I have this exercise from the linear algebra course I'm currently taking. However I'm somewhat stuck.
Let $\alpha$ and $\beta$ be real numbers. Show that the linear equation system \begin{equation*} \begin{pmatrix} 1 & 2\\ 2 & 3\\ 3 & 4 \end{pmatrix}x= \begin{pmatrix} \alpha+2\beta\\ \alpha+2\beta\\ 2\alpha+3\beta \end{pmatrix}, \end{equation*} has solutions if and only if $\alpha+\beta=0$.
Okay, so far I've been able to show, that if $\alpha+\beta=0$, then $\alpha=-\beta$, so the RHS can be written as
$$\begin{pmatrix} -\alpha\\ -\alpha\\ -\alpha \end{pmatrix}$$
I can indeed solve the equation when with gaussian elimination, but I'm not sure how I can show, that the system only has a solution, when $\alpha+\beta=0$. I could come with examples where where is no solution, but I'm unsure how to show it in general for all cases.
You can proceed with the Gaussian elimination keeping the $\alpha$ and $\beta$ in place as variables. The result of the process gives you $$\begin{pmatrix} 1 & 2 \\ 0 & -1 \\ 0 & 0 \end{pmatrix}x = \begin{pmatrix} \alpha + 2\beta \\ -\alpha - 2\beta \\ \alpha + \beta \end{pmatrix}$$ This system clearly has solutions if and only if the last row on the right side is $0$, which is if and only if $\alpha + \beta = 0$.