{pmatrix}\times \begin{pmatrix}4\\3\\12\end{pmatrix}=\begin{pmatrix}4\\3\\12\end{pmatrix}-t\begin{pmatrix}x\\y\\z\end{pmatrix}$$

Otherwise said, with evident notations :

$$X \times V = V - tX \tag{1}$$

(1) looks not to have any solution because $X \times V$ is orthogonal to both $X$ and $V$, therefore cannot lie (as indicated by the RHS) in the plane generated by $V$ and $X$.

But, in fact there is solution iff both sides of (1) are zero, i.e., $V=tX \ \iff X=\tfrac1t V$ :

$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{t}\begin{pmatrix}4\\3\\12\end{pmatrix}=\begin{pmatrix}4/t\\3/t\\12/t\end{pmatrix}\tag{2}$$

It's time to remember that $t=\frac{1}{xyz}$ :

$$t=\frac{1}{(4/t).(3/t).(12/t)}$$

finaly giving $t=\pm 12$.

Having $t$, the solutions are directly obtained from (2).