The original question:
Let $A$ be a $3 \times 3$ matrix with $$ A \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} \quad\text{and}\quad A \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}. $$ Without doing any computations, find $\det A$ and justify your answer.
I know that $\det A = 0$, but I have no idea how to get to that answer. I don't even know where to start.
I noticed that $$ A \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} = A \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} $$ but I don't know what to do with that information.
I can reason some things based on the fact that $\det A = 0$, but that's not a given fact so I don't think that'd be helpful.
$$ A \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix} = A \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}......(*) $$ There are $2$ cases exclusively: Either $A^{-1}$ exists or doesn't.
Whereas the former case leads you (on pre-multiplication to ( * ) by $A^{-1}$) to an absurd equality of two different vectors; the only possibility that (*) holds is the later one i.e. non-existence of $A^{-1}$ which implies det $A=0$.