Finding stationary points of $f(x,y) = x^2+y^2+\beta xy+x+2y$ I am working on a problem which can be seen below. Part i) was fine but part ii) is causing me some issues. Now the First Order Necessary Conditions state that for the interior case, then $$\nabla f(x^*) = 0$$ which I have attempted to use to find $x^*$. Taking the derivatives of the objective function I get $$\begin{bmatrix}2x + \beta y +1 \\
2y + \beta x +2 \end{bmatrix} = \begin{bmatrix}0\\ 0 \end{bmatrix}$$ which I have solved to get $(y-x) = \frac{1}{\beta - 2}$. Now I am convinced that I am wrong here because this does not seem plausible.
I would really appreciate if someone could give me some hints.

Thanks

You got:
$$\begin{bmatrix}2 & \beta \\ \beta & 2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-1\\ -2 \end{bmatrix}$$
Let's perform one step of row reduction:
$$\begin{bmatrix}1 & \frac{1}{2}\beta \\ 0 & 2-\frac{1}{2}\beta^2 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-\frac{1}{2}\\ \frac{1}{2}\beta-2 \end{bmatrix}$$
The system has no solution if $2-\frac{1}{2}\beta^2=0$ but $\frac{1}{2}\beta-2 \neq 0$, so when $\beta=2$ or $\beta=-2$. The system has infinitely many solutions if $2-\frac{1}{2}\beta^2= 0$ and $\frac{1}{2}\beta-2 = 0$, which cannot happen.
When $\beta\neq 2$ and $\beta\neq -2$ we do one more step of row reduction:
$$\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}-\frac{1}{2}-\frac{1}{2}\frac{\beta^2-4\beta}{4-\beta^2}\\ \frac{\beta-4}{4-\beta^2} \end{bmatrix}$$
You can now readily read the solution.
