Say I have two equations:
$$y^3+x-y=0$$ $$x^3-x+y=0$$
These are both partial derivatives of some function, and the condition that they're equal to $0$ implies the values of $x,y,z$ are bound are stationary points. I thought I might try and put these values into a matrix to compute the solution to the system. I think I did something wrong here, because it ended up with a case where my matrix not being in echelon form actually made it easier to solve.
$$\begin{bmatrix} 1 & 0 & 1 & -1 \\ 0 & 1 & -1 & 1\\ \end{bmatrix}$$
Where it is a matrix of coefficients with each row vector with the form $(y^3,x^3,x,y)$. This matrix here, is actually in reduced echelon form, which tells me that there are two free parameters and that the system can be solved as is by back substitution, but if one does..
$$R2 \to R2 + R1$$
One then has (forming the augmented matrix):
$$\begin{bmatrix} 1 & 0 & 1 & -1 &| & 0\\ 1 & 1 & 0 & 0 & | & 0\\ \end{bmatrix}$$
And thus it is shown an implication of the system has been rendered useful to us:
$$x^3 + y^3 = 0$$ $$y^3 +x -y = 0$$
Which is very straightforward to solve as it gives us the insight that $x=-y$. Why, though, did it make sense that doing a row operation after my matrix was in $rref$ helped simply the system further? Also, did what I do make sense? Or is the matrix I made not valid? Just tried applying some linear algebra to a calculus exercise. My hunch is that my matrix is invalid because usually our matrix of coefficients belong to $x,y,z$ terms and for mine, $y^3$ is not independent to $y$ like $z$ is to $y$. But, the vector space of all polynomials with degree at most $2$ have $(a,x,x^2)$ terms which seems to contradict my hypothesis.
adding both equations we get $$x^3+y^3=0$$ and this is equal to $$(x+y)(x^2-xy+y^2)=0$$ can you solve this equation?