I'm asked to to find a $3\times3$ matrix, in which no entry is $0$ but $A^2=0$.
The problem is if I I brute force it, I am left with a system of 6 equations (Not all of which are linear...) and 6 unknowns. Whilst I could in theory solve that, is there more intuitive way of solving this problem or am I going to have to brute force the solution?
Any suggestions would be greatly appreciated.

There's an idea. We can try to put $A = \alpha \beta'$, where $\alpha,\beta$ are $3 \times 1$ matrix.
Then, $A^2 = \alpha \beta' \alpha \beta'$, so if $\alpha,\beta \ne 0$(all the components are not $0$),$\beta' \alpha = 0$ things will be solved.
So,we could let
$\alpha = \begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}, \beta = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$
and $A = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 1 & -2 \\ 1 & 1 & -2 \end{pmatrix}$